basic general topology for graduate studies

Point-set topology with topics

Basic general topology for graduate studies

Robert Andre(Revised: March 14, 2022)

Robert Andre c©2020

(Revised: March 14, 2022)

ToRoberta

∼∫

tion ∼

“I like it when the longing becomes belonging, dance of polarities ultimately be-

comes union, feeling the other like the lost part of oneself.So. . . union becomes “

∫

tion.”

Roberta G

“Je me plaisois surtout aux Mathematiques a cause de la certitude et de l’evidencede leurs raisons. Mais je ne remarquois point encore leur vray usage”

Rene DescartesDiscours de la Methode (1637)

“For though we love both the truth and our friends, piety requires us to honorthe truth first.”

Aristotle

Nicomachean Ethics (350 B.C.)

i

Preface

This text was prepared to serve as an introduction to the study of general topology.Most students in mathematics are required, at some point in their study, to have knowl-

edge of the fundamentals of general topology and master topological techniques that maybe useful in their area of specialization. “Trust us, not only will you be glad to be skilled

at using these tools, but a lot of concepts you study in your field of interest will eventu-ally be seen as specific cases investigated in a more general context in general topology”

they are told. It sometimes occurs that some students develop a particular fascination forgeneral topology in spite of it occasionally being described as being “no longer in fashion

anymore”. Recall that mathematicians such as George Cantor and Felix Hausdorff, werealso told that some of the mathematics they spent time investigating was “not in fashion”.Reasons for the continued study of this topic often go beyond the simple perception that

it is a practical or useful tool. The intricate beauty of the mathematical structures thatare derived in this field become its main attraction. This is, of course, how this writer

perceives the subject and served as the main motivation to prepare a textbook that willhelp the reader enjoy its study. Of course, one would naturally hope that an author would

write a book only about something he or she feels passionate about. I often heard somestudents describe general topology as being “hard”. Well, some of it is. But I have often

thought that maybe this perception was developed because they did not approach it quite inthe right way. In this text, we try to improve on the ways that students are introduced to it.

But first, I should at least write a few words about the mathematical content of thistextbook. The choice of content as well as the order and pace of the presentation of the con-cepts found in the text were developed with senior math undergraduate or math graduate

students in mind. The targeted reader will have been exposed to some mathematical rigorto a level normally found in an introduction to mathematical analysis texts or as presented

in an introduction to linear algebra or abstract algebra texts. The first two sections ofPart I consist mostly of a review in the form of a summarized presentation of very basic

ideas on normed vector spaces and metric spaces. These are meant to ease the reader intothe main subject matter of general topology (in chapter 3 to 20 of Parts II to VI). Parts

II to VI normally form the core material contained in most, one or two semester, BasicGeneral Topology course. Once we have worked through the most fundamental concepts

of topology in chapters one to twenty, the reader will be exposed to brief introductions tomore specialized specialized or advanced topics. These are presented in Part VII in theform of a sequence of chapters many of which can be read or studied, independently, or in

short sequences of two or three chapters, provided the student has mastered chapters 3 to 20.

Each chapter is followed by a list of Concepts review type questions. These questions

highlight for students the main ideas presented in that section and will help test theirunderstanding of these concepts. The answers to all Concept review questions are in the

ii

main body of the text. Attempting to answer these questions will help the student discoveressential notions which are often overlooked when first exposed to these ideas. Reading

a section provides a certain level of understanding, but answering questions, even simpleones, related to its content requires a much deeper understanding. The efforts required in

answering correctly such questions leads the student to the ability to solve more complexproblems in the Exercise sections. If the student desires a more in-depth study of a topic

in Part VII, there are many excellent topology books that can satisfy this need.

Textbook examples will serve as solution models to most of the exercise questions atthe end of each section.

In certain sections, we make use of elementary set theory. A student who feels a bitrusty when facing the occasional references to set theory notions may want to review some

of these. For convenience, a summary of the main set theory concepts appear at the endof the text in the form of an appendix to the book. A more extensive coverage of naive set

theory is offered in the book “Axioms and Set theory” by this writer. It is highly recom-mended and will serve as an excellent companion to this book.

As we all know, any textbook, when initially published, will contain some errors, some

typographical, others in spelling or in formatting and, what is even more worrisome, somemathematical. Critical or alert readers of the text can help weed out the most glaring

mistakes by communicating suggestions and comments directly to the author. This will bemuch appreciated by this writer as well as by future readers.

Robert AndreUniversity of Waterloo, Ontario

[email protected]


ISBN 978-0-9938485-1-3

Contents

I Norms and metrics

1 Norms on vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Metrics on sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

II Topological spaces: Fundamental concepts 27

3 A topology on a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 Set closures, interiors and boundaries. . . . . . . . . . . . . . . . . . . . . . 48

5 Bases of topological spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . 696 Continuity on topological spaces . . . . . . . . . . . . . . . . . . . . . . . . 95

7 Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1168 The quotient topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

III Topological spaces: Separation axioms 169

9 Separation with open sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

10 Separation with continuous functions . . . . . . . . . . . . . . . . . . . . . . 194

IV Limit points in topological spaces 225

11 Limit points in first countable spaces . . . . . . . . . . . . . . . . . . . . . . 22712 Limit points of nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

13 Limit points of filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

V Compact spaces and relatives 275

14 Compactness: Definition and basic properties . . . . . . . . . . . . . . . . . 277

15 Countably compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29916 Lindelof spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

17 Sequentially and feebly compact spaces. . . . . . . . . . . . . . . . . . . . . 31818 Locally compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

v

19 Paracompact topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . 345

VI The connected property 359

20 Connected spaces and properties . . . . . . . . . . . . . . . . . . . . . . . . 361

VII Topics 389

21 Compactifications of completely regular spaces . . . . . . . . . . . . . . . . 391

22 Singular sets and singular compactifications . . . . . . . . . . . . . . . . . . 42223 On C-embedded subsets of S . . . . . . . . . . . . . . . . . . . . . . . . . . 441

24 Realcompact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46525 Perfect functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480

26 Perfect and Freudenthal compactifications . . . . . . . . . . . . . . . . . . . 49327 Spaces whose elements are sequences . . . . . . . . . . . . . . . . . . . . . . 503

28 Completing incomplete metric spaces . . . . . . . . . . . . . . . . . . . . . . 52029 The uniform space and the uniform topology . . . . . . . . . . . . . . . . . 52830 The Stone-Weierstrass theorem . . . . . . . . . . . . . . . . . . . . . . . . . 550

31 Metrizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55832 The Stone space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568

33 Baire spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585

Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593

Bibliography. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 628

Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631

Part I

Norms and metrics

Part I: Norms and metrics 1

1 / Norms on vector spaces.

Summary. In this section, we review a few basic notions about inner productson vector spaces and how they are used as a mechanism to construct a distance

measuring tool called “norm”. We then define “norm on an abstract vectorspace” with no reference to an inner product. We show how to distinguish be-tween norms that are induced by an inner product and those that are not. We

then provide a few examples of norms on vectors spaces of continuous functions,C[a, b]. We end this section with a formal definition of the compact property in

normed vector spaces.

1.1 Review of inner product spaces.

We begin by reviewing a view basic notions about those vector spaces we refer to as

inner product spaces. Recall that the set Rn = (x1, x2, x3, ..., xn) : xi ∈ R equippedwith two operations, addition and scalar multiplication, is known to be a vector space.

We can also equip a vector space with a third operation called “dot product” whichmaps pairs of vectors in Rn to some real number. The dot-product is a specific real-

valued operation on Rn which belongs to a larger family of vector space operationscalled inner products. We briefly remind ourselves of a few facts about inner productson abstract vector spaces.

Definition 1.1 Let V be a vector space over the reals. An inner product is an operationwhich maps pairs of vectors in V to a real number. We denote an inner product on V by

< ~v, ~w >. A real-valued function on V × V is referred to as an inner product if and only ifit satisfies the following 4 axioms :

IP1 The number < ~v, ~v > is greater than or equal to 0 for all ~v in V . Equality holds ifand only if ~v = 0. (Hence, if ~v is not 0, < ~v, ~v > is strictly larger than 0)

IP2 For all ~v, ~w ∈ V , < ~u,~v >=< ~v, ~u > (Commutativity)

IP3 For all ~u, ~v, ~w ∈ V , < ~u+ ~w,~v >=< ~u,~v > + < ~w,~v >

IP4 For all ~u, ~v ∈ V , and α ∈ R, < α~u, ~v >= α < ~u, ~v >

A vector space V is called an inner product space if it is equipped with some specified inner

product.

2 Section 1: Norms on vector spaces

Definition 1.2 Let V be an inner product space. If ~v is a vector in V we define

‖~v‖ =√

< ~v, ~v >

The expression ‖~v‖ is called the norm (or length) of the vector ~v induced by the innerproduct < ~u,~v > on V .

Example 1 . It is easily verified that, for ~x = (x1, x2, x3, ..., xn) and ~y = (y1, y2, y3, ..., yn)in Rn, its well-known dot-product

< ~x, ~y >= ~x · ~y = x1y1 + x2y2 + · · ·+ xnyn

satisfies the four axioms above and so is an inner product on the vector space Rn.This is often referred to as the Euclidean inner product or the standard inner product

on Rn. The norm on Rn induced by the dot product is

‖~x‖ = ‖(x1, x2, x3, . . . , xn)‖=

√

(x1, x2, x3, . . . , xn) · (x1, x2, x3, . . . , xn)

=

√

√

√

√

n∑

i=1

x2i

This particular norm is referred to as the Euclidean norm on Rn. It is also referred to

as the l2-norm on Rn, in which case, it will be represented as ‖~x‖2. We will use thisparticular norm to measure distances between vectors in Rn. That is, the distance

between ~x = (x1, x2, x3, . . . , xn) and ~y = (y1, y2, y3, ..., yn) is defined to be

‖~x− ~y‖ =

√

√

√

√

n∑

i=1

(xi − yi)2

In the case where n = 2 or 3, this represents the usual distance formula betweenpoints in 2-space and 3-space, respectively. In the case where n = 1, it represents the

absolute value of the difference of two numbers.

Example 2 . Consider the vector space, V = C[a, b], the set of all continuous real-valued functions on the closed interval [a, b] equipped with the usual addition and

scalar multiplication of functions. We define the following inner product on C[a, b] as:

< f, g >=

∫ b

af(x)g(x) dx


Showing that this operation satisfies the inner product axioms IP1 to IP4 is left as anexercise. In this case, the norm of f , induced by this inner product, is seen to be

‖f‖ =

√

∫ b

af(x)2 dx

It is also referred to as the L2-norm on C[a, b] and we represent it as ‖f‖2.

The following theorem called the Cauchy-Schwarz inequality offers an important prob-

lem solving tool when working with inner product spaces. The norm which appearsin the inequality is the one induced by the given inner product. This inequality holds

true for any well-defined inner product on a vector space.

Theorem 1.3 Cauchy-Schwarz inequality. Let V be a vector space equipped with an inner

product and its induced norm. Then, for vectors ~x and ~y in V ,

| < ~x, ~y > | ≤ ‖~x‖‖~y‖

Equality holds true if and only if ~x and ~y are collinear (i.e., ~x = α~y or ~y = α~x).

Proof: The statement clearly holds true if ~y = 0.

Let ~x, ~y be two (not necessarily distinct) vectors in V where ~y 6= 0. For any realnumber t,

0 ≤ < ~x− t~y, ~x− t~y >

= ‖~x‖2 − 2t < ~x, ~y > +t2‖~y‖2

Choosing

t =< ~x, ~y >

‖~y‖2

in the above equation we obtain

0 ≤ ‖~x‖2 − < ~x, ~y >2

‖~y‖2

The inequality, | < ~x, ~y > | ≤ ‖~x‖‖~y‖, follows.


We now prove the second part of the statement. If ~x and ~y are collinear, say ~x = α~y,then

| < ~x, ~y > | = | < α~y, ~y > |= |α| < ~y, ~y >

= |α|‖~y‖‖~y‖= ‖α~y‖‖~y‖= ‖~x‖‖~y‖

Conversely, suppose | < ~x, ~y > | = ‖~x‖‖~y‖. If ~y is zero then ~0 = t~y and so ~x and ~y are

collinear. Suppose ~y 6= ~0. Consider t = <~x,~y>2

‖~y‖2 .

< ~x− t~y, ~x− t~y > = ‖~x‖2 − < ~x, ~y >2

‖~y‖2(As described above)

= 0

So, by IP1, < ~x− t~y, ~x− t~y >= 0 implies ~x− t~y = 0 so ~x and ~y are collinear. We have

shown that equality holds if and only if ~x and ~y are collinear.

We now verify that a norm, ‖ ‖, which is induced by an inner product < , > on thevector space V will always satisfy the following three fundamental properties:

1) For all ~x ∈ V , ‖~x‖ ≥ 0, equality holds if and only if ~x = 0.

2) For all ~x ∈ V and scalar α ∈ R, ‖α~x‖ = |α|‖~x‖,3) For all ~x, ~y ∈ V , ‖~x+ ~y‖ ≤ ‖~x‖+ ‖~y‖

The first property follows from the fact that the norm is defined as being the squareroot of a number. The second property follows from the straightforward argument:

‖α~x‖ =√

< α~x, α~x >

=√α2√

< ~x, ~x >

= |α|‖~x‖

The third property is referred to as the triangle inequality. It’s non-trivial proof in-vokes the Cauchy-Schwarz inequality. We prove it below.

Corollary 1.4 The triangle inequality for norms induced by inner products. For any pairof vectors ~x and ~y in an inner product space, ‖~x+ ~y‖ ≤ ‖~x‖ + ‖~y‖. If equality holds thenthe two vectors ~x and ~y are collinear.


Proof:

‖~x+ ~y‖2 = < ~x+ ~y, ~x+ ~y >

= ‖~x‖2 + 2 < ~x, ~y > +‖~y‖2

≤ ‖~x‖2 + 2| < ~x, ~y > | + ‖~y‖2

≤ ‖~x‖2 + 2‖~x‖‖~y‖ + ‖~y‖2(Cauchy-Schwarz)

= (‖~x‖ + ‖~y‖)2

Thus ‖~x+ ~y‖ ≤ ‖~x‖ + ‖~y‖, as required.

In the case where we have equality : Suppose we have ‖~x + ~y‖ = ‖~x‖ + ‖~y‖. Then

‖~x + ~y‖2 = (‖~x‖ + ‖~y‖)2. From the development above, the inequalities must beequalities, so we must have < ~x, ~y >= ‖~x‖‖~y‖. This means | < ~x, ~y > | = ‖~x‖||~y‖. Bythe Cauchy-Schwarz theorem, ~x and ~y are collinear.

1.2 Norms on arbitrary vector spaces.

We now show that norms on a vector space can exist independently from inner prod-ucts.

Definition 1.5 Let V a vector space over the reals. A norm on V is a function, ‖ ‖ : V → R,which satisfies the three following norm axioms:

N1 For all ~v ∈ V , ‖~v‖ ≥ 0. The equality ‖~v‖ = 0 holds true if and only if ~v = 0. Hence if~v is not ~0, ‖~v‖ > 0.

N2 For all ~v ∈ V , α ∈ R, ‖α~v‖ = |α|‖~v‖

N3 For all ~v, ~u ∈ V , ‖~u+ ~v‖ ≤ ‖~u‖ + ‖~v‖ (Triangle inequality)

A vector space V is called a normed vector space if it is equipped with some specified normsatisfying these three norm axioms. It is denoted by (V, ‖ ‖). Note that the vector space

V need not be an inner product space, nor need there be any relationship between ‖ ‖ andsome inner product.

This function, ‖ ‖ : V → R, will be used to measure the “distance”, ‖~x− ~y‖, betweenany two vectors ~x and ~y. Note that ‖~x−~y‖ = ‖(−1)(~y−~x)‖ = |−1|‖~y−~x‖ = ‖~y−~x‖.That is, “the distance between ~x and ~y is the same as the distance between ~y and ~x”.


We provide a few examples of functions known to be norms.

Example 1. The norm ‖ ‖ induced by the inner product of an inner product space

(V, < , >) has already been shown to satisfy the axioms N1, N2 and N3.

Example 2. The 1-norm on Rn is defined as

‖~x‖1 = ‖(x1, x2, x3, . . . , xn)‖1 =

n∑

i=1

|xi|

It can be shown to satisfy the three norm axioms. (Left as an exercise)

Example 3. The ∞-norm on Rn is defined as

‖~x‖∞ = ‖(x1, x2, x3, . . . , xn)‖∞ = max|x1|, |x2|, · · · , |xn|

It can be shown to satisfy the three norm axioms. (Proving that N1 and N2 are

satisfied is left as an exercise)

We prove that the ∞-norm on Rn satisfies the triangle inequality N3.

Proof of ‖~x+~y‖∞ ≤ ‖~x‖∞+‖~y‖∞: Note that |xi| ≤ maxi=1..n|xi| for each i, and|yi| ≤ maxi=1..n|yi| for each i Then for each i = 1 to n,

|xi + yi| ≤ |xi|+ |yi|≤ max

i=1..n|xi| + max

i=1..n|yi|

= ‖~x‖∞ + ‖~y‖∞

Then max1..n

|xi −yi| ≤ ‖~x‖∞+‖~y‖∞ and so ‖~x+~y‖∞ ≤ ‖~x‖∞+‖~y‖∞, as required.

Example 4. Let p ≥ 1. The p-norm on Rn is defined as

‖~x‖p = ‖(x1, x2, x3, . . . , xn)‖p =

(

n∑

i=1

|xi|p)1/p

It can be shown that this function satisfies the three norm axioms. We will not provethis for p in general.

Note that when p = 2 the p-norm is simply the Euclidean norm on Rn. Since theEuclidean norm is induced by an inner product it automatically satisfies the three

norm axioms. Proving that, for any p ≥ 1, the p-norm satisfies N1 and N2 is straight-forward. But proving that the triangle inequality holds true for all p ≥ 1 is not easy.


The interested readers can look the proof up in most Real Analysis texts or find itonline.

A natural question comes to mind . Are all norms on a vector space V induced by

some inner product? The answer is no!

The following theorem tells us how to recognize those norms which are induced bysome inner product:

Theorem 1.5.1 Suppose ‖ ‖ is a norm on a vector space V . There exists an inner

product < , > such that ‖~x‖ =√< ~x, ~x >, for all ~x ∈ V , if and only if, for all ~x, ~y ∈ V ,

the norm ‖ ‖ satisfies the parallelogram identity

2(‖~x‖2 + ‖~y‖2) = ‖~x+ ~y‖2 + ‖~x− ~y‖2

Proof : The proof involves showing that, if ‖ ‖ satisfies the parallelogram identity thenthe identity

< ~x, ~y >=1

4

(

‖~x+ ~y‖2 − ‖~x− ~y‖2)

is a valid inner product on V which induces ‖ ‖. The proof is routine and so is notpresented here.

Examples of norms on C[a, b].

Example 1. The Lp-norm. Let V = C[a, b] denote the family of all real-valuedcontinuous functions on the closed interval [a, b] equipped with the usual + and scalar

multiplication. If p ≥ 1, then we define the Lp-norm on C[a, b] as follows:

‖f‖p =

(∫ b

a|f(x)|p dx

)1/p

In the case where p = 1 we have

‖f‖1 =

∫ b

a|f(x)| dx

which is easily shown to satisfy the three norm axioms. In the case where p = 2, then

this norm is one which is induced by the inner product

< f, g >=

∫ b

a

f(x)g(x) dx

on C[a, b].


It is straightforward to show that, for all p ≥ 1, ‖f‖p satisfies N1 and N2. But provingthat ‖f + g‖p ≤ ‖f‖p + ‖g‖p, for all p ≥ 1, is difficult. This inequality is referred

to as the Minkowski inequality. The interested readers will find a proof in most RealAnalysis texts or online.

Example 2. The sup-norm: We define another norm on the vector space, C∗(S), of allbounded continuous real-valued functions on the space S. Recall that the supremumof a subset A of an ordered set S, written as “sup A” is the least upper bound of A

which is contained in S. Note that sup A may or may not belong to A. We define thesup-norm on C∗(S) as

‖f‖∞ = sup |f(x)| : x ∈ SThe sup-norm is also referred to as the “infinity-norm” or “uniform norm”. Showingthat this is a valid norm is left as an exercise. If a sequence, T = fn : n ∈ N, in

C∗(S) converges to f with respect to the sup-norm, ‖ ‖∞, then we will say that

“T converges uniformly to f ”

and that f is the “uniform limit” of the sequence T .

Suppose C∗(S) is given for some space S and Tx : x ∈ S is a family of sequences inR where,

Tx = fn(x) : n ∈ Neach of which converges (pointwise) to some number ux. Suppose that, for a givenε > 0, there exists an N > 0 such that n > N implies |ux − fn(x)| < ε, independent

of the value of x. Define the function, f : S → R as f(x) = ux. Then, when N > 0,

sup |fn(x) − f(x)| : x ∈ S = ‖fn − f‖∞ ≤ ε

so fn : n ∈ N converges uniformly to f . This provides another way of recognizingthe uniform limit of a sequence of functions. That is, we show that, the existence of

an N depends only on the value of ε and not the particular value of x.

Example 3. Show that the uniform limit, f , of a sequence, fn, in C∗(S), is contin-uous on S and so belongs to C∗(S).

Solution : Let T = fn : n ∈ N be a sequence in C∗(S) which converges uniformly to

f . Let a ∈ S and ε > 0. It suffices to show that there is a δ > 0 such that ‖x−a‖ < δimplies that |f(x) − f(a)| < ε. See that, there is N > 0 such that, n > N implies

|f(x) − f(a)| ≤ |f(x)− fn(x)|+ |fn(x)− fn(a)| + |fn(a) − f(a)|< ‖fn − f‖∞ + |fn(x)− fn(a)|+ ‖fn − f‖∞< ε/3 + |fn(x) − fn(a)|+ ε/3

For n > N , fn is continuous so there exists δ > 0 such that |x − a| < δ implies|fn(x)− fn(a)| < ε/3.

Then |x− a| < δ implies |f(x)− f(a)| < ε. So f is continuous on S.


1.3 Convergence and completeness in a normed vector space.

Convergence of sequences forms and important part of analysis. In what follows,

norms on Rn are always assumed to be the Euclidean norm, unless otherwise stated.An infinite sequence in a normed vector space, V , is a function which maps N (not

necessarily one-to-one) onto a subset, S, of V . This means that a sequence is theindexation of the elements of a countable subset, S, by using the natural numbers.

For example, S = ~x0, ~x1, ~x2, ..., ~xn, ..., . Note that some of these may be repeated(since the function mapping N into S need not be one-to-one). We generalize the

notion of a “convergent sequence and its limit” in R to a “convergent sequence andits limit” in a normed vector space (V, ‖ ‖).

Definition 1.6 Let (V,+, α, ‖ ‖) be a vector space equipped with a norm ‖ ‖. We say asequence of vectors ~xn in V converges to a vector ~a in V with respect to the norm ‖ ‖ if

For any ε > 0, there exists an integer N > 0 such that ‖~xn − ~a|| < ε whenevern > N .

If the sequence of vectors ~xn converges to the vector ~a, then we write

limn→∞

~xn = ~a

The following proposition lists the standard properties respected by limits in normedvector spaces.

Proposition 1.7 Let ~xn and ~yn be two sequences in the normed vector space V .

Suppose ~xn converges to ~a. Let α and β be scalars.

a) limn→∞ (α~xn ± β~yn) = α limn→∞ ~xn ± β limn→∞ ~yn.

b) (limn→∞ ~xn = ~a) ⇔ (limn→∞ ‖~xn −~a‖ = 0)

c) If ~x = (x1, x2, x3, . . . , xn) ∈ Rn then |xi| ≤ ‖~x‖2, for i = 1 to n.

d) A sequence of vectors ~xk : k = 1, 2, 3, . . . , in Rn converges to the vector ~a =

(a1, a2, a3, . . . , an) (with respect to the Euclidean norm) if and only if the componentsof the vectors in ~xk converge to the corresponding components of ~a (with respect

to the absolute value).

Proof: The proof is omitted.


We now define those sequences in a normed vector space called Cauchy sequences. As

we shall soon see all convergent sequences of (V, ‖ ‖) must be Cauchy sequences, butsome “subsets” S of V have Cauchy sequences which do not converge in S.

Definition 1.8 A sequence of vectors ~xn in a normed vector space V is said to be a

Cauchy sequence in V , or simply said to be Cauchy, if for any ε > 0, there exists an integerN > 0 such that ‖~xm − ~xn‖ < ε whenever m, n > N .

At first glance, the reader may feel that a Cauchy sequence is just another way of

referring to a “convergent sequence”. In many situations this is indeed the case. Butthere are sets, S, containing Cauchy sequences which do not converge to a point inside

S. For example, we may refer to the sequence 1, 1/2, 1/3, . . . , 1/n, . . . , as a sequencein the vector space R, or, as a sequence in the subset S = x : 0 < x ≤ 1 where S

inherits the absolute value norm from R. The given sequence certainly converges to0 in R and so can easily be proven to be Cauchy in R. Using the same norm on S

we must conclude that it is also Cauchy in S. However, there is one property of Swhich clearly distinguishes it from R: It is that R contains the limit of this Cauchysequence while S does not contain the limit of this same sequence. So we will use

Cauchy sequences to help us categorize subsets of a vector space based on whetherthey contain the limits of all their Cauchy sequences, or not.

This motivates the following definition of complete subsets of a normed vector space.

Definition 1.9 A subset S of a normed vector space V is a complete subset of V if every

Cauchy sequence in S converges to a vector in S.

Definition 1.9.1 A complete normed vector space is referred to as Banach space.

Definition 1.9.2 If V is a complete normed vector space whose norm is induced by aninner product then V is referred to as a Hilbert space. That is, if (V,+α,<>, ‖ ‖) is

complete then V is a Hilbert space.

Even if all Hilbert spaces are Banach spaces not all Banach spaces are Hilbert spaces.


Theorem 1.10 Completeness of (R, | |). The space (R, | |) is complete with respect to | |.

Proof: Let T = xn : n ∈ N be a Cauchy sequence in R.

We claim that T is bounded: For ε = 1, there exists N such that n,m > N implies|xn − xm| < 1. Then T is bounded by max |a0, |a1|, |a2|, . . . , |aN | + 1. So T is

bounded as claimed.

We are required to show that T converges to a point in R. By the Bolzano-Weierstrass

theorem1 T has a subsequence M = xnk: k ∈ N which converges to, say, L ∈ R.

We claim that T must also converge to L: There exists N such that n,m > N implies|xn − xm| < ε/2. We can pick K > 0 such that |xnk

−L| < ε/2 whenever k > K. We

can choose M = max K,N. Since nk ≥ n (by definition of subsequence), if n > M ,

|xn − L| ≤ |an − ank| + |ank

− L| < ε/2 + ε/2 = ε

So T converges to L, as required.

Theorem 1.11 Completeness of (Cb(S), ‖ ‖∞). Let S be a space and Cb(S) be the set ofall bounded real-valued functions on S equipped with the sup-norm, ‖ ‖∞. Then Cb(S) is

a complete space with respect to ‖ ‖∞.

Proof: Let fn : n ∈ N be a Cauchy sequence in Cb(S). We are required to show thatfn : n ∈ N converges to a function f in Cb(S) with respect to ‖ ‖∞.

Let x ∈ S. Then

|fn(x)− fm(x)| ≤ ‖fn − fm‖∞ < ε for all m, n > N

Then U = fn(x) : n ∈ N is Cauchy in R. Since R is complete (see theorem above) Uconverges to, say, ux. For each x ∈ S define the function, f : S → R, as f(x) = ux ∈ R.

Then fn : n ∈ N converges pointwise to f(x). For m ≥ N ,

|f(x)− fm(x)| = | limn→∞

f(x) − fm(x)|= lim

n→∞|f(x)− fm(x)|

≤ limn→∞

‖fn − fm‖∞≤ ε

N independent of the value of x in S. Then ‖f − fn‖∞ ≤ ε. Therefore, f is the

uniform limit of the Cauchy sequence, fn : n ∈ N. Then f ∈ Cb(S). So Cb(S) iscomplete.

1Bolzano-Weierstrass: Every bounded sequence in R has a convergent subsequence.


We will not dwell on the notions of convergence and completeness of subsets at thistime. We will return to discuss these concepts in the more general contexts of metric

spaces and topological spaces.

1.4 The compact property on normed vector spaces.

We will present immediately the formal definition of the compact property which ap-plies mostly when we confine ourselves to the universe of normed vector spaces. This

is for future reference when, at a point in this book, we will reintroduce the notion ofcompactness, but in a much more general context of topological spaces. This defini-tion will be followed by an important characterization of the compact property which

remains valid as long as we confine ourselves to finite dimensional normed vectorspaces. The proof is non-trivial and so is omitted here. The complete proof is usually

presented in texts which serve as an introduction to Real Analysis courses rather thanin a topology text. One of the main reason for an early introduction is to access an

important “extreme value theore” which states that “A continuous function on a com-pact subset, T , of a normed vectors spaces, V , attains its maximum value at a point

inside T”. The Heine-Borel theorem makes it very convenient, in such situations, torefer to compact subsets simply as being those that are “closed and bounded”. In

abstract topological spaces, we will not have access to this tool, since the Heine-Boreltheorem applies to finite dimensional normed vector spaces.

Definition 1.12 A subset, T , of a normed vector space, V , is said to be compact 1 if and

only if every sequence, xi, in T has a subsequence, xf(i), which converges to some point,p, which belongs to T .

Theorem 1.13 The generalized Heine-Borel theorem. Subsets of a finite dimensionalnormed vector space are compact if and only if they are both closed and bounded.

Proof: The proof is omitted.

EXERCISES

1When we will revisit the property of compactness in our study of topological spaces we will replace theword “compact” with the words “sequentially compact”.


1. Prove that, if ~x and ~y are vectors in Rn, then | ‖~x‖ − ‖~x− ~y‖ | ≤ ‖~y‖.

2. Exercise questions on norms.

a) Consider the vector space, C[a, b], of all continuous functions on the interval

[a, b]. Recall that ‖ ‖1 : C[a, b] → R is defined as

‖f‖1 =

∫ b

a|f(x)|dx

Show that ‖ ‖1 is a valid norm on C[a, b].

b) Let fn : n ∈ N, n 6= 0 be a sequence of constant functions defined as fn(x) =2 + 1

n for each n. That is, f1(x) = 3, f2(x) = 5/2, f3(x) = 7/3, and so on. Iffn is viewed as a subset of C[−1, 2] equipped with the norm ‖ ‖1 determine,

limn→∞

∫ 2

−1|fn(x)| dx

Justify all your steps carefully.

3. Recall that C[a, b] is the vector space of all continuous functions on the interval [a, b].

a) Show that < f, g >=∫ ba f(x)g(x) dx is a valid inner product.

b) Let ‖ ‖ be the norm on C[0, 1] which is induced by the inner product given inpart a) of this question. Compute ‖ex‖.

4. Recall the definition of “p-norm”: ‖~x‖p = (∑n

i=1 |xi|p)1/p, on Rn.

a) Compute ‖(−1, 0, 1)‖3.

b) Holder’s inequality on Rn says that “if p, q ≥ 1 and 1/p + 1/q = 1, then

‖(x1y1, x2y2, . . . , xnyn)‖1 =∑n

i=1 |xiyi| ≤ ‖~x‖p‖~y‖q”. Note that p and q arenot assumed to be integers; Holder’s inequality holds true as long as the condi-

tion “1/p+1/q = 1” is satisfied. Although it is difficult to see why this inequalityis of any interest or is of any value at this point, we will assume it holds true

and use it to practice using simpler concepts. Show that the Cauchy-Schwarzinequality on Rn, | < ~x, ~y > | ≤ ‖~x‖2‖~y‖2, (where < ~x, ~y > is the Euclidean inner

product) follows from the Holder’s inequality on Rn.

c) Holder’s inequality on C[a, b] says that “if p, q ≥ 1 and 1/p + 1/q = 1, then‖fg‖1 ≤ ‖f‖p‖g‖q”. Show that the Cauchy-Schwarz inequality on C[a, b],

| < f, g > | ≤ ‖f‖2‖g‖2

follows from the Holder’s inequality on C[a, b].


5. Invoke the Cauchy-Schwarz inequality to show that

−[√

2 − 1

2

]1/2

≤∫ π/4

0

√cosx

√sinx dx ≤

[√2− 1

2

]1/2

6. Let f be a function in C[a, b]. Show that limp→∞ ‖f‖p ≤ ‖f‖∞.

7. A subset S of a normed vector space is bounded if there exists a positive number M

such that ‖~x‖ < M for all ~x in S. Show that any Cauchy sequence ~xn in a normedvector space (V, ‖ ‖) when viewed as a set is bounded.

8. Suppose (V, ‖ ‖) is a normed vector space and ~xn is a sequence in V which does notconverge to ~0. If δ ∈ R, let Bδ(~0) = ~x ∈ V : ‖~x‖ < δ. Show that there exists some

δ > 0 and a subsequence xni∞i=1 of ~xn such that ~xni ∩Bδ(~0) = ∅.

9. Convergence of vectors in a normed vector space.

a) Review the definition of “convergence of a sequence of vectors in a normed vectorspace”. Let (V, ‖ ‖) be an abstract normed vector. Suppose V contains a sequence

of vectors, ~un : n ∈ N, which converges to some vector in V . Show that

limn→∞

‖~un‖ = ‖ limn→∞

~un‖

Hence, whenever the sequence, ~un : n ∈ N, converges in V then so does thesequence, ‖~un‖ : n ∈ N, of real numbers.

b) Let ~v ∈ R2. Construct a sequence ~un in R2 such that limn→∞ ‖~un‖ = ‖~v‖where limn→∞ ~un 6= ~v.

c) Suppose ~xn and ~un are two sequences in Rn which both converge to the

same point ~y. Show that the sequence ‖~xn − ~un‖ must converge to 0.

10. Let ~xn : n ∈ N and ~yn : n ∈ N be sequences in a normed vector space (V,+, α, ‖ ‖).Let αn : n ∈ N be a sequence of real numbers.

a) Show that, if ~xn : n ∈ N and ~yn : n ∈ N converge to ~x and ~y, respectively,then the sequence ~xn + ~yn : n ∈ N converges to ~x+ ~y.

b) Show that, if ~xn : n ∈ N converges to the vector ~x and αn : n ∈ N convergesto the real number α then the sequence of vectors, αn~xn : n ∈ N, converges to

α~x.

11. Suppose (V, ‖ ‖) is a normed vector space and ~xn is a Cauchy sequence in V . Showthat if ~xn has a convergent subsequence, say ~xni∞i=1, then ~xn converges in V .


12. Let δ > 0. Suppose (V, ‖ ‖) is a normed vector space and ~xn is a sequence in Vsuch that ‖~xn‖ > δ for all n. Show that if ~xn is Cauchy then so is the sequence

xn‖xn‖

on B = ~x ∈ V : ‖~x‖ = 1.

13. Suppose (V, ‖ ‖) is a normed vector space and B = ~x ∈ V : ‖~x‖ = 1.

a) Show that if V is complete with respect to the norm ‖ ‖ then B is complete withrespect to the same norm inherited from V .

b) Show that if B is known to be complete with respect to the norm ‖ ‖, then V

must also be complete with respect to ‖ ‖.

14. Let M be an n-dimensional subspace of the inner product space (V, < , >, ‖ ‖).Prove that M is complete with respect to the norm || ||. Essentially this says that allsubspaces of finite dimensional inner product spaces are complete subsets.

15. Suppose (V, ‖ ‖) is normed vector space whose norm satisfies the parallelogram identity

2(‖~x‖2 + ‖~y‖2) = ‖~x+ ~y‖2 + ‖~x− ~y‖2

a) Show that < ~x, ~y >= 14 (‖~x+~y‖2−‖~x−~y‖2) (referred to as the polarizing identity)

satisfies the three inner product axioms IP1, IP2 and IP4.

b) Show that the polarizing identity in part a) satisfies the inner product property< ~x + ~y, ~z >=< ~x, ~z > + < ~y, ~z > hence the polarizing identity is a valid inner

product.

c) Show that this inner product induces the norm of V .

16. It is known that for any ~x ∈ Rn, ‖~x‖∞ = limp→∞ ‖~x‖p. Prove this for the case wheren = 2.

16 Section 2: Metric on sets

2 / Metrics on sets.

Summary. In this section, we define the concept of a “metric” on a set, giving

rise to “metric spaces”. The metric is defined as a tool for measuring distancesbetween points in the given set. Many metrics can be defined on a set as long

as they each satisfy three metric axioms. Metrics are then used to determinewhether a given sequence converges to a point or not. The subsets of a set called

“open ball” and “open sets” are then defined. Functions are then introduced tomap points from one metric space to another metric space. We are particularly

interested in those functions which are continuous on their respective domains.The topological version of the notion of continuity is presented in terms of “opensets”.

2.1 Measuring distance in arbitrary sets.

We now generalize a few notions of distances between vectors in a normed vectorspace to distances between points in an arbitrary set. Arbitrary sets, in their most

rudimentary form, do not usually appear with some algebraic structure defined onthem (such as vector spaces, for example, on which addition and scalar multiplication

are defined). When a set is equipped with addition and scalar multiplication at theonset, subtraction ~a−~b of two points ~a and ~b is easily given meaning. With a previ-

ously defined norm, we measured the length, ‖~a−~b‖, of the difference ~a−~b to obtainthe number which represents a notion of distance between ~a and ~b. This method is

inspired by the way we normally determine the distance between two real numbers,say −7 and 2.5, in the vector space, R, for example. In R, distance between points is

expressed by referring to the absolute value. To measure distances between points inarbitrary sets we will proceed differently. Given a set S, rather than defining a “norm”function ‖ ‖ : S → R on S, we will define a function, ρ : S×S → R, which maps pairs

of points x and y in S to a number which will represent the distance between thesetwo points in a that particular set. This function is what we will call a “metric” on

S. There will be certain restrictions on the properties possessed by ρ. We would, ofcourse, not want ρ to give us a distance x → y which is different from the distance

y → x. Also, we definitely do not want ρ to give a “negative” distance between twopoints. With different metrics, ρ1 and ρ2, the set, (S, ρ1), may be different in nature

from the set, (S, ρ2).

Definition 2.1 Let S be a non-empty set. A metric, ρ on S, is a function, ρ : S × S → R,which satisfies three metric axioms: .

Part I: Of norms and metrics 17

M1 For every x, y ∈ S, ρ(x, y) ≥ 0. Also,− Identity of indiscernibles: ρ(x, y) = 0 implies x = y.

− Indiscernibility of identicals: x = y implies ρ(x, y) = 0.

M2 Symmetry. For every x, y ∈ S, ρ(x, y) = ρ(y, x)

M3 Triangle inequality. For every x, y, z ∈ S, ρ(x, y) ≤ ρ(x, z) + ρ(z, y) (Triangle equality)

A set S equipped with a metric, ρ, is called a metric space. It is expressed as, (S, ρ)

Many metrics can be defined on a given set S. For a given set, S, if ρ1 and ρ2 aredifferent metrics then (S, ρ1) and (S, ρ2) are considered to be different metric spaces.

We provide a few examples.

Example 1. Let (V, ‖ ‖) be a normed vector space. Define ρ : V × V → R as

ρ(~x, ~y) = ‖~x−~y‖. The function, ρ, is easily seen to satisfy the two first metric axioms,M1 and M2. The triangle inequality for norms guarantees that M3 also holds true.

Then ρ, thus defined on V , is a metric induced by a norm and so (V, ρ) is a metricspace (It can simultaneously be viewed as a normed vector space depending on the

context.) When considering the normed vector space, (R2, ‖ ‖2), the metric inducedby the norm ‖ ‖2 is the distance formula: For ~x = (x1, x2) and ~y = (y1, y2),

ρ(~x, ~y) = ‖~x− ~y‖ =√

(x1 − y1)2 + (x2 − y2)2

This particular example shows that, given any normed vector space (V, ‖ ‖), we can

always express it as a metric space, (V, ρ), by defining ρ(~x, ~y) = ‖~x−~y‖. However, oneshould remember that a metric space (S, ρ) need not be a normed vector space, since

S itself need not be a vector space equipped with addition and scalar multiplication,both of which are necessary to define a norm on S.

Example 2. Let ρ : R2 × R2 → R be defined as:

ρ(~x, ~y) = ρ((x1, x2), (y1, y2)) = sup |x1 − y1|, |x2 − y2|

It is left to the reader to verify that this function satisfies all three metric axioms, andso is a valid metric on R2

Example 3. Let S be any non-empty set. The function ρ : S × S → R defined as

ρ(x, y) =

0 if x = y1 if x 6= y

can be verified to satisfy all three metric axioms and so is a valid metric on S. Anypair of distinct points are at a distance of one from each other while a point is at a


distance zero from itself. This metric is referred to as the discrete metric.

Example 4. Consider the set of all integers Z. Let d : Z × Z → N be defined as

k(x, y) = max 2n : 2n divides x− y

We define ρ : Z × Z → R as follows:

ρ(n,m) =

0 if n = m1

k(m,n) if m 6= n

We verify that ρ is indeed a metric on Z.

– It follows directly from the definition that ρ(n,m) is non-negative and ρ(n,m) =0 if and only if m = n. So ρ satisfies property M1.

– We see that 2n divides x−y if and only if 2n divides y−x. Then k(x, y) = k(y, x).Then, if m 6= n, ρ(m, n) = 1

k(m,n)= 1

k(n,m)= ρ(n,m). So ρ satisfies M2.

– Suppose ρ(m, n) = 1/2d and ρ(n, t) = 1/2e. Then 2d|(m − n) and 2e|(n − t).We will suppose, without loss of generality, that e ≤ d. It is easily verified that

2e|(m− t). It follows that k(m, t) ≥ 2e, hence

ρ(m, t) =1

k(m, t)

≤ 1

2e

= ρ(n, t)

≤ ρ(m, n) + ρ(n, t)

So ρ(m, t) ≤ ρ(m, n) + ρ(n, t). We have shown that ρ satisfies M3.

2.2 Metric subspaces.

We know that certain subsets of a vector space are referred to as “subspaces” pro-

vided they satisfy specific conditions. There are no required conditions on a subset,T , of (S, ρ) to be called a “metric subspace” provided we know what its metric will be.

Definition 2.1.1 Suppose (S, ρ) is a metric space and T ⊆ S. Then T can inherit themetric ρ of its superset, S, and declare itself to be a metric space (T, ρ|T), simply by

restricting the function ρ : S × S → R to ρ|T : T × T → R. In this case, (T, ρT) isreferred to as a metric subspace of (S, ρ). The metric, ρT , is referred to as the subspace


metric.

For example, suppose (R, ‖ ‖) is a normed vector space. Suppose ρ is the metric on R

induced by its norm, ‖ ‖, and T = (−3, 5] ⊂ R. Then (T, ρT) is a metric subspace of R.1

Of course, metric subspaces of a metric space S are themselves metric spaces.

2.3 Convergence and completeness in a metric space.

We now direct our attention to those subsets of the set S which are sequences. The

definition of a sequence, xn, does not involve the notions of “norm” or “metric”and so is precisely as we previously defined it in the section on normed vector spaces.

However convergence or divergence of a sequence depends very much on the tool weuse to measure distances in the set. The following definitions of limits and convergence

in a metric space are in many ways identical to those involving norms.

Definition 2.2 Let (S, ρ) be a metric space. We say that a sequence, xn, of points in Sconverges to the point, a, in S with respect to the metric ρ if and only if

for any ε > 0, there exists an integer N > 0 such that ρ(xn, a) < ε whenevern > N .

If the sequence of points xn converges to the point a (with respect to the metric ρ), then

we writelim

n→∞xn = a

Just as for limits in normed vector spaces, we can say that xn converges to the

point a with respect to a metric ρ in different ways:

limn→∞

xn = a ⇔ limn→∞

ρ(xn, a) = 0

Definition 2.2.1 Suppose (S, ρ) is a metric space, T ⊆ S and xn is a sequence in Swhich converges to the point a with respect to ρ. Then we will say that the point a

is a limit point of T .

1Of course, T is not a vector subspace of the vector space R.


Note that, if xn ⊆ T ⊆ (S, ρ) and a is the limit of this sequence with respect to ρ,this does not guarantee that a also belongs to T . Consider, for example, the metric

space (R, ρ) where ρ(x, y) = |x− y| and T = π + 1/n : n = 1, 2, 3, . . .. Then (T, ρ)is also a metric space. We see that the point π is clearly a limit point of T but π 6∈ T .

Just as in normed vector spaces, the notion of “Cauchy sequence” can also be definedin terms of metrics.

Definition 2.3 A sequence of points xn in a metric space (S, ρ) is said to be a Cauchysequence in S, with respect to the metric ρ (or simply said to be Cauchy) if and only if, for

any ε > 0, there exists an integer, N > 0, such that ρ(xm, xn) < ε, whenever m, n > N .

Example 1. Every convergent sequence is a Cauchy sequence. The following brief

argument confirms what we intuitively feel must be true: If xi converges to thepoint, a, with respect to the metric, ρ, then xi is a Cauchy sequence. To prove this

we let ε > 0. By hypothesis, there exists N > 0 such that n > N ⇒ ρ(xn, a) < ε/2.If m > N , then, invoking the metric axiom M3,

ρ(xm, xn) ≤ ρ(xm, a) + ρ(xn, a) < ε/2 + ε/2 = ε

So xi is Cauchy with respect to the metric ρ.

Definition 2.4 Let (S, ρ) be a metric space. The set S is complete with respect to themetric ρ if every Cauchy sequence in S converges to some point in S.

Example 2. Consider the set S = (0,∞). Let ρ1 denote the discrete metric on S andρ2 be defined as ρ2(x, y) = |x − y|. We compare the two metric spaces (S, ρ1) and

(S, ρ2).

Claim 1. The metric space, (S, ρ2), is not complete. See that the sequence1/n : n = 1, 2, 3, . . . , is a sequence in (R, ρ2) which converges to 0, with re-

spect to ρ2 and so is a Cauchy sequence in (R, ρ2). It is then a Cauchy sequenceinside S. Since the Cauchy sequence 1/n does not converge in S then S is not a

complete subset of S.


Claim 2. The metric space, (S, ρ1), is a complete metric space. Let xi be aCauchy sequence in S with respect to the discrete metric ρ1. Let ε = 1/2. Then

there exists N > 0 such that m, n > N implies ρ1(xm, xn) < ε = 1/2. Then ifm, n > N , ρ1(xm, xn) = 0. This means xn = xN+1, for all n > N . Then for any ε,

n > N ⇒ ρ1(xN+1, xn) = 0 < ε so ~xi converges to xN+1 ∈ S. We conclude that(S, ρ1) is a complete metric space.

Note that a metric space (S, ρ) may be a complete metric space and still have

proper subsets which are not complete metric subspaces.

2.4 Open and closed subsets of a metric space.

We now define a few fundamental subsets of metric spaces.

Definition 2.5 Let (S, ρ) be a metric space.

a) If y ∈ S, and ε > 0 we define an open ball of radius ε of center y as being the set

Bε(y) = x ∈ S : ρ(y, x)< ε

b) Let U ⊆ S. We say that U is an open subset in S if U is the union of open balls

each of which is entirely contained in U . That is, for each x ∈ U , there exists a realnumber number εx > 0 such that

U = ∪Bεx(x) : x ∈ U

Instead of saying “U is an open subset of S” we often simply say “U is open in S”.

c) Let F ⊂ S. We say that F is a closed subset of S if and only if every limit point of

F belongs to F . That is,

F = x ∈ S : x = limn→∞

xn for some sequence xn ⊆ F

Instead of saying “F is a closed subset of S” we often simply say “F is closed in S”.

Example 1. We define the metric ρ on R as ρ(x, y) = |x− y|. If U is the the interval(−5, 3) = x ∈ R : −5 < x < 3, then U is an open subset in R since, for every point

x ∈ U , we can find ε > 0 such that (x− ε, x+ ε) ⊆ (−5, 3). However, V = (−4, 3] isnot an open subset in R since V contains a point 3 such the open ball (3−ε, 3+ε) 6⊆ V

no matter how small we choose ε to be.


Example 2. We define the metric ρ on the set U = (−5, 3] as ρ(x, y) = |x−y|. Considerthe subset S = (0, 3]. The subset S is open in the metric space U since, for each y ∈ S

there is εy such that, S = ∪Bεy(y) : y ∈ S. Some readers might object, stating thatthere can be no ε such that Bε(3) ⊆ S. But we must be careful and see that we are

viewing S as a subset of the metric space U = (−5, 3]. See that, if we choose ε = 1/2,Bε(3) = x ∈ U : (3 − ε, 3 + ε) = (3− ε, 3] ⊆ (2, 3] is indeed entirely contained in S.

Example 3. We define the metric ρ on R2 as ρ(~x, ~y) = ‖~x − ~y‖2. The x-axis, U =(x, 0) : x ∈ R is closed in R2 since if ~a = (a, b) is a limit point of U then there must

exist a sequence (xn, 0) such that (a, b) = limn→∞(xn, 0) = (limn→∞ xn, limn→∞ 0);this implies b = 0, hence (a, b) ∈ U . It then follows that U is closed in R2

Example 4. Let S be an arbitrary non-empty set equipped with the discrete metricρ. For any x ∈ S, B1/2(x) = x ⊆ x hence x is an open subset of S. Let x be a

limit point of x. The only sequence in x is the constant sequence x, x, x, . . . , which converges to x ∈ x. Hence x is closed in S.

Remark . We can similarly define the notion of an “open subset S” of a normed vectorspace (V, ‖ ‖) by defining ρ(~x, ~y) = ‖~x− ~y‖, in which case U is open in (V, ‖ ‖) if and

only if U is open in (V, ρ). In (V, ‖ ‖)), the open ball, Bε(~y), centered at ~y is definedas Bε(~y) = ~x ∈ V : ‖~y − ~x‖ = ρ(~y, ~x) < ε. Similarly, if F ⊆ (V, ‖ ‖), F is said to

be closed in V if and only if F contains all its limit points with respect to the norm ‖ ‖.

Theorem 2.6 Let (S, ρ) be a metric space.

a) Both the empty set, ∅, and S are open subsets of S.

b) Both the empty set, ∅, and S are closed subsets of S.

c) Finite intersections of open subsets of S are open subsets of S.

d) Arbitrarily large unions1 of open subsets of S are open in S.

Proof: The proofs of all four parts are left as an exercise.

Suppose (S, ρ) is a metric space and T ⊆ S. Let ρT : T × T → R be the restriction of

the function ρ to T × T . Then (T, ρT) is a metric space. Or we simply say that T isa subspace of (S, ρ). One may wonder what relationship exists between the open sets

of the subspace T and the open sets of the space S. The following theorem answersthis question.

1The words “arbitrarily large unions” include the notion of the union of a countably or uncountablyinfinite number of sets. For example, ∪An : n ∈ N


Theorem 2.7 Suppose (S, ρ) is a metric space and T ⊆ S equipped with the metric ρT

inherited from the set S. Also suppose U ⊆ T ⊆ S. Then U is open in T (with respect tothe metric ρT ) if and only if there exists an open subset U∗ in S (with respect to the metric

ρ) such that U = U∗ ∩ T .

Proof: We are given that (S, ρ) is a metric space and T ⊆ S equipped with the metric ρT

inherited from the set S. Suppose U ⊆ T ⊆ S.

( ⇐ ) Suppose U∗ is open in S and U = U∗ ∩ T . If x ∈ U , then there exists ε > 0such that Bε(x) ⊆ S. Now

Bε(x) ∩ T = y ∈ T : ρ(x, y) < ε) = y ∈ T : ρT (x, y) < ε) = B′ε(x)

by definition, an open ball radius ε center x, in T . Since U = ∪B′ε(x) : x ∈ U thenU is open in T .

( ⇒ ) Suppose U is an open subset of T ⊆ S. If x ∈ U then there exists ε > 0 suchthat Bε(x) ⊆ U . Let B′ε(x) = y ∈ S : ρ(x, y) < ε. Then B′ε(x) is an open subset of

S such that B′ε(x) ∩ T = y ∈ T : ρT (x, y) < ε = Bε(x). Then

[∪B′ε(x) : x ∈ T] ∩ T = ∪B′ε(x) ∩ T : x ∈ T= ∪Bε(x) : x ∈ T= U

If U∗ = ∪B′ε(x) : x ∈ T, then U∗ is an open subset of S such that U = U∗ ∩ T .

Theorem 2.8 A subset F is a closed subset of a metric space (S, ρ) if and only if itscomplement, S\F , is an open subset of S.

Proof: Recall that closed subsets of S are those subsets F which contain all their limitpoints. So the statement says that F is closed in S if and only if S \F contains no

limit point of F . The proof is left as an exercise.

2.5 Characterizations of continuous functions on a metric space.

In what follows we study continuous functions mapping one metric space, (Sa, ρa) toanother metric space (Sb, ρb).


Definition 2.9 Let (Sa, ρa) and (Sb, ρb) be two metric spaces.

a) We say that a function f : Sa → Sb is continuous at the point u ∈ Sa if and only if itsatisfies the following condition:

For every ε > 0, there exist δ > 0 such that ρa(u, x) < δ ⇒ ρb(f(u), f(x))< ε.

b) If A ⊆ Sa, we say that a function f : A→ Sb is continuous on the set A if and only ifit is continuous at each point in A.

There are various ways of recognizing continuous functions on a subset T of a metricspace (S, ρ). The following theorem illustrates the most important ones.

Theorem 2.10 Let (Sa, ρa) and (Sb, ρb) be two metric spaces and f : Sa → Sb be a functionmapping the Sa into the set Sb. Then the following are equivalent:

1) The function f is continuous on S.

2) Whenever a sequence xn in Sa converges to a point u ∈ Sa, then the sequence

f(xn) in Sb converges to f(u) in Sb.

3) Suppose f [Sa] denotes the range of f on the domain Sa. For any open subset U off [Sa] ⊆ Sb, the subset f←[U ] = x ∈ Sa : f(x) ∈ U is open in Sa.

Proof: We are given two metric spaces, (Sa, ρa) and (Sb, ρb) and a function and f : Sa → Sb.

( 1 ⇒ 2 ) Suppose f : Sa → Sb satisfies the formal definition of continuity on Sa. Let

u ∈ Sa and ε > 0. There exists δ such that ρa(u, y) < δ ⇒ ρb(f(u), f(y)) < ε. Sup-pose xn is a sequence in Sa such that limn→∞ xn = u. Then the sequence f(xn)and the point f(u) are defined in Sb. We claim that f(xn) converges to f(u). Sincexn converges to u, there exist an N > 0 such that n > N implies ρa(u, xn) < δ.

Then, for n > N , ρb(f(u), f(xn)) < ε. It then follows that limn→∞ f(xn) = f(u), asrequired.

( 2 ⇒ 1 ) Suppose the function f : Sa → Sb is such that, for u ∈ Sa and xn ⊆ Sa,

limn→∞

xn = u ⇒ limn→∞

f(xn) = f(u)

Let ε > 0. Suppose f is not continuous at u. Then there exists ε > 0 such that forany δ there exists x ∈ Sa such that ρa(u, x) < δ and ρb(f(u), f(x))> ε. Then, for any

δn = 1n , there exists xn ∈ Sa such that ρa(u, xn) < δn = 1

n and ρb(f(u), f(xn)) > ε.We can then construct a sequence xn in Sa such that limn→∞ (xn − u) = 0 and


limn→∞ f(xn) 6= f(u), contradicting our hypothesis.1 The source of our contradictionis our supposition that f is not continuous at u. So f must be continuous at u, as

required.

( 1 ⇒ 3 ) Suppose f : Sa → Sb satisfies the formal definition of continuity on Sa. Let

u ∈ Sa and ε > 0. Let U be an open subset of f [Sa] ⊆ Sb. Let u ∈ f←[U ]. Thenf(u) ∈ U . Since U is open in f [Sa] there exists ε > 0 such that Bε(f(u)) ⊆ U where

Bε(f(u)) = x ∈ f [Sa] : ρb(f(u), x) < ε. Since f is continuous at u, there existsδ > 0 such that ρa(u, x) < δ ⇒ ρb(f(u), f(x)) < ε. This means that, if x ∈ Bδ(u),

then f(x) ∈ Bε(f(u)). Then

u ∈ Bδ(u) ⊆ f←[Bε(f(u))] ⊆ f←[U ]

So f←[U ] is open in Sa, as required.

( 3 ⇒ 1 ) Suppose f : Sa → Sb is a function on Sa such that whenever U is open in

f [Sa], f←[U ] is open in Sa. Let u ∈ Sa and ε > 0. By hypothesis, f←[Bε(f(u))] is open

in Sa and contains u. Then there exists δ > 0 such that u ∈ Bδ(u) ⊆ f←[Bε(f(u))].This means that if ρa(u, x) < δ then ρb(f(u), f(x))< ε, as required.

Note that the third characterization of a continuous function on a set, “U is open

⇒ f←[U ] is open”, is normally not used to determine continuity of f at a pointin the domain. This characterization “U is open ⇒ f←[U ] is open” is referred to

as the topological definition of continuity on a set while the second characterization“xn → a⇒ f(xn) → f(a)” is referred to as the sequential definition of continuityat a point.

EXERCISES

1. Prove parts a) to d) of theorem 2.6.

2. a) Show that the function which appears in Example 2 on page 17 is a valid metric.

b) Show that the function which appears in Example 3 on page 17 is a valid metric.

3. Define ρ : R×R → R as ρ(x, y) = |x−y|. Prove that, thus defined, ρ is a valid metricon R.

4. Let ρ1 and ρ2 be two metrics on the set S.

1Note that the axiom of choice is invoked here.


a) For a given k > 0 we define the function kρ : S×S → R as (kρ)(x, y) = k×ρ(x, y).Show that kρ is valid metric on S.

b) If ρ1 +ρ2 : S×S → R is defined as (ρ1 +ρ2)(x, y) = ρ1(x, y)+ρ2(x, y) show that

ρ1 + ρ2 is a valid metric on S.

c) If ρ : S × S → R is defined as ρ(x, y) = min1, ρ1(x, y) show that ρ is a validmetric on S.

5. Suppose xn is a sequence in a metric space (S, ρ). Show that if limn→∞ xn = a and

limn→∞ = b where both a and b belong to S then a = b.

6. We define two functions ρ1 : R2 × R2 → R and ρ2 : R2 × R2 → R as follows:

ρ1((a1, a2), (b1, b2)) = |a1 − b1| + |a2 − b2|ρ2((a1, a2), (b1, b2)) = max |a1 − b1|, |a2 − b2|

a) Show that ρ1 thus defined is a valid metric on R2.

b) Show that ρ2 thus defined is a valid metric on R2.

c) Show that limn→∞~xn = ~x ∈ R2 with respect to ρ1 if and only if limn→∞~xn =~x ∈ R2 with respect to ρ2.

7. Let a, b be distinct points in the metric space (S, ρ). Find disjoint open sets A and

B such that a ∈ A, b ∈ B and A ∩B = ∅.

8. Suppose g : S → R and h : S → R are two continuous functions on the metric space

(S, ρ). Show that the set U = x ∈ S : h(x) < g(x) is an open subset of M .

9. Prove that a finite subset of a metric space (S, ρ) is always closed in S.

10. Prove theorem 2.8: A subset F is a closed subset of a metric space (S, ρ) if and only

if its complement, S\F , is an open subset of S.

Part II

Topological spaces: Fundamentalconcepts

27

Part II: Topological spaces: Fundamental concepts 29

3 / A topology on a set

Summary. In this section we define the notion of a topological space. We

will begin by describing those families of subsets of a set which form a topology,τ , on a set, showing along the way how to recognize“open subsets” and “closedsubsets”. There can be many topologies on a set. Given a pair of topologies τ1and τ2 on S one can sometimes be seen as being “weaker” or “stronger” thanthe other. We have provided many examples both in the main body of the text

as well as in the given exercises.

3.1 Introduction.

In our review of metric spaces we realized, in theorem 2.10, that a function, f : S → Y ,from a metric space S to another metric space T is continuous whenever the following

condition is satisfied: “The set, f←[U ], is open in S whenever the set, U , is openin T”. We learned that, some sets are described as being “open”, others as being

“closed”, some are both open and closed, while some are neither. We then learnedhow to distinguish one from the other. For example, we recognized closed sets as

being those whose complement is open. We can recognize a closed set, F , as beingone that “contains the limit point of every convergent sequence in F”. It seems thatknowing various characterizations for each type of set is not only useful, but impor-

tant. How did this notion of an “open set” originate? Recall that we constructedopen sets in a set S with the help of a norm or metric previously defined on S. These

two distance measuring tools defined on S allowed us to define the notion of an “openball” in S which, in turn, allowed us to recognize those subsets, T , of S which are

open. We would now like to define “open set” in a much more abstract context, ina way that is independent of any distance measuring tool such as a norm or metric.

We will gradually add more structure to what we will call, topological spaces. Wewill the proceed to classify the family of all topological spaces according to some of

their intrinsic properties. Thus gathering together those spaces which are similar withrespect to these properties. It may occur that a particular class of topological spaces,is actually a subclass of some other class.

Definition 3.1 Let S be a non-empty set.

a) A topology on S is a collection, τ , of subsets of S which possesses the following prop-

erties:

O1. Both the empty set, ∅, and S belong to τ .

O2. If C ⊆ τ , then ∪C ∈ C is a set which also belongs to τ .

O3. If F is a finite subset of τ , then ∩C ∈ F is also a set which belongs to τ .

30 Section 3: A topology on a set

We will refer to O1, O2 and O3 as the open set axioms.1

b) If τ is a topology on a set S (as defined above) then each member, U ∈ τ , is called anopen subset of S. We often simply say that “U is open in S”.

c) Suppose we have defined a topology, τ , on some set S. Then this set, S, whenconsidered together with this topology τ , is called a topological space and can be

represented as (S, τ). The condition, O2, can also be expressed by the phrase “τis closed under arbitrary unions”. The condition, O3, can also be expressed by thephrase “τ is closed under finite intersections”.

It is important to remember that the elements of a topology τ on S are all subsets ofS. To say that S is an open subset of S it is to assume that a topology τ has been

previously defined; that is, that τ satisfies the three properties stated above. Therecan be many topologies defined on a given set. If τ1 and τ2 are two different topologies

on a set S, then (S, τ1) and (S, τ2) are considered to be distinct topological spaceseven though they contain the same points. When we view (S, τ1) and (S, τ2) simply

as sets they are of course equal. We speak of U as being an open subset of S when itis clearly understood from the context that U is a member of a predefined topology τon S.

Example 1. Suppose (S, ρ) is a metric space and Bε(y) = x ∈ S : ρ(x, y) < εrepresents an open ball of radius ε and center y in S. Let

τρ = ∅ ∪ U ⊆ S : U is the union of open balls in S

Then τρ is a topology on S. (Showing that τρ satisfies the three open set axioms is

left as an exercise.) We say that τρ is the topology induced by the metric ρ on S.

If a metric, ρ, is defined on S, a set A belongs to τρ if and only if A is open in themetric space (S, ρ). In this sense, every metric space is a topological space.

Note : There exist topological spaces whose topology is not derived from a metric ρ.

Those topological spaces whose open sets can be derived some metric have a specialname.

Definition 3.2 Let (S, τ) be a topological space. We will say that (S, τ) is metrizable if

there exists a metric, ρ, on S such that τ = τρ (where τρ is induced on S by ρ).

1Some readers may notice that, if C = ∅ ⊆ τ , then, by O2, ∪C ∈ ∅ = ∅ ∈ τ . Also, if the finite subset,F , is empty then, by O3, ∩C ∈ ∅ = S ∈ τ . So, theoretically, it would be sufficient to axiomatize “opensets” with O2 and O3 where O1 would logically flow from these two. The axiom O1 is normally includedfor convenience to make it easier to identify a topology.


From this definition, we can clearly state that the class of all topological spaces canbe subdivided into a subfamily of all metrizable spaces and the subfamily of non-

metrizable ones.

For example, the usual metric ρ(x, y) = |x − y| on R induces a topology, τρ, on R.We normally refer to this topology as being the usual topology on R. In many text-

books it is also referred to as the Euclidean topology on R. In this case, the opensubsets of R are the sets which are unions of open intervals Bε(y) = (y − ε, y + ε).

For example, the sets U1 = ∪(n, n+ 1) : n ∈ Z and U2 = (−4, 9) can be shown tobelong to τρ, but the subset U3 = [−7,−3) 6∈ τρ. (It is left to the reader to verify this.)

Example 2. A non-metrizable space. Consider the set S = R2 equipped with thetopology

τ = B ⊆ R2 : R2\B is countable1 ∪ ∅,R2

a) Verify that the family, τ , of subsets of R2 is indeed a topology on R2.

b) Verify that the topological space, R2, equipped with the topology, τ , describedabove is not metrizable.

Solution : Given: τ = B ⊆ R2 : R2\B is countable ∪ ∅,R2.

a) Verification that the sets in τ satisfy the three open sets axioms O1, O2, and O3

is left as an exercise.

b) Suppose ρ is a metric on R2 such that τ = τρ. We will show that, given τ = τρ,

then ρ(~x, ~y) > ρ(~x, ~z) + ρ(~y, ~z), contradicting the fact that ρ is a valid metric onR2.

Suppose ~x and ~y are distinct points in R2 and ρ(~x, ~y) = α 6= 0. Consider the openballsBα/2(~x) and Bα/2(~y). Since both balls belong to τ , each ball has a countable

complement; so each ball is an uncountable subset. Hence Bα/2(~y) cannot beentirely contained in R2\Bα/2(~x). That means that Bα/2(~y)∩Bα/2(~x) 6= ∅. Let

~z ∈ Bα/2(~y) ∩ Bα/2(~x). Then

ρ(~x, ~z) + ρ(~y, ~z) < α/2 + α/2

= α

= ρ(~x, ~y)

We see that ρ(~x, ~y) > ρ(~x, ~z) + ρ(~y, ~z) and so ρ does not satisfy the triangleinequality, contradicting the fact that ρ is a metric on R2. So there can be nometric ρ on R2 such that τ = τρ. We conclude that the topological space, (R2, τ),

1If A is a subset of a space S, then S\A is the complement of A. It denotes the set of all points in Swhich don’t belong to A.


is not a metrizable space.2

For convenience, we introduce the following supplementary definition.

Supplementary definition: If (S, τ) is a topological space and x ∈ U where U is anopen subset of S, then we say that U is an open neighbourhood of x. Furthermore,

if U is open and T is a subset of S such that x ∈ U ⊂ T , then we say that T is aneighbourhood of x.

Note that it is required that U be open in S for T to be called a “neighbourhood of x”.A point, x, in a topological space, (S, τ), always has at least one open neighbourhood,namely S. If T = (−2, 4)∪ (4, 7] in R, equipped with the usual topology, we see that

T is a neighbourhood of 3 but not a neighbourhood of 4 nor of 7 (since there is noopen set U such that 7 ∈ U ⊆ T ).

Definition 3.3 Suppose τ1 and τ2 are two topologies on a given set S. If τ1 ⊆ τ2, then we

say that τ1 is a weaker topology than τ2 on S or that τ2 is a stronger topology than τ1 onS. We can also say that τ1 is a coarser topology than τ2 on S, or that τ2 is a finer topology

than τ1 on S. We will say that the two topological spaces (S, τ1) and (S, τ2) are equivalentif and only if τ1 = τ2.

Example 3. Suppose S is a non-empty set and P(S) denotes the power set of S (that

is, P(S) denotes the collection of all subsets of S). Let τd = P(S). Then τd is atopology on S. It is left to the reader to verify this. In this case, for every single point

x ∈ S, x is open in S.

Supplementary definition: For a given space S, if

τd = P(S)

the topology, τd, is referred to as the discrete topology on S.

2Later, once we covered the concept of “Hausdorff” this problem will be more easily solved by statingthat “this space is not metrizable because it not Hausdorff”.


If S is equipped with the discrete topology then every non-empty subset of S is anopen neighbourhood of the elements it contains.

Example 4. For a non-empty set S, if τi = ∅, S, then the two-element set, τi, is atopology on S.

Supplementary definition: If τi = ∅, S, the topology, τi, is normally referred to as

the indiscrete topology on S. In this case, for any x ∈ S, S is its only neighbourhood.

Since all topologies on S must at least contain ∅ and S, τi is the weakest (coarsest)

of all topologies on S. On the other hand, τd = P(S) is the strongest (finest) of alltopologies on S. For any topology τ on S, we then have

τi ⊆ τ ⊆ τd

It is left for the reader to verify the following important fact.

Fact : Given a family τk : k ∈ I of topologies on a set S, it is an easy sxercise to

show that the family ∩k∈Iτk is also a topology on S.1 This fact can also be expressedas: “The set of all topologies is closed under arbitrary intersections.”

But ∪τk : k ∈ I need not necessarily be a topology on S. For example, verify that

τ1 = ∅, S, b, a, b and τ2 = ∅, S, c, a, c are two topologies on S; but theirunion, τ1 ∪ τ2, is not a topology on S. (Since a, b, c 6∈ τ1 ∪ τ2. Verify this.)

3.2 Closed subsets of a topological space.

In our brief overview of metric spaces, (S, ρ), we defined a closed subset, F , of Sas one which contains all its limit points. We then saw that, in a metric space S,

a subset, F , of S is closed if and only if its complement, S \F , is open.2 Metricspaces were equipped with a “metric” so that we could discuss the much needed no-

tions of “convergence of a sequence” and “closed set” in a set S. We now formallydefine the analogous notions of “closed subsets in a topological space”. We will not

discuss the concepts of “convergence” and “limit points” in a topological space imme-diately. We will be doing an in-depth study of this topic in a section later in this book.

1Caution: However, arbitrary unions of topologies may not form a topology on a set.2S\F = x ∈ S : x 6∈ F


Definition 3.4 Let F be a subset of a topological space (S, τ). If the complement, S\F ,

of F is an open subset in S then we say that F is a closed subset in S.

The definition of “closed” states that “(S\F is open) ⇒ (F is closed”). Conversely,

A is closed ⇒ A = S \ [S\A] is closed

⇒ [S\A] is open

We can then actually write

(F is closed) ⇔ (S\F is open)

Suppose (S, τ) is a topological space and F = A ⊂ S : A is closed in S. Then wecan define the topology in terms of F ,

τ = A : S\A ∈ F and F = A : S\A ∈ τ

Theorem 3.5 Let (S, τ) be a topological space and I be any indexing set. Then,

a) Both ∅ and S are closed in S.

b) If Fi : i ∈ I is a family of closed subsets of S then ∩Fi : i ∈ I is a closed subset

of S.

c) If Fi : i = 1, 2, 3, . . . , k is a finite family of closed subsets of S then ∪Fi : i =1, 2, 3, . . . , k is a closed subset of S.

Proof:

a) Since ∅ is open, then S = S\∅ is closed. Since S is open, ∅ = S\S is closed.

b) Let Fi : i ∈ I be a family of closed subsets of S. Then, for each i ∈ I , S\Fi is

open. Since (by De Morgan’s law) S\∩Fi : i ∈ I = ∪S\Fi : i ∈ I is open (beingthe union of open sets) then ∩Fi : i ∈ I is a closed subset of S.

c) This part is left as an exercise.


Supplementary proposition: Defining a topology on S in terms of “closed sets”. Sup-

pose we are given a set S and family, F = F : F ⊆ S of elements from P(S) whichsatisfies the following three conditions:

F1. The sets ∅, S both belong to F .

F2. If Fi : i ∈ I ⊆ F then ∩Fi : i ∈ I ∈ F .

F3. If Fi : i = 1, 2, 3, . . . , k ⊆ F then ∪Fi : i = 1, 2, 3, . . .k ∈ F .

Then the family, τ = S\F : F ∈ F, forms a topology on S.

Proof : The proof showing that τ is a topology is left to the reader. It easily followsfrom an application of De Morgan’s law1.

Supplementary definition: The conditions F1, F2 and F3 described above are referredto as the...

“closed set axioms”

Example 5. Suppose S is a non-empty set and

F = F : F is a finite subset of S ∪ ∅, S

Then the set F satisfies the three closed sets axioms F1, F2 and F3. (Verify this!)Then τ = S\F : F ∈ F forms a topology on S. The elements of τ are

τ = ∅, S ∪ U ⊂ S : S\U is finite

Given a set S, the family of subsets,

τ = A : A ⊆ S and S\A is finite ∪ S,∅1

forms a topology of S called the cofinite topology on S or the Zariski topology on S.

1That is, S \ [∪i∈IFi] = ∩i∈I [S\Fi] and S \ [∩i∈IFi] = ∪i∈I [S\Fi]1If S\U is finite we say U is cofinite.


3.3 Subspace topology on a subset.

We previously defined the notion of a “metric subspace”, T , as being a subset of ametric space, (S, ρ), equipped with the subspace metric, ρT .

In the more general case of a topological space, (S, τ), any subset T can be declared to

be a “topological subspace” provided the reader understands what topology is definedon T . Suppose H is a non-empty subset of a topological space (S, τ). Then H can

inherit its topology from τ , in a natural way, as shown the following theorem.

Theorem 3.6 Let (S, τ) be topological space and H ⊆ S.

a) Let

τH = U ⊂ H : U = K ∩H, for some K ∈ τThen τH is a topology on H .

b) Let

FH = F ⊂ H : F = M ∩H where M = S \K,K ∈ τThen FH represents all closed subsets of the topological space (H, τH)

Proof: The proof is left as an exercise.

Definition 3.7 If (S, τ) is topological space and H ⊆ S and

τH = U ⊂ H : U = K ∩H, K ∈ τ

then τH is called the subspace topology or relative topology on H induced by S, or inherited

from S. In such a case, we will say that (H, τH) is a subspace of S.

Some subsets of a topological space (S, τ) can be both open and closed. For example,if (R, τi) is equipped with the indiscrete topology, τi = ∅,R, both ∅ and R are the

only open subsets of R and so both ∅ and R are the only closed subsets of R. That is,∅ and R are simultaneously open and closed in R. We consider a less trivial example.

Example 6. Let H = [3, 5). Consider the subset

T = H ∪ 9 = [3, 5)∪ 9


of (R, τ) where R is equipped with the usual topology, τ . Suppose the subspace (T, τT)is equipped with the subspace topology inherited from τ . Now H is a subset of both

T and R. Since H = T ∩ [2, 6], then H is closed in T with respect to the subspacetopology τT . Since H = T ∩ (2, 5), then H is open in T with respect to the subspace

topology τT . So the subset, H , is both open and closed in the subspace, T . Note,however, that H is neither open nor closed in S.

There is a special adjective used to refer to those subsets which are both open andclosed. Since its use is fairly common we formally define it below.

Definition: If T is a subset of a topological space (S, τ), and T is both open and closedwith respect to τ , we say that...

“T is clopen in S.”

Suppose (S, τ) is a topological space which contains the non-empty subset (F, τF )equipped with the subspace topology, τF , inherited from S. If one speaks of an opensubset, U , of F , it may sometimes not be obvious whether U ∈ τ or U ∈ τF . To be

more specific we can sometimes write “U is an S-open subset in F” to mean U ∈ τand “U is an F -open subset of F” to mean U ∈ τF . 1

3.4 Other examples.

We provide a few more examples of topological spaces.

Example 7. Let S be a set and B ⊆ S. Let τB = A ∈ P(S) : B ⊆ A ∪ ∅.

a) Verify that τB is indeed a topology.

b) Describe the closed subsets of (S, τB).

Solution : We are given that B ⊆ S and τB = A ∈ P(S) : B ⊆ A ∪ ∅.

a) We verify that τB is a topology on S by confirming that it satisfies the open setaxioms O1, O2 and O3.

− By definition, ∅ ∈ τB; also, since B ⊆ S then S ∈ τB .

− Suppose U is a non-empty subset of τ . Then B ⊆ U for each U ∈ U . SoB ⊆ ∪U : U ∈ U . Hence ∪U : U ∈ U ∈ τB.

− Suppose F is a finite subset of τB . Then B ⊆ F for each F ∈ F . So

B ⊆ ∩F : F ∈ F. Hence ∩F : F ∈ F ∈ τB.

1Note that if F is S-open and U is an F -open subset of F then U is also S-open.


b) We now describe the closed subsets of S. We consider the sets, A, such thatA ∩B = ∅, and those satisfying A ∩B 6= ∅.

− If A ⊆ S such that A ∩ B = ∅, then B ⊆ S \A; this means S \A is open,hence A is a closed subset of S.

− Suppose, on the other hand, that for A ⊆ S and A∩B 6= ∅. We claim thatS is the only closed subset which contains A. To see this, note that if F isclosed then S \F is open and so either contains B or is ∅. If A ∩ B 6= ∅

and A ⊆ F then S\F cannot containB so S\F = ∅, which means that F = S.

So the closed subsets of S are the family

F = F ⊂ S : F ∩B = ∅ ∪ S

Example 8. Let (S, τ) be a topological space and suppose B is a fixed subset of S.

Let

τB = A ∈ P(S) : A = C ∪ (D ∩ B) where C,D ∈ τ

a) Verify that τB is another topology on S.

b) Is one of the two topologies, τB, τ , stronger than the other? Are these two

topologies equivalent topologies?

Solution : We are given that (S, τ) is a topological space and B is a fixed subset of S.

a) We begin by showing that τB satisfies the three open set axioms O1, O2 and O3.

− We know ∅, S ∈ τ . So we have ∅ = ∅∪ (∅∩B) ∈ τB and S = S ∪ (S∩B) ∈τB.

− Suppose U = Ci ∪ (Di ∩ B) : i ∈ I ⊆ τB, where Ci, Di ∈ τ for all i ∈ I .

Then ∪i∈ICi, and ∪i∈IDi both belong to τ . Then

⋃

i∈I

[Ci ∪ (Di ∩B)] = [∪i∈ICi] ∪ ([∪i∈IDi] ∩B) ∈ τB

Thus τB is closed under arbitrary unions.

− Suppose F = Ci ∪ (Di ∩B) : i = 1, 2, . . . , n ⊆ τB. Then

⋂

i=1,...,n

[Ci ∪ (Di ∩B)] =⋂

i=1,...,n

[(Ci ∪Di) ∩ (Ci ∪B)]

= [∩i=1,...,n(Ci ∪Di)] ∩ [(∩i=1,...,nCi) ∪ B]

= [∩i=1,...,n(Ci ∪Di)] ∩ [(∩i=1,...,nCi)]

∪ [∩i=1,...,n(Ci ∪Di) ∩B]

∈ τB

So τB satisfies the three open set axioms O1, O2 and O3.


b) We claim that τ ⊆ τB : See that, if C ∈ τ , then C = C ∪ (∅ ∩ B) ∈ τB, soτ ⊆ τB . So τB is a topology on S which is finer (stronger) than τ .

We claim that these two topologies are not equivalent: Suppose B 6∈ τ , D ∈ τand B ⊂ D. Then ∅∪(D∩B) = B 6∈ τ . Since B ∈ τB then τB contains elements

which are not in τ . So τ ⊂ τB.

In the above example, we say that “the topology τB extends τ over B”.

3.5 Free union of topological spaces.

Suppose we are given a family of topological spaces. There is a way to unite theminto one single new larger topological space without altering their individual topology.

This is referred to a being the “free union” of these topological spaces. We define thisconcept.

Definition 3.8 Let Si : i ∈ I be a family of topological spaces. For each space, Si, weassociate a space, S∗i = i × Si in such a way that S∗i and Si are identical except for the

fact that i × Si has a label i attached to Si. This is to guarantee that if i 6= j then S∗iand S∗j are entirely different sets and so have empty intersection. This allows us to view

the family, S∗i : i ∈ I, as being pairwise disjoint, in the sense that no two spaces haveelements in common. We define the free union of the family Si : i ∈ I,1 denoted as,∑

i∈I S∗i , as being the topological space

∑

i∈I

S∗i = ∪S∗i : i ∈ I

in which U is open in∑

i∈I S∗i if and only if U ∩ S∗i is open for each i ∈ I . This topology,

thus defined, is referred to as the disjoint union topology.

Example 9. For each n ∈ N\0, let Ln denote the set,

Ln =

(x, yn) : yn = gn(x) =

(

sin ( π4n)

cos ( π4n)

)

x, x 6= 0

Let L0 = (x, 0) : x ∈ R and

T = ∪Ln : n ∈ N\0∪ L01Some texts may refer to this set as direct sum or free sum or topological direct sum.


It is possible to view T as a single subset of R2 and equip it with the subspace topology,in which case open neighbourhoods of points on L0 would intersect points on other

lines. For a different topology, we could view T as a free union of disjoint subspacesof R2, where each line is equipped with the subspace topology. With the free union

topology, an open neighbourhood of a point on a line is restricted to the line itself.Each line would be clopen in T . For example, (x, 0) : 1 < x < 7 would be an open

neighbourhood of (5, 0) in T .

3.6 Topics: G-delta and F -sigma sets.

Besides the fundamental open and closed subsets of a topological space introducedearlier, there are other subsets defined in terms of open and closed sets with special

properties. Two of these are called G-delta’s, F -sigma’s. Developing some familiaritywith these now will be good practice. As well, it will allow us to freely refer to them

in various examples, theorems and exercise questions to come.

The G-delta and F -sigma sets in a topological space. We have seen that arbitraryunions of open subsets of a topological space are open. However, only the intersectionof (at most) finitely many open sets are guaranteed to be open. Similarly, arbitrary

intersections of closed sets are closed, but the union of (at most) finitely many closedsets are guaranteed to be closed. The intersection of countably many open sets and

the union of countably many closed sets may be relevant in the study of certain typesof spaces.

Before we continue, we remind the reader that a non-empty set S is said to be “count-

able” if it is finite or, in the case where it is infinite, the elements of S can be indexedby the natural numbers. That is, S = xi : i = 0, 1, 2, 3, . . .. We can also say that S

is countable if there exists a function f : N → S mapping the natural numbers onto S.We now formally define those special subsets of a topological space we call G-delta’s

and F-sigma’s.

Definition 3.9 The sets in a topological space, (S, τ), which are the intersection of at mostcountably many open sets are called Gδ-sets (or simply Gδ). Those sets in S which are the

union of at most countably many closed sets are called Fσ-sets (or simply Fσ). Neither ofthese special sets need be open or closed.

Trivially, if F is closed in (S, τ) then F is an Fσ and if U is open then U is a Gδ.


Example 10. If R is equipped with the usual topology, the set T = [2, 7] is obviouslya Fσ. It is also a Gδ since

[2, 7] = ∩(2 − 1/n, 7 + 1/n) : n = 1, 2, 3, . . .

So some sets can be both a Gδ and an Fσ with respect to the same topology τ .

Example 11. On the other hand, suppose (S, τi) is equipped with the indiscrete topol-

ogy, τi = ∅, S. If T is a proper non-empty subset of S, we see that T 6⊆ ∅ andT 6= S; so T is neither a Gδ nor an Fσ with respect to τi.

Example 12 . We consider the set of all rationals, Q, as a subset of R equipped withthe usual topology. It is known that Q is countably infinite and so can be expressedin the form Q = xi : i = 1, 2, 3, . . .. Then Q = ∪xi : i = 1, 2, 3, . . . where each

xi is a closed subset of R. So Q is an Fσ .1

The following theorem exhibits properties respected by each Fσ and Gδ and the fam-ilies of all Gδ’s and Fσ’s of a topological space.

Theorem 3.10 Suppose F is an Fσ and G is a Gδ in S.

a) The complement of F in S is a Gδ and the complement of G in S is an Fσ .

b) There exists a sequence, Fi : i = 1, 2, 3, . . ., of closed subsets of S such that

Fi ⊆ Fi+1 for all i = 1, 2, 3, . . ., and F = ∪Fi : i = 1, 2, 3, . . .

c) There exists a nonincreasing sequence, Gi : i = 1, 2, 3, . . ., of open subsets of S such

that

Gi+1 ⊆ Gi for all i = 1, 2, 3, . . ., and G = ∩Gi : i = 1, 2, 3, . . .

d) Suppose F denotes the family of all Fσ ’s in S and Fi : i = 1, 2, 3, . . . represents atmost countably many elements in F . Then

∪Fi : i = 1, 2, 3, . . . ∈ F and ∩ Fi : i = 1, 2, 3, · · · , k ∈ F for any k

e) Suppose G denotes the family of all Gδ’s in S and Gi : i = 1, 2, 3, . . . represents atmost countably many elements in G . Then

∩Gi : i = 1, 2, 3, . . . ∈ G and ∪ Gi : i = 1, 2, 3, · · · , k ∈ G for any k

1Once we have the necessary tools we will prove that Q is not a Gδ


Proof: Given: (S, τ) be topological space; F is an Fσ and G is a Gδ in S.

a) Suppose F = ∪Ki : i = 1, 2, 3, . . . , where each Ki is closed in S

S\F = S\(∪Ki)= ∩S\Ki (By De Morgan’s rule)

= a Gδ-set

The proof of the second part of a) follows by a similar application of De Morgan’srule.

b) The proof is left as an exercise for the reader.

c) The proof is left as an exercise for the reader.

d) For countable unions of Fσ ’s:

∞⋃

j=1

[∪F(i,j) : i = 1, 2, 3, . . . , ] = ∪F(i,j) : (i, j) ∈ N × N

where N × N is known to be countable.

For finite intersections of Fσ ’s:

k⋂

i=1

[∪∞j=1F(i,j)] =⋃

(j1,··· ,jk)∈N×N×···×N

F(1,j1) ∩ · · · ∩ F(k,jk)

where N × N × · · · × N is known to be countable.

e) This part is proved similarly to part d).

We summarize two of the results in the above theorem:

“The family, F , of all Fσ’s of a topological space is closed under countableunions and closed under finite intersections.”

“The family, G , of all Gδ’s of a topological space is closed under countable

intersections and closed under finite unions.”


We will see a bit later that, in certain specific classes of topological spaces, Fσ’s areclosed and Gδ’s are open. We have to do some groundwork before we can discuss these.

3.7 Topics: The family of Borel sets.1

Topological spaces each contain a particular family of subsets of P(S) which playsa role in certain fields of study where topology is applied. Particularly in analysis.Before we formally define it, we begin by defining a special type of subset of P(S)

called a “σ-ring”.

A subset, K , of P(S) is called a σ-ring if:

1) For any A ∈ K , S\A ∈ K

2) Whenever Ai : i ∈ N ⊆ K then ∪Ai : i ∈ N ∈ K .

To summarize, a σ-ring is simply a family of sets which is “closed under complements”

and “countable unions of its sets”. See that P(S), itself a σ-ring, may contain manyσ-rings. Given a topological space, (S, τ), we will consider all those σ-rings in P(S),

which contain τ . To obtain the unique smallest σ-ring, B, in P(S) that contains τ ,we then take the intersection of all σ-rings in P(S) which contain τ ,

B = ∩K ⊆ P(S) : K is a σ-ring, τ ⊆ K

The reader should verify that this “intersection of all σ-rings in P(S) which containτ” is itself a σ-ring containing τ ; we emphasize, that this intersection is the unique

and smallest such σ-ring. We have a name for this particular set.

Definition 3.11 Given a topological space (S, τ), we call the smallest σ-ring, B, in P(S)which contains τ ,...

“the family of Borel sets in S”

Each member, A ∈ B, is referred to as a Borel set. That is,

A = “a Borel set in S” ⇔ A ∈ B

Every topological space, S, has its unique family, B, of Borel sets. We identify aBorel set by confirming that it belongs to B. To help us identify Borel sets we list a

few properties of B. The reader is left to verify that B . . .

– is closed under complements,

1This is a more specialized topic. It can be omitted without loss of continuity.


– is closed under countable unions and countable intersections

– contains all Gδ’s and all Fσ ’s.

The definition of a “Borel set” in S makes it difficult to recognize such a subset of S.

But it also makes it easy to recognize a large subfamily of B: “All open subsets (in-cluding ∅ and S itself), all Gδ’s and all Fσ’s are Borel sets”. But there may be others.

Theorem 3.12 Let (S, τ) be a topological space. The family, B, of Borel sets is the uniquesmallest subfamily of P(S), that

a) contains τ ,

b) is closed under complements

c) is closed under countable unions.

Furthermore, B satisfies the following three properties:

1. B contains all Fσ ’s of S,

2. B is closed under countable intersections,

3. B contains all Gδ’s of S.

Proof: Given: (S, τ) is a topological space.

Suppose B is the family of Borel sets in P(S). By definition, B is the intersectionof all σ-rings that

a) contain τ ,

b) that are closed under complements and

c) closed under countable unions.

So B is itself the unique smallest σ-ring of subsets of S which satisfies these three

properties.

Since B contains τ , it contains all open subsets of S and since it is closed undercomplements, it contains all closed subsets of S. Since it is closed under countable

unions then it must contain all Fσ ’s. This establishes property 1.

We now verify that B is closed under countable intersections: Let Ai : i = 1, 2, 3, . . .be a countable family of subsets in B. Then

∩Ai : i = 1, 2, 3, . . . = S\S\(∩Ai : i = 1, 2, 3, . . .)= S\∪[S\Ai : i = 1, 2, 3, . . .]∈ B


This establishes property 2.

It then follows that, since B contains all open sets, it follows from property two thatit must also contain all Gδ’s of S. This establishes property 3.

The above theorem guarantees that every open set, closed set, Gδ and Fσ in a topo-

logical space S can be referred to as a Borel set in S. It is sometimes difficult toidentify subsets of a topological space (S, τ) which are not Borel sets (with respect to

τ). Consider for example, the topological space (S, τi) equipped with the indiscretetopology. If A is a non-empty proper subset of S then A is not a Borel set since

∅, S = τi is the smallest σ-ring which contains τi and does not contain the elementA. On page 61 of this text we provide another example.

Concepts review:

1. Given a set S, what does a topology τ on S represent? How does one verify whethera family of subsets is a topology?

2. Given an open subset, U , of a topological space, (S, τ), what is the relationship

between U and τ?

3. What are the three open set axioms of a topological space (S, τ)?

4. Given a metric space, (S, ρ), describe a topology on S which is induced by ρ.

5. Describe the usual topology on R.

6. Given a point, x, in a topological space, (S, τ), what is a neighbourhood of x?

7. Given two topologies, τ1 and τ2, what does it mean to say that τ1 is weaker than τ2?

8. Given two topologies, τ1 and τ2, what does it mean to say that τ1 is finer than τ2?

9. Given a non-empty set, S, describe the discrete topology on S.

10. Given a non-empty set, S, describe the indiscrete topology on S.

11. Suppose F is a closed subset of the topological space (S, τ). What is the relation

between F and τ?

12. What are the three closed set axioms, F1, F2, and F3, of a topological space (S, τ).

13. If S is a non-empty set what do we mean by the cofinite or Zariski topology on S?


14. If T is a subset of the topological space (S, τ) what is the subspace topology on T ?

15. Suppose B is a subset of the topological space (S, τ) such that B 6∈ τ . Describe thetopology τB which extends τ over B?

16. What does it mean to say that a set is metrizable?

17. What is a Gδ of a topological space? What is an Fσ of a topological space?

18. Describe the family of Borel sets in a topological space (S, τ).

19. Provide a few examples of Borel sets in R equipped with the usual topology. Is Q aBorel set? Why?

20. Define the topological space called the “free union” of the spaces, Si : i ∈ I.

EXERCISES

1. Prove the statement in theorem 3.5.

2. Consider the open interval S = (−3, 7) in R.

a) Construct a topology τ on S which contains five elements.

b) Consider the subset T = (−2, 4] ⊂ S. For the topology τ constructed in part a)what is the subspace topology τT on T inherited from S.

c) Are the open subsets of T necessarily open subsets of S?

3. Consider R equipped with the usual topology τ (induced by the Euclidian metric).

Let (Q, τQ) be the set of all rational numbers equipped with the subspace topologyinherited from R. Consider the subset T = [−π, π)∩Q. Determine whether T is open

in Q, closed in Q, both open and closed in Q, or none of these.

4. Construct a topology other than the discrete or indiscrete topology on the set S =

4,♦,.

5. Let F = A ⊆ R : A is countable ∪ ∅,R. Show that F satisfies the three condi-tions F1, F2 and F3 described on page 35. Then use this to construct a topology on

R. (This is referred to as being the cocountable topology.

6. Suppose τA and τB are two topologies on a set S. Determine whether τA ∩ τB is atopology on S.

7. If R is equipped with the usual topology and Z represents the set of all integersdetermine whether Z is open or closed (or both or neither) in R.


8. Suppose R is equipped with the usual topology and T = [1, 4]∪ (6, 10) ⊂ R where Tis equipped with the subspace topology. Determine whether [1, 4] is open in T , closed

in T or both open and closed in T . Determine whether (6, 10) is open in T , closed inT or both open and closed in T .

48 Section 4: Set closures, interiors and boundaries.

4 / Set closures, interiors and boundaries.

Summary. In this section, we introduce the notions of closure and interior

of subsets of a topological space. The concept of the boundary of a set is thendefined in terms of its interior and closure. Based on their properties, we derive

the “closure axioms” and “interior axioms”. We then begin viewing closure andinterior of sets as being operators on P(S). From this perspective, we better

see how closure and interior operators on P(S) can be used to topologize a set,providing examples on how this can be done.

4.1 The closure of a set.

If T = (2, 7] is viewed as a subset of the topological space R (equipped with theusual topology), we easily see that it is not closed since its complement, R\T =

(−∞, 2] ∪ (7,∞), is not open in R. And yet we feel that it wouldn’t take very muchfor us to “make it closed”: We need only add the element, 2, to T to obtain the closed

subset T ∗ = [2, 7]. Adding the fewest number of points possible to a set T to obtaina closed set is what we will refer to as obtaining the closure of T . The key words

here are “fewest number” of points, and no more. In this case, we would say thatthe “closure of T = (2, 7] is the set T ∗ = [2, 7], the smallest closed subset of R whichcontains all the elements of T . With this example in mind, we will now formally define

a concept called “closure of a subset”.

Definition 4.1 Let S be a topological space and T ⊆ S. We define the closure of T in S,

denoted by, clST (or by clS(T )), as

clST = ∩F : F is closed in S and T ⊆ F

The reader should first be aware of the following verifiable facts for any subset, T , ofthe topological space S.

1) The closure of T , clST , is closed in S: This follows from the fact that arbitrary

intersections of closed sets are closed.

2) The set T ⊆ clST : This follows from the definition of closure of T .

3) The set, clST , is the smallest closed set which contains T : Suppose A is a closed

set containing T . Then A ∈ F : F is closed in S and T ⊆ F. Hence clST ⊆ A.

4) If T is closed then T = clST : This is true since T is the smallest closed set

containing T .


Definition . Let A be a subset of a topological space S.

If x is a point in S such that, for every open neighbourhood U of x, U ∩ A contains

some point other than x, then we say that...

x is a cluster point of A

The set of all cluster points of A is called the derived set of A.

This definition provides us with another way of describing a closed set.

“The set B is closed if and only if it contains all its cluster points.”

Verification of this fact is left to the reader.

For example, if B = (1, 3) ∪ (3, 5] ∪ 6 in R, the derived set, (that is, the set of allcluster points of B) is [1, 5]. The element, 6, is not a cluster point of B since there

is an open neighbourhood, (5.5, 7) of 6 which does not meet other elements of B.The set B is not closed since it doesn’t contain the cluster points 1 and 3.

Example 1. Let T be the open interval, (0, 1), viewed as a subset of R equipped withthe usual topology. Then clRT = [0, 1]. To prove this we must show that:

1) [0, 1] is closed by showing that R\[0, 1] is open.

2) 0, 1 ⊆ A for any closed set A containing the interval (0, 1).

This is left as an exercise.

Example 2. If Q is the set of all rational numbers then clR(Q) = R.

Proof : To prove this we will show that, if F is a closed subset of R such that Q ⊆ F ,then F = R.

Suppose F is a closed subset of R such that Q ⊆ F . Then R\F is open. SupposeR\F 6= ∅. Then R\F is the union of open intervals each of which must contain a

rational number. Since Q ∩ (R\F ) 6= ∅, this contradicts Q ⊆ F . Then R\F = ∅.This means that the only closed set containing Q is R. So clR(Q) = R.

Example 3. Suppose S is a topological space induced by the metric ρ (that is, the

elements of τ are unions of open balls of the form Bε(x) = y : ρ(x, y)< ε). SupposeF is a non-empty subset of S. We define

ρ(x, F ) = inf ρ(x, u) : u ∈ F


Show that clSF = x : ρ(x, F ) = 0.

Solution : To do this we must show

1. F ⊆ x : ρ(x, F ) = 0,2. S \ x : ρ(x, F ) = 0 is open in S,

3. If F ⊆ A where A is a closed subset of S, then x : ρ(x, F ) = 0 ⊆ A.

The details are left as an exercise.

We now list and prove a few of the most fundamental closure properties.

Theorem 4.2 Let A and B be two subsets of a topological space (S, τ). Then,

1) clS(∅) = ∅.

2) If A ⊆ B then clS(A) ⊆ clS(B)

3) clS(A ∪B) = clS(A) ∪ clS(B) (Closure “distributes” over finite unions.)

4) clS(clS(A)) = clS(A)

Proof:

1) Since ∅ is closed clS(∅) ⊆ ∅. Since ∅ ⊆ clS∅, then clS(∅) = ∅.

2) We are given that A ⊆ B. If F is closed in S and B ⊆ F then A ⊆ B ⊆ F . Then

A ⊆ ∩F : F is closed in S and B ⊆ F = clS(B)

By 1) of the facts above, clS(B) is closed in S and so

clS(A) = ∩F : F is closed in S and A ⊆ F ⊆ clS(B)

We have shown that clS(A) ⊆ clS(B).

3) Since A ⊆ A∪B and B ⊆ A∪B then clS(A) ⊆ clS(A∪B) and clS(B) ⊆ clS(A∪B)

(by parts 1) and 2)). So

clS(A) ∪ clS[B) ⊆ clS(A ∪B)

Since A ⊂ clS(A) and B ⊂ clS(B), A ∪ B ⊆ clS(A)∪ clS(B), a closed subset in S.Since clS(A∪ B) is the smallest closed set containing A ∪ B then

clS(A ∪B) ⊆ clS(A)∪ clS[B)

We conclude that clS(A ∪ B) = clS(A)∪ clS(B).


4) By part 1)A ⊆ clS(A) ⊆ clS(clS(A)). Since clS(A) is closed (see fact 1)), clS(clS(A)) ⊆clS(A). It then follows that clS(clS(A)) = clS(A).

Example 4. Closure does not distribute over intersections. Suppose A = (2, 5) andB = (5, 7). Then clR(A ∩ B) = clR(∅) = ∅. On the other hand, clR(A) = [2, 5]

and clR(B) = [5, 7] hence clR(A)∩ clR(B) = 5. This shows that clS(A)∩ clS(B) 6=clS(A ∩B) may sometimes occur.

It is however possible to prove that

clS(A ∩B) ⊆ clS(A) ∩ clS(B)

Proving this is left as an exercise.

Remark on closures of arbitrary unions. We have seen in theorem 4.2 that clS(A∪B) =

clS(A)∪ clSB, so the “closure distributes over finite unions”.

This does not hold true for arbitrary unions. Consider the sets of the form

Ai =

(

1

i, 3

]

where clRAi =

[

1

i, 3

]

for i = 1, 2, 3, . . . Verify that ∪clR(Ai) : i = 1, 2, 3, . . . = (0, 3] (left as an exercise).

Since clR[∪Ai : i = 1, 2, 3, . . .] = [0, 3] (left as an exercise), then

∪clR(Ai) : i = 1, 2, 3, . . . 6= clR[∪Ai : i = 1, 2, 3, . . .]

4.2 Closure viewed as an operator on P(S).

The closure of a set can be viewed as an action

clS : P(S) → P(S)

performed on a set. It takes an arbitrary set, A, and associates to it another set,

clS(A), obtained by adding sufficiently many points (but no more) so as to producea “closed set”. It can then be viewed as a function. With this in mind, we define the

Kuratowski closure operator.

Definition: Kuratowski closure operator. Suppose S is a non-empty set and K :

P(S) → P(S) is a function which satisfies the four conditions:


K1. K(∅) = ∅ and A ⊆ K(A)

K2. If A ⊆ B then K(A) ⊆ K(B)

K3. K(A∪ B) = K(A) ∪K(B)

K4. K(K(A)) = K(A)

A function, K : P(S) → P(S), satisfying these four properties is referred to as a Ku-ratowski closure operator where K1 to K4 are the Kuratowski closure operator axioms.

Topologizing a set S by using a closure operator. The reader should notice that, in

the above definition, the set S is not described as being a “topological space“ sinceno topology is defined on it. It is just a set. The following theorem shows that, if we

are given a Kuratowski operator, K : P(S) → P(S), on P(S) then we can use K togenerate a topology, τK , on S such that

clSA = K(A)

for all A ∈ P(S) .

Theorem 4.3 Let S be a set and suppose K : P(S) → P(S) satisfies the four Kuratowskiclosure operator axioms. Define F ⊆ P(S) as

F = A ⊆ S : K(A) = A

a) Then F , is the set of all closed subsets of some topology, τK , on S. That is,

τK = S\A : where A ∈ F

b) Furthermore, in (S, τK), clS(A) = K(A), for any A ⊆ S.

Proof: Given: The operator K : P(S) → P(S). Also, F = A ⊆ S : K(A) = A.a) To prove the statement a) it will suffice to show that F satisfies the three “closedsets” conditions F1, F2 and F3 on page 35. If so, then we can define τK by invoking the

statement of the proposition on page 35.

− Note that ∅, S ⊆ F . To see this note that, by K1, S ⊆ K(S) ⊆ S ⇒ K(S) = S.

Both ∅ and S belong to F . The set F satisfies condition F1.

− Suppose U = Fi : i ∈ I is an arbitrarily large family of sets in F . By K1,

∩U ⊆ K(∩U ). Also, by K2, K(∩U ) ⊆ K(Fi) = Fi for each i ∈ I . So K(∩U ) ⊆∩Fi : i ∈ I = ∩U . Hence K(∩U ) ⊆ ∩U . So K(∩U ) = ∩U . We then have

K(∩U ) ∈ F . The set F is then closed under arbitrary intersections. The set Fsatisfies condition F2.


− We now show that F is closed under finite unions. We must show that if A andB ∈ F then K(A∪ B) = A ∪B. Consider A,B ∈ F . By K2,

A ∪B ⊆ K(A ∪ B)

A ∪ B ⊆ K(A) ∪K(B) ⇒ K(A ∪B) ⊆ K(K(A) ∪K(B)) (By K2.)

⇒ K(A ∪B) ⊆ K(K(A)) ∪K(K(B)) (By K3.)

⇒ K(A ∪B) ⊆ A ∪B (By K4.)

We conclude that K(A ∪B) = K(A) ∪K(B). The set F satisfies condition F3.

So F is the set of all closed sets in S. This means the topology τK on S is

τK = S\A : where A ⊆ F

Hence the set, F = A ⊆ S : K(A) = A, represents all closed subsets of S (with respectto τK).

b) We now prove the second statement, clS(A) = K(A).

Let A ⊆ (S, τK). Then clS(A) ∈ F . So K(clS(A)) = clS(A). We claim that, from this

we can obtain clS(A) = K(A).

Proof of claim : Let A ⊆ S.

K(K(A)) = K(A) (By K4.)

⇒ K(A) ∈ F

⇒ S\K(A) ∈ τK⇒ K(A) is closed with respect to τK

⇒ clS(A) ⊆ K(A) (By K2, A ⊆ K(A).)

A ⊆ clS(A) ⇒ K(A) ⊆ K(clS(A)) (By K2.)

⇒ K(A) ⊆ clS(A) (Since K(clS(A)) = clS(A))

We conclude that clS(A) = K(A), as claimed.

We have shown that any Kuratowski closure operator K : P(S) → P(S) can be usedto construct a topology, τK , on S in such a way that, for any A ⊆ S, K(A) = clS(A).

We illustrate this in the following example.


Example 5. We consider the set R2, a set with uncountably many elements. We definea function K : P(R2) → P(R2) as follows:

K(A) = A if A is countable

K(A) = R2 if A is uncountable

K(∅) = ∅

Show that K is a Kuratowski operator. Then find the topology on R2 induced by theoperator K.

Solution. Proving that K satisfies the properties K1 to K4 is routine and so is left asan exercise. Then, thus defined, K is a Kuratowski closure operator. By theorem 4.3

the set

F = A ⊆ R2 : K(A) = A = A ⊆ R2 : A is countable ∪ ∅,R2

represents the set of all closed subsets of the topological space (R2, τK). We deduce

that

τK = B ⊆ R2 : R2\B is countable ∪ ∅,R2

Definition We will refer to τK in this example as the cocountable topology on R2.

4.3 The interior of a set.

Given a non-empty set A the “closure of A” has been defined as being the intersectionof all closed subsets of S which contain A. We now wish to consider the union of all

open subsets of S which are entirely contained in A.

Definition 4.4 Let A be a non-empty subset of the topological space (S, τ). We say that a

point x is an interior point of A if there exist an open subset, U , of S such that x ∈ U ⊆ A.We define the interior of A, denoted intSA (or as intS(A)) as follows:

intSA = x ∈ S : x ∈ U ⊆ A for some open U in S.

If A contains no interior points then we will say that the interior, intSA, of A is empty.


Clearly intSA ⊆ A. For example, if A = (2, 4] ∪ 5, intRA = (2, 3). The element 5does not belong to intRA since there is no open interval U in R such that 5 ∈ U ⊆ A.

The fact that 5 is open in A is irrelevant.

The reader is left to verify that intSA can equivalently be described as being...

“the largest open subset of S which is entirely contained in A”

The following theorem shows a relationship between the interior and closure of a set.It also proposes a method to determine the interior of a set, A, by considering the

closure of its complement, S\A.

Theorem 4.5 Let (S, τ) be a topological space and A be a subset of S. Then,

S\intS(A) = clS(S\A)

Proof : Given: (S, τ) is a topological space and A is a subset of S.

Since intS(A) ⊆ A then S\A ⊆ S\intS(A). But S\intS(A) is closed in S, hence

clS(S\A) ⊆ S\intS(A)

Also,

S\A ⊆ clS(S\A) ⇒ S\(clS(S\A)) ⊆ A

⇒ S\(clS(S\A)) ⊆ intS(A)

⇒ S\intS(A) ⊆ (clS(S\A))

We thus obtain S\intS(A) = clS(S\A)

Using this theorem, we let the reader verify that the following three statements are

equivalent:

a) intS(A) = S\clS(S\A),

b) intS(S\A) = S\clS(A),

c) clS(A) = S\(intS(S\A))

Just as for the closure of a set we have four basic similar properties for the interior of sets.


Theorem 4.6 Let (S, τ) be a topological space and A and B be subsets of S.

a) The set, intS(A), is open in S. Also, intS(A) is the largest open subset of S which isentirely contained in A.

b) If B ⊆ A then intS(B) ⊆ intS(A).

c) The set intS(A ∩B) = intS(A)∩ intS(B). (IntS “distributes” over finite intersections.)

d) The set intS(intS(A)) = intS(A).

Proof: The proofs of statements a), b) and d) are left as an exercise.

Proof of intS(A ∩B) = intS(A)∩ intS(B):

S\intS(A ∩B) = clS(S\(A∩ B))

= clS [(S\A)∪ (S\B)]

= clS(S\A)∪ clS(S\B)

= [S\intS(A)] ∪ [S\intS(B)]

= S\[intS(A) ∩ intS(B)]

⇒intS(A ∩B) = intS(A) ∩ intS(B)

Example 6. Given that R is equipped with the usual topology what is intR(Q)?

Solution. We consider the subset, Q, of all rationals in R. By theorem 4.6 part a),intR(Q) ⊆ Q. If intR(Q) is non-empty, it should be a union of non-empty open inter-

vals. But every open interval in R contains an irrational; then intR(Q) = ∅.

Example 7. If R is equipped with the usual topology, then intR([0, 1]) = (0, 1).

Example 8. The set intR(A ∪ B) need not be equal to intR(A)∪ intR(B): If R is

equipped with the usual topology, then

intR ( [0, 1]∪ [1, 2] ) = (0, 2)

while

intR[0, 1] ∪ intR[1, 2] = (0, 1)∪ (1, 2)


4.4 The interior viewed as an operator intS : P(S) → P(S).

Just as for closures of sets we can view “intS” as a function, I : P(S) → P(S).

Definition Let S be a non-empty set and I : P(S) → P(S) be a function satisfying

the properties:

I1. I(S) = S

I2. I(A) ⊆ A, for all A ⊂ S,

I3. I(I(A)) = I(A), for all A ⊂ S,

I4. I(A∩ B) = I(A) ∩ I(B), for all A,B ∈ P(S) (I distributes over finite intersections.)

The function I : P(S) → P(S) satisfying the listed properties is called an interior

operator. We refer to I1 to I4 as being the interior operator axioms.

Topologizing a set S by using an interior operator. Again, just as for the closure oper-

ator, the definition of the function, I : P(S) → P(S), doesn’t refer to any topologyon S. But we will show that the function, I , can be used to define a topology on S by

choosing appropriate sets in its range. We desire a topology, τ , such that A is openin S if and only if A = intSA = I(A).

Theorem 4.7 Let S be a set and suppose I : P(S) → P(S) satisfies the four interior

operator axioms. Define U ⊆ P(S) as

U = A ⊆ S : I(A) = AThen,

a) The set U forms a topology on S.

b) Furthermore, if S is equipped with topology U , intS(A) = I(A), for any A ⊆ S.

Proof: Let S be a set and I : P(S) → P(S) be a function satisfying the four interioroperator axioms I1 to I4 listed above. Let U = A ∈ P(S) : I(A) = A.

a) We are required to prove that U forms a topology on S.

We see that:

− By I1, I(S) = S, so S ∈ U . By I2, I(∅) ⊆ ∅. Since ∅ ⊆ I(∅), thenI(∅) = ∅ and so ∅ ∈ U

− Suppose A and B belong to U . By property I4, I(A ∩B) = I(A) ∩ I(B) =A ∩B. So U is closed under finite intersections.


− To show that U is closed under arbitrary unions we first verify that( A ⊆ B ) ⇒ ( I(A) ⊆ I(B) ) (∗)

A ⊆ B ⇒ A = B ∩A⇒ I(A) = I(B ∩ A) = I(B) ∩ I(A)

⇒ I(A) ⊆ I(B)

Let Aii∈M be a collection of sets in U . It suffices to show I(∪Aii∈M) =

∪Aii∈M .

I(Ai) ⊆ ∪I(Ai)i∈M ⇒ I(I(Ai)) ⊆ I(∪I(Ai)i∈M) (By *.)

⇒ I(Ai) ⊆ I(∪I(Ai)i∈M) (Since Ai ∈ U for all i ∈M .)

⇒ ∪I(Ai)i∈M ⊆ I(∪I(Ai)i∈M)

By I2, I(∪I(Ai)i∈M) ⊆ ∪I(Ai)i∈M . Then I(∪I(Ai)i∈M) = ∪I(Ai)i∈M

so ∪I(Ai)i∈M ∈ U .

Then set U satisfies the three open set axioms O1, O2, and O3. So U is a

topology on S, as required.

We will denote the topology U on S induced by the operator I , by τI .

b) We are now required to show that I(A) = intS(A) with respect to τI .

Suppose A ⊆ S.

By I3, I(I(A)) = I(A), so I(A) ∈ U = τI ; so I(A) is open. Since, intS(A) is thelargest open subset of S contained in A, and I(A) ⊆ A (by I2),

I(A) ⊆ intS(A)

Also see that, since intS(A) ∈ τI = U and intS(A) ⊆ A,

intS(A) = I(intS(A))

⊆ I(A) (By * above A ⊆ B ⇒ I(A) ⊆ I(B))

then intS(A) ⊆ I(A).

We conclude that I(A) = intS(A).

We provide a few examples.

Example 9. Let I : P(R) → P(R) be a function defined as

I(R) = R

I(A) = A\Q otherwise.


a) Show that I : P(R) → P(R), thus defined, is an interior operator on P(R).

b) Use the interior operator described in part a) to define a topology, τI , on R.

c) For the topology, τI, on R shown in part b), describe the open subsets, the closedsubsets of R, the closure of sets and the interior of sets.

d) A Borel sets example: If F represents the set of all closed subsets with respect

to the topology, τI , on R, show that

B = τI ∪ F

is the smallest σ-ring containing τI and so is a family of Borel sets.

Solution.

a) We show I is an interior operator.

1. By definition, I(R) = R so I1 is satisfied.

2. Also, if A 6= R, I(A) = A\Q ⊆ A. So I2 is satisfied.

3. If A 6= R

I(I(A)) = I(A)\Q

= (A\Q)\Q

= A\Q

= I(A)

So I(I(A)) = A. So I3 is satisfied.

4. If neither A nor B is R,

I(A∩ B) = (A∩ B)\Q

= (A\Q)∩ (B\Q)

= I(A) ∩ I(B)

So I(A∩ B) = I(A) ∩ I(B). Then I4 is satisfied.

This means that I : P(R) → P(R) is an interior operator.

b) We now use the interior operator described in part a) to topologize R.

Since I : P(R) → P(R) has been shown to be an interior operator then

τI = A : I(A) = A= A : A\Q = A ∪ R= A : A does not contain any points of Q ∪ R

is a topology on R.


c) For the topology, τI , on R we now describe the open subsets, the closed subsetsof R, the closure of sets and the interior of sets .

Open sets in R. Open sets in R, are R itself and all sets which do not contain

any rationals, including ∅.

For example, if J is the set of irrationals and r ∈ J, then, since r contains norationals, r is an open singleton set. Also, if q ∈ Q, R is the only open set

containing q. So q is not an open singleton set.

Closed sets in R. Suppose B is not R. We claim that B is closed in R withrespect to τI if and only if Q ⊆ B:

Q ⊆ B ⇔ R\B = (R\B)\Q

⇔ R\B = I(R\B)

⇔ R\B is open (with respect to τI )

⇔ B is closed (with respect to τI )

For example, if r ∈ J, since Q 6⊆ r, the singleton set, r, is not a closed set.

Closure of a set . Then taking the closure of a subset A of R comes down to

adding all of Q to A. That is, if A 6= R, clRA = A ∪ Q. For example, if q ∈ Q,since clRq = Q, q is not closed.

Interior of a set in R. IntRA = A ∩ J. Finding the interior of A comes down to

removing any trace of Q in A.

d) We are required to show: The set, B = τI ∪F , is the smallest σ-ring containingτI and so is a family of Borel sets of R.

Given: F is the set of all closed subsets in R with respect to τI . We claim that

B is a σ-ring.

Closure of B under countable unions. If U ∈ τI and V ∈ F then U ∪ V ∈ F(since an open subset union a closed subset containing Q contains Q. So U ∪ Vis a closed subset.)

The set F is closed under countable unions (since closed subsets are those sub-

sets of R which contain all of Q, arbitrary unions of elements of F are closedwith respect to τI). This actually show that F contains all its Fσ’s. Since all

Fσ’s are closed then all Gδ’s are open and so belong to τI .

Trivially, τI is closed under arbitrary unions and so is closed under countable

unions.

Closure of B under “complements”. Let U ∈ B. If U ∈ τI then R\U ∈ F . IfU ∈ F then R\U ∈ τI .

So B is a σ-ring, as claimed. Since it must contains τI and all complements it isthe smallest σ-ring containing τI . By definition, it is a family of Borel sets of R


with respect to τI.1

4.5 The boundary of the subset of a space.

We have seen that, for any subset, A, of a topological space (S, τ),

intS(A) ⊆ A ⊆ clS(A)

Often, clS(A) \ intS(A) 6= ∅. We will now briefly discuss those sets whose pointsbelong to clS(A)\intS(A).

Definition 4.8 Let A be a subset of a topological space (S, τ). We define the boundary ofA, denoted as bdS(A), as

bdS(A) = clS(A)\intS(A)2

The expressions, FrS(A), ∂S(A), BdS(A) are also commonly used to represent the“boundary of A”. It is easily verified that

bdS(A) = clS(A)∩ clS(S\intSA)

Since the finite intersection of closed sets is closed, we see that the boundary, bdS(A)

of a set A, is always closed. Furthermore, for any set A in S, both A and S \A sharethe same boundary (like adjacent neighbours who share the same fence). It is always

the case that intS(A)∩ bdS(A) = ∅ and that intS(A), bdS (A) and intS(clS(S \ A))are pairwise disjoint sets whose union is S. The reader is left to verify this.

Example 10. If Q is viewed as a subset of R equipped with the usual topology then

bdR(Q) = clR(Q)\ intR(Q)

= R\∅

= R

Example 11. If B = [0, 1] is a closed interval viewed as a subset of R equipped with

the usual topology then bdR(B) = 0, 1. It is left to the reader to verify this.

1Note that, if T is the closed interval [1, 3] in R then T 6∈ τI ∪ F (since [1,3] contains some elements ofQ, but does not contain all of Q) and so is not an element of the unique family, B, of all Borel sets, withrespect to τI . We can then refer to it as a “non-Borel set”.

2The word “Frontier” is also sometimes used instead of “boundary”. When the term “Frontier of A” isused, it is denoted by FrSA.


Example 12. Let B = (x, y) ∈ R2 : x2 + y2 < 1be a subset of R2 equipped with thetopology induced by the Euclidean metric. Then

bdR2(B) = clR2(B)\intR2(B) = (x, y) ∈ R2 : x2 + y2 = 1

The reader is left to verify the details.

Example 13. Suppose the set of natural numbers N is equipped with the cofinite

topology.1 Let E denote the set of all even natural numbers. Then

bdN(E) = clN(E)\intN(E)

= N\∅

= N

Example 14. Suppose B = (0, 1) ∪ (1, 2] viewed as a subset of R equipped with theusual topology. We compute the boundary to be,

bdR(B) = clR(B)\intR(B)

= [0, 2] \ (0, 1)∪ (1, 2)

= 0, 2

It is interesting to note that

intR(clR(B)) = intR([0, 2])

= (0, 2) 6= B

4.6 On dense subsets of a topological space.

Suppose B and A are subsets of (S, τ) such that A ⊆ B and B \A contains onlyboundary points of A. When two sets relate to each other in this way we say that “A

is dense in B”. We define this formally.

Definition 4.9 Suppose A and B are subsets in a topological space (S, τ). If

A ⊆ B and B ⊆ clS(A)

then we will say that“A is a dense subset of B”

In the case where B = S, then A is dense in S if and only if clSA = S. If A ⊆ S issuch that intSclSA = ∅ then we say that A is nowhere dense in S.

1The open subsets are those whose complement is finite.


Example 15. Suppose

A = (x, y) : x2 + y2 < 1 and B = A ∪ (0, 1), (1, 0), (0,−1), (−1, 0)

Since

A ⊆ B and B ⊂ clR2A = (x, y) : x2 + y2 ≤ 1then A is dense in B.

Example 16. A nowhere dense subset. Let A = 6 be a subset of R. Then clRA = 6and intT clTA = ∅. So A is nowhere dense in the space R.

Another example: The set of all integers, Z, is nowhere dense in R, since intRclRZ = ∅.

Example 17. Verify that the set C = (x, y) : x2 + y2 = 1 is nowhere dense in R2.

On countable dense subsets of a space.

The property in the following definition is one which refers to an upper bound for the

cardinality of a dense subset of a topological space. Hence, in a way it expresses arestriction on its size.

Definition 4.10 We will say that a topological space (S, τ) is separable if and only if S

contains a countable dense subset.

For example, the topological space, (R, τ), equipped with the usual topology is a sep-

arable space since, clRQ = R, and the subset of all rational numbers is a countablesubset of R. Notice how the border of Q is much larger than Q itself.

We will subdivide the class of all topological spaces into two subclasses: Those thatare separable and those that are not. Determining in which one is not always a trivial

task, specially in the less familiar and complexly defined spaces.

4.7 Topic: Regular open sets and regular closed sets.

Suppose A is a non-empty open subset of a topological space (S, τ). We know thatA ⊆ clSA is always true. Also, it is always true that, if A is open, A = intSA ⊆intSclSA (by part b) of theorem 4.6). So it is always true that

A ∈ τ ⇒ A ⊆ intSclSA (†)


One may wonder whether, when A is open, must we have equality between A andintRclRA. The following simple example will help answer this question.

Let A = (1, 3) ∪ (3, 7), an open subset of R. Then by applying the above reasoning

we have,A = (1, 3)∪ (3, 7) ⊆ intRclRA = intR[1, 7] = (1, 7)

So in the case where A = (1, 7) we do have equality; but if A = (1, 3)∪ (3, 7) we havean open subset, A, such that A 6= intRclRA. Then the best we can then do is affirm

that, when A is open, A is a subset of intSclSA. We have a special name for thoseopen subsets where equality holds.

Definition 4.11 An open subset, A, of a topological space (S, τ) is called a regular opensubset of S if and only if A = intSclSA. In this book, we denote the set of all regular opensubsets of S by Ro(S)

We know that for an open set A, it is always true that, A ⊆ intSclSA. Then

S\( intSclSA) ⊆ S\A ⇒ clS [ S\(clSA) ] ⊆ S\A⇒ clS intS(S\A) ⊆ S\A

So, it is always true that,

F closed ⇒ clSintS(F ) ⊆ F

Then, if A is regular open, A = intSclSA, so for the complement, F = S\A, we cansay that

F = clS intS(F )

So F = clSintS(F ) if and only if the complement of F is regular open.

We have a name for the complements of regular open subsets.

Definition 4.12 A closed subset, F , of a topological space (S, τ) is called a regular closedsubset of S if and only if

F = clS intSF

Hence the regular closed subsets of S are precisely the complements of the regular opensubsets. We denote the set of all regular closed subsets of S by Rc(S).


Example 18. Let S be a topological space and Ro(S) be the set of all regular open

sets in S.

a) Verify that Ro(S) is closed under finite intersections but is not closed under finite

unions.

b) We know that, if A = intSclSA, then A belongs to Ro(S). Suppose A is somenon-empty subset of S which does not belong to Ro(S). Verify that intSclSA

belongs to Ro(S).

c) If Rc(S) denotes the set of all regular closed sets in S, show that Rc(S) is closed

under finite unions.

Solution : Given: Ro(S) is the set of all regular open sets in S.

a) Suppose A and B are open subsets. Then A∩B is open. Suppose A = intSclSAand B = intSclSB. Then

intSclS(A ∩B) ⊆ intS(clSA ∩ clSB) By theorem 4.6

= intSclSA ∩ intSclSB

= A ∩ B

Since A∩B ⊆ intSclS(A∩B) and intSclS(A∩B) ⊆ A∩B then A∩B is regular

open.

So the family of all regular open sets is closed under finite intersections.

On the other hand, (1, 5) and (5, 9) are both regular open but (0, 5) ∪ (5, 9) is

not. So the family of all regular open sets is not closed under finite unions.

b) We are given A ⊆ S. We are required to show that intSclSA is regular open. Seethat

intSclS(intSclSA) ⊆ intSclS(clSA) (Since intSclSA ⊆ clSA)

= intSclSA

Since intSclSA ⊆ intSclS(intSclSA) then intSclSA is regular open.

c) This part is left an exercise.


Concepts review:

1. Given a topological space (S, τ) and T ⊆ S, define the closure, clS(T ), in S, with

respect to the topology τ .

2. Does clS “distribute” over finite unions? How about finite intersections?

3. List the four Kuratowski closure operator axioms, K1 to K4.

4. If K : P(S) → P(S) satisfies the four Kuratowski closure operator axioms describe

a topology on S which can be constructed from K.

5. Define the cocountable topology on R2. Describe the Kuratowski operator used todevelop this topology.

6. Given a topological space (S, τ) and A ⊆ S, define “interior point” of A with respectto τ . Define the interior, intS(A), of A with respect to τ .

7. Does intS distribute over finite unions? How about finite intersections?

8. State the four interior operator axioms I1 to I4 for I : P(S) → P(S).

9. Given a set S, describe a topology that can be constructed on S by using the operatorI : P(S) → P(S).

10. Describe the topology induced on R by the operator I(A) = A\Q.

11. Define the boundary of a set with respect to a topology τ on S.

12. Define what we mean when we say that “A is dense in B”.

13. Define what we mean when we say that “A is nowhere dense in B”.

14. Define a regular open subset and a regular closed subset of a space.

15. Show that for any subset U of S, intSclSU is regular open in S.


EXERCISES

1. Let T = (0, 1), viewed as a subset of R equipped with the usual topology. Show thatclR(T ) = [0, 1]. This is Example 1 on page 49.

2. Suppose S is a topological space induced by the metric ρ (that is, the elements ofτ are unions of open balls of the form Bε(x) = y : ρ(x, y) < ε). Suppose F is

a non-empty subset of S. We define ρ(x, F ) = infρ(x, u) : u ∈ F. Show thatclS(F ) = x : ρ(x, F ) = 0. This is Example 3 on page 49.

3. Consider R equipped with the usual topology τ (induced by the Euclidian metric).Let (Q, τQ) be the set of all rational numbers equipped with the subspace topology

inherited from R. Consider the subset T = [−π, π)∩Q. Determine whether T is openin Q, closed in Q, both open and closed in Q, or none of these.

4. In Example 4 on page 51 it is shown that clS(A)∩ clS(B) 6= clS(A ∩ B) sometimesoccurs. Show that clS(A∩ B) ⊆ clS(A)∩ clS(B) is always true.

5. Let (S, τ) be a topological space and A and B be subsets of S. Show that:

a) intS(A) is open in S. Also, intS(A) is the largest open subset of S which isentirely contained in A.

b) If B ⊆ A then intS(B) ⊆ intS(A).

c) intS(A ∩ B) = intS(A)∩ intS(B).

d) intS(intS(A)) = intS(A).

(This is theorem 4.6. Part c) is already proven.)

6. Let S be a set and I : P(S) → P(S) be an interior operator satisfying the fourconditions I1 to I4 listed on page 57. Show that, if τ = A ∈ P(S) : I(A) = A,(S, τ) is a topological space such that I(A) = intS(A) for all A ∈ P(S).

7. If B = [0, 1] is a closed interval viewed as a subset of R equipped with the usualtopology show that bdR(B) = 0, 1.

8. Let B = (x, y) ∈ R2 : x2 + y2 < 1, x > 0 ∪ (x, y) ∈ R2 : x2 + y2 ≤ 1, x ≤ 0 be asubset of R2 equipped with the topology induced by the Euclidean metric. Show that

bdR2(B) = clR2(B)\intR2(B) = (x, y) ∈ R2 : x2 + y2 = 1

9. Show that A is both open and closed in the topological space (S, τ) if and only ifbdS(A) is empty.

10. Let A and B be two subsets of the topological space (S, τ). Show that if bdS(A)∩bdS(B) = ∅ then intS(A ∪ B) = intS(A)∪ intS(B).


11. Let B = [0, 7)∪ 9 where B is equipped with the subspace topology inherited fromR itself equipped with the usual topology. If A = 0∪ 3 ∪ (5, 7), find i) clB(A), ii)

bdB(A), iii) intB(A).


5 / Bases of topological spaces.

Summary. In this section we define a neighbourhood system of x ∈ S withrespect to a given topology on S. Given a topology, τ , on S, we define a “base

for the topology τ”. We deduce a set-theoretic property called the “base prop-erty” possessed by any base. Those subsets of P(S) which satisfy the described

property will be shown to be a base for some topology on S. We introduce thenotion of a “subbase for the topology τ” by describing its properties. We then

show how to topologize a set both from a collection of sets which possesses the“base property” and also from an arbitrary collection of subsets.

5.1 Neighbourhoods of points.

It is not always easy to confirm that a given subset, S , of P(S) is a well-definedtopology, τ , on a set S. Ultimately, we would prefer a topology on S to be describedin a way that provides some intuitive idea about what its open subsets are like. In

this section we describe a subfamily, B, of open sets from which every open set in τis constructed. With this objective in mind, we begin by introducing the concept of

“neighbourhoods” of a point in (S, τ).

Definition 5.1 Let (S, τ) be a topological space and x ∈ A ⊆ S. We will say that A is a

neighbourhood of x with respect to τ if x ∈ intS(A). For a given x ∈ S, the set

Ux = A ∈ P(S) : A is a neighbourhood of x

is called a neighbourhood system of x with respect to τ .

A subfamily, Bx, of a neighbourhood system, Ux, such that, for any open set, A, containing

x there exists Ux ∈ Bx such that x ∈ Ux ⊆ A, is called a neighbourhood base at x. If everyelement of Bx is open then we can say it is an open neighbourhood base at x.

Note that, since its definition refers to the “interior” of sets, a neighbourhood system is

always expressed with respect to some topology, τ , defined on the whole set S. Also,see that we define two concepts in the above statement. We see that a neighbourhoodbase, Bx, at x is a subset of a neighbourhood system at x. The set, Bx, must be

such that, any open neighbourhood of x, contains an element of Bx. While there canbe only one neighbourhood system at x we will see that there can be more than one

neighbourhood base at this point.

70 Section 5: Bases of topological spaces.

For example, B = [−1, 5) ∪ [6,∞) is a neighbourhood of 0 with respect to the usualtopology of R. The set

U3 = U ∈ P(R) : 3 ∈ intR(U)

is a neighbourhood system of 3. If A belongs to U3, the element “3” must belong tothe interior of A in the sense that 3 cannot be sitting on A’s boundary. For example,[0, 4] ∈ U3, but [1, 3] 6∈ U3. Notice that, for a given x ∈ S, its neighbourhood system,

Ux, with respect to τ is, by definition, unique.

Based on the definition, we can make the following comments about neighbourhoodsof a point in a topological space, (S, τ).

− The empty set, ∅, is not the neighbourhood of any point, so ∅ 6∈ Ux for any

x ∈ S.

− In general, a neighbourhood, U , of a point x is not necessarily open, but it mustcontain x in its interior, intS(U).

− If x ∈ S, then Ux is not empty since S is a neighbourhood of x. If τ is the indis-

crete topology, ∅, S, then S is the only neighbourhood of every point x ∈ S.

The “neighbourhood of a point” concept allows us to come up with another charac-terization of open sets provided we have predefined “interior of a set”. We propose:

“ [ A is open in S ] ⇔ [ x ∈ A⇒ ∃ neighbourhood, Ux, where x ∈ Ux ⊆ A ]”

This will work since this would imply that

A = ∪x∈AintS(Ux) : Ux ∈ Ux and Ux ⊆ A

Or, we could simply say that “A is open in S provided A contains an x-neighbourhood,

Ux, for each point x in A”. This definition of “open set” (in terms of neighbourhoods)is equivalent to its formal definition of “open set”.

5.2 A base for a topology.

In our study of normed vector spaces (as well as of metric spaces) we have seen that,

introducing the notion of “open ball,Bε(x), center x with radius ε”, led to a convenientway of constructing a subfamily, Bx = Bε(x) : ε > 0, of open neighbourhoods of a

point x. Every open subset of S can then simply be described as being the union ofopen neighbourhoods of the form Bε(x) contained in the set. The collection of opensets,

B = ∪Bx : x ∈ Sis sufficient to generate all open subsets of S.


Definition 5.2 Let (S, τ) be a topological space. Suppose B is a subset of τ satisfying the

property:

“For any A ∈ τ , A is the union of elements of a subset, C , of B. ”

We call the subset, B, a base for open sets or an open base for the topology τ (often abbre-viated by simply saying a base for S). The word basis is sometimes used instead of “base”.1

The elements of a base are referred to as basic open sets.2

If F is a family of closed subsets of S satisfying the property:

“For any closed subset B in S, B is the intersection of elements of a subset, C ,of F . ”

we call the subset, F , a base for closed sets of S.

A base for open sets is generally not unique. Given the topological space (R, τ), where

τ is the usual topology, both τ and B = (a, b) : a < b are bases for R.

It is easily verified that if F is a base for closed subsets of S then the family, B, ofall complements of the elements of F will form a base for open sets.

How does one go about constructing a useful base for a topology? A good way tostart is to determine some properties possessed by a known useful open base. The

following theorem characterizes a base for a topology on a set S.

Theorem 5.3 Let (S, τ) be a topological space and B ⊂ τ . Then the following are equiv-alent:

1. The family B is a base for τ .

2. For any x ∈ S, B contains an open neighbourhood base, Bx, of sets at x.

3. Whenever x ∈ U ∈ τ , there exists B ∈ B such that x ∈ B ⊆ U .

1Both words “base” and “basis” are commonly used; thus the reader can assume these have the samemeaning.

2Note that, if A = ∅ ∈ τ , then A is the union of all elements from C = ∅ = ⊆ B. So B alsogenerates the empty set.


Proof: We are given that (S, τ) is a topological space and B ⊂ τ .

(1 ⇒ 2) Suppose the set B is a base for a topology τ on S and x ∈ S. Let

Bx = B ∈ B : x ∈ B

Suppose A is an open subset of S which contains x. It suffices to show that Bx

contains an open neighbourhood U of x such that x ∈ U ⊆ A. By hypothesis and

definition 5.2, A = ∪B : B ∈ B. Then, there exists some Bx ∈ B such thatx ∈ Bx ⊆ A. By definition, Bx ∈ Bx. So Bx is an open neighbourhood base of xcontained in B.

(2 ⇒ 3) This follows immediately from the definition of “neighbourhood base”.

(3 ⇒ 1) Suppose A ∈ τ . If A is empty then A is the union of all open sets in∅ = ∈ B. Suppose x ∈ A. By hypothesis, there exists B ∈ B such that

x ∈ B ⊆ A. Then A is the union of sets from B. So B is a base for open subsets ofS.

5.3 The “base property”.

We have seen how an arbitrary set can be topologized by two different techniques.

One by using a Kuratowski closure operator, the other by using an interior operator.These operators must satisfy certain axioms. If they do, then we can define a par-

ticular topology corresponding to the operator. In this section, we will propose twoother methods for topologizing a set. We will first discuss another characterization ofa base for a topology on a set S.

Theorem 5.4 Let S be a non-empty set and B ⊆ P(S). The set B is a base for a

topology τ on S if and only if S = ∪B : B ∈ B and, if x ∈ A ∩ B for some A,B ∈ B,then there exists C ∈ B such that x ∈ C ⊆ A ∩B.

Proof: We are given that S is a non-empty set and B ⊆ P(S).

(⇒) Suppose the set B is a base for a topology τ on S.Since S ∈ τ then S = ∪B : B ∈ C where C ⊆ B. Since ∪B : B ∈ B ⊆ S then

S = ∪B : B ∈ B.Suppose x ∈ A ∩ B for some A,B ∈ B. Then A,B ∈ τ and so A ∩ B ∈ τ . Since

A ∩ B is open and (by theorem 5.3) B contains an open neighbourhood base, Bx,there exists Bx ∈ Bx ⊆ B such that x ∈ Bx ⊆ A ∩ B. So Bx plays the role of C is

the statement, as required.


(⇐) We are given that B ⊆ P(S) and satisfies the two property: S = ∪B : B ∈ Band if x ∈ A∩B for some A,B ∈ B then there exists C ∈ B such that x ∈ C ⊆ A∩B.

Let T = A ⊆ S : A = ∪C : C ∈ C for some subset C of B.We are required to show that T is a topology on S and that B is base for T .

If we show that T is a topology on S then, by definition, B is a base for T .

O1 The set S belongs to T : Since S = ∪B : B ∈ B then S ∈ T . We claim theset ∅ belongs to T : The union of all elements in ∅ ⊂ B is ∅. So ∅ ∈ T .

O2 Claim : The set T is closed under unions. Let Aα : α ∈ Γ ⊆ T . For α ∈ Γ,

Aα = ∪B : B ∈ Bα ⊆ B. Then ∪α∈ΓAα = ∪α∈Γ∪B∈BαB, a union ofelements in B. So ∪α∈ΓAα ∈ T .

O3 Claim : The set T is closed under finite intersections. Let A,C ∈ T . It suffices

to show that A ∩C ∈ T . Let x ∈ A ∩ C. See that

A = ∪B : B ∈ BA ⊆ B and C = ∪B : B ∈ BC ⊆ B

There exist BA ∈ BA and BC ∈ BC such that x ∈ BA∩BC . By hypothesis, thereexists Bx ∈ B such that x ∈ Bx ⊆ BA ∩BC ⊆ A∩C. Then for every x ∈ A∩Cthere exists Bx ∈ B such that x ∈ Bx ⊆ A ∩B. Then A∩B = ∪x∈A∩CBx; soA ∩C ∈ T , as required.

So T is a topology on S. By definition of T , every element of T is a union ofelements of B so, by definition of “open base”, B is base of T .

In this text, we will call the special property which is satisfied by B ⊆ P(S) in thetheorem statement, the “base property”. We formally define this below.

Definition 5.4.1 : Let S be a non-empty set and let B be a subset of P(S).

Base property : The set B satisfies the base property, if S = ∪B : B ∈ B and, if

x ∈ A ∩ B for some A,B ∈ B, then there exists C ∈ B such that x ∈ C ⊆ A ∩B.

The theorem 5.4 guarantees that, given any non-empty set S, if a subset, B, of P(S)satisfies the base property then B forms a base for some topology, τ , on S. The family,

τ , is made precisely of the arbitrary unions of the elements of subsets of B. In thiscase, we say that B generates the topology τ . The above statement is a powerful tool

for topologizing sets.

Throughout our study of general topology it will be very useful to have a variety ofexamples of topological spaces at hand. The reader is encouraged to take note of

these, or at least bookmark them, for future reference in this book. For this purpose,most are given a particular name so that they can be more easily be referred to in the


index of this book. The following examples, illustrate how the above result is used totopologize a subset of R2.

Example 1. Recall that a Gδ-set is a subset of a space, (S, τ) which is a countableintersection of elements of τ . Let

G = G ∈ P(S) : G is a Gδ-set ⊆ P(S)

(the set of all Gδ-sets in a topological space, (S, τ)). Show that G is an open base forsome topology, τG , on S.

Solution : Since every element of τ is an element of G then S = ∪G : G ∈ G .Suppose A,B ∈ G and x ∈ A ∩ B. Then there exists open subsets, Ui : i ∈ N andVi : i ∈ N, such that A = ∩Ui : i ∈ N and B = ∩Vi : i ∈ N. Then

x ∈ (∩Ui : i ∈ N) ∩ (∩Vi : i ∈ N)= ∩Ui ∩ Vi : i ∈ N (A Gδ.)

⊆ A ∩ B

Since ∩Ui ∩ Vi : i ∈ N is a Gδ, then G satisfies the base property and so generates

a topology, τG .

We will refer to τG as the

“Gδ-topology generated by τ”

Example 2. The Moore plane. Also called, Niemytzki’s topology.

Let S = (x, y) ∈ R2 : y ≥ 0. The set Bε(a, b) represents the usual open ball of

center, (a, b), and radius ε. Let

A = Bε(x0, y0) : x0 ∈ R, y0 > 0, and ε < y0

For each x ∈ R and ε > 0, let

Dε(x, 0) = (x, 0) ∪ Bε(x, ε)

That is, Dε(x, 0) is an open ball, Bε(x, ε), of radius ε tangent to the horizontal axis

at (x, 0) with the point (x, 0) attached to it. Let

D = Dε(x, 0) : x ∈ R, ε > 0

Let B = A ∪ D . Show that B is the base for some topology on S.

Solution : If we show that B satisfies the “base property” then B is a base which gen-

erates a topology, τ , on the half-plane S. It is first easily seen that S = ∪B : B ∈ B.


We consider case 1: Suppose A,B ∈ B and (x, y) ∈ A ∩ B where y > 0. Then thesituation is analogous to what occurs in R2 with the usual topology, and so

(x, y) ∈ Bε(x, y) ⊆ A ∩ B

for some ε > 0.

We consider case 2: Suppose A,B ∈ B and (x, 0) ∈ A ∩ B. Then A = (x, 0) ∪Bε1(x, ε1) and B = (x, 0) ∪ Bε2(x, ε2). Let ε = minε1,ε2

2 and C = (x, 0) ∪Bε(x, ε). Then

(x, 0) ∈ C ⊆ A ∩B

By invoking theorem 5.4, we conclude that B is a base for a topology, τM , on S.

This well-known topological space, (S, τM), is referred to as the Moore plane. Someauthors also refer to this topology, τM , on S, as the Niemytzki’s tangent disc topology .

The topology τM is strictly stronger than τ , the usual topology. To see this, let τrepresent the subspace topology on the set S, inherited from R2 equipped with the

usual topology. We verify that τM is strictly stronger than τ . Let B(0, 0) denote abasic open neighbourhood of the point (0, 0) in τM . Then B(0, 0)∩(0, 0) = (0, 0).The set Bε(0, 0) ∩ S is a half-open disc which contains the interval, (−ε, ε), and socannot be contained in B(0, 0). Then B(0, 0) 6∈ τ so TM 6⊆ τ . On the other hand,

given (u, 0) ∈ Bε(0, 0) ∩ S one can construct a neighbourhood base element, B(u, 0),of radius small enough so that it is contained in Bε(0, 0)∩ S. So τ ⊆ τM .

So the Moore plane topology is strictly stronger than the usual topology on S.

Example 3. The radial plane. Consider the set R2. We will equip R2 with what is

called the radial plane topology. If (x, y) ∈ R2, we define a “radial ball centered at(x, y)”, B(x,y), in R2, as follows:

The singleton set, (x, y), union the set of all open line segments originat-ing at (x, y), precisely one in each direction.

The line segments need not be of the same length. For each (x, y), let

B(x,y) = B(x,y) : B(x,y) is a radial ball centered at (x, y)

Show that the collection, B = ∪B(x,y) : (x, y) ∈ R2, forms a valid base for a topol-ogy, τ∗, on R2. Then show that the topology , τ∗, generated by B is strictly stronger

than the usual topology, τ .

Solution : First part: We claim that B satisfies the “open base property” and

so generates a topology on R2. Clearly, R2 = ∪B(x,y) : (x, y) ∈ R2. Suppose(x, y) ∈ U∩V where U , V ∈ B(x,y). Let KU be a line segment in a particular direction


originating at (x, y) such that KU ⊂ U and KV be a line segment originating at (x, y)pointing in the same direction as KU but KV ⊂ V . Let KU∩V = KU ∩ KV . Then

KU∩V is an open line segment originating at (x, y) which is contained in U ∩V . Thiscan be repeated for all open lines in all directions originating at (x, y) to obtain a

radial ball, M(x,y). So there exists M(x,y) ∈ B(x,y) such that

(x, y) ∈M(x,y) ⊆ U ∩ V

So B satisfies the open base property.

Hence, by theorem 5.4, B is an open base which generates a topology on R2, say τ∗.We are done with the first part of the question.

Second part: We claim that the usual topology, τ , is contained in τ∗. Suppose(x, y) ∈ U where U is an open base element of τ centered at (x, y). Then U = Bε(x, y)

for some ε. Let (a, b) ∈ Bε(x, y). Then there exists δ > 0 such thatBδ(a, b) ⊆ Bε(x, y).Let V be a set such that all open line segments originating at (a, b) are of length less

than δ/2. Then V ∈ B and V ⊆ Bδ(a, b) ⊆ Bε(x, y). So U ∈ τ∗. Then τ ⊆ τ∗, asclaimed.

We claim that τ∗ is strictly stronger than τ . To show this it suffices to show that τ∗

contains an element which does not belong to τ . Consider the sets U = B1(0, 1) and

V = B1(0,−1).1 Then, if L = (x, 0) : x ∈ R, |x| ≤ k the set,

W = U ∪ L ∪ V

representing the two balls U and V with a straight horizontal line through (0, 0) addedto them does not belong to τ . However, the set M(0, 0) which has rays emanating

from (0, 0), in all directions, each of which is contained inW is an open neighbourhoodof (0, 0) in τ∗. So τ is a proper subset of τ∗, as claimed.

5.4 The subbase of a topology.

We have seen that, when given an arbitrary set S, a subfamily of P(S) which satisfiesthe “base property” can be used to generate a topology on S. We shall soon see thatany subfamily family of P(S) can be used to generate a base which, in turn, will

generate a topology, τ , on S. We will refer to such a family as a “subbase” for τ .Even if this may, at first, appear to be a very convenient way to generate topologies

on a set, it does make it more difficult to predict, from a subbase, what the topologygenerated by such a collection of sets will be like. So, for this convenience, there is a

price to pay. We know that we can obtain a topology from a collection of sets which

1The set B1(0, 1) refers to the open ball center (0, 1) and radius 1 while B1(0,−1) refers to the open ballcenter (0,−1) with radius 1.


satisfies the “base property” in a single step: base → τ . But to obtain a topology, τ ,from a subbase requires two steps:

subbase → base → τ

But the notion of a subbase is nevertheless interesting to investigate since, if a spaceis already equipped with a topology, τ , it can be useful to understand what kind of

subbase generates τ . It may allow us to dig deeper into the background of τ to bettersee why τ satisfies certain properties.

Definition 5.5 Let (S, τ) be a topological space. A subbase for the topology τ is a non-

empty subfamily, S , of τ such that the set, B, defined as

B = B : B = ∩U : U ∈ F where F is a finite subset of S

forms a base for τ .

The definition states that, if a base of τ can be obtained from the set of all finite

intersections of the elements of a collection, S , then S , is called a subbase of τ . Thisof course means that every element of a subbase of τ belongs to τ and so is open.

In the following two examples we are given a particular topology on a set. For eachtopological space we identify a subbase for the given topology.

Example 4. A subbase for the usual topology on R. Consider the sets

(a,∞) = x ∈ R : a < x(−∞, b) = x ∈ R : x < b

and the family

S = (a,∞) : a ∈ R ∪ (−∞, b) : b ∈ RWe see that S forms a subbase for the usual topology τ on R since the set of all finiteintersections of elements of S form the set, B = (a, b) : a < b, known to be a base

of τ .

Example 5. Consider (N, τd) where τd is the discrete topology (see the definition on

page 32). Then τd = P(N).

The set of all singleton sets, B = n : n ∈ N, forms a base for τd since every subset

of N is the union of elements from some subset C ⊆ B. If

Na = n ∈ N : n ≤ aMb = n ∈ N : n ≥ b


andS = Na : a ∈ N ∪ Mb : b ∈ N

then B is a subset of the family of finite intersections of elements of S . So S is a

subbase for the topology τd on N.

Example 6. Consider the sets Xa = x ∈ R : a < x and Yb = x ∈ R : x ≤ b and

the family S = Xa : a ∈ R ∪ Yb : b ∈ R. We see that the intersections of finitesubsets of S are either ∅ or of the form (a, b]. So

B = (a, b] : a < b ∪ ∅

forms a base for a topology, τS , on R generated by the subbase S . This topology isreferred to as the

“upper limit topology on R”

or the Sorgenfrey line.1

5.5 A subbase as a generator of a topology.

We now show how any non-empty subset, S , of P(S) will generate a topology, τSon S. This topology, will have S as “subbase”.

Suppose S is a set and S is a non-empty subset of P(S). Let,

J = τj : j ∈ I,S ⊆ τj

denote all topologies on S which contain S . The set J is non-empty since it at leastcontains the discrete topology, τd = P(S), which, by definition, contains S . Sincethe family of all topologies on a set is closed under intersections (see page 33), (but

not necessarily under unions) then the family

τS = ∩τj : j ∈ I, S ⊆ τj

is also a topology on S which contains all elements of S . In fact, it is easily verifiedthat τS is the smallest possible topology on S which contains S . We are now leftwith the task of showing that S indeed satisfies the definition of “subbase” for the

constructed topology, τS , on S.

Theorem 5.6 Let S be a non-empty set and S ⊆ P(S). Suppose

τS = ∩τj : j ∈ I, τj is a topology on S, S ⊆ τj

where I is an indexing set. Then S is a subbase for the unique smallest topology, τS , onS which contains S .

1If B = [a, b) : a < b ∪ ∅, this topology is referred to as the “lower limit topology on R”


Proof: Given: τS = ∩τj : j ∈ I, τj is a topology on S, S ⊆ τj.We have already seen that τS is the smallest topology on S with contains S . If

S ⊆ P(S) let

B = B : B = ∩M∈FM where F is a finite subset of S

To prove that S is a subbase of τS , it suffices to show that B is a basis for a topology

τ on S.

Then to prove that B is a basis for a topology τ on S it now suffices to show that,∪B : B ∈ B = S and that B satisfies the “base property”: That is, if A,D ∈ Band x ∈ A ∩D, there exists E ∈ B such x ∈ E ⊆ A ∩D.

We first claim that ∪B : B ∈ B = S: It suffices to show that S ∈ B. To do this

we must show that S is the intersection of all elements of a finite subset of S . Notethat, the empty set, ∅, is a finite subset of S . Then

∩M∈∅M = x ∈ S : x ∈M for every M ∈ ∅ = S

So S ∈ B, as claimed.

Claim: If A,D ∈ B and x ∈ A ∩D, there exists E ∈ B such that x ∈ E ⊆ A ∩ D.Let A,D ∈ B and A and D be finite subfamilies of S such that A = ∩M∈AM andD = ∩M∈DM . Then A ∪ D is a finite subfamily of S . Suppose x ∈ A ∩D. Then

x = (∩M∈AM) ∩ (∩M∈DM) = ∩M∈A∪DM = A ∩D

If E = ∩M∈A∪DM , x ∈ E ⊆ A ∩D.

Then B is a base for some topology τ on S and so S is a subbase for the topology τ .

Since τS is the smallest topology which contains S then τS ⊆ τ . On the other hand,if U ∈ τ then U is the union of elements of B ⊆ τS . So U ∈ τS and τ ⊆ τS . ThenτS = τ and so S is a subbase of τS .

Example 7. Consider the set, S = [a, b] : a < b, of all non-empty closed andbounded intervals in P(R).

We see that S does not satisfy the “base property” (since there does not exist [x, y]in S such that [x, y] ⊆ [a, b]∩ [b, c]) and so cannot form a base for a topology on R.

However, the theorem guarantees that S is a subbase for some topology on R. De-scribe this topology.

Solution : We see that non-empty finite intersections of elements of S are of the form[c, d] where c ≤ d. In particular, [u, x] ∩ [x, v] = x is an open base element for all


x ∈ R. So the subbase S will generate an open base, B = x : x ∈ R, for thediscrete topology, τd.

We see that the subbase, S , holds more sets then is really required to generate τd.We can reduce its size to

S ∗ = [x− 1, x] : x ∈ R ∪ [x, x+ 1] : x ∈ R

Since, for each x ∈ R, [x− 1, x] ∩ [x, x+ 1] = x, S ∗ still generates τd.

Example 8. Consider two topological spaces (S, τS) and (T, τT). Using the setsS and T we can construct a new set, the Cartesian product S × T , defined as

S × T = (x, y) : x ∈ S, y ∈ T. We can topologize the set S × T by defining asuitable subbase.

We will proceed as follows. Define the two projection functions πS : S × T → S and

πT : S × T → T as πS(x, y) = x and πT (x, y) = y, respectively. We will define assubbase for S × T

S = π−1S (U) : U ∈ τS ∪ π−1

T (V ) : V ∈ τT

where π−1S (U) = U × T and π−1

T (V ) = S× V . By referring to the principle (A×B) ∩(C×D) = (A∩C)× (B ∩D) (verify this!), the basis B induced by this subbase is of

the form

B = U × V : U ∈ τS, V ∈ τT

The topology on S×T whose base is B is called the product topology on S×T . Such

topological spaces created by combining known topological spaces are important andso will be discussed in depth in the next few chapters.

Supplementary definition : Two bases B and B∗ are said to be equivalent bases for a

particular set S if they generate the same topology.

Note that the usual basis for R and the basis for the upper limit topology described inthe example above are not equivalent bases since (a, b] does not belong to the usual

topology on R. However, since

(a, b) = ∪(a, b− 1/n] : n = 1, 2, 3, . . . ∈ τS

then every base element for the usual topology belongs to τS and so . . . “the usual

topology is weaker than τS ”.


In the following example, we illustrate a different topology, τscat on R. The topologicalspace, (R, τscat), is known as the “scattered line” or the “discrete irrational extension

of R”.

Example 9. Consider the real line, R. Let τ denote the usual topology on R and J

denote the subset of all irrationals. Let

τscat = U ∪ V : U ∈ τ and V ⊆ J

Verify that τscat is indeed a topology on R. Find a base for this topology.

Solution. O1. Since ∅ ∈ τ and ∅ ⊆ J, ∅ = ∅ ∪ ∅ ∈ τscat. So ∅ ∈ τscat. SinceR = R ∪ J, then R ∈ τscat.

O2. Let A = Ui ∪ Vi : i ∈ I be a subfamily of τscat. Then

∪A = ∪Ui ∪ Vi : i ∈ I=

⋃

i∈I

Ui ∪⋃

i∈I

Vi

where⋃

i∈I Ui ∈ τ and⋃

i∈I Vi ⊆ J. So ∪A ∈ τscat.

O3. Let D = Ui ∪ Vi : i = 1 to n be a finite subfamily of τscat. Then

(U1 ∪ V1) ∩ (U2 ∪ V2) = (U1 ∩ U2) ∪ [(U1 ∩ V2) ∪ (V1 ∩ (U2 ∪ V2))]

The right-hand side is an element of τscat (where U1 ∩ U2 ∈ τ and (U1 ∩ V2) ∪ (V1 ∩(U2 ∪ V2)) ⊆ J). Proceeding by finite induction on n, we conclude O3 is satisfied.

So τscat is indeed a topology on R.

For neighbourhoods, let A ∈ τscat which contains the rational number q. ThenA = U ∪ V for some U ∈ τ and V ⊆ J. Since q is not irrational, q ∈ U . Then thereexists ε > 0 such that q ∈ (q− ε, q+ ε) ⊆ U ⊆ A. So the family (q− ε, q+ ε) : q > 0forms a neighbourhood base for q in τscat.On the other hand, if r ∈ J, and A ∈ τscat contains r, then r = ∅ ∪ r ∈ τscat. So

r : r ∈ J forms neighbourhood base of r.

Thenr : r ∈ J ∪ (q − ε, q + ε) : ε > 0, q ∈ Q

forms a base for the scattered line.

5.7 Topic: Spaces with countable bases.

Some properties of a topological space, S, may depend on the cardinality of an open

base on S. We will see that some spaces have a countable open base while otherssimply have a countable neighbourhood base at each point. It will be useful to discuss

these notions as soon as possible in the text.


Definition 5.7 Let (S, τ) be a topological space. The topological space, S, is said to be

“first countable”

if and only if every point, x ∈ S, has a countable neighbourhood base. The topologicalspace, S, is said to be

“second countable”

if and only if the smallest of its open bases is a countable base.

Countability properties. We have discussed a countability property which is indirectlyrelated to the topology of a space before. Recall (from definition 4.10) that a space is

“separable”

if it has at least one countable dense subspace. These properties, separable, first count-

able and second countable are often referred to as “the three countability properties”of a topological space. When introduced to an unfamiliar topological space we often

try to determine which property (or properties) apply to the space in question. Webegin by discussing in which ways these properties relate to each other.

Theorem 5.8 A second countable topological space is a first countable space.

Proof: Suppose (S, τ) is a second countable space. Then S has a countable base, B. Then,

for each x ∈ S, Bx = B ∈ B : x ∈ B is a neighbourhood base at x. Since B iscountable and Bx ⊆ B, Bx cannot be uncountable. So (S, τ) is first countable.

The class of all first countable spaces is a subclass of all second countable spaces.

However there are some uncountably large topological spaces which are first count-able but not second countable. The following example illustrates such a space.

Example 10. Suppose τS represents the upper limit topology on R (the Sorgenfreyline) (see example 4 on page 78). Recall that, when R is equipped with this topology,B = (x, y] : x, y ∈ R is an open base for R. Show that (R, τS) is first countable but

not second countable.

Solution : The space, (R, τS) is first countable: For each x ∈ R,

Bx = (x− 1/n, x] : n ∈ N


is a countable neighbourhood base at x. So (R, τS) is first countable.

The space, (R, τS) is not second countable: Let the function φ : R → B be defined as

φ(x) = (x− 1, x]

So φ[R] ⊆ B. If x 6= y, then

φ(x) = (x− 1, x] 6= (y − 1, y] = φ(y)

Since R is uncountable and φ is one-to-one on its domain, R, then φ[R] is an uncount-able subset of B. Also (∞, b ] : a ∈ R ∪ (c,∞) : c ∈ R forms a subbase for τSgenerating a base which contains every element of φ[R]. So B is uncountable.

So (R, τS) is not second countable. As required

Metrizable spaces turn out to be first countable spaces. That is, every point of a

metrizable space has a countable neighbourhood base. We already have the toolsneeded to prove this immediately.

Theorem 5.9 Any metrizable space is first countable.

Proof: Suppose (S, τ) is a metrizable space whose open sets are generated by the metric

ρ. Then the set

Bx = B1/n(x) : n ∈ N, n > 0

(an open ball center x with radius 1/n) forms a countable neighbourhood base at x.

Hence S is a first countable space.

Example 12. Since R (equipped with usual topology) is metrizable then it is first

countable.

Example 13. Verify that the space, R, when equipped with the usual topology, is

second countable.

Solution : We are required to show R has a countable open base.

Suppose U is an open subset of R and x ∈ U . Then there exists an open interval,Bε(x) = (x− ε, x+ ε), such that x ∈ Bε(x) ⊆ U .

We consider two cases.

− If x ∈ Q, then there exists integer m such that x ∈ B 1m

(x) ⊆ Bε(x) ⊆ U .


− If x is an irrational, we know that there is a sequence of rationals which convergesto x, so we can find a rational y in B 1

4m(x), such that

x ∈ B1/4m(y) ⊆ Bε(x)

We have just shown that B = Bm(y) : y,m ∈ Q forms a base for the open sets in

R. Since Q is countable, then so is Q × Q†2 so

|B| = |Bm(y) : (y,m) ∈ Q × Q| = |Q × Q| = ℵ0

We have shown that, when equipped with the usual topology, R has a countable base

for open sets.

Example 14. Recall that the radial plane, (R2, τr), is equipped with what is calledthe radial plane topology, τr, (see the example on page 75). The open sets in the

radial plane are defined as follows: the subset U is an open neighbourhood of q ifand only if q ∈ U , and U is the union of a set of open line segments, precisely one in

each direction, each one originating at q. It is shown on page 75, that this is a validtopology and that it is strictly stronger than the usual topology on R2. Verify thatthe radial plane is not first countable.

Solution : Let p ∈ R2. Suppose (R2, τr) is first countable. Then p has a countableopen neighbourhood base,

Bp = Bi : i ∈ N\0We will construct an element, V , of τr which contains p but does not contain anyelement of Bp.

For each radial set, Bi ∈ Bp, let ri be the length of the longest ray originating at

p which appears in Bi and |ri| denote its length. We then inductively construct asequence of rays, R = r∗i : i ∈ N\0 each originating at p in the direction of ri and

of length|r∗i | = |ri/2i|

Since we have only countably many rays the set,

r∗i : i ∈ N\0 ∪ p

is not yet a complete radial set centered at p. We will add uncountably many morerays of arbitrary length to the ones we have to obtain a complete radial open set, V ,centered at p. No open neighbourhood Bi can be contained in this V (since the rays

become arbitrarily small, shrinking down to p). So Bp cannot be an open neighbour-hood base for p. So p must have an uncountable open neighbourhood base.

2†See appendix on cardinalities.


5.8 Topic: Relating the countable base property with the separable property.

Recall that in definition 4.10, we defined a separable topological space as being a spacewhich has a countable dense subset. We saw, for example, that, since Q is a countable

dense subset of the reals, R is separable.1

In the following theorem, we see that all second countable spaces are guaranteed to

have a countable dense subset. The proof of this statement follows from a more gen-eral result (in part a) ) which states that, given any space S, given any open base, B,

S has a dense subset D whose cardinality, |D|, is less than or equal to the cardinalityof B.

Theorem 5.10 “Second countable ⇒ separable” theorem. Suppose B is an open base ofthe space (S, τS) such that, amongst all bases of S, it is one of least cardinality.

a) There exists in S a dense subset, D such that |D| ≤ |B|.

b) Any second countable topological space is a separable space.

Proof: We are given that B = Bi : i ∈ I is a base such that if any other base hascardinality J, J ≥ |I |.

a) For each i ∈ I , choose xi ∈ Bi. (Choice!)1 Let D = xi : i ∈ I. Then |D| ≤ |B|.We claim that D is dense in S: Let U be a non-empty open subset of S. Then,

for x ∈ U there exists Bi ∈ B such that x ∈ Bi ⊆ U . Then xi ∈ Bi ⊆ U . ThenU ∩D 6= ∅. So every open set in S intersects D. This means that D is dense in S,as claimed.

Then D is a dense subset of S such that |D| ≤ |I |. Since I is the smallest indexingset for any open base, then D has a cardinality which is less than or equal to the

cardinality of any base.

b) It follows immediately from part a) that “second countable property” implies the

“separable property”. For, if |B| = ℵ0, then D is a countable dense subset of S. SoS is separable.

We have just shown that those spaces that have a countable base of open sets musthave a countable dense subset. In general, the converse does not hold true. That is,

there are spaces that have a countable dense subset which do not have a countablebase of open sets. But if S is a metrizable space then the converse holds true, as weshall now see.

1Where R is equipped with the usual topology.1Existence theorems often (but not always) suggest an application of the Axiom of choice in the proof.

Keep an eye open for it.


Theorem 5.11 Separable metrizable spaces are second countable.

Proof: We are given that S is a metrizable separable topological space. Then there is a

metric, ρ, on S such that the space, S, is equivalent to (S, ρ).

Since S is separable, then S has a countable dense subset, D = xi : i ∈ N\0.

For each i and n in N\0 let B(i,n) = B1/n(xi), be an open ball center xi and radius1/n. Consider

B = B(i,n) : i, n ∈ N\0

For x ∈ S, let Ux be an open neighbourhood of x. Then there exists, j ∈ N\0 such

that B1/j(x) ⊆ Ux. Since D is dense in S, B1/j(x) ∩D is non-empty. Say,

xk ∈ B 12j

(x) ∩D

Then

x ∈ B 12j

(xk) ⊆ B 12j

(x) ⊆ Ux

Then B is a countable base for open sets in S. So the separable metric space, S, issecond countable.

The following diagram summarizes the above results.

metrizable + separable metrizable

⇓ ⇓second countable =⇒ first countable

⇓separable

Example 15. From the above statements, we have another proof that the set of all

reals equipped with the usual topology is both separable and second countable.

5.9 Topic : Hereditary topological properties.

Some properties on spaces are carried over from the whole space to their subspaces,

while others are not. Those properties that do are called “hereditary properties”.


Definition 5.12 A topological property, say P , of a space (S, τS) is said to be a hereditarytopological property provided every subspace, (T, τT), of S also has P .

Example 16. Show that metrizability is a hereditary property.

Solution : Suppose (S, τ) is metrizable. Then there exists a metric ρ such that (S, τ)

and (S, ρ) have the same open sets. Suppose T ⊆ S has the subspace topology andρt : T × T → R is the subspace metric on T . Then (T, τt) and (T, ρt) have the same

open sets and so T is metrizable. So subspaces of metrizable spaces are metrizable.

“First countable” is another example of a hereditary topological property. However,

“separable” is not a hereditary topological property. Witness the separable space,R with the usual topology; its subspace of all irrationals, J, has no countable densesubset.

However, the reader may want to verify that, if S is separable and V is an open sub-space of S, then V is separable.

We now show that the “second countable” property is hereditary.

Theorem 5.13 Suppose (S, τS) is a second countable topological space. Then any non-empty subspace of S is also second countable. So “second countable” is a hereditary prop-

erty.

Proof : Suppose (S, τS) has a countable base B = Bi : i ∈ N. Suppose (T, τ) is a non-empty subspace of S. Let U be an open subset of T . Then there exists an open subset

U∗ of S such that U = U∗∩T . Then there existsN ⊆ N such that U∗ = ∪Bi : i ∈ N.Then U = ∪Bi ∩ T : i ∈ N. So BT = Bi ∩ T : i ∈ N is a countable basis of T .Hence T inherits the second countable property from its superset S.

5.10 Topic : Ordinal space.

Ordinal numbers play a useful role in general topology. This “linearly well-ordered”set often serves as a counterexample to certain conjectures . Recall that a linearly

ordered set is said to be “well-ordered” if,

“...every non-empty subset contains its least element.”


We state some of the most fundamental facts about ordinal numbers.

Definition. An ordinal number is a set whose elements are themselves sets.1 We say

that the set, α, is an ordinal number if it satisfies the two properties:

− If u, v are elements of α, they are themselves “sets of sets” such that either u ∈ v,v ∈ u, or u = v

− If u ∈ v, and v ∈ α, then u ∈ α.

The elements of an ordinal number, α, are linearly ordered by “∈ or =”. The first

countable ordinal numbers are in fact the natural numbers. Each ordinal is the setwhose elements are all of its predecessors.

∅ = 0

∅ = 0 = 1∅, ∅ = 0, 1 = 2∅, ∅, ∅, ∅ = 0, 1, 2 = 3

∅, ∅, ∅, ∅, ∅, ∅, ∅, ∅ = 0, 1, 2, 3= 4...

...n ∪ n = 0, 1, 2, . . . , n = n+ 1

......

0, 1, 2, 3, . . . = ω

ω ∪ ω = ω + 1...

...

See that n + 1 = 0, 1, 2, 3, . . . , n is both a subset of n + 2 and an ordinal con-tained in the next ordinal, n+ 2 = 0, 1, 2, 3, . . . , n, n+ 1. Also see that any n+ 1 is

both a subset of ω and an element of ω (where ω is the first countably infinite ordinal).

All elements of an ordinal are themselves ordinals (a statement which requires proof).Those ordinals that do not contain a maximal ordinal (such as ω, for example) are

referred to as limit ordinals. These can also be recognized by the fact that limitordinals don’t have an immediate predecessor. The limit ordinal, ω, can be expressed

as a half-open interval,

ω = [0, ω) = ordinal α : α < ω

where “α < ω” is interpreted as α ∈ ω. A non-limit ordinal, β + 1 = 0, 1, 2, . . . , βcan be expressed as an interval

β + 1 = [0, β] = [0, β + 1) = α : α ≤ β1The family of all ordinal numbers is too large to be called a “set”. So we refer to it as the class of all

ordinal numbers.


where “α ≤ β” is interpreted as “α ∈ ω or α = β”.

In the case of non-limit ordinals, β + 1 = [0, β],

∪ (β + 1) = ∪[0, β] = ∪α : α “∈or=” β = β

sup (β + 1) = sup [0, β] = supα : α “∈or=” β = β

Limit ordinals are quite different in nature. We can recognize non-zero limit ordinalsby the following characterization. The following three statements are equivalent:

− The ordinal, γ = [0, γ) = α : α ∈ γ, is a limit ordinal.

− For the ordinal γ, ∪γ = ∪[0, γ) = γ.

− For the ordinal γ, sup γ = sup[0, γ) = γ.

From this characterization, we can eventually argue (this is not immediate) that theunion of all countable ordinals must be an uncountable ordinal, ω1, which we refer to

as the first uncountable ordinal. The ordinal ω1 ∪ ω1 = ω1 + 1 is its successor.

5.11 Topic : Topologizing an ordinal set.

We will topologize an initial segment of ordinals, S = [0, ωα], (either with an open orclosed right end) by defining an appropriate subbase, S , which will generate a col-

lection, B, satisfying the “base property”. The collection, B, will, in turn, generatea topology on S.

Definition 5.14 Let ωγ be an ordinal and S = [0, ωγ] = ordinals α : α ≤ ωγ.Suppose β and µ are both ordinals which belong to S. Let

Sµ = (µ, ωγ] = α ∈ S : α > µSβ = [0, β) = α ∈ S : α < β

The standard subbase of S is defined as,

S = Sµ : µ ∈ S ∪ Sβ : β ∈ S = (µ, ωγ] : µ ∈ S ∪ [0, β) : β ∈ S

This subbase will generate a base, B, which in turn will generate the topology, τω, of S.

When the space S is equipped with the topology τω, (S, τω), is referred to as an

“ordinal space”

The topology that is generated by this subbase is called the

“interval topology on the set of ordinals”


A few facts about an ordinal space. When we say “ordinal space” we mean a setof ordinals with the topology generated by the described subbase S . But the best

way to memorize the topology of the ordinal space is to remember what the elementsof its base for open sets look like. Remember that there are two types of ordinals:

limit ordinals and successor (non-limit) ordinals. Every ordinal number, α, withoutexception, has an immediate successor, α + 1, by definition. Some ordinals, γ, have

an immediate predecessor, say β, provided

γ = β + 1

In this case, sup [0, γ) = sup [0, β + 1) = sup [0, β] = β.

While some ordinals, µ don’t have an immediate predecessor (limit ordinal). In thiscase,

sup δ : δ < µ = sup [0, µ) = µ

So when we consider the intersection of two elements, (µ, ωγ] and [0, β), of the subbase,S , we get the open-ended interval (µ, β). At this point, we see only two possible cases

for β:

– Case 1 : Suppose β has an immediate predecessor, say γ (because β = γ + 1).

In this case, we can express (µ, β) as the half-open interval, (µ, γ].

– Case 2 : Suppose β doesn’t have an immediate predecessor. Then β = δ : δ <β. In this case, (µ, β) can be expressed as the half-open interval, (µ, β].

In both cases we have an open base element which is a half-open interval. So weconclude that a base, B, for open sets in the ordinal space, S = [0, ωγ], is the set

B = (α, β] : α, β ∈ S, α < β

This is easier to remember.1

5.12 Topic : On bases generated by regular open sets.

Suppose we are given a topological space (S, τ). Recall that, in 4.12, we defined:

An open subset, U , of S is called a regular open subset if it satisfies the

property,U = intSclSU

1A more detailed study of the ordinals in the context of set theory is found in Axioms and set theory, byRobert Andre (can be found online)


In the expression, intSclSU , the interior and closure are with respect to the topologyτ on S. The symbol

Ro(S) = U ∈ τ : U is regular open

represents the set of all regular open subsets of S. So Ro(S) ⊆ τ . But since Ro(S)

is not closed under unions (see the example on page 65) then it is not, by itself, atopology on S.

Clearly, ∅ and S belong to Ro(S). On page 65, we showed that Ro(S) is closed underfinite intersections. Then, if x ∈ S and x ∈ A ∩ B where A,B ⊆ Ro(S), since

A ∩B ∈ Ro(S), then Ro(S) satisfies the “base property”. Then, by theorem 5.4,

Ro(S) is an open base for some topology, say τs, on S

The elements of Ro(S) are all open with respect to τ but not all elements in τs arenecessarily regular open in S. Furthermore there may be some open sets in τ whichare not unions of elements in Ro(S) and so these are not in τs. So we have possibly

distinct topologies, τs and τ , where,

Ro(S) ⊆ τs ⊆ τ

Note that the topology generated by Ro(S) is weaker than τ , but under certain con-ditions, τ and τs may be equivalent topologies.

Definition 5.15 Let (S, τ) be a topological space and τs denote the topology whose base isRo(S). We have seen that τs and τ may be distinct topologies for the same underlying set, S.

If τs is a proper subset of τ , the topological space, (S, τs), is called the semiregularizationof S with respect to τ .

If τ = τs, then we will call (S, τ) a semiregular topological space.1 That is, a semiregular

topological space is a space such that for all U ∈ τ , U is the union of regular open sets.2

The semiregularization of, (S, τ), is akin to running τ through a strainer where only

the elements of τs successfully make it through. The reader should verify that the setof real numbers, R, equipped with the usual topology is an example of a semiregular

topological space. Hence it is its own semiregularization. We will refer to variousproperties of semiregular spaces further on in the text.

1We will specify later in this text that, in the literature, semiregular spaces are assumed to be Hausdorff.But for now this is irrelevant.

2The choice of the name “semiregular space” suggests that we will be introduced to a “regular space” atsome point. This is the case. We will show in chapter 9 that “regular spaces” (to be defined in that chapter)are semiregular spaces, but there exists semiregular spaces which are not regular spaces.


5.13 Topic : Spaces with a base of clopen sets.

We now consider the set, B(S) = U ∈ P(S) : U is clopen, of all clopen sets in thetopological space, (S, τ). The set B(S) is never empty since ∅ and S are elements of

that set. Furthermore, if U and V belong to B(S) and x ∈ U ∩ V , given that U ∩ Valso belongs to B(S) then B(S) satisfies the “base property”.

This means that B(S) forms a base for some topology, τb, on S. Since B(S) ⊆ τ , then

the topology, τb, is weaker than τ , but may, under certain conditions, be equivalentto it.

Definition 5.16 Let (S, τ) be a topological space and τb denote the topology whose baseis

B(S) = U ∈ P(S) : U is clopenIf τ = τb, then we say that (S, τ) is a zero-dimensional topological space.

Example 17. Consider the subspace, (Q, τ), of rational numbers with the subspace

topology of R itself equipped with the usual topology. Verify that Q is a zero-dimensional topological space.

Solution : Consider U = (a, b) : a < b, a and b irrationals ⊆ R. This forms a basefor open sets in R. Then UQ = U ∩ Q : U ∈ U forms a base for open sets for Q.

Each element of UQ is clopen in Q so Q is zero-dimensional.

This example confirms that there are non-discrete zero-dimensional spaces.

Concepts review:

1. Define a neighbourhood system of x with respect to τ .

2. Define a neighbourhood base of x with respect to the topology τ .

3. Is ∅ a neighbourhood of a point x? Is a neighbourhood of x necessarily open?

4. Define a base for a topology τ .

5. Find a base for the usual topology τ on R.

6. Give a characterization of a base of τ in terms of “neighbourhoods”.


7. What does it mean to say that a subset P(S) satisfies the “base property”?

8. Describe the Moore plane and the base for its topology.

9. Given an arbitrary subset S of P(S) explain how it can be used to construct atopology on S.

10. Define a subbase for a topology.

11. Describe a subbase for the usual topology on R.

12. Describe a topology generated by a subbase S in terms of other topologies on S.

13. Given two topological spaces S and T and a Cartesian product S × T . Find a useful

subbase involving projection maps that can be used to generate a topology on S× T .

14. Describe the upper limit topology (or Sorgenfrey topology) on R in terms of its baseand subbase.

15. Is R with the upper limit topology first countable? Is it second countable.

16. What does it mean to say that two subsets of P(S) are equivalent topologies for the

set S.

17. What can we say about the size of the neighbourhood bases at the points of R withrespect to the usual topology?

18. What can we say about the size of the base of R with respect to the usual topology?

19. Define first countable topological space.

20. Define second countable topological space.

21. Is R equipped with the usual topology first countable? What about second countable?

22. Describe a topological space which is first countable but not second countable.

23. State a relationship between “second countable” and “separable”.

24. Describe the topology on an ordinal space.

25. Metrizable space are necessarily first countable. Describe a neighbourhood base.

26. Describe how the set of regular open subsets can be used to generate a topology on aspace.

27. What does it mean to say that a property is hereditary?

28. Define a semiregular space.


29. Given a topological space, (S, τ), what is the semiregularization of (S, τ)?

30. Define a zero-dimensional space. As an example provide a subspace of R which iszero-dimensional.

EXERCISES

1. Show that, if F is a closed subset of the topological space (S, τ) and x 6∈ F , then x

has a neighbourhood which does not intersect F .

2. Let ~x = (a, b) ∈ R2 equipped with the usual (Euclidean) topology. For each q ∈ R, let

B~x(q) = (x, y) : max|a− x|, |b− y| < q. Show that B~x = B~x(q) : q ∈ R, q ≥ 0forms a neighbourhood base at ~x.

3. Suppose τ1 and τ2 are two topologies on the set S with respective bases B1 and B2.

Show that τ1 ⊆ τ2 if and only if whenever x ∈ B1 there exists B2 ∈ B2 such thatx ∈ B2 ⊆ B1.

4. Let A and B be two infinite sets each equipped with the cofinite topology. Describea basis of A× B equipped with the product topology.

5. Let (S, τ) be a topological space with base B. If A ⊆ S, show that BA = B ∩ A :

B ∈ B is a base for the open sets of A.

6. Let (S, τS) and (T, τT ) be two first countable topological spaces. Show that S × T

equipped with the product topology is first countable.

7. Prove that, if (A, τ1) is a subspace of (B, τ2) and (B, τ2) is a subspace of (C, τ3) then(A, τ1) is a subspace of (C, τ3).


6 / Continuity on topological spaces.

Summary. In this section we formally define the notion of a continuous func-

tion mapping one topological space into another. We discuss various charac-terizations of these. An important subclass of continuous functions is the one

called “homeomorphisms”. We will see why these are fundamental in the studyof topological spaces. Finally, we examine how a continuous function and a

topological space can be used to topologize set.

6.1 Basic notions and notation associated to functions mapping sets to sets.

We begin by establishing the notation and terminology we will use in our discussion

of functions. Suppose f : S → T is a well-defined function mapping a topologicalspace, S, into another topological space, T . For f : S → T the expressions, “f is

one-to-one and onto” and “f is a bijection between S and T” are simply differentways of conveying the same idea.

The function, f : S → T , induces another function, f : P(S) → P(T ), where

f [A] = y ∈ T : y = f(x) for some x ∈ A ⊆ S

Essentially, f [A] is the image of the set A under the function f : S → T . The functionf : S → T also induces the function f← : P(T ) → P(S) where

f←[B] = x ∈ S : where f(x) ∈ B

We can also say that, if f←[B] = D, then D is the “inverse image” or “pre-image”of B under f , or that the function f← “pulls back” the set B onto the set D inside

the domain S of f . In the case where f←[y] = x, if there is no risk of confusion,we will simply write f←(y) = x.1 In the case where f : S → T is one-to-one and

onto T then f← can itself be seen as a well-defined function, f← : T → S, and sowe can write “f←(x) = y if and only if f(y) = x” (without using the square brackets).

In this book, if f : S → R is a function and 0 6∈ f [S], f−1 will be interpreted as

follows:

f−1(x) =1

f(x)

In the following theorem statement, we review some basic principles on how functionsact on sets. In particular, we are reminded of the following three principles:

1) A function “respects arbitrary unions of sets”.

1Note that f need not be one-to-one on all of the domain in order for us to speak of f← in this way.

96 Section 6: Continuity on topological spaces

2) If a function is one-to-one it will “respect arbitrary intersections”. Otherwise, afunction “does not always respect intersections of sets”.

3) An inverse function, f←, “always respects unions, intersections and complementsof sets”.

Theorem 6.1 Let f : A → B be a function mapping the set A to the set B. Let A be a

set of subsets of A and B be a set of subsets of B. Let D ⊆ A and E ⊆ B. Then:

a) f[⋃

S∈AS]

=⋃

S∈Af [S]

b) f[⋂

S∈AS]

⊆ ⋂S∈Af [S] where equality holds true only if f is one-to-one.

c) f [A\D] ⊆ B\f [D]. Equality holds true only if f is one-to-one and onto B.

d) f←[⋃

S∈BS]

=⋃

S∈Bf← [S]

e) f←[⋂

S∈BS]

=⋂

S∈Bf← [S]

f) f← [B\E] = A\f← [E]

Proof:

a) x ∈ f

[

⋃

S∈A

S

]

⇔ x = f(y) for some y ∈⋃

S∈A

S

⇔ x = f(y) for some y in some S ∈ A

⇔ x = f(y) ∈ f [S] for some S ∈ A

⇔ x ∈⋃

S∈A

f [S]

b) It will be helpful to first prove this statement for the intersection of only two setsU and V . The use of a Venn diagram will also help visualize what is happening.

So we first prove the statement: f [U ∩ V ] ⊆ f [U ] ∩ f [V ] with equality only if f

is one-to-one on U ∪ V .

Case 1: We consider the case where U ∩ V = ∅.

Then f [U ∩ V ] = ∅ ⊆ f [U ] ∩ f [V ]. So the statement holds true.

Case 2: We now consider the case where U ∩ V 6= ∅.

x ∈ f [U ∩ V ] ⇔ x = f(y) for some y ∈ U ∩ V⇔ x = f(y) for some y contained in both U and V

⇒ x = f(y) ∈ f [U ] and f [V ]

⇔ x ∈ f [U ] ∩ f [V ]

We now show that if f is one-to-one on U ∪V , then f [U ]∩f [V ] ⊆ f [U ∩ V ] andso equality holds true.


− Suppose x = f(y) ∈ f [U ] ∩ f [V ]. Then there exists u ∈ U and v ∈ V suchthat f(u) = f(v) = f(y). Since f is one-to-one, u = v = y. This implies

y ∈ U ∩ V . Hence, f [U ∩ V ] = f [U ] ∩ f [V ].

The proof of the general statement is left as an exercise.

c) Proof is left as an exercise.

d) x ∈ f←

[

⋃

S∈B

S

]

⇔ x = f(y) for some y ∈⋃

S∈B

S (By definition of f←.)

⇔ x = f(y) for some y in some S ∈ B

⇔ x ∈ f←[y] ⊆ f←[S] for some S ∈ B

⇔ x ∈⋃

S∈B

f← [S]

Thus, f←(⋃

S∈BS)

=⋃

S∈Bf← (S).

e) Proof is left as an exercise.

f) Proof is left as an exercise.

6.2 Continuous functions on topological spaces.

Given two topological spaces, (S, τS) and (T, τT), we will discuss various types of

functions, f : S → T , which map S into T . The reader is already familiar with thosefunctions called “continuous functions” mapping R to R. We will generalize this no-

tion of continuity to topological spaces.

Our formal definition of a continuous function mapping a topological space into an-

other is presented below. Those readers who are familiar with the “epsilon-delta”definition of a continuous function (normally presented in any Introduction to anal-

ysis course) will notice the analytical approach cannot be used in topology sincetopological spaces, in their most rudimentary form are not equipped with distance

functions such as absolute values, norms or metrics.

Definition 6.2 Let f : S → T be a function mapping (S, τS) into (T, τT). We say that

“f : S → T is continuous on S if, for any open subset U ∈ T , f←[U ] is open in S”

If x ∈ S, we will say that “f is continuous at the point x” if, for any neighbourhood, U , off(x) (inside T ) there exists a neighbourhood V of x such that f(x) ∈ f [V ] ⊆ U .


So a function f : S → T is continuous if f pulls back open sets in T to open sets in S”.

The above definition has two parts to it. The first describes continuity of f on a setwhile the second describes continuity of f at a point x. Clearly, we cannot apply

the first definition to determine continuity of f at a point. But a set A is simply acollection of points. If a function f can be shown to be continuous at every point x

in a set A we would hope that, we would obtain continuity on the set as defined inpart a). The next theorem confirms that this is the case.

Theorem 6.3 Let (S, τS) and (T, τT) be two topological spaces and f : S → T be a

function. Then f is continuous on S if and only if f is continuous at every point of S.

Proof :

(⇒) Suppose f is continuous on S, x ∈ S and y = f(x) ∈ f [S]. We are required to

show that f is continuous at the point x.

Suppose U is a neighbourhood of y. By definition of continuity on a set, f←[intTU ]

is an open neighbourhood of x ∈ f←[y] ⊆ f←[intTU ] in S. By definition ofneighbourhood, there exists an open set, V ⊆ f←[intTU ], such that x ∈ V . Then

y = f(x) ∈ f [V ] ⊆ U . So we have found the required neighbourhood, V , of x. So fis continuous at the point x.

(⇐) Suppose that f is continuous at every point of S. Let U be a non-empty opensubset of f [S] ⊆ T . We are required to show that f←[U ] is open in S.

Let x ∈ f←[U ]. Then f(x) ∈ U . Since f is continuous at x then there exists a neigh-

bourhood V of x such that f [V ] ⊆ U . Now x ∈ intTV ⊆ f← [f [V ]] ⊆ f←[U ]. Thenf←[U ] is an open subset of S. So f is continuous on S.

There are other ways of recognizing those functions which are continuous on a set.

For example, if f : S → T satisfies the property,

“f←[F ] is closed in S whenever F is closed in the codomain T”

then, when U is open in T , S \f←[U ] = f←[T \U ] is closed in S. Then f←[U ] is open

in S and so f is continuous on S. The reader is left to verify that the converse alsoholds true.

Other useful characterizations of continuity on a topological space are given below.


Theorem 6.4 Let (S, τS) and (T, τT) be two topological spaces and f : S → T be afunction.

a) The function f is continuous on S if and only if f pulls back subbase elements of thetopological space T to open sets in S.1

b) The function f is continuous on S if and only if f pulls back open base elements of

the topological space T to open sets in S.2

c) The function f is continuous on S if and only if, for any subset U of S,

f [clSU ] ⊆ clT f [U ]

Proof : The proofs of parts a) and b) are left as an exercise.

c) ( ⇒ ) Suppose f is continuous on S. To show that f [clSU ] ⊆ clT f [U ] it will sufficeto show that x 6∈ clTf [U ] implies x 6∈ f [clSU ].

Suppose x 6∈ clT f [U ]. Then there musts exist an open V in T such that x ∈ V ⊆T \clT f [U ]. Then f←(x) ⊆ f←[V ] ⊆ S \U . By continuity of f , f←[V ] is open in Sand so clSU ∩ f←[V ] = ∅. Then f [f←(x)] = x 6∈ f [clSU ]. So f [clSU ] ⊆ clT f [U ].

The proof of ( ⇐ ) is left as an exercise.

Suppose A is a subspace of the topological space, (S, τ), equipped with the subspace

topology, τA. Suppose f : S → T is known to be continuous on its domain. Then,when f is restricted to f |A on A, f |A preserves the continuity property on A. This is

confirmed by the following theorem statement.

Theorem 6.5 Let (S, τS), (T, τT) and (Z, τZ) be topological spaces.

a) If f : S → T and g : T → Z are both continuous on their domains then g f : S → Z

is continuous on S. (That is, the composition of continuous functions is continuous.)

b) Suppose f : S → T is a continuous function on S and A ⊆ S. Let f |A : A→ T denotethe restriction of f to the subset A. Then f |A is continuous on A.

1That is, f←[B] is open in S whenever B is a subbase element of T .2That is, f←[B] is open in S whenever B is a base element of T .


Proof : a) The proof of part a) is left as an exercise.

b) Let U be an open subset of f [A] with respect to the subspace topology τf [A] in T .Then there exists U∗ ∈ τT such that U = U∗ ∩ f [A]. Now

x ∈ f |←A [U ] ⇒ x ∈ A : f(x) ∈ U⇒ x ∈ x ∈ S : f(x) ∈ U∗ ∩ A⇒ x ∈ f←[U∗] ∩A

Since f is continuous on S, f←[U∗] ∩ A is an open subset of A with respect to the

subspace topology. So f |←A [U ] is open in A.

We have seen that continuous functions are those functions which “pull back” opensets to open sets, or, equivalently, “pull back” closed sets to closed sets. We will

encounter at least two other similar types of functions, which are not necessarily con-tinuous.

Definition 6.6 Let f : S → T be a function mapping the topological space, (S, τS), into

the space, (T, τT). We say that f : S → T is an open function on S, if for any open subsetU of S, f [U ] is open in the space, (T, τT). We say that f : S → T is a closed function on

S, if for any closed subset F of S, f [F ] is closed in the space, (T, τT).1

The reader should be alerted to the fact that an open function need not be a contin-uous function; similarly, a closed function need not be continuous. Also, we caution

the reader by pointing out that, even if f : S → T is an open function, it does notnecessarily follow that f← : T → S is a continuous function on T . (See one of the

examples that will follow.) However, for one-to-one functions f , “f is closed if andonly if f is open” is true, as we shall now prove.

Supplementary theorem. Suppose f : S → T is one-to-one and onto T . Then f is anopen function if and only if f is a closed function.

Proof : ( ⇒ ) Suppose f : S → T is a one-to-one and onto open function. We are

required to show that f is also a closed function. Let F be a closed subset of S.Then U = S\F is open in S and so f [U ] is open in T . Then

f [U ] = f [S\F ]

= f [S]\f [F ] (Since f is one-to-one.)

= T \f [F ]

1Whether the function f : S → T is open or closed depends on the topology defined on S and T .


Since T \f [F ] is open in T then f [F ] is closed in T . We conclude that f is aclosed function, as required.

Proof of the converse ( ⇐ ) is left to the reader.

Example 1. Suppose (R, τ) denotes the real line with the usual topology, τ , and (R, τS)

denotes the real line with the upper limit topology (Sorgenfrey line), τS .

Let i : (R, τ) → (R, τS) denote the identity map, i(x) = x. Verify that this identity

map is open on (R, τ) but is not continuous on its domain.

Solution : Since the function i(x) = x maps the open base element, (a, b), of τ to the

open set (a, b) ∈ τS (as we have seen earlier, τ ⊂ τs) then i maps open sets to opensets hence, i is an open function. But (a, b ] 6∈ τ . So i←[(a, b ]] = (a, b ] 6∈ τ . So the

open identity map i is not continuous on R with respect to τ .

This example illustrates that, if the codomain has more open sets then the domain,

then the identity function will not pass the test of continuity.

Example 2. Let (R2, τ) be equipped with the usual topology. Suppose the open ball

center (0, 0) of radius 1,

B = (x, y) ∈ R2 : x2 + y2 < 1

is equipped with the subspace topology, τB. We define the function f : B → R2 asfollows:

f(x, y) = (x, y)

If U is an open subset of B then f [U ] = U ∩ B an open subset of R2. So f is anopen map. But f is not a closed map. To see this note that B is a closed subset of it-self with respect to τB. On the other hand, f [B] = B is not closed in the codomain R2.

If f is one-to-one and onto, we arrive at a different conclusion, as we shall now see.

Theorem 6.7 Let (S, τS) and (T, τT) be topological spaces. For a one-to-one and onto

continuous function f : S → T ,

f← : T → S is continuous ⇔ f : S → T is open

Proof : The proof is left as an exercise.


6.3 Homeomorphic topological spaces.

For a continuous function, f : S → T , its inverse function, f← : T → S, may, or may

not, be continuous, even if f is one-to-one. Those one-to-one continuous functionsf : S → T where f← is continuous on f [S] ⊆ T are fundamental in the study of

topology. They have a special name.

Definition 6.8 Let (S, τS) and (T, τT) be topological spaces and f : S → T be a function.If f simultaneously satisfies all three of the following conditions,

1. f is one-to-one on S and onto T

2. f is continuous on S

3. f← is continuous on T

then the function, f , is called a homeomorphism from S onto T . If f : S → T is ahomeomorphism then S and T are said to be homeomorphic topological spaces, or f is said

to map S homeomorphically onto T .

Example 3. Let the open interval, S = (−π/2, π/2), be equipped with the usual sub-space topology. The one-to-one and onto function, tan : S → R, is well-known to

be continuous on its domain, S. Similarly its inverse (arctan) , tan← : R → S, iscontinuous on its domain, R. By definition, tan is a homeomorphism. We can then

say that S = (−π/2, π/2) and R are homeomorphic topological spaces. In fact, as weshall see, any non-empty open interval of R is homeomorphic to R.

Theorem 6.9 Let (S, τS) and (T, τT) be topological spaces. Suppose f : S → T is one-to-

one and onto S. The following are equivalent:

1. The function f is a homeomorphism.

2. The function f is both continuous and open.

3. The function f is both continuous and closed.

Proof : The proof follows from the statement in theorem 6.7. It is left as an exercise.


When the function f : S → T is one-to-one and open but is not onto T , the pair Sand T cannot be said to be homeomorphic spaces. In this case S is homeomorphic

to the proper subspace, f [S], of T . We often express this situation by saying that “fembeds S in T”. We define this formally.

Supplementary definition. Let (S, τS) and (T, τT) be topological spaces and f : S →T be a function mapping S into T (which may or may not be “onto”). If f is a

homeomorphism then we say that

“f embeds S into T”

or, simply that “f is an embedding”. When the specific homeomorphism in question

is not explicitly described, we might simply say that

“T contains a homeomorphic copy of S”

This homeomorphic copy is, of course, the image of the hypothesized homeomorphism.

Also note that, if f : S → T embeds S into T , then f← : f [S] → S is also a homeo-morphism.

We have seen that, if f : S → T is a homeomorphism between the two topologicalspaces (S, τS) and (T, τT), the one-to-one function, f , maps each open base element

in τS to a unique set in τT , and vice-versa. Since f is one-to-one it respects botharbitrary unions and arbitrary intersections; then every element of τS will be paired,under f , to exactly one element in τT . This suggests that some properties in S which

involve open sets will be “mirrored” inside T . We will refer to such properties as“topological properties”. We formally define this notion.

Definition 6.10 Suppose P is property which is satisfied on a topological space S. If P

is satisfied in any homeomorphic copy of S, then P is called a topological property or atopological invariant.

If f : S → T is known to be a homeomorphism and P is known to be a topologicalproperty, then, by definition,

P is satisfied in f [S] if and only if P is satisfied in S

Example 4. Recall that in a metric space, (S, ρ), a sequence, xn : n ∈ N, is referred

to as being a Cauchy sequence if and only if, for every ε > 0, there exists N such thatwhenever n,m > N then ρ(xn, xm) < ε. Being a “Cauchy sequence” is a metric space


property which does not translate easily to a topological space. To see this considerthe function g(x) = 1/x on R\0. The function g maps R\0 homeomorphically

onto itself. The sequence T = 1, 1/2, 1/3, . . . , is a Cauchy sequence in the domainof g. But g[T ] = 1, 2, 3, . . . , is not Cauchy in the image of g in R\0. 1

Example 5. Consider the following property, P , on a space S:

“Every real-valued continuous function on S assumes its maximum value on S”

Verify that this is a topological property.

Solution: The negation, ¬P , of the property P is: “There is a continuous real-valuedfunction on S which does not attain its maximum value in S”. We will show that ¬Pis a topological property. The desired result will follow.

Suppose S is a space which satisfies ¬P . Then there exists a continuous function,f : S → R, which does not attain its maximum value in S.

Let h : S → T be a homeomorphism mapping S onto T . We claim that T also satisfies¬P .

Proof of claim. Consider the continuous function, fh← : T → R. If b ∈ T , thenh←(b) ∈ S. Then, by hypothesis, there must exist a ∈ S such that f(a) > (fh←)(b).

Note that (fh←)(h(a)) = f(a) > (fh←)(b). Thus fh← does not assume a maxi-mum value on T . Then T also satisfies ¬P , as claimed.

We conclude that S satisfies ¬P if and only if T satisfies ¬P . Then S satisfies P ifand only if T satisfies P . So P is a topological property.

It is interesting to note that the closed interval [0, 1] cannot be homeomorphic to R

since elementary calculus shows that [0, 1] satisfies property P and R does not.

6.4 Continuity and countability properties.

We have previously defined the three countability properties,

separablefirst countable

second countable

We would like to verify whether or not a continuous function will always carry over

each of these properties from its domain into its codomain.

Recall that a topological space, (S, τ), is separable if and only if S contains a count-

able dense subset. We expect that “separable” is a topological property. This fact

1We will eventually see that the notion of uniform continuity of a function on a subset of a metric spacealso does not translate automatically to a topological space.


is confirmed by the following result which shows that separability is carried over bycontinuous functions. Hence, if S is separable, every topological space which is home-

omorphic to S is also separable.

Theorem 6.11 Suppose (S, τS) is a separable topological space. Then any continuousimage of S is also separable.

Proof : Suppose (S, τS) is separable and f : S → T is a continuous function mapping Sonto the topological space (T, τT).

We claim that the image, T , of S under f is also separable. Given that S is separable,then it contains a countable dense subset, say D. Since the cardinality of the image

of a function is less than or equal to the cardinality of its domain, then f [D] is acountable subset of T . It now suffices to show that f [D] is dense in T . Suppose U is

a non-empty open subset of T . Continuity of f guarantees that f←[U ] is open and sothere must exist x ∈ f←[U ] ∩D. Then

f(x) ∈ f [f←[U ] ∩D]

⊆ f [f←[U ]] ∩ f [D]

= U ∩ f [D]

Hence, every open subset U of T intersects f [D] in a non-empty set. We concludethat f [D] is dense in T . So T is separable.

A simple continuous function is (by itself) not quite strong enough to carry over

the second countable property or the first countable property from its domain to itscodomain. We suspect that homeomorphisms can certainly do so. But, as we shallsee, we don’t need the full power of a homeomorphism for this. The following theorem

shows that a continuous open function will suffice. If so, since homeomorphisms areboth continuous and open, then “second countable” and “first countable” are topo-

logical properties.

Theorem 6.12 Let f : S → T be a continuous open function mapping S onto a space T . If

the space S is second countable then T is second countable. If the space S is first countablethen T is first countable.


Proof : Let (S, τS) and (T, τT) be topological spaces and f : S → T be a continuous openfunction mapping S onto T .

If S is second countable then S has a countable open base, B. Consider the family,

BT = f [B] : B ∈ B. By hypothesis, BT is a set of countably many open subsetsof T . We claim that the countable set, BT , is a base for open sets in T .

Proof of claim : Let U be a non-empty open subset of T and let y ∈ U . Let

x ∈ f←(y) ∈ f←[U ]. Since f is continuous, f←[U ] is open in S, so there ex-ist an open V ∈ B such that x ∈ V ⊆ f←[U ]. Since f is declared to be open,

f(x) = y ∈ f [V ] ∈ BT , an open subset of U . We can then conclude that U is theunion of elements from BT . So BT forms a base for T , so T is second countable.

The proof that the “first countable” property is carried over by continuous open func-

tions is left as an exercise.

6.5 The weak topology induced by a family of functions.

Suppose (T, τT) is a space and we are given a function, f : S → T , mapping the set

S onto the topological space, T (where S is not yet topologized). With this premisealone, we would normally not discuss the continuity of f since “continuity” is defined

in terms of a topology on both the function’s domain and codomain. But we couldfirst hypothesize “continuity of f” and then force on S enough open sets so that this

family of open sets would support the continuity of f .

Is this hard to do? It is quite easy. We need only equip the domain, S of f , with thediscrete topology, τd. Then, no matter how f is defined, since every subset of S is

open then f is, by definition, continuous on S. But this is not entirely satisfactory −nor an interesting problem − since the discrete topology on S doesn’t depend on thefunction, f , at all. We can tighten up our question a bit so that the topology on S

will relate to the specific function f .

Let f : S → T be some function mapping the set S into the topological space T . Wewill topologize S so that, not only is the function f guaranteed to be continuous on

S, but that it is the smallest such topology. To do this we will define

Sf = f←[U ] : U ∈ τT

a subset of P(S), declaring it to be a subbase for a topology. By taking all finite

intersections of elements from Sf we will generate the subset, Bf , of P(S) whichsatisfies the “base property”. By taking all unions of elements from Bf we will gener-

ate the smallest topology, τf , with subbase Sf . The family τf is the weakest topologypossible that will guarantee the continuity property on f . We will refer to τf as being


the weak topology on S induced by f . Eliminating just one element from this topologywould be sufficient to prevent f from being continuous on S. Also the topology we

obtain is directly related to the function f . If we consider a different function we mayobtain a different topology.

If we want the two functions f : S → T and g : S → T to be continuous on S, wewould require more open sets on S. That is, the required subbase would have to be,

Sf,g = f←[U ] : U ∈ τT ∪ g←[U ] : U ∈ τT

In this case the topology generated by the subbase Sf,g would be called the weaktopology induced by f, g.

We can generalize this even more. Rather than restrict ourselves to only two functionseach with the same domain S, we will consider the more general “weak topology on S

induced by a family of functions fα : S → Tαα∈I”, formally defined below.

Definition 6.13 Let S be a non-empty set and (Tα, τα) : α ∈ Γ denote a family of

topological spaces. For each α ∈ Γ, suppose

fα : S → Tα

is a function mapping S onto Tα. Let

S = f←α [Uα] : Uα ∈ ταα∈Γ

Then S can be declared to be the subbase of a topology, τS , on S. The topology τS iscalled the

“weak topology induced by the family of functions fα : α ∈ Γ on S”

The reader is encouraged to keep this definition in mind. We will refer to the weak

topology induced by a set of functions when we will discuss the question of a suitabletopology on the Cartesian product of a family of topological spaces.

6.6 Topic: Continuous functions on a dense subset.

We examine particular properties of a continuous function on a dense subset of (S, τ).

Note that, if A is dense in B and A ⊆ D ⊆ B, then D must be dense in B. Thereader is left to verify this. It is also easy to verify that, if D is dense in S, then S\D


cannot contain any non-empty open subsets of S.

Before we state the next theorem we provide the following definition. If f : S → Tand g : S → T are two functions such that f(x) = g(x) for all x ∈ A ⊆ S then we will

say that

“f and g agree on A”

We present the following theorem concerning continuous functions which agree on a

dense subset of topological space.

Theorem 6.14 Suppose (S, τ) is a topological space and (T, τρ) is a metrizable topologicalspace induced by the metric ρ. Suppose f : S → T and g : S → T are two continuous

functions which agree on some dense subset D of (S, τ). Then f and g must agree on all ofS.1

Proof : Let (T, ρ) be the metric space which is equivalent to (T, τρ). Let

D = x ∈ S : f(x) = g(x)

By hypothesis, f and g agree on D. We are required to show that D = S.

Suppose a ∈ S\D; then f(a) 6= g(a). Then there exists in T , disjoint basic open balls,Bε(f(a)) and Bε(g(a)), of S of radius

ε =ρ( f(a), g(a) )

3

with center f(a) and g(a), respectively. Since both f and g are continuous on S, bothf←[Bε(f(a))] and g←[Bε(g(a))] are open in S each containing at least the point a.

So, the open subset

f←[Bε(f(a))] ∩ g←[Bε(g(a))] 6= ∅

in S. We claim that f←[Bε(f(a))] ∩ g←[Bε(g(a))]∩D = ∅. For if,

x ∈ f←[Bε(f(a))]∩ g←[Bε(g(a))]

f(x) ∈ Bε(f(a)) and g(x) ∈ Bε(g(a)) so f(x) 6= g(x) (since Bε(f(a)) and Bε(g(a))are disjoint). So a 6∈ S\D. So S\D is empty. So D = S, as required.

1Once we have introduced the concept of “Hausdorff” this statement generalizes from “metrizable topo-logical spaces” to “Hausdorff topological spaces”.


6.7 Topic: On algebras of continuous real-valued functions.

We now take a quick glance at the algebra1, (C(S),+, ·, scalar mult)”, of those con-tinuous functions on a topological space, S, with range in R. If S is a topological

space, we denote the set of all continuous functions, f : S → R, mapping S into R byC(S). The set C(S) is normally considered equipped with the algebraic operations

+, · and scalar multiplication defined as

(f + g)(x) = f(x) + g(x)

(f · g)(x) = f(x)g(x)

(αf)(x) = αf(x)

We will assume that the reader is able to write proofs showing that C(S) is closed

under sums, multiplication and real scalar multiplication and that |f | defined as

|f |(x) = |f(x)|

also belongs to C(S) whenever f is in C(S). We will also refer to (C(S),+, ·) as aring of continuous functions. The elements of C(S) can be partially ordered with “≤”where

f ≤ g if and only if f(x) ≤ g(x)

for all x ∈ S. We also define the operations ∨ and ∧ on C(S) as,

(f ∨ g)(x) = max f(x), g(x) for x ∈ S

(f ∧ g)(x) = minf(x), g(x) for x ∈ S

Verification of the following formulas is left to the reader.

f ∨ g =f + g + |f − g|

2

f ∧ g =f + g − |f − g|

2

From these we can conclude that both f ∨ g and f ∧ g are continuous whenever f andg are continuous.

6.8 Topic: The Pasting lemma and a generalization.

The continuity of a function, f , on a given space can sometimes be more easily con-

firmed by determining the continuity of f on some of its subspaces. The “pastinglemma” is a statement which says that two continuous functions can be ”pasted to-gether” to create some other continuous function. The lemma is implicit in the use of

piecewise functions.

1An algebra is a set with + and · and scalar multiplication by elements of a field.


Lemma 6.15 Let A and B, be closed subsets of a topological space S such thatS = A ∪ B. Then f : S → T is continuous if and only if f |A and f |B are both

continuous.

Proof : Let f : S → T be a function and A and B are both closed in S.

( ⇒ ) Suppose f is continuous on S. Then, by theorem 6.5 both f |A and f |Bare continuous on A and B, respectively.

( ⇐ ) Suppose both f |A and f |B are continuous on the closed subsets A and

B, respectively. Let F be a closed subset of T . Then both f |←A [F ] and f |←B [F ]are closed since each is the pre-image of f when restricted to A and B respec-

tively. Then their union, is also closed, being a finite union of closed sets. Sof←[F ] = f |←A [F ] ∪ f |←B [F ], a closed subset of S. So f is continuous on S.

We generalize the pasting lemma to the case where f acts on an infinite collection

of closed sets. The statement in the following theorem involves a family (possiblyinfinite) of subsets referred to as being a “locally finite collection”.

Definition. A locally finite collection, C , of subsets of a topological space, S, is onefor which every point in S has at least one neighbourhood which meets at mostfinitely many elements of C .

Such families of subsets satisfy an interesting property, presented in the followinglemma.

Lemma 6.16 Suppose (S, τ) is a topological space. Let U = Ui : i ∈ I be a locally

finite collection of sets in S. Then U ∗ = clSUi : i ∈ I is also locally finite.

Furthermore,

clS [∪Ui : i ∈ I] = ∪clSUi : i ∈ I

Proof : Let U = Ui : i ∈ I be a locally finite collection of sets in S.

Then, for x ∈ S, there is an open neighbourhood B such that B ∩ U 6= ∅ for finitelymany members of U and B ∩U = ∅ for all others. For these “other”, U ’s, since B is

open in S, B∩clSU = ∅. In the case where B∩U 6= ∅ then B∩clSU 6= ∅ for finitelymany U ’s. The existence of this open neighbourhood B of x is all that is required toprove the local finiteness property of clSUi : i ∈ I.We now prove the second part. Let

M = ∪Ui : i ∈ I

We are required to prove clSM = ∪clSUi : i ∈ I.


Clearly, ∪clSUi : i ∈ I ⊆ clSM . So we need only prove clSM ⊆ ∪clSUi : i ∈ ILet p ∈ clSM . We claim that p ∈ ∪clSUi : i ∈ I. There exists an open neighbour-hood B of p such that B ∩ clSUi 6= ∅ for finitely many elements, F , in U ∗.

See that, for p ∈ clSM ,

p ∈ clSM = clS

[

∪Ui ∈ U : clSUi 6∈ F⋃

∪Ui ∈ F]

= clS [∪Ui ∈ U ∗ : clSUi 6∈ F]⋃

clS [∪Ui ∈ F]

= clS [∪Ui ∈ U ∗ : clSUi 6∈ F]⋃

∪clSUi ∈ F (clSUi is closed andF is finite)

Now p cannot belong to clS [∪Ui ∈ U ∗ : clSUi 6∈ F] since B ∩ clSUi = ∅ for allclSUi 6∈ F . Then p ∈ ∪clSUi ∈ F ⊆ ∪clSUi : i ∈ I, as claimed.

So clSM = ∪clSUi : i ∈ I, as required.

It follows from the lemma that, if F = F : F ∈ F is any locally finite collection ofclosed sets, then

clS[∪F : F ∈ F = ∪F : F ∈ F

Theorem 6.17 Suppose (S, τ) is a topological space. Let F be a locally finite family ofclosed subsets which covers all of S (i.e., S = ∪F ). Let f : S → T be a function mapping

S into a space T . Then f is continuous on S if and only if f |F : F → T is continuous oneach F ∈ F .

Proof : By theorem 6.5, b), we need only prove ( ⇐ ).

( ⇐ ) Suppose F is a locally finite collection of closed subsets of S which covers all

of S. Also suppose f : S → T is a function mapping S into a space T such thatf |F : F → T is continuous on each F ∈ F . We are required to show that f is contin-

uous on all of S. To prove this it suffices to show that, if A is closed in T , then f←[A]is closed in S.

Let A be a closed subset of T . Then,

f←[A] = ∪F ∩ f←[A] : F ∈ F= ∪f |←F [A] : F ∈ F

Since f |F is continuous for each F ∈ F then f |←F [A] is a closed subset of F . Then

there exists a closed subset G of S such that f |←F [A] = G ∩ F , a closed subset of S.


Consider the collection

C = f |←F [A] : F ∈ Fof closed subsets of S. Note that f←F [A] = ∪C . We will show that this set is closed in S.

Claim. We claim that C is locally finite.

Proof: Let p ∈ S. Then there exists an open neighbourhood, V , of p and a finitesubcollection, G , of F such that V ∩ F 6= ∅ if F ∈ G and V ∩ F = ∅ for F ∈ F \G .

Note that for F \G ,

V ∩ f |←F [A] = V ∩ (F ∩ f←[A])

= (V ∩ F ) ∩ f←[A]

= ∅ ∩ f←[A]

= ∅

Then at most finitely many elements of C intersect the neighbourhood V of p. So Cis locally finite collection of closed subsets, as claimed.

Then f←[A] = ∪C is closed (see paragraph preceding the theorem).

Then f : S → T is continuous on S.

In the above theorem, if the members of the collection F are all open (rather than

all closed as hypothesized in the theorem) the family F need not be locally finite andso the statement may not hold true.

Corollary 6.18 Suppose (S, τ) is a topological space. Let F = Fi : i ∈ F be a finite

family of closed subsets which covers all of S (i.e., S = ∪U ). Let f : S → T be a functionmapping S into a space T . Then f is continuous on S if and only if f |Fi : Fi → T is

continuous on each Fi ∈ F .

Proof : See that, if F = Fi : i ∈ F be a finite family of closed subsets which coversall of S, then F is a σ-locally finite cover of closed sets on S. The result follows

immediately from the theorem.

Concepts review:

1. Given the function f : S → T and A ⊆ S, define f [A].


2. Given the function f : S → T and B ⊆ S, define f←[B].

3. State the formal topological definition of “f is continuous function on the set A”.

4. State the formal topological definition of “f is continuous function at the point x”.

5. Give a formal theorem statement which link the above two definitions of continuity.

6. A continuous function f : S → T pulls back open subbase elements in P(T ) to whatkind of set in S?

7. A continuous function f : S → T pulls back open base elements in P(T ) to what

kind of set in S?

8. If f : S ⇒ T is continuous on S and A ⊆ S, show that f |A is continuous on A.

9. Is it correct to say “f : S → T is continuous if and only if f pulls back closed sets toclosed sets”?

10. What does it mean to say f : S → T is an open functions?

11. What does it mean to say f : S → T is a closed functions?

12. Is it okay to say that “open functions are always closed”? What about “continuous”?

13. What does it mean to say that f : S → T is a homeomorphism?

14. What does it mean to say that S and T are homeomorphic spaces?

15. Give two characterizations of homeomorphic functions.

16. What is a topological property?

17. Is “first countable” a topological property? What about “second countable”?

18. Do continuous functions necessarily carry over the second countable property to itscodomain? What about the first countable property?

19. What does it mean to say that A is a dense subset of B?

20. What can we say about two continuous functions which agree on a dense subset of a

metrizable topological space?

21. What does it mean to say that a topological space is separable?

22. What can we say about continuous images of separable spaces?

23. What can we say about second countable spaces in reference to the “separable prop-erty”?


24. Provide examples of hereditary and non-hereditary topological properties.

25. Define the weak topology induced by a family of functions fα : S → Tαα∈I.

26. Provide two characterizations of “f : S → T is continuous on S” involving families ofsubsets which cover S.

EXERCISES

1. Suppose S = a, b is a set with topology τS = ∅, a, S. Let T = a, b where Tis equipped with the discrete topology τd. Let i : S → T denote the identity map.

What can we say about the continuity or non-continuity of the functions i : S → Tand i← : T → S?

2. Let f : S → T be a continuous function mapping (S, τS) onto (T, τT). If U is a Gδ inT , is f←[U ] necessarily a Gδ in S? Show that f pulls back Fσ ’s in T to Fσ ’s in S.

3. Let f : (S, τS) → (T, τT) be a continuous function mapping S onto T . Show that f iscontinuous on S if and only if f [clS[U ]] ⊆ clT f [U ] for any U ∈ P(S).

4. Recall that infinite countable sets are those sets which can be mapped one-to-one andonto the natural numbers N. Suppose X and Y are both countable dense subsets of(R, τ) where τ is the usual topology. Show that X and Y must be homeomorphic

subspaces of R.

5. Consider the topological spaces S = (R2, τ1) and T = (R, τ) where τ1 and τ represent

the usual topology. (The open base of τ1 are the open balls, Bε(x, y), with center(x, y) and radius ε). Consider the function f : S → T defined as f(x, y) = x.

a) Show that f is an open function.

b) Show that f is not a closed function. (Hint: See that F = (x, y) : xy = 1 is aclosed subset of S. Consider the image of F under f .)

6) Prove: The function f : S → T is a closed function if and only if whenever F is aclosed subset of S then t ∈ T : f←[t] ∩ F is non-empty is a closed subset of T .

7) Suppose f : S → T is a closed function. Show that whenever V is an open subset of

S and f←[x] is a subset of V then x ∈ intT (f [clSV ]).

8) Suppose f : S → T is a closed function and F is closed subset of S. Show that the

restriction, f |F : F → f [F ], is also a closed function.

9) Suppose U and V are subsets of S such that U ∪ V = S and x ∈ U ∩ V . Suppose

f : S → T is a function such that both f |U and f |V are continuous on U and V ,respectively. Show that f is continuous at x.


10) Suppose that f : S → T is a one-to-one and onto function. Show that f is a homeo-morphism if and only if, for any U ∈ P(S), f [clSU ] = clT f [U ].

116 Section 7: Product spaces

7 / Product spaces.

Summary. In this section we will review some fundamental facts about those

sets that are “Cartesian products of sets”. We will then consider two topolo-gies on a Cartesian product of topological spaces and study some of their most

fundamental properties. In the last half of the chapter, we will look at some ap-plications where product spaces play an important role. In particular, we prove

the existence of a continuous function which maps the closed interval [0, 1] ontothe product space, [0, 1]× [0, 1]× [0, 1], a cube in a three dimensional space.

7.1 Fundamentals of Cartesian products.

In topology, Cartesian products are a rich and important source of examples of vari-

ous known topological properties. Given a family of known topological spaces, we canconstruct a new larger topological space with it’s own particular properties. For thisreason, it is crucial to understand them well. The best way to do this is to practice

using them in various contexts.

Most students are exposed to the notion of a Cartesian product with two or threefactors, in some form or other, at the high school level. For example,

A× B ×C = (a, b, c) : a ∈ A, b ∈ B, c ∈ C

When discussing infinite Cartesian products of sets we must proceed cautiously while

deciding which topology we will adopt to best suit our purposes. We thought it wouldbe best if we begin by presenting a formal definition of products of sets. Those readers

already well familiar with these concepts can skim through this section and go directlyto section 7.2.

Definition 7.1 Let Sα : α ∈ I be an indexed family of sets. The Cartesian product ofthese sets, denoted by

∏

α Sα, or in more detail as,

∏

α∈ISα = f | f : I → ∪α∈ISα

is the set of all functions f mapping the index set, I , into the union, ∪α∈ISα, such that, forβ ∈ I , f(β) ∈ Sβ.2 So, if u is an element of

∏

α Sα and yα = f(α), then we can express it

in the form

u = f(α) : α ∈ I = yα : α ∈ I = <yα>α∈I

2We will assume that a verification involving a combination of the Axiom of union and the Axiom of

power set guarantees thatQ

α Sα is indeed a “set”.


or we can write it more simply as yα =< yα >. If β ∈ I , the set,

Sβ , is called the βth factor

of the Cartesian product,∏

α Sα, while

yβ is called the βth coordinate of the element, < yα >α∈I

For example, if I = 1, 2, 3∏

α∈ISα = (a, b, c) : a ∈ S1, b ∈ S2, c ∈ S3 = S1 × S2 × S3

The size of∏

α∈I Sα depends on the size of the respective sets, Sα, and the size of theindex set.

If we are given an indexed family of sets, say Sα : α ∈ J, where all factors, Sα,

represent the same set, S, the reader should be familiar with the following notation:

SJ =∏

α∈JS

The expression, SJ , is normally interpreted as meaning “the set of all functions map-ping J into S”. For example, if J = N and Sα = R for all α ∈ J, then

RN =∏

α∈NR

It thus represents all countably infinite sequences of real numbers, a0, a1, a2, a3, . . .,or equivalently, the set of all functions mapping N into R. Another example is, RR,

which represents the set of all functions mapping R into R. Since R is not normallyviewed as an index set, the set, RR, of all functions mapping R into R is not often

expressed as a Cartesian product.1

A few words on the definition of Cartesian product. In our definition of Cartesian

product we refer to an indexed family, Sα : α ∈ I, of sets. We didn’t state explic-itly that each one of these is non-empty. Should we include this requirement in the

definition? What happens if, say Sβ = ∅, for some β ∈ I? Since there is nothing inSβ then there cannot exist a function which will map β to some element in Sβ and so

the cautious reader will conclude that∏

α Sα must be empty. This is not catastrophicsince to say the product is non-empty is to assume that each factor is non-empty.

So we can leave it as is. On the other hand, if no Sα is empty, are we guaranteedthat there exists at least one function, f : I → ∪α∈ISα, such that f assigns β in I

to a particular element in Sβ? If so, is there a way to decide which element shouldbe selected by f? The assumption that at least one such function exists invokes thestatement in the Axiom of choice:

1If we write the product,Q

α∈RSα, then the reader can assume that the Well-ordering theorem (equivalent

to the Axiom of Choice) has been invoked to linearly order R.


“Given any set A of non-empty sets, there is a rule f which associates toeach set A in A some element a ∈ A”.

So the Axiom of choice grants us permission to assume that at least one function f

will select a point f(β) = mβ in Sβ for us. However, other than this guarantee thatat least one f exists, we have no way of ever determining what that function f is. We

are assuming the existence of a mathematical entity we will never ever see. Invokingthe Axiom of choice is not ideal, but it is the best we can do. For most people, the

fact that the Cartesian product of non-empty sets is non-empty is obvious and sothey don’t lose any sleep over it. Throughout this section, we will, as a rule, assume

the Axiom of choice holds true, and not point out it’s application at each place it isinvolved, unless it is of particular interest to do so.

In this book, we declare that the “Cartesian product of topological spaces, S =∏

i∈I Si”, we will always mean “the non-empty Cartesian product, S, of spaces.”

A few set-theoretic properties of Cartesian products.

We have yet to topologize Cartesian products of spaces. But before we do so, wepresent a few of the Cartesian product’s most fundamental properties.

Definition 7.2 Let S =∏

i∈I Si be a Cartesian product of sets. The family of functionsπi : i ∈ I where πi :

∏

i∈I Si → Si, is defined as

πj(< xi >i∈I) = xj

We will refer to πj as being the jth projection which maps∏

i∈I Si onto Sj.

Theorem 7.3 Let Si : i ∈ I and Ti : i ∈ I be two families of sets and S =∏

i∈I Si

and T =∏

i∈I Ti be their corresponding Cartesian products.

a) If∏

i∈I Si ⊆∏

i∈I Ti then Si ⊆ Ti for each i ∈ I .

b) For Ui, Vi, both subsets of Si,

i)∏

i∈I Ui ∩ ∏

i∈I Vi =∏

i∈I(Ui ∩ Vi)

ii)∏

i∈I Ui ∪ ∏

i∈I Vi =∏

i∈I(Ui ∪ Vi)


Proof : a) Let β ∈ I . Since the βth projection map, πj , is onto the (Sj)th factor,

Sj = πj

(∏

i∈ISi

)

⊆ πj

(∏

i∈ITi

)

= Tj

The proofs of part b) are left for the reader.

7.2 Topologizing the Cartesian product of topological spaces.

Taking Cartesian products of large numbers of topological spaces is a powerful way to

construct new topological spaces from “old” ones. There is more than one topologyfor us to choose from, some of which will eventually be more useful than others. For

finite products there is one which most users would consider as being the most intu-itive and as being the “standard” one. For infinite products the choice of topology

may depend on the context.

The third example on page 80, illustrates how one might proceed to topologize theCartesian product of two spaces (S, τS) and (T, τT). In that example, the topology, τ ,

on S × T was generated by choosing, as subbase for τ , the family of sets

S = π←S [U ] : U ∈ τS ∪ π←T [V ] : V ∈ τT

where πS : S × T → S and πT : S × T → T are projection maps. Recall that

πS(a, b) = a for all b ∈ T and πT (a, b) = b for all a ∈ S and so π←S (a) = a × T andπ←T (b) = S × b. So the set of all finite intersections of the elements in S forms a

base, B, for this topology. Thus, the base elements of τ are of the form

π←S [U ] ∩ π←T [V ] = (U × T ) ∩ (S × V )

= (U ∩ S)× (V ∩ T )

= U × V

where U ∈ τS and V ∈ τT . That is, B = U × V : U ∈ τS, V ∈ τT.Note that, in this particular example, the topology constructed for S × T is the

“weak topology induced by the projection functions πS, πT”

This “weak topology” on S × T comes with the guarantee of continuity for each pro-

jection map, πα, on S × T . This motivates the choice of the “standard topology” forarbitrary products described in the following definition.


Definition 7.4 Let (Sα, τα) : α ∈ I be an indexed family of non-empty topologicalspaces and

S =∏

α∈ISα

be the Cartesian product of these spaces. Let πα : α ∈ I be the family of the associated

projection maps

πβ :∏

α∈ISα → Sβ

We define the product topology or the standard topology as being the weak topology on∏

α∈I Sα induced by the family of functions πα : α ∈ I. The Cartesian product of

topological spaces, when equipped with this weak topology, is referred to as a productspace. The family of all sets of the form,

π←α [U ] : U ∈ τα

is the subbase, S , for the product space. While each element of the base of open sets, B,

is of the form

∩α∈Fπ←α [Uα] : Uα ∈ τα

where F is a finite subset of I .

Since π←α [Uα] ∩ π←α [Vα] = π←α [Uα ∩ Vα], we can assume that all the α’s in the finiteset, F , are distinct. This assumption does not alter the definition of product topology.

Also, π←β [Uβ] is a subset of∏

α∈I Sα where, if α 6= β, the αth factor is, Sα, itself, andonly the βth factor is Uβ . Hence, for any open base element, every factor is Sα itself

except for finitely many factors, Uα, as proper subsets of Sα.

It is also worth noting that the product topology is the absolute smallest topology onS =

∏

α∈I Sα which guarantees that each and every projection map in

πα :∏

α∈ISα → Sα

is continuous on its domain, S.

Finally, note that the product topology depends on the topology of each of the factors.

It does not depend on some topology defined on the index set, I .

Projection maps are open.

In the following proposition we will show that each map, πα, is also open. This is a

useful fact to remember.


Proposition 7.5 Given a product space, S =∏

α∈I Sα, the projection map,

πβ : S → Sβ

is an open map.

Proof : Given: A product space, S =∏

α∈I Sα, and a projection map, πβ : S → Sβ .

Since functions respect arbitrary unions, it suffices to show that the projection mapsends basic open sets to open sets.

Let

∩π←αi[Uαi ] : i = 1, 2, . . . , k

be a basic open set in S.

If β = αj ⇒ πβ[∩π←αi[Uαi] : i = 1, 2, . . . , k] = Uαj

If, for i = 1, 2, . . . , k, β 6= αi ⇒ πβ[∩π←αi[Uαi] : i = 1, 2, . . . , k] = Sβ

So πβ is an open map, as required.

Theorem 7.6 Let (Sα, τα) : α ∈ I be an indexed family of non-empty topological spacesand

∏

α∈I Sα be the Cartesian product space equipped with the product topology, τ . For

each α ∈ I , Sα has an open base represented by, Bα. Then the set

B∗ = ∩α∈Fπ←α [Bα] : Bα ∈ Bα α∈I

where F is finite, forms a base for τ .

Proof : Let V = ∩α∈Fπ←α [Uα] : Uα ∈ τα (with F finite) be an open base element of the

product topology, τ . Suppose <xα>α∈I ∈ V . Then, for each α ∈ F , there exists anopen base element, Bα ∈ Bα, such that xα ∈ Bα ⊆ Uα. Then

<xα>α∈I ∈ ∩α∈Fπ←α [Bα] : Bα ∈ Bα ⊆ V

Then every open base element of τ is the union of elements from B∗. So B∗ forms a

base for the product topology.


The “box topology” on a Cartesian product.

The product topology is not the only topology we can define on (infinite) productsof topological spaces. Some readers may have noticed that, for an infinite product,

S =∏

α∈I Sα, the collection

B∗ = ∩α∈Iπ←α [Uα] : Uα ∈ τα

(note the intersection of infinitely many sets) satisfies the “base property”. Hence,

this set will be a base for a topology τ∗ which is different from the product topology,τ , on S. In the literature, it is referred to as the box topology. Notice that every open

base element, B ∈ B, for the product topology is an open base element in B∗. Hencethe “box topology” is stronger (finer) than the “product topology”. In fact, if every

factor of∏

α∈I Uα is a proper open subset of Sα and <xα>α∈I ∈∏α∈I Uα ∈ τ∗, thenthere does not exist a V ∈ B such that <xα>α∈I ∈ V ⊆∏α∈I Uα. Can you see why?

So, B∗ 6⊆ B.

Note that, for the Cartesian product of finitely many spaces, the product topologyand the box topology are equivalent topologies.

Example 1. Show that the product space R3 = R × R × R is metrizable.

Solution : This can be seen since the basic open sets in (R3, τ) equipped with theproduct topology are 3-dimensional open rectangular boxes. While the basic open

sets of (R3, ρ), where ρ(~x, ~y) is the distance between the two given points, are openballs. Since the rectangular boxes can be filled with open balls and vice versa the

metric space (R3, ρ) and the product space R3 = R×R×R (equipped with the prod-uct topology) are equivalent topological spaces. So (R3, τ) is metrizable.

Example 2. The product space [0, 1]3 is metrizable. Since [0, 1]3 is a subspace of themetrizable, R3, then by the example on page 87, [0, 1]3 is metrizable.

Example 3. Let S =∏

j∈J [0, 1]. We define the following metric on S,

ρ(<xj>j∈J , <yj>j∈J) = sup |xj − yj| : j ∈ J

The verification that this is a valid metric on S is left to the reader. Clearly, the

induced metric topology on S is not equivalent to the product topology (since, forsmall ε, no j-factor of the open base element, Bε(<xj >j∈J), is equal to [0, 1]). The

metric, ρ, is referred to at theuniform metric

on a product on S =∏

j∈J [0, 1]. The topology induced by the uniform metric is

referred to at the uniform topology on∏

j∈J [0, 1]. A word of caution: Theorems inwhich the a Cartesian product equipped with the product topology is involved maynot hold true if that product is equipped with the uniform topology.


7.3 On interiors, closures and boundaries of product spaces.

It is sometimes useful to know how to handle the interior and the closure of a product.

The interior of a product is well-behaved in the case of a finite product space, whilethe closure of a product distributes nicely over its factors even for infinite products.

Theorem 7.7 Let (Sα, τα) : α ∈ I be an indexed family of non-empty topological spacesand S =

∏

α∈I Sα be a Cartesian product equipped with the product topology. Let T =∏

α∈I Aα where each Aα is a subspace of Sα. Then,

clST =∏

α∈IclSαAα

Proof : The Cartesian product space S =∏

α∈I Sα is given.

1)∏

α∈IclSαAα ⊆ clST : Let <xα> ∈ ∏α∈IclSαAα and let U =∏

α∈I Uα

be a basic open neighbourhood of <xα>. Since xα ∈ clSαAα, there exists <yα> such

that yα ∈ Uα ∩Aα, for all α. So <xα> ∈ clS∏

α∈IAα

2) clST ⊆∏α∈IclSαAα : Let <xα> ∈ clS∏

α∈IAα and Uβ be an open neighbourhood

of xβ in Sβ . Then π←β [Uβ ] is open in∏

α∈ISα. The set π←β [Uβ] contains some point

<yα> ∈ ∏α∈IAα. Then yβ ∈ Uβ ∩Aβ . So xβ ∈ clSαAβ .

Note that the above statement holds true also if the product is equipped with the box

topology.

Theorem 7.8 Let S and T be topological spaces which contain subsets A and B, respec-

tively. Then,intS×T (A× B) = intSA × intTB

Proof : We are given the Cartesian product M = S × T and A ⊆ S, B ⊆ T . Since intSA

and intTB are both open in S and T , respectively, then intSA× intTB is open in M .Since intSA× intTB ⊆ A× B, then intSA × intTB ⊆ intM (A×B).

Conversely, let (a, b) ∈ intM(A×B). Then there is an M -open neighbourhood, U , of(a, b) such that U ⊆ intM(A × B) ⊆ A × B. Then U = ∪i∈I(Vi ×Wi), where Vi isopen in S and Wi is open in T . Let j ∈ I such that (a, b) ∈ Vj ×Wj ⊆ U ⊆ A × B.

Then a ∈ intSA and b ∈ intTB. Then (a, b) ∈ intSA × intTB. We conclude thatintM (A× B) ⊆ intSA× intTB.

So intM (A×B) = intSA× intTB.


The reader should resist the temptation of generalizing the statement in the theoremimmediately above to arbitrary products. Consider, M =

∏

n∈N R equipped with

the product topology, and the subset,∏

n∈N(0, 1), of M . Then∏

n∈NintR(0, 1) =∏

n∈N(0, 1). On the other hand consider,

W = intM[∏

n∈N(0, 1)]

Let U =∏

n∈N Yn be a basic open subset (with respect to the product topology) ofM , where Yn = R for all but finitely many n’s. Since

∏

n∈N(0, 1) cannot contain even

a single basic M -open set,∏

n∈N(0, 1) can have no interior.

We conclude that intM

[∏

n∈N(0, 1)]

= ∅ 6=∏n∈NintR(0, 1).

Example 4. Let S and T be topological spaces which contain subsets A and B,

respectively. Verify that

bdS×T (A×B) = [clSA × bdTB] ∪ [ bdSA × clTB]

Solution. See that

bdS×T (A× B) = clS×T (A×B)\intS×T (A×B)

= (clSA× clTB)\(intSA× intTB)

= [clSA\intSA× clTB)] ∪ [clSA× clTB\intTB]

= [ bdSA× clTB] ∪ [clSA × bdTB]

7.4 Countability properties involving product spaces.

In this text, when we say “product space”, we will always mean the Cartesian product

equipped with the product topology. The product topology is to be considered as thedefault topology unless stated otherwise.

In the following two theorems we investigate whether the countably properties carryover from factors to product and vice versa. Recall that a topological space, (S, τS), is

second countable if it has at least one countable base for open sets. It is first countableif each point, x ∈ S, has a countable open neighbourhood base at x.

Theorem 7.9 Let (Sα, τα) : α ∈ N be an indexed countable family of non-empty topo-logical spaces and S =

∏

α∈N Sα be the corresponding product space.

a) Then the product space, S, is second countable if and only if each Sα is secondcountable.


b) Then the product space, S, is first countable if and only if each Sα is first countable.

Proof : We are given that S =∏

α∈N Sα is a product space of countably many factors.

We will prove part a).

( ⇒ ) Suppose S is second countable. We claim that each Sα is second countable.

Recall (from proposition 7.5) that the projection map πα is continuous is and openfor each α ∈ N. Hence, by theorem 6.12, given that πα is continuous each Sα = πα[S]is second countable. We are done with this direction. (Note that the countability of

the index set plays no role in this direction.)

( ⇐ ) We are given that, for each α ∈ N, (Sα, τα) is a second countable topological

space, and∏

α∈N Sα is a countable product equipped with the product topology, τ .

By hypothesis, for each α ∈ N, Sα has a countable base, Bα, of open sets. Then, bytheorem 7.6, the elements of τ of the form

∩α∈Fπ←α [Bα] : Bα ∈ Bα

when gathered together form a base, B, for open sets in τ . Since each Bα is countablethen so is the subfamily

π←α [Bα] : Bα ∈ Bα, α ∈ N

of the base, B, of S. Let F = F : F ⊂ N, F is finite. Note that the set, F , hascountably many elements. (See footnote1) (This is where the countable the number

of factors is countable is taken into consideration.)

Since the cardinality of F × Bα × N is ℵ0, then the base

B = ∩α∈Fπ←α [Bα] : Bα ∈ Bα F finite, α∈N

is countable. This means∏

α∈N Sα is second countable.

The proof of part b) is similar and so is left as an exercise.

In an analogous theorem, the “separable property” will carry over from factors to the

product and vice-versa, provided the number of factors in the product is restricted tono more than c = |R| = 2ℵ0. The provided proof is notationally involved, but exhibitsan interesting technique.

1The number of finite subsets of N is countable: For each n ∈ N, let Un = A ∈ P(N) : maxA ≤ n,the set of all subsets of 0, 1, 2, . . . , n. Then, for each n, |Un| = 2n+1. The countable union of countablesets is countable (See 19.3 of Appendix B.), so ∪n∈NUn is countable. If F is a finite subset of N, F ∈ Um

for some m. Then ∪n∈NUn contains all finite subsets of N. Then the number of finite subsets of N iscountable.


Theorem 7.10 Let (Sα, τα) : α ∈ I be a family of sets indexed by the set I , where

|I | ≤ c = 2ℵ0. Also suppose that Sα is a non-singleton set, for all α ∈ I . Let S =∏

α∈I Sα,be the corresponding product space. Then S is separable if and only if each Sα is separable.


α∈I Sα is a product space of non-empty non-singleton setswhere | I | ≤ c = 2ℵ0.

( ⇒ ) Suppose S is separable. Recall from the proof that appears on page 121, the

projection map πα is continuous for each α ∈ N. Hence by theorem 6.11, the contin-uous image of a separable space is separable, so each Sα = πα[S] is separable. We are

done with this direction. The cardinality of I is irrelevant for this direction.

( ⇐ ) Suppose each Sα is separable and |I | ≤ c. Then |I | ≤ | [0, 1] |. So there is aone-to-one function f : I → [0, 1] which maps I into [0, 1]. We can then index the

product space S with f [I ]. That is, S =∏

α∈I Sf(α). So, if x ∈ S then x is of theform < xf(α) >α∈I.

Since each factor of S is separable, each Sf(α) contains a countable dense subset

Df(α) = df(α)n: n = 1, 2, 3, . . . ,

We will now construct a countable subset, D∗, of S which is dense in S.

For each k ∈ N\0 we can construct in [0, 1] a finite sequence of pairwise disjoint

closed intervalsAk = [aj, bj]k : j ∈ 1, 2, . . . , k

such that aj, bj ∈ Q ∩ [0, 1] and aj+1 > bj ≤ 1.

For each k, if α ∈ I , f(α) ∈ [aj, bj]k for, at most, one j ∈ 1, 2, . . . , k.See that W = ∪Ak : k = 1, 2, 3, . . . is a countable set of intervals. We can then

index the elements of W as

W = [aj, bj]k,n : j ∈ 1, 2, . . . , k, k ∈ 1, 2, 3, . . . , , n ∈ 1, 2, 3, . . . ,

where, for each n, [aj, bj]k,n = [aj, bj]k ∈ Ak.

We define a function h : W →∏

α∈I Df(α) ⊆∏

α∈I Sf(α) as follows:

h( [aj, bj]k,n ) = < yf(α) >α∈I where

yf(α) = df(α)nif f(α) ∈ [aj, bj]k,n

yf(α) = df(α)1otherwise

Is h : W → S a well-defined function? Yes it is, since (as mentioned earlier) for each

k, if α ∈ I , f(α) ∈ [aj, bj]k for, at most, one j ∈ 1, 2, . . . , k.


Also note that every component of h( [aj, bj]k,n ) = < yf(α) >α∈I belongs to someDf(α) (which is dense in Uf(α)).

Let D∗ = h[W ].

Claim 1 : That D∗ is countable in S.

Since the domain of h, W , is countable then so is its range, D∗ = h[W ], in∏

α∈I Sf(α),

as claimed.

Claim 2 : That D∗ is dense in S.

Let B = π←α1[Uf(α1)]∩ · · ·∩π←f(αk)[Uf(αk)] be a basic open neighbourhood of a point in

S =∏

α∈I Sf(α).

We need to verify that B ∩D∗ 6= ∅.

Since Df(α) = df(α)n: n = 1, 2, 3, . . . , is dense in Sf(α), then for each f(αi) there

exists df(αi)ni∈ Uf(αi), for i = 1 . . . , k.

If df(αi)m∈ Sf(αi), then

πf(αi)←(df(αi)m

) = < yf(α) >: f(αi) ∈ [aj, bj]k,m = h([aj, bj]k,m)] ∈ h[W ]

πf(αi)←(df(αi)1

) = < yf(α) >: f(αi) 6∈ [aj, bj]k,m = h([aj, bj]k,m) ∈ h[W ]

Then, h[W ] ∩ π←α1[Uf(α1)]∩· · ·∩π←f(αk)[Uf(αk)] 6= ∅. Then B∩D∗ 6= ∅, so D∗ is dense

in S, as claimed.

So h[W ] is a countable dense subset of∏

α∈I Sf(α). Then∏

α∈I Sf(α) is separable, as

required.

One should exercise caution when transferring a topological property possessed by allfactors of a product to the product itself. The statements of the above two theorems

were proven with particular restrictions on the number of factors.

7.5 On continuous functions involving product spaces.

The choice of the “product topology” on a Cartesian product,∏

α∈I Sα, is is due

mostly to the fact that it guarantees the continuity on a product space of all its pro-jection maps, πα : α ∈ I. It is the smallest topology that will guarantee this.


Lemma 7.11 Let S =∏

α∈I Sα be a product space and (T, τ) be a space. Suppose g :T → S is a function mapping the space T to the product space, S. For each α ∈ I , let

fα = πα g. Then

[ g is continuous on T ] ⇔ [ fα : T → Sα is continuous for all α ]

Proof : Recall that, by definition of “product space”, each πα is guaranteed to be continu-ous on S.

( ⇒ ) If g : T → S is continuous then, for all α ∈ I , so is (πα g) : T → Sα (since thecomposition of continuous functions was shown to be continuous).

(⇐ ) We fix β ∈ I . We are given that (πβ g) : T → Sβ is continuous. We are requiredto show that g is continuous. If Vβ is open in Sβ, then

g←[

π←β [Vβ]]

= (πβ g)←[Vβ]

is open in S =∏

α∈I Sα. So g pulls back an open subbase element, π←β [Vβ], to an

open set. By theorem 6.4, g is continuous on T , as required.

The following example may also serve as more motivation for the choice of the producttopology on infinite products over the box topology. 1

Example 5. Consider the Cartesian product∏

n∈N R. Let i : R → R denote theidentity function i(x) = x. Let g : R → ∏

n∈N R be the function where, for each

n ∈ N,

(πng)(x) = i(x)

That is, g(x) =< i(x)>= (x, x, x, . . .). Then for each n ∈ N, πng = i, is clearly acontinuous function on R. If

∏

n∈N R is equipped with the product topology, then by

lemma 7.11, g is a continuous function. Show that if∏

n∈N R is equipped with thebox topology, g is not continuous.

Solution : If∏

n∈N R is equipped with the box topology, consider the open subset

B = (−1, 1)× (−1/2, 1/2)× (−1/3, 1/3)× (−1/4, 1/4) · · ·

with respect to the box topology. Suppose g←[B] was open in R. Then there would

exist ε, such that g[(−ε, ε)] ⊆ B. Then (πng)[(−ε, ε)] = i[(−ε, ε)] = (−ε, ε) ⊆(−1/n, 1/n) for all n. This can’t possibly be. So g←[B] is not open.

1This do not mean that we will always choose the product topology over the box topology. Later, whenwe study spaces of sequences, for example, we may decide that it makes more sense to work with the boxtopology on such spaces.


Theorem 7.12 Let Sαα∈I and Tαα∈I be two sets of topological spaces and let fαα∈I

be a family of functions, fα : Sα → Tα. Then the function, g :∏

α∈I Sα → ∏

α∈I Tα, definedas

g(<xα>α∈I) = <fα(xα)>α∈I

is continuous if and only if fα is continuous on Sα for each α ∈ I .

Proof : Suppose S =∏

α∈I Sα, T =∏

α∈I Tα and fαα∈I is a family of functions,fα : Sα → Tα. Let πβS

: S → Sβ and πβT: T → Tβ be βth projection maps.

Let g :∏

α∈I Sα →∏

α∈I Tα be defined as g(<xα>α∈I) = <fα(xα)>α∈I.

( ⇐ ) For this direction we are given that each function fα is continuous on Sα. Wewant to show that the continuity of g follows.

To show continuity of g it will suffice to show that g pulls back subbase elements of∏

α∈I Tα to open sets in∏

α∈I Sα. Let Uβ be an open subset of Tβ . Then π←βT[Uβ] is

a subbase element for open sets in T =∏

α∈I Tα.

By lemma 7.11, if παg is continuous for all α, then g is continuous. It then sufficesto show that g←[π←βT

[Uβ]] is open in S.

See that,

π←βS[f←β [Uβ]] = π←βS

[x ∈ Sβ : fβ(x) ∈ Uβ]= < xα >α∈I ∈ S : fβ(xβ) ∈ Uβ= < xα >α∈I ∈ S : fα(xα) ∈ π←βT

[Uβ]= < xα >α∈I ∈ S : g(< xα >α∈I ) ∈ π←βT

[Uβ]= g←

[

< xα >α∈I ∈ S : xα ∈ π←βT[Uβ]

]

= g←[π←βT[Uβ ]]

so,g←[π←βT

[Uβ]] = π←βS[f←β [Uβ]]

Since both πβSand fβ are continuous then the right-hand side is open; so g←[π←βT

[Uβ]]

is open. So g is continuous, as required for ⇐.

( ⇒ ) Suppose g :∏

α∈I Sα →∏

α∈I Tα is continuous, where

g(<xα>α∈I) = <fα(xα)>α∈I

We are required to show that each fα : Sα → Tα is continuous.

Then for each β, (πβg)(< xα >α∈I) = πβ(<fα(xα)>α∈I) = fβ(xβ) ∈ Tβ. Since bothπβ and g are continuous on their domain then so is fβ on Sβ, as required. This proves

⇒.


We expect that, if two product spaces, S and T , have corresponding factors which

are homeomorphic then S and T are homeomorphic. In the proof of the followingtheorem, we confirm this by explicitly exhibiting the required homeomorphism.

Theorem 7.13 Let S =∏

α∈I Sα and T =∏

α∈I Tα be two product spaces. Suppose that,

for each α ∈ I , Sα and Tα are homeomorphic. Then the spaces S and T are homeomorphic.

Proof : For each α ∈ I , let fα : Sα → Tα be a homeomorphism mapping Sα onto Tα.

Let g :∏

α∈I Sα →∏

α∈I Tα be defined as

g(<xα>α∈I) = <fα(xα)>α∈I ∈ ∏

α∈ITα

Since each fα is continuous and one-to-one then so is g (by the theorem 7.12 above).

We claim that g is open.

By hypothesis, each fα is an open map. Let π←α1[Uα1] ∩ π←α2

[Uα2] · · · ∩ π←αk[Uαk

] be anopen base element in

∏

α∈I Sα. Since g is one-to-one, see that

g[

π←α1[Uα1] ∩ π←α2

[Uα2] · · · ∩ π←αk[Uαk

]]

= g[

π←α1[Uα1]

]

∩ g[

[π←α2[Uα2]

]

· · · ∩ g[

[π←αk[Uαk

]]

= fα1

[

π←α1[Uα1]

]

× · · · × fαk

[

π←αk[Uαk

]]

with all other factors equal to Tγ .

Since each fαiis open RHS is open in

Q

α∈I Tα.

Since the right-hand side is open, then g is open. So g is a homeomorphism.

We have shown that g(<xα>α∈I) = <fα(xα)>α∈I is the required homeomorphism

mapping S onto T .

We now show that every product space contains a homeomorphic copy of each of its

factors.


α∈I Sα be a product space. Then, for each α ∈ I , S contains a

subspace which is a homeomorphic copy of Sα.


Proof : Let β ∈ I . For α 6= β, we choose and fix kα ∈ Sα. Let

T =∏

α∈ITα = < xα >i∈I : Tβ = Sβ and [α 6= β] ⇒ [Tα = kα]

So every T is a subspace of S. We claim that T is homeomorphic to Sβ.

We define g : Sβ → T as g(x) =< xα >α∈I, where xβ = x, and if α 6= β, xα = kα.Then g maps Sβ one-to-one and onto T ⊆ S. Then the function, (πβ g) : Sβ → Sβ

is the continuous identity map onto Sβ. Invoking theorem 7.11, we conclude that g iscontinuous on Sβ . So g embeds Sβ in the proper subset T of

∏

α∈I Sα and so embedsSβ in

∏

α∈I Sα, as required.

Example 5. Let (Sn, ρn) : n ∈ N be a countable family of metric spaces. Suppose

that, for each n ∈ N, supρn(xn, yn) : xn, yn ∈ Sn ≤ 1. Let S =∏

n∈N Sn andρ : S × S → R be defined as

ρ(x, y) =∑

n∈N

ρn(xn, yn)

2n

Verify that ρ : S × S → R is a metric on S. Furthermore, verify that the topology onS =

∏

n∈N Sn derived by the metric ρ is the product topology.

Solution : We are given that S =∏

n∈N Sn and , for each n,

supρn(xn, yn) : xn, yn ∈ Sn ≤ 1

We begin by showing that ρ : S × S → R thus defined, is a valid metric on S.

For x, y ∈ S let x =<xn>n∈N and y =<yn>n∈N.

For M1, ρ(x, y) =∑

n∈Nρn(xn,yn)

2n ≤∑n∈N12n = 2 ≥ 0.

If ρ(x, y) =∑

n∈Nρn(xn,yn)

2n = 0. Then ρn(xn, yn) = 0 for all n. Since x =<xn>n∈N

and y =<yn>n∈N then x = y.

For M2, ρ(x, y) =∑

n∈Nρn(xn,yn)

2n =∑

n∈Nρn(yn,xn)

2n = ρ(y, x).

For M3, we compare ρ(x, z) and ρ(x, y) + ρ(y, z). Since, for each n, ρn(xn, zn) ≤


ρn(xn, yn) + ρn(yn, zn), then

ρ(x, z) =∑

n∈N

ρn(xn, zn)

2n

≤∑

n∈N

ρn(xn, yn) + ρn(yn, zn)

2n

=∑

n∈N

[

ρn(xn, yn)

2n+ρn(yn, zn)

2n

]

=∑

n∈N

ρn(xn, yn)

2n+∑

n∈N

ρn(yn, zn)

2n(All terms are positive)

= ρ(x, y) + ρ(y, z)

So ρ is a valid metric on S =∏

n∈N Sn.

We will now compare the the metric topology on (S, ρ) to the product topology on S.

Let x =< xn >∈ S. Let V = B1/2p(x) be an open ball in S with respect to ρ. Weclaim that V is open in S with respect to the product topology.

Proof of claim. Let

U = y ∈ S : y =<yn>, ρ(xn, yn) < 2−p−n−2 for n ≥ p+ 2

Note that U is a union of open base elements and so is an open subset of S withrespect to the product topology. It suffices to show that U ⊂ V . If y ∈ U , then

ρ(x, y) =∑

n∈N

ρn(xn, yn)

2n

=

(

ρ0(x0, y0)

20+ρ1(x1, y1)

21+ · · ·+ ρp+2(xp+2, yp+2)

2p+2

)

+ρp+3(xp+3, yp+3)

2p+3+ · · ·

<

(

2−p−0−2

20+

2−p−1−2

21+

2−p−2−2

22+ · · ·+ 2−p−(p+2)−2

2p+2

)

+1

2p+3+

1

2p+4+ · · ·+

=

(

1

2p+2+

1

2p+4+

1

2p+6+ · · ·+ 1

23p+6

)

+

(

1

2p+3+

1

2p+4+

1

2p+5+ · · ·+

)

<1

2p+1+

1

2p+1

=1

2p

So y ∈ V = B1/2p(x).

So V is open in S with respect to the product topology.


Suppose now that U is an open subbase element in S with respect to the producttopology. Suppose U contains the point a =< an >n∈N. It will suffice to find an ε

such that Bε(a) (with respect to ρ) is contained in U .

Now U is of the form x : xn ∈ Wn where Wn = Sn for all n except, possibly, for

some k ∈ N where Wk is open in Sk. Then there is an ε1 such that

Ba = y ∈ S : ρk(ak, yk) < ε1

is a subset of U containing a. Then

ρk(ak, yk)

2k<ε12k

Choose ε = ε1/2k. We then claim thatBε(a) ⊆ Ba ⊆ U (with respect to the metric, ρ).

Proof of claim. Let y ∈ Bε(a). Then

ρ(a, y) =∑

n∈N

ρn(an, yn)

2n< ε =

ε12k

Sinceρk(ak, yk)

2k≤∑

n∈N

ρn(an, yn)

2n= ρ(a, y)< ε = ε1/2

k

then this y ∈ Ba. So Bε(a) ⊆ Ba ⊆ U , with respect to ρ. We conclude that U is openin (S, ρ).

Then, the metric space, (S, ρ), and the product space, (S, τ), have the same topology.

7.6 Topic: Product spaces are “commutative”.

What effect does it have on the topology of a product space if we alter the order ofits factors? We investigate this question by first considering a particular case in the

following example.

Example 6. Show that for topological spaces S1, S2 and S3,

S = S1 × S2 × S3 and T = S3 × S1 × S2

are homeomorphic.

Solution : Let q : 1, 2, 3 → 3, 1, 2 be the one-to-one function where q(1) = 3,

q(2) = 1, q(3) = 2. Let q∗ : S1 × S2 × S3 → S3 × S1 × S2 be defined as

q∗< x1, x2, x3 >=< x3, x1, x2 >


We claim that πq(3)q∗ is continuous. If so we can apply theorem 7.11, to conclude q∗

is continuous. Let U be an open subset of Sq(3). Then U ×S1 ×S2 is a basic open set

in T . Then

(πq(3)q∗)←[U ] = q∗←[πq(3)

←[U ]]

= q∗←[U × S1 × S2]

= S1 × S2 × U (Since q∗ [S1 × S2 × U ] = U × S1 × S2 )

(A basic open subset of S).

So πq(3)q∗ is continuous, as claimed. Similarly, πq(2)q

∗ and πq(1)q∗ are continuous.

By theorem 7.11, q∗ : S → T is continuous. Similarly q∗← : T → S is continuous. So

q∗ : S → T is a homeomorphism, as required.

In the next theorem we generalize the proof in the example to one involving largerproduct spaces.


α∈I Sα and let q : I → J be a one-to-one function mapping Ionto an indexing set, J. Then there is a homeomorphism mapping S onto T =

∏

α∈I Sq(α).

Proof : Let S =∏

α∈I Sα and let q : I → J be a one-to-one function mapping the indexingset, I , onto the indexing set, J. We are required to show that T =

∏

α∈I Sq(α) and

S =∏

α∈I Sα are homeomorphic.

Let q∗ : S → T be defined as q∗(< xα >α∈I) =< xq(α) >α∈I.

Since q is one-to-one and onto J, then both q∗ : S → T and q∗← : T → S are

one-to-one and onto. Then πq(α)q∗ : S → Sq(α), where

(πq(α)q∗)(< xα >α∈I) = πq(α)(< xq(α) >α∈I) = xq(α)

Let Uq(β) be an open neighbourhood of xq(β) in Sq(β).

Then

(πq(β)q∗)←[Uq(β)] = q∗←[πq(β)

←[Uq(β)] ]

= q∗←[· · · × Sq(i) × · · · × Uq(β) × · · · × Sq(i) × · · · ]= · · · × Si × · · · × Uβ × · · · × Si × · · ·

(A basic open subset of S).

So πq(3)q∗ is continuous, as claimed. By theorem 7.11, q∗ : S → T is continuous.

Similarly q∗← : T → S is continuous. So q∗ : S → T is a homeomorphism, as required.


The above theorem confirms that “altering the order of the factors of a product space

produces another product space which is homeomorphic to the original one”. We cansummarize the statement by saying that . . . ‘

“product spaces are commutative.”

7.7 Example: The Tychonoff plank

We now briefly discuss the product of two ordinal spaces, for future reference. Theordinal space was introduced in definition 5.14.1 In the following example, ω1 rep-

resents the first uncountable ordinal, while, ω0 represents the first countable infiniteordinal. Let W represent the ordinal space, [0, ω1], and T represent the ordinal space,

[0, ω0]. Recall from the discussion which appears following the definition 5.14, thatthe elements of the open base for an ordinal space are of the form (α, β].

Let

S = W × T = [0, ω1] × [0, ω0]

be the product space of the two given ordinal spaces. Then the elements of S canbe viewed as ordered pairs (α, β) ∈ W × T . Since both sets are linearly ordered, it

doesn’t hurt to represent the product space, S, as a Cartesian plane of numbers whereW represents the horizontal axis and T represents the vertical axis. We would thenhave (0, 0) in the lower left corner and (ω1, ω0) in the top right corner. The topological

space S = [0, ω1] × [0, ω0] equipped with this topology is commonly referred to bytopologists as the

“Tychonoff plank”

The subspace S∗ of S defined as

S∗ = S\(ω1, ω0)

simply obtained by deleting the top right corner from the Tychonoff plank is appro-priately referred to as the

“Deleted Tychonoff plank”

As an open neighbourhood base of the point, (β, µ) ∈ S, we can use elements of theform

B(β,µ) = (α, β]× (γ, µ] : α < β and γ < µ1A more detailed study of the ordinals in the context of set theory is found in Axioms and set theory, by

Robert Andre (can be found online)


The Tychonoff plank may appear to be a topological space that is, in many ways,similar to the the product space R×N but, as we will eventually see, has quite differ-

ent properties. Both the Tychonoff plank and the Deleted Tychonoff plank are usefultopological spaces to remember.

7.8 Topic: Embedding a topological space in a product space.

In this section we will show how the family, F , of continuous functions on a topologicalspace, (S, τ), can be used to embed S in a product space whose factors are the range

of the functions in F . This method is exhibited in a theorem titled “The EmbeddingTheorem I”. The Embedding theorem I applies only to topological spaces, S, in which

singleton sets are closed. In order to understand this theorem one must be familiarwith two very important notions in topology:

− Family of functions which separates points and closed sets of S,

− Evaluation map with respect to a family, F , of functions.

Definition 7.16 Let (S, τS) be a topological space and (Sα, τα) : α ∈ Γ be an indexedfamily of non-empty topological spaces. Let F = fα : α ∈ Γ be a set of continuous

functions, fα : S → Sα, each one mapping S onto its range fα[S] ⊆ Sα.

a) We say that F separates points and closed sets if, whenever F is a closed subset of S

and x 6∈ F , then there exists at least one function fβ ∈ F such that fβ(x) 6∈ clSβfβ [F ].1

b) We define a function, e : S →∏

α∈Γ Sα, as follows:

e(x) =<fα(x)>α∈Γ ∈∏α∈Γfα[S] ⊆∏α∈ΓSα

We refer to the function e : S →∏

α∈Γ Sα as the

“evaluation map of S into∏

α∈Γ Sα with respect to F”

Theorem 7.17 The embedding theorem I. Let (S, τS) be a topological space in which every

singleton set in S is a closed subset of S. Given an indexed family of non-empty topologicalspaces, (Sα, τα) : α ∈ Γ, let F = fα : α ∈ Γ be a set of continuous functions where

each fα maps S onto its range, fα[S], inside Sα.1Similar expressions such as “the subsets A and B are completely separated” and “the real-valued func-

tion, f : S → R separates A and B if A ⊆ Z(f) and B ⊆ Z(f − 1) for some f ∈ C(X)” will be defined whendiscussing normal spaces in definition 10.2.


a) If F separates points and closed sets of its domain, S, then the evaluation map,

e(x) =< fα(x)>α∈Γ∈∏

α∈Γfα[S] ⊆∏α∈ΓSα

with respect to F , is both continuous and one-to-one on S.

b) Furthermore, the function, e : S →∏

α∈Γ Sα, maps S homeomorphically onto

e[S] ⊆∏α∈Γfα[S] ⊆∏α∈ΓSα

Hence this evaluation map embeds a homeomorphic copy of S into the product space,∏

α∈Γ Sα.

Proof : We are given that (S, τS) is a topological space in which all singleton sets, x, areclosed in S and a family of topological spaces, Sα : α ∈ Γ. For the set,F = fα : S → Sαα∈Γ, of continuous functions on S, we define

e : S →∏

α∈Γfα[S] ⊆∏α∈ΓSα

as an evaluation map with respect to F .

a) Note that, for each α ∈ Γ and x ∈ S, (πα e)(x)) = πα(< fα(x)>α∈Γ ) = fα(x).Since, for each α ∈ Γ, fα is continuous then so is πα e : S → fα[S]. By lemma 7.11,

e : S →∏

α∈Γ fα[S] is continuous.

We claim that e : S → ∏

α∈Γ fα[S] is one-to-one on S. Suppose a and b are distinctpoints in S. Then, since the single set b is closed and F separates points and closed

sets, there exists β ∈ Γ such that fβ(a) 6∈ clSβfβ [b]. Then the βth component of

e(a) =< fα(a)>α∈Γ and e(b) =< fα(b)>α∈Γ are distinct and so e(a) 6= e(b). We

conclude that the evaluation map e : S →∏

α∈Γ fα[S] is one-to-one on S, as claimed.

b) To prove that the evaluation map e : S →∏

α∈Γ fα[S] embeds S into∏

α∈Γ Sα, itwill suffice to show that it is an open function and then invoke theorem 6.9.

Let U be a non-empty open subset of S with the point u ∈ U . Then F = S \U is

closed in S. Since F separates points and closed sets, there exists β ∈ Γ such thatfβ(u) 6∈ clSβ

fβ[F ]. That means, fβ(u) ∈ Sβ\[

clSβfβ[F ]

]

.

We now show that e[U ] is open in∏

α∈Γ Sα.

Note that,

(πβ e)(u) = πβ(e(u))

= πβ(<fα(u)>α∈Γ)

= fβ(u)

∈ Sβ\[

clSβfβ [F ]

]


Since e(u) ∈ π←β[

Sβ\[

clSβfβ [F ]

]]

and since πβ is continuous then π←β[

Sβ\[

clSβfβ [F ]

]]

is an open neighbourhood of e(u) in∏

α∈ΓSα. It now suffices to show that

π←β[

Sβ\[

clSβfβ [F ]

]]

⊆ e[U ]

Suppose e(a) ∈ π←β[

Sβ\[

clSβfβ [F ]

]]

.

e(a) ∈ π←β[

Sβ\[

clSβfβ[F ]

]]

⇒ (πβ e)(a) ∈ πβ

[

π←β[

Sβ\clSβfβ [F ]

]]

⇒ fβ(a) ∈[

Sβ\clSβfβ [F ]

]

⇒ fβ(a) 6∈ clSβfβ [F ]

⇒ a ∈ S\ F = S\(S\U) = U

⇒ e(a) ∈ e[U ]

So π←β[

Sβ\[

clSβfβ [F ]

]]

is an open neighbourhood of e(u) which is entirely containedin e[U ]. We conclude e[U ] is open in

∏

α∈Γ Sα and so e : S → ∏

α∈Γ Sα is a homeo-

morphism.

We have thus shown that e : S →∏

α∈Γ Sα embeds S into the product space,∏

α∈Γ Sα.

7.9 Topic: The Cantor Set: Viewed a the continuous image of a product space.

The Cantor set is a notion which is part of general mathematical culture. We often

refer to it because of its interesting set-theoretic and topological properties. It can bedescribed in various ways. We will formally provide a set-theoretic definition of the

Cantor set and then study it from a topological point of view.

A prologue to a definition of the Cantor set.

Recall that Z+ = 1, 2, 3, . . . , . Let D = 0, 1, 2, . . . , 8, 9. So∏

n∈Z+ D = DZ+

represents the set of all functions mapping Z+ into D. The set,∏

n∈Z+ D can bedescribed as the set of all countably infinite ordered strings, of the form,

< mn >n∈Z+ , where mn ∈ 0, 1, 2, . . . , 9

Suppose we define the function, ϕ :∏

n∈Z+ D → [0, 1] (the closed interval in R from 0to 1) as follows:

ϕ(< m1, m2, m3, . . . , mn, . . . , >) =

∞∑

n=1

mn

10n

Note that every number x ∈ [0, 1] (represented in its infinite decimal expansion

0.m1m2m3 · · · ) can be expressed in the form,

x =

∞∑

n=1

mn

10n= 0.m1m2m3 · · ·


For example,

ϕ(< 0, 0, 0, , . . . >) =

∞∑

n=1

0

10n= 0

ϕ(< 9, 9, 9, . . . >) =

∞∑

n=1

9

10n= 0.9999 . . .= 1

ϕ(< 2, 9, 9, . . . >) = 0.2999 . . .= 0.3000 . . .= ϕ(< 3, 0, 0, . . . >)

and so ϕ maps∏

n∈Z+ D onto [0,1] but is not necessarily one-to-one (since an endlessstring of 9’s and an endless string of 0’s may be mapped to the same element in [0,

1]). But the entire set [0,1] is, indeed, the image of∏

n∈Z+ D under ϕ.

However, if we reduce the size of D to say, D∗ = 0, 1, 2, 5, 7, 9, then the function,ϕ :

∏

n∈Z+ D∗ → [0, 1], defined similarly, would produce a range which is a proper

subset of [0, 1] which would contain multiple gaps in it (since the digits 3, 6, and 8are lacking in the ordered strings of the domain).

If we set D = 0, 1, 2 and every number x in [0, 1] is expressed in its triadic expansion

form1 then the set

ϕ[∏

n∈Z+0, 1, 2]

= [0.000 . . .3 , 0.2222 . . .3] = [0, 1]

would look like,

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n= 0.m1m2m3 · · ·3 ∈ [0.000 . . .3 , 0.2222 . . .3] = [0, 1]

For example,

ϕ(< 0, 0, 0, . . . >) = 0.0000 . . .3 =∞∑

n=1

0

3n

ϕ(< 2, 2, 2, . . . >) = 0.2222 . . .3 =

∞∑

n=1

2

3n= 1

ϕ(< 1, 1, 1, . . . >) = 0.1111 . . .3 =

∞∑

n=1

1

3n=

1

2(Verify!)

ϕ(0, 1, 2, 2, . . .) = 0.0122223 . . . = 0.0200000000 . . .3 = ϕ(0, 2, 0, 0, 0, . . .)

Note that, ϕ is not one-to-one onto∏

n∈Z+0, 1, 2.

1. . . , 0, 1, 2, 10, 11, 12, 100, 101, 102, 110, . . .


If we shrink the set 0, 1, 2 to D = 0, 2, with mn ∈ 0, 2 and ϕ :∏

n∈Z+0, 2 →[0, 1] is defined as,

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n= 0.m1m2m3 . . .3 ∈ [0.000 . . .3 , 0.2222 . . .3] = [0, 1]

where 0.m1m2m3 . . .3 contains only 0’s and 2’s. The set, ϕ[∏

n∈Z+0, 2]

is now a

proper subset of [0.000 . . .3 , 0.2222 . . .3] = [0, 1].

In this case, ϕ is indeed onto [0, 1] its domain. For example, even if 0.02000 . . .3 =0.01222 . . .3

ϕ←(0.012222 . . .3) = ϕ←(0.02000 . . .3) = 0, 2, 0, 0, 0, . . .

So every point in [0, 1] can be expressed with simply 0’s and 2’s in ternary expansionform.

The “Cantor set” involves the particular function

ϕ :∏

n∈Z+0, 2 → [0, 1]

who’s range, ϕ[∏

n∈Z+0, 2]

, is a proper subset of [0, 1].

Definition 7.18 The Cantor set: A definition. Suppose Z+ = N\0 and∏

n∈Z+0, 2 =

0, 2Z+, the set of all countably infinite sequences of 0’s and 2’s. Let the function

ϕ :∏

n∈Z+0, 2 → [0, 1]

be defined as

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n

where ϕ(< mn >n∈Z+) = 0.m1m2m3 . . .3 is an element of [0, 1] expressed in its triadic ex-pansion form. Since the digit 1 is lacking in the ordered strings, ϕ is one-to-one on its domain

and the range, ϕ [∏

n∈Z+0, 2 ], is a proper subset of the interval, [0.000 . . .3 , 0.2222 . . .3] =[0, 1], with multiple gaps in it. The one-to-one image,

C = ϕ [∏

n∈Z+0, 2 ]

of∏

n∈Z+0, 2 is a proper subset of [0.000 . . .3 , 0.2222 . . .3] = [0, 1]. The set, C, is referred

to as the Cantor set.


Then the Cantor set can be seen as

C =

∞∑

n=1

mn

3n: mn ∈ 0, 2

all possible values of the infinite sum∑∞

n=1mn3n where mn = 0 or 2.

What does the Cantor set look like?

The definition of the “Cantor Set”, C, states that C is a particular subset of the

closed unit interval, [0, 1]. The Cantor set is the range, ϕ [∏

n∈Z+0, 2 ], under the awell-defined function, ϕ, with domain,

∏

n∈Z+0, 2. This definition, by itself, is not

very useful when trying to visualize what kind of subset of [0, 1] it represents. If thereader is willing to go through the trouble of verifying this, C does not contain any of

the points in the open intervals, (1/3, 2/3), (1/32, 2/32), (7/32, 8/32). Since ϕ maps∏

n∈Z+0, 1, 2 onto [0, 1] then ϕ will map∏

n∈Z+0, 2 only onto a proper subset of

[0, 1]. Such gaps will occur. It doesn’t take long before one sees a pattern emerge. Wenormally view the Cantor set as a set which is inductively constructed in stages bysuccessively defining a nested sequence, C0, C1, C2, . . ., of subsets of [0,1]. Ultimately

we define the Cantor set as being the intersection, C = ∩n∈NCn, of all of these (afterconvincing ourselves that C would not be empty). We can visualize this procedure as

follows.

C0 = [0, 1]

C1 = C0\(1, 3)C2 = C1\( 1

32232 ) ∪ ( 7

32832 )

C3 = C2\...open middle thirds in C2...

Cn+1 = Cn\...open middle thirds of the remaining intervals in Cn...

The construction of Cn is normally described by saying “ . . . to obtain Cn, subtract

open middle thirds from Cn−1” After inductively obtaining an infinite sequence ofnested sets, Cnn∈N, in this way we define the Cantor set as being the infinite

intersection

C = ∩∞n=0Cn

We visualize the Cantor set geometrically (up to the construction of C5) as follows.


In the definition 7.18 of the Cantor set C = ϕ[∏

n∈Z+0, 2] → [0, 1] where

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n

the topology of the sets involved was not discussed. So we didn’t speak of the “conti-

nuity” of ϕ on the product∏

n∈Z+0, 2. We will define topologies of all sets involvedin the most natural way. We will equip the set 0, 2 with the discrete topology, the

set∏

n∈Z+0, 2 with the product topology, and finally, the Cantor set C, with thesubspace topology inherited from R. We will now show that, the Cantor set, C, is a

homeomorphic copy of∏

n∈Z+0, 2.

Theorem 7.19 The one-to-one function, ϕ :∏

n∈Z+0, 2 → [0, 1], defined as,

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n

which maps the product space∏

n∈Z+0, 2 homeomorphically onto the Cantor set,

C = ϕ[∏

n∈Z+0, 2]

a subset of [0, 1]. So the Cantor set is a topological copy of the product space,∏

n∈Z+0, 2,contained in [0, 1].

Proof : Given the one-to-one function ϕ :∏

n∈Z+0, 2 → [0, 1] where

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n

Let ε > 0. First note that, for all < mn >n∈Z+∈∏

n∈Z+0, 2,

|ϕ(< mn >n∈Z+)| ≤∞∑

n=1

2

3n= 1


Then the sequence,∑∞

n=k23k

n∈Z+ , converges to zero. So there exists N such that

∞∑

n=N+1

2

3n< ε (∗)

Claim #1. The function ϕ is continuous. If < kn >n∈Z+ ∈ ∏n∈Z+0, 2 then

ϕ(< kn >n∈Z+) =

∞∑

n=1

kn

3n= x ∈ C

Let Bε(x) be and open interval in R with center x and radius ε. Then Bε(x)∩C is an

open neighbourhood of x in C. To show continuity of ϕ at < kn >n∈Z+ , it will sufficeto find an open neighbourhood, U , of < kn >n∈Z+ such that ϕ[U ] ⊆ Bε(x) ∩C.

Since the series,∑∞

n=1kn3n , converges to x then the sequence

x−∑m

n=1kn3n

=∑∞

n=m+1kn3n

converges to zero as m tends to infinity. Then,

∣

∣

∣

∣

∣

∞∑

n=N+1

kn

3n

∣

∣

∣

∣

∣

≤∞∑

n=N+1

2

3n< ε (By *)

Let U = π←1 (k1)∩ · · ·∩π←N (kN) be an open base neighbourhood of kn in∏

n∈Z+0, 2and suppose < bn >n∈Z+ is some element in U . This implies b1 = k1, b2 = k2, . . . , bN =

kN . So

|x− ϕ(< bn >n∈Z+)| =

∣

∣

∣

∣

∣

x−∞∑

n=1

bn3n

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∞∑

n=1

kn

3n−∞∑

n=1

bn3n

∣

∣

∣

∣

∣

≤ 0 +

∞∑

n=N+1

|kn − bn|3n

≤∞∑

n=N+1

2

3n

< ε

Then ϕ[U ] ⊆ Bε(x)∩C. Then ϕ is continuous at < kn >n∈Z+ , and so at all points of∏

n∈Z+0, 2. Then ϕ is continuous on∏

n∈Z+0, 2, as claimed.

Claim #2. The function ϕ is homeomorphic. To prove that ϕ is a homeomorphism itwill suffice to show it is an open function.

For y =< kn >n∈Z+∈∏n∈Z+0, 2,∑∞

n=1kn3n converges to ϕ(y).


Let U = π←j1 (kj1) ∩ · · · ∩ π←jm(kjm) be an arbitrary open neighbourhood base element

of < kn >n∈Z+ .

To show that ϕ is open, it will suffice to find some ε such that Bε(ϕ(y)) ∩C ⊆ ϕ[U ].

Let N = max j1, j2, . . . , jm and let ε = 13N+1 . (**)

We claim that, if |ϕ(< kn >n∈Z+) − ϕ(< zn >n∈Z+) | < ε, then ϕ(< zn >n∈Z+) ∈ϕ[U ].

|ϕ(< kn >n∈Z+) − ϕ(< zn >n∈Z+) | =

∣

∣

∣

∣

∣

∞∑

n=1

kn

3n−∞∑

n=1

zn3n

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∞∑

n=1

kn − zn3n

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

N∑

n=1

kn − zn3n

+

∞∑

n=N+1

kn − zn3n

∣

∣

∣

∣

∣

< ε

Then

−ε =−1

3N+1<

N∑

n=1

kn − zn3n

+

∞∑

n=N+1

kn − zn3n

<1

3N+1= ε

−1

3N+1−

∞∑

n=N+1

kn − zn3n

<

N∑

n=1

kn − zn3n

<1

3N+1−

∞∑

n=N+1

kn − zn3n

After some computation we obtain,

−4

3N+1<

N∑

n=1

kn − zn3n

<4

3N+1(Verify this!)

To show that ϕ(< zn >n∈Z+) ∈ ϕ[U ], it suffices to show that, yn − zn = 0 for n = 1to N .For m = 1 to N , let

Sm =m∑

n=1

kn − zn3n

Suppose k1 − z1 = 2 or −2; then S1 = ±2/3 6∈ (− 43N+1 , 4

3N+1 ). So k1 − z1 = 0.

Suppose km − zm = 0 for m < N and km+1 − zm+1 = 2 or −2. Then Sm+1 = ±2/3 6∈(− 4

3N+1 , 43N+1 ). So km+1 − zm+1 = 0.

We then have yn − zn = 0 for n = 1 to N .


This is precisely what was needed to show that ϕ(zn) ∈ ϕ[U ]. This means that, forour choice of ε = 1

3N+1 (at (**)), Bε(ϕ(kn) ∩ C ⊆ ϕ[U ]. So ϕ is an open map on∏

n∈Z+0, 2. So ϕ is a homeomorphism, as claimed.

The theorem is proved, as required

The Cantor set was defined as being the range ϕ [∏

n∈Z+0, 2 ] of∏

n∈Z+0, 2. Butthe above statement shows it is much more. Since we have shown

∏

n∈Z+0, 2 is a

homeomorphic copy of C then, topologically speaking, the subspace, C, of [0, 1] wecall the Cantor set “is” the product space,

∏

n∈Z+0, 2. The topological point of

view certainly provides much more insight on the nature and various properties of Cas well as those sets that are linked to it via continuous functions. The set-theoretic

definition of C is blind to this aspect.

About the cardinality of C.

Our geometric description of the construction of the Cantor set, C, showed that Cwas obtained by successively removing open middle-third interval from a previous

set, leaving behind, at least, the endpoints of countably many closed intervals. Theendpoints, all of them of the form, mn

3n, (mn ∈ 0, 2) are rationals and are never

removed and so must belong to C. This may lead one to believe that C is countablyinfinite. But our formal definition of C (as well as the previous theorem) shows that

this cannot be so. It states that C can be mapped homeomorphically onto the un-countable set

∏

n∈Z+0, 2. Then, |C| = 2|Z+| = 2ℵ0 = c. So C is uncountable. We

leave the reader to reflect on the more challenging question: If x is a point in C whichis not an endpoint of a middle third, what does it look like? How can it be that a

“non-endpoint” is left behind?

Is the Cantor topological space discrete?

By this question, we are wondering whether every single point of C is both openand closed. We have shown that the Cantor set, C, is the homeomorphic image of

the product space,∏

n∈Z+0, 2. If every point of C was open then every point of∏

n∈Z+0, 2 would be open. But this is easily verified not to be the case. Since

no point of the product space can contain a basic open set of the product space,∏

n∈Z+0, 2.So C can be viewed as a sprinkling of an uncountable set of points in [0, 1]. It inter-

esting to note that C can continuously be mapped onto [0, 1] as the following theoremshows.1

1The rest of this section can be omitted without loss of continuity.


Theorem 7.20 There is a continuous function, δ : C → [0, 1], which maps the Cantor set,C, onto the closed interval [0, 1].

Proof : Recall that the function, ϕ :∏

n∈Z+0, 2 → C, maps∏

n∈Z+0, 2 homeomorphi-cally onto C. It is defined as

ϕ(< mn >n∈Z+) =

∞∑

n=1

mn

3n

We begin by defining a similar function, ψ :∏

n∈Z+0, 2 → [0, 1] which maps thesame set,

∏

n∈Z+0, 2, onto [0,1]. It is defined as,

ψ(< mn >n∈Z+) =

∞∑

n=1

mn

2n+1

We claim that ψ is continuous on its domain and onto the closed interval [0, 1]. To

see that ψ is onto [0, 1] simply note that the expression

∞∑

n=1

mn

2n+1=

∞∑

n=1

mn/2

2n=

∞∑

n=1

sn2n

= 0.s1s2s3s4 . . .2 ∈ [0, 1]

where each sn is 0 or 1 and 0.s1s2s3s4 . . .2 is simply the dyadic expansion for an ele-

ment in [0, 1].1

To prove continuity of ψ proceed precisely as for ϕ in the previous theorem.

We define δ : C → [0, 1] as

δ = ψϕ←

where ϕ← continuously maps C one-to-one and onto∏

n∈Z+0, 2 and ψ continuouslymaps

∏

n∈Z+0, 2 onto [0, 1].2 So δ continuously maps C onto [0, 1], as required.

1For example, if y =< mn >n∈Z+ is the string 2, 0, 2, 0, 2, 0, . . . , , then ψ(y) = 0.10101010 . . . a pointin [0, 1] in dyadic expansion form. The function ψ is easily seen to be onto [0, 1]. For example, givenx = [0.001001001 . . . , ], ψ(0, 0, 2, 0, 0, 2, . . .) = x.

2Note that ψ is not one-to-one since ψ maps the two distinct strings 0, 0, 2, 2, 2. . . . , and0, 1, 0, 0, 0. . . . , to the same point 0.001111 . . . = 0.01000000 . . .


7.10 Topic: Peano’s space filling curve.1

As a final example of an application of product spaces we show the somewhat surpris-

ing result which states that a cube

[0, 1]3 = (x, y, z) : x, y, z ∈ [0, 1]

equipped with the product topology, in R3 is the continuous image of the closed in-

terval [0, 1] with the usual topology inherited from R. In the first example, we showthat there is a continuous function which maps the Cantor set onto the cube, [0, 1]3.

Example 7. Find a function, t : C → [0, 1]3, which maps the Cantor set, C, continu-ously onto the cube [0, 1]3.

Solution : We first list a few relevant functions presented in this chapter up to now.

− ϕ :∏

n∈Z+0, 2 → ϕ[∏

n∈Z+0, 2]

= C :

This function maps∏

n∈Z+0, 2 homeomorphically onto C. (Theorem 7.19).

− ϕ← : C →∏

n∈Z+0, 2 :

The function ϕ← maps C homeomorphically onto∏

n∈Z+0, 2.− δ :

∏

n∈Z+0, 2 → [0, 1] :

There exists a continuous function δ mapping∏

n∈Z+0, 2 onto [0,1]. (By the-

orem 7.20.)

− q∗ :∏

n∈Z+0, 2 →∏

n∈Z+A∪Z

+B∪Z

+C0, 2 :

Since Z+ = N\0 is countably infinite it has the same cardinality as 3Z+. Then

there is a one-to-one function mapping Z+ onto 3Z+ = Z+A∪Z+

B∪Z+C (the disjoint

union of three copies of Z+).

Then there is a homeomorphism, q∗ mapping∏

n∈Z+0, 2 onto∏

n∈Z+A∪Z

+B∪Z

+C0, 2.

(By theorem 7.15)

− f :∏

n∈Z+A∪Z

+B∪Z

+C0, 2 → ∏

n∈Z+A0, 2×∏n∈Z

+B0, 2×∏n∈Z

+C0, 2 :

There is an onto homeomorphism f with domain∏

n∈Z+A∪Z

+B∪Z

+C0, 2 defined as

f(<xα>α∈3Z+) = 〈 fA(<xα>α∈3Z+), fB(<xα>α∈3Z+), fC(<xα>α∈3Z+) 〉

where fA, fB and fC are defined in proposition 7.21 as,

fA(< xα >α∈3Z+) = < xα >α∈Z+A

fB(< xα >α∈3Z+) = < xα >α∈Z+B

fC(< xα >α∈3Z+) = < xα >α∈Z+C

1Pronounced: pay-an-o. This section can be omitted without loss of continuity.


− h :∏

n∈Z+A0, 2×∏n∈Z

+B0, 2 ×∏n∈Z

+C0, 2 → [0, 1]× [0, 1]× [0, 1]:

If TA =∏

n∈Z+A0, 2, and h[TA × TB × TC ] = δ[TA] × δ[TB] × δ[TC] then h con-

tinuously maps TA × TB × TC onto [0, 1]× [0, 1]× [0, 1]. (By 7.12 and 7.20).

By appropriately composing these functions, we have,

C → 1-1 onto ϕ←→ ∏

n∈Z+0, 2→ 1-1 onto q∗→ ∏

n∈Z+A∪Z

+B∪Z

+C0, 2

→ 1-1 onto f→∏

n∈Z+A0, 2×

∏

n∈Z+B0, 2×

∏

n∈Z+C0, 2

→ onto h→ [0, 1]× [0, 1]× [0, 1]

Lett = hfq∗ϕ←

Then the function, t : C → [0, 1]3, continuously maps C onto [0, 1]3, as required.

Example 8. Show that there exists a function which continuously maps [0, 1] onto thecube [0, 1]3.

Solution : We adopt the notation introduced in the previous example.

Each of the three functions

fA = πAt = πAhfq∗ϕ← : C → [0, 1]

fB = πBt = πBhfq∗ϕ← : C → [0, 1]

fC = πC t = πChfq∗ϕ← : C → [0, 1]

continuously maps the points in the Cantor set, C, onto the points of [0, 1].

Recall that C is a subset of [0, 1] with multiple open intervals missing. Also recallthat a function g : [0, 1] → [0, 1] is a continuous extension of fA if g is continuous on

[0, 1] and g|C = fA. We can then extend the continuous function fA : C → [0, 1] to acontinuous function FA : [0, 1] → [0, 1] so that, on each missing open interval (a, b) in

C, FA is defined as the being the function which is linear from (a, FA(a)) = (a, fA(a))to (b, FA(b)) = (b, fA(b)). The continuous extension, FA, of fA then maps [0, 1]

continuously onto [0, 1]. The same holds true for the similarly defined functions,FB : [0, 1] → [0, 1] and FC : [0, 1] → [0, 1].

We then have the three continuous onto functions,

FA : [0, 1] → [0, 1]

FB : [0, 1] → [0, 1]

FC : [0, 1] → [0, 1]


We define the function F : [0.1] → [0, 1]3, as

F (x) = 〈FA(x), FB(x), FC(x) 〉 ∈ [0, 1]× [0, 1]× [0, 1]

Since all three functions, FA, FB, and FC are continuous then by theorem 7.12, F iscontinuous on [0, 1]. So F continuously maps the closed interval [0, 1] onto [0, 1]3.

We normally refer to the continuous image of the closed interval [0, 1] as a

“curve”

In this sense, saying that “there is a curve that can fill the cube [0, 1]3” or “there is acurve that goes through every point of the cube [0, 1]3” is another way of saying that

“[0, 1] can be mapped continuously onto the cube [0, 1]3”. The reader who followsthrough the proof carefully will notice that it can be generalized to the statement

“For any integer n there is a curve which goes through each point of the cube [0, 1]n”.The set

[0, 1]n

is referred to as a cube, for any natural number, n ≥ 1. The curve is referred to as

Peano’s space filling curve. A quick internet search will lead to many illustrations ofcurves which fill [0, 1]2 or [0, 1]3. The figure below which can be graphed by math

software such as Wolfram or Maple illustrates part of a curve gradually filling up acube.

Figure 1: Part of a space filling curve in [0, 1]3.


7.11 Topic: Product spaces are “associative”.

We address the question: How can we simplify the product of a product spaces? We

begin by illustrating what we mean by this question.

Let I = 1, 2, . . . , 12, A = 1, 2, 3, 4, B = 5, 6, and C = 7, 8, . . . , 12. The set I

is the disjoint union the of sets A, B and C. Let (Sα, τα) : α ∈ 1, 2, . . . , 12 be aset of twelve topological spaces. Consider the two product spaces S and Y defined as

follows:

S =∏

α∈ISα and Y =∏

α∈ASα ×∏

α∈BSα ×∏

α∈CSα

We wonder, “How do the two product spaces compare?”. We see that S is a product

space with twelve factors, while Y is a product space of only three factors. This ob-servation is sufficient to conclude that S 6= Y . On the other hand, we see that each

of the three factors of Y are themselves product spaces, all three with factors, Sα,identical to the ones found in S. While we have excluded, the possibility that S equals

Y , it would be reasonable to suspect that these two spaces are homeomorphic copiesof each other. We will present this conjecture as a proposition (in just a slightly more

general form) followed by its proof.

Proposition 7.21 Let (Sα, τα) : α ∈ I be a set of non-empty topological spaces. Suppose

I is the disjoint union of three non-empty subsets, A, B and C where the elements inA ∪ B ∪C respect the order in which they appear in I . Then

S =∏

α∈ISα and Y =∏

α∈ASα ×∏α∈BSα ×∏α∈CSα

are homeomorphic product spaces.

Proof : We define three functions, fA : S → ∏

α∈A Sα, fB : S → ∏

α∈B Sα and fC : S →∏

α∈C Sα as follows:

fA(< xα >α∈I) = < xα >α∈A

fB(< xα >α∈I) = < xα >α∈B

fC(< xα >α∈I) = < xα >α∈C

All three functions are easily seen to be onto∏

α∈A Sα,∏

α∈B Sα and∏

α∈C Sα, re-spectively.

Note that,

For β ∈ A, (πβfA)(< xα >α∈I) = xβ = πβ(< xα >α∈I)

For β ∈ B, (πβfB)(< xα >α∈I) = xβ = πβ(< xα >α∈I)

For β ∈ C, (πβfC)(< xα >α∈I) = xβ = πβ(< xα >α∈I)


By theorem 7.11, the functions fA, fB , and fC are all continuous on S.

We now consider the function

f :∏

α∈ISα →∏


defined as

f(< xα >α∈I) = 〈 fA(< xα >α∈I), fB(< xα >α∈I), fC(< xα >α∈I) 〉

We claim f is one-to-one and continuous on∏

α∈ISα.

Since fA, fB and fC are all three continuous then, by theorem 7.12, f : S → Y is con-

tinuous on its domain S. Also, since I is the disjoint union of A, B and C, then f mapsS one-to-one and onto Y . (Verify this!). So f is one-to-one and continuous as claimed.

To show that f is a homeomorphism it now suffices to show that it is open.

Let V = π←α1[Uα1]∩π←α2

[Uα2]∩· · ·∩π←αk[Uαk

] be an open base element in S =∏

α∈ISα.

Then

f [V ] = 〈 fA[V ], fB [V ], fC [V ] 〉For fA[V ] = fA[π←α1

[Uα1] ∩ π←α2[Uα2] ∩ · · · ∩ π←αk

[Uαk]]:

fA[π←αi[Uαi]] = Uαi if αi ∈ A

=∏

α∈ASα if αi 6∈ A

The same holds true for fB and fC .

So f [V ] = 〈 fA[V ], fB[V ], fC [V ] 〉 is open.

The given homeomorphism, f , confirms that the product spaces

∏

α∈A∪B∪C Sα and∏


are homeomorphic.

Concepts review:

1. Give a general definition of a Cartesian product of sets.


2. Provide two definitions of the Cantor set.

3. Define the product topology on a Cartesian product.

4. Define the box topology on a Cartesian product.

5. Define what we mean by product space.

6. What do we mean by “partitioning” a product space and how does it change its

topology.

7. What effect does changing the order of the factors have on the topology of a productspace.

8. Describe the Tychonoff plank and its topology.

9. What does it mean to say that a set of functions F separates points and closed setsof a set S.

10. Define evaluation map with respect to a set of functions F .

11. State the Embedding theorem.

12. The embedding theorem describes a homeomorphism between a topological space and

a product space. What are these spaces? What is the homeomorphism?

13. The Cantor set equipped with the usual topology is shown to be homeomorphic to aproduct space. Which one? Describe the homeomorphism.

14. Describe a continuous function which maps the Cantor set, C, continuously onto theclosed interval [0, 1].

15. Is the Cantor set a countable subset of R?

16. Is the Cantor set, equipped with the subspace topology, discrete? Is it second count-

able?

17. Does there exist a continuous function which maps C onto the cube, [0, 1]3?

18. Does there exist a continuous function which maps [0, 1] onto the cube, [0, 1]3?


EXERCISES

1. Show that∏

α∈Γ Sα is dense in∏

α∈Γ Tα if and only if, for each α ∈ Γ, Sα is dense inTα.

2. Suppose that, for each α ∈ Γ, Sα ⊆ Tα. Then∏

α∈Γ Sα is a subset of the product

space,∏

α∈Γ Tα. Show that the product topology on∏

α∈Γ Sα is the same as thesubspace topology

∏

α∈Γ Sα inherits from∏

α∈Γ Tα.

3. Given the product space∏

α∈Γ Sα, show that the projection map πα :∏

α∈Γ Sα → Sα

is an open function. Is it a closed function?

4. If in the product space, S =∏

α∈Γ Sα, each Sα is discrete, describe the open subsets

of S.

5. If in the product space, S =∏

α∈Γ Sα, each Sα is indiscrete, describe the open subsets

of S.

6. Let X =∏

α∈Γ Sα and Y =∏

γ∈Φ Tγ be two product spaces where Γ and Φ have thesame cardinality confirmed by the one-to-one function q : Γ → Φ. For each α, Sα and

Tq(α) are homeomorphic topological spaces. Show that X and Y are homeomorphic.

7. Suppose we are given a topological space (S, τS) and a family of topological spaces,

(Tα, τα) : α ∈ Γ. Suppose F = fα : α ∈ Γ is a family of functions, fα : S → Tα,where each fα maps its domain S into Tα. Let

B = f←α [U ] : (α, U) ∈ Γ × τα

Show that B is a base for open sets of S if and only if F separates points and closed

sets in S.

8. Suppose U ⊆ S and V ⊆ T . Show that intS×T (U × V ) = intSU× intTV .

9. Show that, if the product space,∏

α∈Γ Sα, is first countable, then Sα is first countablefor each α ∈ Γ.

154 Section 8: The quotient topology

8 / The quotient topology.

Summary. In this section we will present a method to topologize the range, T,of a function, f, whose domain, S, is a topological space. The topology on therange is referred to as the “quotient topology induced by f”. When the function

f is used for this purpose, it is referred to as the associated quotient map.

8.1 The strong topology induced by a function.

We have previously discussed a function, f , with an untopologized domain, S, which ismapped to a topological space (T, τT). We topologized S in such a way that guaranteesthe continuity of the function f : S → T on S. To do this we must be sure that S is

provided with enough open sets so that f : S → T is continuous. The easiest way todo this is to assign to S the discrete topology, since, on such a space any function will

be continuous on S. But, ideally, we preferred a topology which is custom-made forf and τT . We then opted for the weak topology induced by f and τT . Namely,

τS = U ⊆ S : U = f←[V ], for some open V in T

In this section, we will work the other way around. We wish to assign to T a topology,

τT that will guarantee that f is continuous on S. Again, we could take the easy wayout by assigning to T the indiscrete topology, ∅, T. It only has one non-empty

open set, T , pulled back by f to the open set, S ∈ τS. We again obtain continuity,but one which is independent of the function, f . We opt for choosing a topology on

T which is custom made for the given function, f and the topology on the domain.That is, we will choose the strongest topology on T that will guarantee continuity of

f on the given topological space (S, τS). With this mind, we present the followingformal definition.

Definition 8.1 Let f : S → T be a function mapping the topological space (S, τS) onto

the set T . We assign to the set T the following topology

τf = U ⊆ T : f←[U ] is open in S

The set, τf is referred to as the quotient topology induced by f and τS. When T is equippedwith the quotient topology, then (T, τf) is referred to as the quotient space of S and

f : S → T is referred to as its associated quotient map.1

1Some authors use the terms “identification topology” instead of quotient topology and “identificationmap” instead of quotient map.


Since f← respects both infinite unions and intersections of sets then τf is a well-definedtopology on T . Also since τf contains precisely those sets which are pulled back to

some set in τS , and no other sets, then τf is indeed the largest topology that guaran-tees continuity for f on S. The quotient topology, τf , induced by f is then unique.

It is important for the reader to notice that, in the above definition, we declare thefunction f : S → T to be onto T . This fact may be relevant in some of the proofs

that follow. If f : S → T was not declared to be onto T , we would at least have tomodify our definition of quotient topology to τf = U ⊆ f [S] : f←[U ] ∈ τS.Suppose we are given two topological spaces (S, τS) and (T, τT) and a continuous func-

tion, f , mapping S onto T . If we are given no more information about τT , it may ormay not be the quotient topology induced by f . The following theorem shows that

there may be various ways of recognizing a quotient topology. By definition, quotientmaps must at least be continuous.

Theorem 8.2 Suppose (S, τS) and (T, τT) are topological spaces and f : S → T is acontinuous function onto T .

a) If f : S → T is an open map, then τT is the quotient topology on T induced by f .

That is, τT = τf .

b) If f : S → T is a closed map, then τT is the quotient topology on T induced by f .

That is, τT = τf .

c) If there is a continuous function g : T → S such that (f g)(x) = x on T , then τT = τf .

Proof : We are given that f : S → T is a continuous function mapping S onto T withtopology τT . To show that τT is the quotient topology it suffices to show that τT is

the strongest topology, τf that will guarantee continuity to this function, f . Sincef : S → T is continuous, and τf is the largest topology on T for which f is contin-

uous, then τT ⊆ τf . For each of the three parts it will then suffice to show that τf ⊆ τT .

a) We are given that f : S → T is open.

We claim that τf ⊆ τT . Let U ∈ τf . By definition of τf , f←[U ] is open in S. Since

f is both open and onto, then f [f←[U ]] = U ∈ τT . So τf ⊆ τT . Hence τT = τf .

b) The proof of part b) is similar to a) and so is left to the reader.

c) We are given that g : T → S is continuous such that (f g)(x) = x (where

fg : T → T ).


We claim that τf ⊆ τT . Suppose U ∈ τf . Then, by definition of τf , f←[U ] isopen in S. We have

U = x ∈ T : (f g)(x) = x ∈ U= (f g)←[U ]

= g←[f←[U ]]

⇒ U is open in T since g and f are continuous.

⇒ U ∈ τT

Hence U ∈ τT . So τf ⊆ τT , as claimed

Then τf = τT , as required.

Since there can be many different open maps from (S, τS) into (T, τT), it is possibleto have distinct maps f and g, associated to the same quotient topology, τT = τf = τg.

8.2 An equivalence relation induced by a function f .

Any function, f , mapping a set S onto a set T can be used to partition the domain

into subsets we call “fibres in S induced by the function f”. We will first formallydefine this set theoretic notion.

Definition 8.3 Let f : S → T be a function mapping a set S onto a set T . If w ∈ T ,we will refer to f←[w] as the fibre of w under the map f . Fibres in the domain are the

preimages of singleton sets with respect to f . 1

Let f : S → T be a function which maps the topological space (S, τS) onto the quotient

space, (T, τf). Using this function we will construct a new set by defining a relationRf on S as follows:

[u is related to v ] ⇔ [ f(u) = f(v) in T ]

The phrase, “u is related to v” in S, can be more succinctly expressed as,

uRfv or (u, v) ∈ Rf

1The word “fibre” is also written as “fiber”. It is usually interpreted in this way in the field of set theory.But it can have an entirely different meaning in other mathematical fields. Its interpretation is determinedby the context.


This essentially means that u and v are related if and only if they both belong to the

same fibre under the map f : S → T .2 The relation, Rf , on S is easily seen to bereflexive, symmetric and transitive, and so Rf is an equivalence relation on S. We will

denote an equivalence class of x under Rf by

Sx = y ∈ S : xRfy

and the set of all such equivalence classes in S by

S/Rf = Sx : x ∈ S

Each element, Sx, is a fibre under, f , and the set, S/Rf , can be viewed as the set ofall fibres in S under f . In set theory, S/Rf , is normally called the quotient set of S

induced by Rf . Just like the set of all fibres of a function, f : S → T , partitions thedomain, we see that the set of all equivalence classes partitions the set S. By this

we mean that the elements of S/Rf are pairwise disjoint subsets which cover all of S.One should remember that, when Sx is in S, it is a subset of S, but, whenever Sx is

in S/Rf , it is viewed as one of its elements.

There is a “natural” function, θ : S → S/Rf , defined as

θ(x) = Sx

mapping the points of the topological space (S, τS) onto the “elements” of the set

S/Rf . One might more figuratively say that θ collapses each fibre, f←(x), in S downto a unique element, Sx, in S/Rf , or, equivalently, maps fibres in S to singleton sets,

Sx, in S/Rf .

As long as no topology is declared on S/Rf our discussion remains within the boundsof set theory. We will topologize S/Rf as being the quotient space induced by the

function, θ : S → S/Rf , a function we can now refer to as a quotient map. So S/Rf

will be equipped with the quotient topology, τθ. More specifically,

τθ = U : θ←[U ] is open in S

The topology on S/Rf will be the strongest topology that guarantees the continuityof θ : S → S/Rf .

We now have two quotient spaces induced by two functions with domain (S, τS):

f : S → (T, τf)

θ : S → (S/Rf , τθ)

An insightful reader may already have a feeling that the two quotient spaces aretopologically the same. To confirm this, we have to show that they are linked by

2In this context a fibre is a set theoretic concept and is independent of the topologies involved.


some homeomorphism. We will connect S/Rf to T by defining a third function,φf : S/Rf → T as

φf(Sx) = f(x) (Where Sx = θ(x).)

which maps points of S/Rf to points of T (remembering that f is onto T and so every

element of T can be represented by f(x) for some x in S.) The following diagramillustrates the relationship between S, S/Rf and T .

S - T@

@@R

S/Rf

f

θ φf

This allows us to express the function, f , as a composition of functions φf θ = f .The following definition refers to expressions of the form f← [f [U ] ]. One should keep

in mind that, even though it may occur that U = f← [f [U ] ], it is always the case thatU ⊆ f← [f [U ] ].

Definition 8.4 Let g : S → T be a function mapping the topological space (S, τS) onto

a topological space (T, τT). A subset, U ⊂ S, is said to be g-saturated whenever U is thecomplete inverse image of some subset V in T . Equivalently, U is g-saturated if and only if

U = g←(g(U)).

Theorem 8.5 Suppose f : S → T is continuous on S and U is a non-empty open subsetof T . Then f←[U ] is θ-saturated.

Proof : It suffices to show that f←[U ] = θ←[ θ [ f←[U ] ] ]. Since f←[U ] ⊆ θ←[ θ [ f←[U ] ] ]

we need only show inclusion in the opposite direction.

x ∈ θ←[ θ [ f←[U ] ] ] ⇒ θ(x) ∈ θ [ f←[U ] ]

⇒ θ(x) ∈ θ(y) : y ∈ f←[U ]⇒ Sx ∈ Sy : f(y) ∈ U⇒ f(x) ∈ U

⇒ x ∈ f←[U ]

Then θ←[ θ [ f←[U ] ] ] ⊆ f←[U ]. Then f←[U ] is θ-saturated. That is,

θ←[ θ [ f←[U ] ] ] = f←[U ]


Theorem 8.6 Suppose (S, τS) is a topological space. Let (T, τf) and (S/Rf , τθ) be two

quotient spaces induced by the quotient maps f : S → T and θ : S → S/Rf , whereθ(x) = Sx. Then the function, φf : S/Rf → T defined as φf(Sx) = f(x), maps S/Rf

homeomorphically onto T .

Proof : We are given that f : S → T and θ : S → S/Rf and are quotient maps, hence arecontinuous on S. We are required to show that the function, φf : S/Rf → T , definedas, φf(Sx) = (φf θ)(x) = f(x), is a homeomorphism.

We claim that φf is one-to-one and onto T : See that, since f is onto T , φf is onto T .Also, if Sx 6= Sy, then f(x) 6= f(y), which implies φf(x) 6= φf(y); so φf is one-to-one

and onto.

We claim that φf is continuous on S/Rf : Let U be open in T . Since f is continuousf←[U ] is open in S.

φ←f [U ] = Sx : φf(Sx) ∈ U= Sx : φf(θ(x)) ∈ U= θ(x) : f(x) ∈ U= θ(x) : x ∈ f←[U ]= θ[f←[U ]]

We have shown above that f←[U ] is θ-saturated hence θ←[ θ [ f←[U ] ] ] = f←[U ].Since θ : S → S/Rf is a quotient map and f←[U ] is open then θ [ f←[U ] ] is open. So

φ←f [U ] is open. This establishes the claim that φf is continuous.

We claim that φf is open on S/Rf : From the diagram above we know that f = φf θ.

So θ = φ←f f . Since θ is continuous then (φ←f f) : S → S/Rf is continuous on S. LetU is an open subset of S/Rf .

θ←[U ] = (φ←f f)←[U ] ∈ τS

⇒ f←[(φ←f )←[U ]] ∈ τS

⇒ f←[φf [U ]] ∈ τS

Since T has the quotient topology induced by f , φf [U ] is open in T . We have shownthat φf is an open map, as claimed.

We have shown that the one-to-one onto function, φf : S/Rf → T , is both continuous

and open, hence it maps S/Rf homeomorphically onto T .


We summarize the main ideas behind the above result. If we are given a topological

space, (S, τS), and a function f mapping S onto a set T we have the two main ingre-dients necessary to topologize T with the quotient topology, τf . We do so in a way

that guarantees continuity of the function, f : S → T . The fibres of the function, f ,covers its domain. By collapsing, via the map θ, each fibre down to a point we create

another set, S/Rf , and topologize it with the quotient topology, τθ. The expression“identifying the points of the fibres” is also of common usage. This new topologicalspace, (S/Rf , τθ) has been proven to be a homeomorphic copy of (T, τf).

Example 1. Let J denote the set of all irrational numbers in the closed interval [0, 1]

and let T = 1 ∪ J, a proper subset of [0, 1]. We define a function f : [0, 1] → T asfollows

f(x) =

x if x is irrational in [0, 1]1 if x is rational in [0, 1]

Let the domain, [0, 1], be equipped with the usual topology and the range, T = 1∪J,be equipped with the quotient topology induced by the function f . Describe the open

subsets of the quotient space, T , induced by the function f on [0, 1].

Solution. The open subsets of T are defined as

τf = U ⊆ T : f←[U ] is open in [0, 1]

We have f←[∅] = ∅ and f←[T ] = [0, 1] so ∅, T ⊆ τf .

Consider, for example, the open interval U = (1/5, π/4). Then [U ∩ J] ∪ 1 ⊆ T . So

f← [ [U ∩ J] ∪ 1 ] = f←[U ∩ J] ∪ f←[1]= [U ∩ J ] ∪ [ [0, 1]∩ Q ]

Since [U ∩ J ]∪ [ [0, 1] ∩ Q ] is not open in [0, 1] then [U ∩ J ]∪ 1 is not open in T .

It seems that the complete, open, preimage, f←[V ], of any non-empty subset, V , ofT should contain open intervals and, at least, both 0 and 1. Are there such proper

subsets of [0, 1]? We test the following subset:

f←[V ] = [0, π/7)∪ (π/7, π/5) ∪ (π/5, π/4) ∪ (π/4, 1]

is open in [0, 1]. We seek a set V in T that f pulls back to the subset, f←[V ]. We test

f [f←[V ]] = T \π/7, π/5, π/4

to find

f←[T \π/7, π/5, π/4] = [0, π/7)∪ (π/7, π/5) ∪ (π/5, π/4) ∪ (π/4, 1]

An open subset of [0, 1].


So T \π/7, π/5, π/4 ∈ τf .

Then, if K is any finite subset of irrationals in [0, 1], T \K is open in T . So

τf = ∅, [0, 1] ∪ T \K : K ⊂ J ∩ [0, 1], K finite

8.3 Decomposition spaces.

We have seen that the quotient space induced by θ : S → S/Rf involves the decom-position of the set S into non-overlapping sets, Sx. We will now slightly generalize

this notion of “decomposition of a set”. Suppose we are given a topological space(S, τS). Rather than partition S into equivalence classes, each of which contains all

elements which belong to a particular fibre, Sx, of f , we can also partition the set S inan arbitrary way. By this we mean that we express S as the union of non-intersecting

subsets. We will denote this set of subsets by, DS . Each element x of S belongs tosome element, labeled Dx, of DS . Note that, if y ∈ Dx then Dx and Dy are simply

different labels for the same element of DS . The set

DS = Dx : x ∈ S

is reminiscent of a quotient set whose elements are equivalent classes of some relation

R on S. We can similarly define a function θ : S → DS , on the topological space,S, where θ(u) = Du, the unique element of DS which contains u. Then θ←(Du) =

x ∈ S : x ∈ Du = Du ⊆ S. The set, DS , is not yet topologized, but we know of aprocedure to topologize it, since it is, after all, the range of a function θ on S. We

can equip DS with the quotient topology,

τθ = U ⊆ DS : θ←[U ] is open in S

induced by θ. We formally define the concepts we have just presented.

Definition 8.7 Let (S, τS) be a topological space. If DS is collection of pairwise disjointsubsets of S such that every element of S belongs to some element of DS , then we say DS

is a“decomposition of S”

The function, θ : S → DS , defined as, θ(u) = Dx if and only if u ∈ Dx, mapping every

element of S onto DS is called the decomposition map of S onto DS. The function, θ, isalso referred to as the identification map. The function, θ, is said to identify the elements

of the subset Dx ⊆ S. If τθ is the quotient topology induced on DS by θ, then we say that(DS, τθ) is a

“decomposition space or quotient space”


Why does all of this sound familiar? Recall that, if f : S → T , then the function, f ,

decomposes its domain, S, into fibres. In this case, (DS , τθ) is a copy of (S/Rf , τθ).The following theorem will help recognize the open subsets of the decomposition space,

DS. It states that open subsets U = Dx : Dx ∈ U of DS correspond to open θ-saturated subsets ∪Dx : Dx ∈ U of S.

Theorem 8.8 Suppose (DS, τθ) is a decomposition space of the topological space (S, τS)and let U = Dx : x ∈ I ⊆ S be a subset of DS . Then U is open in DS if and only if

∪Dx : x ∈ I

is open in S.

Proof : We are given that U = Dx : x ∈ I ⊆ S be a subset of DS.

( ⇒ ) Suppose U is open in DS .

Then θ←[U ] = x ∈ S : θ(x) = Dx ∈ U = x ∈ S : x ∈ I is open in S. We arerequired to show that ∪Dx : x ∈ I is open in S.

θ←[U ] = θ←[∪Dx : x ∈ I]= ∪θ←[Dx] : x ∈ I= ∪Dx ⊆ S : x ∈ I

Is open since θ is continuous.

( ⇐ ) Suppose ∪Dx ⊆ S : x ∈ I is open in S. We are required to show that

U = Dx : x ∈ I ⊆ S is open in DS. To show that U is open in DS it sufficesto show that θ←[U ] is open in S (since DS is equipped with the quotient topology

induced by θ).

∪Dx ⊆ S : x ∈ I = ∪θ←[Dx] : x ∈ I= θ←[∪Dx : x ∈ I]= θ←[U ]

Since ∪Dx : x ∈ I = θ←[U ] is open then U is open.


Example 2. Suppose we are given the subset, S = [0, 2π] × [0, 2π] of R2. We willdecompose S as follows: For each point (x, 0) on the line [0, 2π]× 0 let

D(x,0) = (x, 0), (2π− x, 2π)

So for each x ∈ [0, 2π], D(x,0) and D(2π−x,2π) represent the same element of the de-

composition, DS . For example, D(0,0) = D(2π,2π) = (0, 0), (2π, 2π).For each (x, y) 6∈ [0, 2π]× 0 ∪ [0, 2π]× 2π let

D(x,y) = (x, y)

Each subset of the form D(x,0) is a subset of S which contains two points. All other

subsets of the decomposition are singleton sets. The decomposition space, DS, of thesubspace S obtained is referred to as the Mobius strip.1 See the figure.

Figure 2: Topological representation of the mobius strip

The two lines L1 = (x, 0) : 0 ≤ x ≤ 2π and L2 = (x, 2π) : 0 ≤ x ≤ 2π arecollapsed together (after inverting one of the lines) to form the mobius strip.

Without the inversion of one of the lines, the decomposition space becomes a topo-logical representation of a cylindrical shell.

Moore plane decomposition

Example 2. Let (S, τS) denote the Moore plane. (Review the description of the

Moore plane by looking over the example on page 74.) Let F = (x, 0)|x ∈ R andW = S \F . We can then express S as the disjoint union of the sets F and W .

Let D = x : x ∈ W ∪ F, be the decomposition space of S which results fromcollapsing the subset, F , down to a point and identifying all other points to themselves.

1In reference to August Ferdinand Mobius, 1790-1868. A German mathematician and astronomer


Describe an open neighbourhood base of each point in the decomposition space D .

Solution : We are given the Moore plane, (S, τS), and the decomposition D = x :

x ∈W ∪ F. It easily verified that points in S are closed and so S is T1.

If θ : S → D is the corresponding quotient map, then θ(a, b) = (a, b) if (a, b) ∈W and

θ(u, v) = F when (u, v) ∈ F . The quotient topology on D is, by definition,

τθ = U ⊆ D : θ←[U ] is open in S

The open neighbourhood base of points, x ∈ D for x ∈W is the same as for τW . IfτW is the subspace topology on the open subspace, W , inherited from S, then τW ⊆ τθ.

We now describe an open neighbourhood base for the element, F ∈ D .

Let N = N\0. Consider the set, NR, of all functions which map R into N .

We now construct a basic open neighbourhood of the element F . For f ∈ NR, let

Uf = ∪

B 1f(x)

(

x,1

f(x)

)

: (x, 0) ∈ R × 0

where 1f(x) is the radius of each open ball of center

(

x, 1f(x)

)

tangent to the x-axis.

Since, for each f , θ←[Uf ∪ F] = Uf ∪ F is open in S, then

BF = Uf ∪ F : f ∈ NR

is an uncountably large base of open neighbourhoods for F , since for any open neigh-

bourhood V of the x-axis, there exists f ∈ NR such that (x, 0) : x ∈ R ⊆ BF ⊆ V .

On subspaces of a decomposition space.

Example 3. Let (S, τS) be a topological space. Suppose (D , τD) is a decompositionspace induced by some decomposition of S. Let T be a non-empty subset of S. Then

the decomposition D of S will induce a decomposition space (DT , τT ),

DT = D ∩ T : D ∈ D

of T . We can identify another decomposition space (D |T , τ|T ) associated to T :

D |T = D ∈ D : D ∩ T 6= ∅

The space, (D |T , τ|T ) is equipped with the subspace topology, τ|T , inherited from

(D , τD).

Provide an example that shows that (D |T , τ|T ) and (DT , τT ) are generally not home-

omorphic spaces.


Solution . Consider the decomposition, D = x × R : x ∈ R, of R2. LetT = (x, 0) ∈ R2 : x < 0 ∪ (0, 1), a subset of R2, with the subspace topology.

Let DT = D ∩ T : D ∈ D. Then DT = (x, 0) : x < 0 ∪ (0, 1).Let θT : T → DT be the quotient function which maps a point x in T to the uniqueelement of DT which contains it. Then θT induces a the quotient topology, τD on DT .

That is, U is open in DT if and only if, θ←T [U ] is open in T . The function θT is clearlyone-to-one on T . Also, since θ←T [(0, 1)] = (0, 1) is open in T (with respect to the

subspace topology), the point (0, 1) is a clopen element of DT . The point (0, 1)in DT is referred to as an isolated point.

We now consider the decomposition,

D |T = D ∈ D : D ∩ T 6= ∅

We claim that D |T does not contain an isolated point, and so cannot be homeomor-

phic to DT . Note that D |T = x × R : x ≤ 0. Since every neighbourhood ofa vertical line in ∪D |T intersects another vertical line then D |T cannot contain an

isolated point. So D |T and DT cannot be homeomorphic.

Example 4. Let Sn denote the n-sphere in the space, T = Rn+1\~0 (equipped with

the usual topology). That is, Sn = ~x ∈ Rn+1\~0 : ‖~x‖ = 1. For example, S1 is theunit circle in R2\(0, 0), S2 is the sphere surface in R3\(0, 0). Let

Lλ = λ~x : ~x ∈ T

represent a line in T passing through the point, ~0, and ~x with “slope” λ. Then the

collection of lines, Lλ : λ ∈ R>0, constitutes a decomposition, DT , of the space T .The elements of DT = Dλ : λ ∈ R>0, are of the form Dλ = Lλ for λ ∈ R>0.

Let f : T → Sn be defined as

f(~x) =~x

‖~x‖a continuous function on T . Then

f [Lλ] = f [λ~x : ~x ∈ T] = λ~x/‖λ~x‖ ∈ Sn ∩ Lλ

So f is a function which collapses each set Lλ down to the singleton set Lλ ∩ Sn.Then, f induces a map

f∗ : DT → Sn

which is one-to-one on DT onto Sn. If U is open in Sn, f∗←[U ] is open in DT .

If V is open in DT , then f∗[V ] is open in Sn. So f∗ is a homeomorphism mappingDT = T/Rf onto Sn. So Sn is homeomorphic to Rn+1/Rf .

Note: See that Sn cannot be homeomorphic to Rn+1 nor to Rn since Sn is closed andbounded (compact) and neither Rn+1 nor Rn are.


Concepts review:

1. Define the quotient topology induced by a function.

2. If τT is a topology on T for which f : S → T is continuous, how does τT compare with

the quotient topology, τf , induced by f?

3. If τT is a topology on T for which f : S → T is a continuous open map, how does τT

compare with the quotient topology, τf , induced by f?

4. If τT is a topology on T for which f : S → T is a continuous closed map, how does τT

compare with the quotient topology, τf , induced by f?

5. What is a fibre of a point under a map f?

6. Define the set S/Rf of all equivalence classes induced by a function f : S → T by

referring to a function θ : S → S/Rf .

7. What topology is defined on S/Rf?

8. How is the function φf : S/Rf → T defined?

9. If g : S → T is a function and U ⊆ S, what does it mean to say that U is g-saturated?

10. If g : S → T is a function and U ⊆ S is g-saturated, is there a way to describe the

subset U in terms of g?

11. Given f : S → T and θ : S → S/Rf describe the homeomorphism which links S/Rf

to T .

12. Describe how to construct a decomposition space, DS , from a topological space (S, τS).

EXERCISES

1. Let (S, τS) be a topological space. We define a relation, R, on S as follows: (u, v) ∈ Rif and only if clSu = clSv.

a) Show that R is an equivalence relation on S.

b) Let (S/R, τθ) denote the quotient space induced by the natural map θ : S → S/R.

i. If Sx is an element in S/R, is Sx necessarily a closed subset of S/R?


ii. If Sx and Sy are distinct elements of S/R, does there exist an open set in τθwhich contains Sx but not Sy?

2. Describe the quotient space, R/R, if (u, v) ∈ R if and only if x− y is an integer.

3. Suppose DR2 is a decomposition of R2 (with the usual topology) whose elements are

circles with center at the origin. Show that the corresponding decomposition space,(DR2, τθ), and the set of all non-negative real numbers, x ∈ R : x ≥ 0, equippedwith the usual topology are homeomorphic topological spaces.

Part III

Topological spaces: Separationaxioms

169

Part III: Topological spaces: Separation axioms 171

9 / Separation with open sets.

Summary. In this section we will present five classes of topological spaces

which, together, form a hierarchy of separation axioms. They differ by the waytheir open sets can be used to separate their points and their closed sets. For eachof these spaces, we present a few of their characterizations and basic properties.

These five types of topological spaces are called, Ti-spaces, where i = 0, 1, 2, 3, 4.

9.1 The T0 separation axiom.

In this chapter we will introduce five other classes of topological spaces. We will also

discuss various characterizations and properties of these spaces. We distinguish thesefive classes by the separation axioms they satisfy. The first one is referred to as the

T0 axiom.

Definition 9.1 Let (S, τS) be a topological space. We say that S is a T0-space if, for every

pair of distinct points u and v, there exists at least one open set in τS, which contains oneof these two points, but not the other.1

Most of the topological spaces we have been exposed to are T0-spaces, so finding anexample of a T0-space is easy, as long as we can justify why a particular space is

declared to be T0. The space, R, with the usual topology, is T0 since, for any a, b ∈ R

where a < b, (a−1, b) contains a but not b. We will soon see that the Euclidean space,

R, can separate points in much more intricate ways. For many, the topological spacesof interest are, at the very least, T0. Are there any non-T0-spaces? Let’s consider

the space which has the fewest open sets possible: The indiscrete space (S, τi) withtopology, τi = ∅, S. If a, b ∈ S, since S is the only non-empty open set, it contains

all points and so is unable to separate a point a from a point b in the way prescribedby the T0-property. So, for this reason, there is simply not much else to say about the

indiscrete spaces.

Combining the T0 separation property with other topological properties.

Even though T0-spaces by themselves may be considered by some as being of little

interest, it is good practice to see what happens when we combine the T0-propertywith other topological properties. Consider the following example in which we show

1A T0-space is also known as a Kolmogorov space (named after the mathematician Andrey Kolmogorov).

172 Section 9: Separation with open sets.

that infinite second countable T0-spaces cannot be arbitrarily large.

Example 1. Suppose (S, τS) is an infinite second countable T0-space. Show that thecardinality, |S|, of S is less than or equal to 2ℵ0 (where, ℵ0 = |N|).

Solution: Since S is an infinite second countable space, then, by definition, it has an

open base, B, such that |B| ≤ ℵ0. For each x ∈ S, let

Bx = B ∈ B : x ∈ B ⊆ B

If y 6= x then Bx cannot equal By, for, if it did, then y ∈ Bx for all Bx ∈ Bx, and

there would be no Bx (By) which can separate x from y, a contradiction of the T0

property. So there is a one-to-one function, f : S → P(B), defined as,

f(x) = Bx

Then |S| = |Bx : x ∈ S|. Recall that the cardinality of the power set, P(T ), of a

set T is 2|T |. Since |B| ≤ ℵ0, then |P(B)| = 2|B| ≤ 2ℵ0.

Since, for each x ∈ S, Bx ∈ P(B) then |Bx : x ∈ S| ≤ |P(B)|.

We conclude that

|S| = |Bx : x ∈ S| ≤ |P(B)| ≤ 2ℵ0

so the cardinality of a second countable T0-space, S, is less than or equal to 2ℵ0, asrequired.

9.2 The T1 separation axiom.

In general, we prefer topological spaces whose points are closed. The next separationaxiom called, T1, requires that τS has sufficiently many open sets to guarantee that

its singleton sets, x, must be closed subsets.

Definition 9.2 Let (S, τS) be a topological space. We say that S is a T1-space if, for everypair of distinct points u and v, there exists an open set, U , which contains u but not v, and

an open set, V , which contains v but not u.1

1The T1-spaces are also referred to as Frechet spaces, named after the mathematician Maurice Frechet.


Of course, every T1-space is T0. Also, Rn, when equipped with the usual topology, is aT1-space. Are there T0-spaces which are not T1? In the following example we exhibit

a non-T1-space which is T0.

Example 2. Let S be a set which contains more than a single point. Suppose u ∈ S

and F ⊆ P(S) where

1. For non-empty F , [F ∈ F ] ⇔ [u ∈ F ].

2. ∅ ∈ F .

Verify that the two given conditions on F together satisfy the closed sets axioms andso define a topology, τS , on S. Then show that (S, τS) is T0, but not T1.

Solution. We claim that F describes all closed subsets with respect to the topologyτS = U : u 6∈ U ∪ S. Clearly, S,∅ ⊆ F . Finite unions of sets each of which

contains the point u will contain the point u and arbitrary intersections of sets eachof which contain u will also contain u. Then F satisfies the closed set axioms. Then

the set, F , represents all closed sets for the topology

τS = U : u 6∈ U ∪ S

We see that, if x 6= u, then S\u is an open neighbourhood of x and does not contain

u. So S is T0. But since the only open neighbourhood of u is all of S, then S cannotbe T1. So (S, τS) is T0 not T1.

Characterizations of T1-spaces.

The most useful characterization of T1-spaces is given below. This characterization issuch that some may use it as a definition of T1.

Theorem 9.3 Let (S, τS) be a topological space. The space S is T1 if and only if everysingleton set, x, of S is a closed subset of S.

Proof : Let (S, τS) be a topological space. If S is a singleton set then, trivially, S is T1 and

its only singleton set is S, which is closed. Suppose S contains more than one point.

( ⇒ ) Suppose S is T1. Let u ∈ S. We claim that u is closed. For each, x 6= u,

there exists open, Ux such that, x ∈ Ux and u ∈ S\Ux. Then u = ∩S\Ux : x 6= u,the intersection of closed subsets of S. So u is closed in S.

( ⇐ ) Suppose every point of S is closed. Then, then for any u, y ∈ S, if y 6= u,S\u is an open neighbourhood of y which does not contain u and S\y is an open

neighbourhood of u which does not contain y. Hence S is T1.


Theorem 9.4 Let (S, τS) and (T, τT) be topological spaces.

a) If S is a T1-space then so are all its subspaces. So T1 is a hereditary property.

b) Suppose S is a T1 space and f : S → T is a closed function. Then f [S] is a T1 subspace

of T .

c) Suppose S =∏

i∈I Si is a product space. The space S is a T1 product space if andonly if each of its Si-factors is a T1-space.

Proof : Let (S, τS) and (T, τT) be topological spaces.

a) Suppose S is T1 and u ∈W ⊆ S. Since u is a closed subset of S it is a closedsubset of its subspace W (with respect to the subspace topology). So W is T1

(by theorem 9.3).

b) Suppose S is T1 and f : S → f [S] is a closed map. Let v ∈ f [S]. Then thereexists x ∈ S such that f [x] = v. By hypothesis, x is closed and f is closed

so, v is closed. Hence f [S] is T1.

c) ( ⇒ ) Suppose S =∏

i∈I Si is a T1 product space. Recall that each factor, Si, ishomeomorphic to a subspace, Ui, of S. By part a), Ui is T1 and so Si is T1.

( ⇐ ) Suppose each factor, Si, of S =∏

i∈I Si is T1. Let q = < ai >i∈I be apoint in S. Then no Si is empty since, for each i ∈ I , ai ∈ Si. Since each Si

is T1, for each coordinate ai, the singleton set, ai, is a closed subset of Si.

For all j ∈ I , πj : S → Sj is continuous; then π←j (aj) is closed in S. Then

q = ∩j∈I

π←j (aj)

is closed in S. So S is T1.

In the following example, we show that the decomposition space, D , whose elements

are the fibres of f : S → T , is T1 if and only if the fibres of f are closed subsets of S.

Example 3. Let f : S → T be a function. Suppose τf is the quotient topology on S

induced by the quotient map f . Let DS = Dx : x ∈ T be a decomposition space ofS, with elements, Dx = f←(x). Show that DS is T1 if and only if each of its elements,

Dx = f←(x), is a closed subset of S.

Solution : Let f : S → (T, τT) be a function. f←(x) = Dx for each x ∈ T andDS = Dx : x ∈ T be a decomposition space of S induced by f , equipped with the

quotient topology τf . Let q : S → DS be the quotient map q(x) = Dx.


( ⇐ ) We are given that, for each x ∈ T , Dx = f←(x) is a closed subset of S. We arerequired to show that DS is T1. Let Sy ∈ DS. It suffices to show that Sy is closed

in DS. Consider

q←[DS\Sy] = S\f←[y]

where by hypothesis, S\f←[y] is open in S. Since DS is equipped with the quotienttopology,

q←[DS\Sy] open in S ⇒ DS\Sy open in DS

So Sy is closed in DS. This implies DS is T1.

( ⇒ ) We are given that DS is a T1 quotient space. Then, for each x ∈ T , Dx is aclosed subset of DS . We are required to show that Dx = f←(x) is a closed subset of

S. Since q : S → DS is continuous, then q←[Dx] = f←(x) is closed in S, as required.

9.3 The T2 separation axiom or Hausdorff topological spaces

We will now define a T2-space. The terminology “T2-space” helps remind us that

T2 ⇒ T1 ⇒ T0. Although both T2 and Hausdorff mean the same thing, the word“Hausdorff” is more commonly used.1

Definition 9.5 Let (S, τS) be a topological space. We say that S is a T2-space or Haus-dorff space if, for every pair of distinct points u and v, there exists non-intersecting open

neighbourhoods, U and V such that u ∈ U and v ∈ V .

Clearly, any Hausdorff space, (S, τS), satisfies the T1 axiom. Hence,

“If S is Hausdorff then, for each x ∈ S, x is a closed subset of S”.

Example 4. An infinite set, S, with the cofinite topology (also referred to as the Zariskitopology, see page 35) is T1 but not Hausdorff since every pair of open neighbourhoods

will intersect with infinitely many points in their intersection. However, it is easilyverified to be T1.

Example 5. Any metrizable topological space, (S, τρ), is Hausdorff since, given dis-

tinct points a and b, Bε(a) ∩ Bε(b) = ∅ where ε < ρ(a,b)3 . A standard approach to

verifying whether a topological space is metrizable is to first check if it is Hausdorff.

1Sadly, the talented mathematician, Felix Hausdorff, (along with many other very talented Europeanmathematicians) did not survive persecution by the Nazi regime in Germany.


If it is not Hausdorff then the question is settled.

Hausdorff characterizations

There are various equivalent ways of expressing the Hausdorff separation property.We present a few of the more common characterizations.

Theorem 9.6 Let (S, τS) be a topological space. The following statements are equivalent.

a) The space, S, is Hausdorff.

b) If u and v are distinct points in S then there exists an open neighbourhood, U , of u

such that v 6∈ clSU .

c) If u ∈ S, then ∩clSU : U is an open neighbourhood of u = u

d) The set D = (u, u) : u ∈ S is closed in the product space, S × S.

Proof : Let (S, τS) be a topological space.

(a ⇒ b) We are given that S is Hausdorff. Suppose u and v are distinct points inS. Then there exists disjoint open neighbourhoods, U and V , that contain u and v,

respectively. See that U ⊆ S \V , a closed subset of S; hence clSU ⊆ S \V . Thenv 6∈ U ⊆ clSU .

(b ⇒ c) Suppose that, if u 6= v in S, there exists an open neighbourhood, U , of u

such that v 6∈ clSU . Then v ∈ ∩clSU : U is a neighbourhood of u is impossible.So ∩clSU : U is a neighbourhood of u = u.

(c ⇒ d) Suppose that ∩clSU : U is a neighbourhood of u = u, for all u ∈ S.

Let D = (a, a) ∈ S × S : a ∈ S. We are required to show that D is closed in S × S.Let (u, v) ∈ (S×S)\D. Then u and v are distinct elements of S. By hypothesis, thereexists an open neighbourhood, U , of u such that v 6∈ clSU . Then U × (S\clSU) is an

open neighbourhood of (u, v). If a ∈ U , then a 6∈ S\clSU so (a, a) 6∈ U × (S\clSU).So U × (S\clSU) does not intersect D. So (S × S)\D is open. Hence D is closed.

(d ⇒ a) Suppose D is closed in S × S. Let u and v be distinct points in S. Then

(u, v) ∈ (S ×S)\D. Since D is closed there exists and open neighbourhood, U × V of(u, v) which does not intersect D. Then U and V are disjoint open neighbourhoods

of u and v, respectively. So S is Hausdorff.


In the following theorem, we show that a subspace will always inherit the Hausdorffproperty from its topological space. Also, the Hausdorff property is carried over from

a set of factors, Sii∈I, to the Cartesian product space,∏

i∈I Si, they generate. TheHausdorff property is also carried over from the domain, S, of a one-to-one closed

function, f , to its range, f [S].

However, continuous images of Hausdorff spaces need not necessarily be Hausdorff, asthe example following the theorem below will show.

Theorem 9.7 Let (S, τS) be a Hausdorff topological space and Sii∈I be a family of

Hausdorff topological spaces.

a) If U is a subspace of S, then U is Hausdorff. That is, “Hausdorff” is a hereditaryproperty.

b) If f : S → T is a one-to-one closed function onto T , then T is Hausdorff.

c) The product space,∏

i∈I Si, generated by the Hausdorff factors, Sii∈I, is Hausdorff.

Proof : Let (S, τS) be a Hausdorff topological space.

a) Suppose u and v are distinct points in the subspace, W , of S. Then there exists

disjoint open neighbourhoods, U and V , which respectively contain the points u andv. Then U ∩W and V ∩W are disjoint open neighbourhoods in W , which respectively

contain the points u and v.

b) Suppose f : S → T is a one-to-one closed function mapping S onto T . Recall that

a one-to-one closed function is also an open function. (See the corresponding theoremon page 100). Then disjoint open neighbourhoods U and V of u and v are mapped to

disjoint open neighbourhoods of f [U ] and f [V ] of f(u) and f(v).

c) We are given that Sii∈I is a family of Hausdorff topological spaces. Let uii∈I

and vii∈I be distinct points of∏

i∈I Si. Then, for some j ∈ I , uj 6= vj. Let Uj andVj be disjoint open neighbourhoods of uj and vj in Sj. Then π←j [Uj] and π←j [Vj] are

disjoint open base elements containing uii∈I and vii∈I. So∏

i∈I Si is Hausdorff.

Examples involving the Hausdorff property.

Example 6. Consider the space S = 0, 1 equipped with the discrete topology,

τd = ∅, 0, 1, 0, 1

and T = 0, 1 equipped with the topology,

τT = ∅, 0, 0, 1


Let f : S → T be a one-to-one onto function defined as f(x) = x on S. Since Sis discrete, each point can serve as it own open neighbourhood and so is Hausdorff.

Also, every function is continuous on a discrete space so f is continuous on S. Use thisexample to show that the continuous image of a Hausdorff space need not be Hausdorff.

Solution: In τT , there is no open neighbourhood that contains 1 but not 0. So T is

not Hausdorff. In fact, T is not even T1. So continuous images of Hausdorff spacesneed not be Hausdorff.

Example 7. Show that, if a continuous function, f : U → V , maps U one-to-one intoa Hausdorff space V then its domain, U , must also be Hausdorff.

Solution. Let f : U → V be a one-to-one continuous function mapping U into a Haus-

dorff space, V . Then, since Hausdorff is a hereditary property, f [U ] is a Hausdorffsubspace of V . Also, since f is continuous, f← : f [U ] → U is a one-to-one closedfunction onto U . By the theorem 9.7, part b), the space U must be Hausdorff. This

is what we were required to show.

In the following two examples we show how the Hausdorff characterization in theorem9.6 part d)

“[S is Hausdorff ] ⇔ [D = (u, u) : u ∈ S is a closed subset of S × S ]”

can serve as a useful tool for solving certain types of problems.

Example 8. Show that, if f and g are two continuous functions each mapping a topo-

logical space S into a Hausdorff space T , then the set, U = u ∈ S : f(u) = g(u), isa closed subset of S. That is, the set on which f and g agree is closed in S.

Solution. Let h : S → T × T be a function defined as, h(x) = (f(x), g(x)), and let

U = u ∈ S : f(u) = g(u). By theorem 9.6, the statement,

“T Hausdorff” ⇔ “D = (u, u) : u ∈ T is a closed subset of T × T ”

Since both f and g are continuous on S then so is h (see theorem 7.12). Hence h←[D]is closed in S. Since

h←[D] = u ∈ S : h(u) = (f(u), g(u)) ∈ D= u ∈ S : f(u) = g(u)= U

the set, U , is closed in S, as required.

Example 9. Show that, if f : S → T is continuous, where T is Hausdorff then the

graph of f ,G = (u, f(u)) : u ∈ S ⊆ S × T


is closed in S × T .

Solution. Let t : S × T → T × T be defined as

t(u, v) = (f(u), v)

Then, since f and the identity map are continuous on S, t is continuous on S × T .

Hence, since T is Hausdorff, by theorem 9.6, D = (u, u) : u ∈ T is a closed subsetof T × T . Thus t←[D] is closed in S × T . But

t←[D] = (u, v) ∈ S × T : t(u, v) ∈ D= (u, v) ∈ S × T : (f(u), v) ∈ D= (u, v) ∈ S × T : f(u) = v= (u, f(u)) ∈ S × T : u ∈ S= G

So the graph G is closed in S × T .

Theorem 9.8 Let S and T be two spaces where T is Hausdorff. If D is a dense subset ofS and f : S → T and g : S → T are continuous on S which agree on D, then f = g on S.

Proof : Let D be a dense subset of S such that f(x) = g(x) for all x ∈ D.

Suppose a ∈ S \D. Then f(a) 6= g(a) in T . Since T is Hausdorff then there exists

disjoint basic open sets, Bf(a) and B(g(a) with empty intersection.

Both f←[B(f(a)] and g←[B(g(a)] are open in S and each contains the point a. But they

should have empty intersection. Contradiction. So S\D must be empty. So D = S.

9.4 Completely Hausdorff spaces

We will now slightly strength the Hausdorff property to one called the completelyHausdorff property. So that we understand why this stronger property is significant

we will then show that there are Hausdorff spaces which are not completely Hausdorff.

Definition 9.9 Let (S, τS) be a topological space. We say that a space is completely Haus-dorff if and only if for any pair of points a and b in S, there exist open neighbourhoods U

and V of a and b, respectively, such that

clSU ∩ clSV = ∅

The completely Hausdorff property is also represented by T2 12. It is also sometimes referred

to as a Urysohn space.


Clearly every completely Hausdorff space is Hausdorff. But,

“Hausdorff 6⇒ completely Hausdorff”

as the following example shows. Equivalently, T2 6⇒ T2 12.

Example 10. LetT = intR2( [0, 1]× [0, 1] )

andS = T ∪ (0, 0), (1, 0)

The points in T have basic neighbourhoods which are the usual Euclidean open balls.The basic open neighbourhoods of the point (0, 0) are defined to be open rectangles

of the formB(0,0) = (0, 1/2)× (0, 1/n)∪ (0, 0) : n > 0

where (0, 1/2) and (0, 1/n) are open intervals. The basic open neighbourhoods of the

point (1, 0) are defined to be open rectangles of the form

B(1,0) = (1/2, 1)× (0, 1/m)∪ (1, 0) : m > 0

Verify that the basic open neighbourhoods of (0, 0) and (1, 0) can have empty inter-section but that their closures will always have a point of the form (1/2, x) in common.

So this topological space is Hausdorff but not completely Hausdorff.

9.5 T3-spaces and “regular spaces”.

The separation axioms, T0, T1, T2 and T2 12

illustrate four different ways that open sets

can separate two sets of points. The completely Hausdorff (T2 12-space) space imposes

the strongest conditions since T2 12⇒ T2 ⇒ T1 ⇒ T0.

We now want to use open sets to separate a non-empty closed set, F , from a point,x, such that x ∈ S\F . To do this we will define a T3-space by using the definition ofa T2-space as a model.

Definition 9.10 Let (S, τS) be a topological space.

a) We say that S is a T3-space if, for any non-empty closed subset, F and point v ∈ S\F ,there exists non-intersecting open subsets, U and V such that F ⊆ U and v ∈ V .

b) If (S, τS) is both T1 and T3 then we will say that S is a “regular space”.


Let’s consider the most trivial of topological spaces, the indiscrete space, (S, τi), where|S| > 1. There does not exist a “closed F and x ∈ S \F” in S and so S is (vacu-

ously) a T3-space. But distinct points u and v in S are not contained in disjointopen sets and so, S is not Hausdorff. So we have a situation here which needs fixing:

T3 6⇒ T2. We can avoid this situation by defining “regular space = T1 + T3”. If S isboth T1 and T3 and u, v are distinct points then, since S is T1, they are both closed,

and, since S is T3, they are, respectively, contained in disjoint open sets and so S is T2.

Then “[T3 + T1] ⇒ T2 ⇒ T1 ⇒ T0”. Equivalently,

“Regular ⇒ Hausdorff ⇒ T1 ⇒ T0”

In the following example we produce a topological space which is Hausdorff but not

regular.

Example 11. Let (R, τ) we the space of real numbers equipped with the usual topology,

τ . Then B = (a, b) ⊂ R : a < b denotes the standard open base of R. Let

S = B ∪ Q

the standard open base increased by one element, Q. That is, the set of all rationals

is a subbase element and so is declared to be an open set. We declare S to be a subbasefor some topology τS on R. Show that the topological space, (R, τS), is Hausdorff butnot regular.

Solution. Note that Q belongs to τS but is not an element of the usual topology, τ ,and so S = B ∪ Q 6⊆ τ . Since every element of B is a base element of τ , then

τ ⊂ τS . So τS is strictly stronger then the usual topology on R. Since (R, τ) isHausdorff then so is (R, τS ).

We now claim that (R, τS ) is not regular. Since R is Hausdorff, R is T1. So we arerequired to show that the topological space generated by S is not T3. Let J = R\Q.Since Q ∈ τS , Q is an open subset of R with respect to τS . So J is a closed subset of

R with respect to τS .

Then 1 is a point which does not belong to the closed subset, J, of R. Suppose U is

any open neighbourhood of 1. Then there exists and open interval (a, b) such that1 ∈ (a, b) ⊆ U . Since (a, b) ∩ J 6= ∅ then there can be no pair of disjoint open sets

which contain 1 and J, respectively. So (R, τS ) is not regular.

We now present two useful characterizations of “regular”.

Theorem 9.11 Let (S, τS) be a T1 topological space. The following statements about S

are equivalent.

a) The space S is regular.


b) For every x ∈ S and open neighbourhood, U of x, there exists an open neighbourhood,V of x, such that x ∈ clSV ⊆ U .

c) For every x ∈ S and closed subset F disjoint from x, there exists an open neigh-bourhood, V of x, such that clSV ∩ F = ∅.

Proof : Let (S, τS) be a T1 topological space.

(a ⇒ b) We are given that S is regular and that U is an open neighbourhood ofx ∈ S. Then S\U and x are disjoint closed subsets of S. Since S is regular, there

exists disjoint open sets, V and W , containing x and S\U , respectively. Then S\Wis a closed set, entirely contains V and is entirely contained in U . So clSV ⊆ S\W ⊆ U .

(b ⇒ c) We are given that for every x ∈ S and open neighbourhood, U of x, there

exists an open neighbourhood, V of x, such that x ∈ clSV ⊆ U . Let x ∈ S and F be aclosed subset disjoint from x. By hypothesis, there exists an open neighbourhood,

V , of x such that clSV ⊆ S\F .

(c ⇒ a) We are given that, for every x ∈ S and closed subset F disjoint from x,there exists an open neighbourhood, V of x, such that clSV ∩ F = ∅. Let x ∈ S and

F be a closed subset disjoint from x. By hypothesis, there exists an open neigh-bourhood, V , of x such that clSV ∩ F = ∅. Then V and S\(clSV ) are disjoint open

sets of containing x and F respectively. So S is T3. Since S is Hausdorff, it is T1,hence S is regular.

Example 12. Show that any metrizable space is regular.

Solution: Suppose (S, τρ) is a metrizable space whose open sets are generated by ametric, ρ. Metrizable spaces are known to be Hausdorff and so are T1.

Let y ∈ S and U be an open neighbourhood of y. Then there exists ε > 0 such that theball, Bε(y), (center y and radius ε) is entirely contained in U . Then, if V = Bε/3(y),clSV = x : ρ(x, y) ≤ ε/3 ⊆ Bε(x) ⊆ U . By theorem 9.11, S is regular.

Basic properties of regular spaces.

In the following theorems we confirm that “regular” is a hereditary property and thatit carries over products, in both directions.

Theorem 9.12 Subspaces of regular spaces are regular.


Proof : Let (S, τS) be a regular topological space and (T, τT) be a subspace of S.

Any point, x, in T is a point in S and so is closed in S. Then x is closed in T . Wemust conclude tha T is T1.

Let x and F be disjoint closed subsets of T . Then there exists a closed subset, F ∗,

of S such that F = F ∗ ∩ T . By hypothesis, there exists disjoint open subsets, U andV , of S which contain x and F ∗, respectively. Then the disjoint open subsets, U ∩ Tand V ∩T , of T contain x and F , respectively. We can then conclude that T is regular

and so “regular” is a hereditary topological property.

Semiregular spaces revisited.

Recall that a set, U , is regular open in S, if

U = intSclSU

Also, from definition 5.15,

. . . a space (S, τ) is said to be semiregular if the set, Ro(S), of all regular

open subsets of S forms a base for open set of S.

Although it is not explicitly stated in the original definition 5.15, semiregularity is tra-

ditionally defined on spaces which are assumed to be Hausdorff only. We will adoptthis tradition here. We will show that the “regular” separation axiom on a space, S,

forces semiregularity on S.

Theorem 9.13 A regular space is semiregular.

Proof : Let (S, τ) be a regular space. To prove that S is semiregular it will suffice, by

definition, to show that Ro(S) is a base for the open sets of S.

Suppose u ∈ S and B is an open neighbourhood of u. By theorem 9.11, there exists

an open subset, U , of S such that u ∈ clSU ⊆ B. Then u ∈ intSclSU ⊆ B. SinceintSclSU ∈ Ro(S), then Ro(S) is a base for open sets of S, as required.

Theorem 9.14 A regular space is completely Hausdorff.

Proof : The proof is straightforward and so is left an exercise.


In example 10 on page 180, we presented a topological space which is easily verified to

be semiregular but is not completely Hausdorff (since the rectangular neighbourhoodsdo not have closures which separate the points (0, 0) and (1, 0). From this examplewe can conclude that

“semiregular” does not imply “completely Hausdorff”

We then have the followings diagram of implications,

completely Hausdorff

6⇑regular ⇒ semiregular ⇒ Hausdorff ⇒ T1 ⇒ T0

⇓completely Hausdorff

We will now examine regular separation properties involving products.

Theorem 9.15 Let Sii∈I be a family of topological spaces and S =∏

i∈I Si be a corre-sponding product space. Then S is regular if and only if each factor, Si, is regular.

Proof : In theorem 9.4, it is shown that the T1-property carries over both from products tofactors and from factors to products. It will then suffice to show that this holds true

for the T3 property.

( ⇒ ) We are given that∏

i∈I Si is a T3-space. We have shown that∏

i∈I Si containsa homeomorphic copy of each Si. Since “regular” is a hereditary property each Si isT3 and so is regular.

( ⇐ ) We are given that S =∏

i∈I Si is a product space and each Si is T3.

Let <ui>i∈I ∈ S and W be any open neighbourhood of <ui>i∈I. Then there exists

a base element, ∩π←i [Ui] : i ∈ Ffinite ⊆ I, of S such that

<ui>i∈I ∈ ∩π←i [Ui] : i ∈ Ffinite ⊆ I ⊆W

Since each Si is regular there exists open Si-neighbourhoods, Vii∈F⊆I , such thatui ∈ Vi ⊆ clSiVi ⊆ Ui for i ∈ Ffinite. Then

<ui>i∈I ∈ ∩π←i [Vi] : i ∈ F ⊆ clS [∩π←i [Vi] : i ∈ F]= ∩clSπ←i [Vi] : i ∈ F= ∩π←i [clSiVi] : i ∈ F⊆ ∩π←i [Ui] : i ∈ F⊆ W


Since<ui>i∈I ∈ ∩π←i [Vi] : i ∈ F ⊆ clS [∩π←i [Vi] : i ∈ F] ⊆W

then, by the characterization theorem above, S is T3 and so is regular.

A few more examples involving the regular separation axiom.

Example 13. Show that the Moore plane is regular. (For a description of the Moore

plane see the example on page 74.)

Solution : Let (S, τS) denote the Moore plane. The proof that S is T1 is straightfor-

ward and so is left as an exercise. We now show that S is T3.

In the Moore plane there are two types of points. Those points, (x, y), whose basic

open neighbourhoods are of the form Bε(x, y) and the points of the form, (x, 0), whichhave basic open neighbourhoods of the form

“Bε(x, y) ∪ (x, 0) where ε = y”

So we consider these two cases separately.

Let U be an open base neighbourhood, Bε(x, y), of (x, y), y 6= 0. Then

(x, y) ∈ clSBε/3(x, y) ⊂ U

Let V be an open base neighbourhood of (x, 0) of the form, Bε(x, y)∪(x, 0), whereε = y. Then

(x, 0) ∈ clSBε/3(x, y/3)∪ (x, 0) ⊂ V

We conclude that the Moore plane is T3. So S is regular.

Example 14. Recall from definition 5.16 that a space is said to be zero-dimensionalif it has a base of clopen sets. Show that zero-dimensional spaces which are T1 are

regular spaces.

Solution : Let F be a closed subset of a zero-dimensional T1-space S. Then S has a

base of clopen subsets. If x ∈ S \F , then there is a subset, B, which is both openand closed such that x ∈ B ⊆ S\F . Then B is an open neighbourhood of x which is

disjoint from the open neighbourhood, S\B of F . So S is regular.

9.6 T4-spaces and normal spaces.

Our final axiom of separation “by open sets” involves, what we will call, “T4-space”’.

“Normal” spaces will be those T4-spaces which are also T1.


Definition 9.16 Let (S, τS) be a topological space. We say that S is a T4-space if, for any

pair of non-empty disjoint closed subsets, F and W , in S, there exists non-intersecting opensubsets, U and V containing F and W , respectively.

Example 15. As our first example of a T4 space we consider the topological space,

(R, τu), where τu = (x,∞) : x ∈ R ∪ ∅,R. Since the closed subsets of R are ofthe form (−∞, y], disjoint pairs of non-empty closed sets do not exist in R and so R is

vacuously T4. Consider the subsets, F = (−∞, 5] and 6. Any open subset U whichcontains F will contain the point 6. So (R, τu) is not T3. Then, T4 6⇒ T3.

We would prefer that spaces satisfying our next level of closed set separation be, at

least, regular spaces. We rectify this situation by defining normal spaces as onesrespecting the rule,

“normal” ⇔ T4 + T1

Let us verify why this will work. If S is T1 +T4 and F is a non-empty closed subset of

S, for u 6⊂ F , we would be guaranteed the existence of disjoint open subsets, U anV , which, respectively, contain the closed subsets u and F , hence S would be both

T1 and T3; by definition, S would be regular. So we would have the desired chain ofimplications,

normal ⇒ regular ⇒ completely Hausdorff ⇒ Hausdorff ⇒ T1 ⇒ T0

We will then proceed with this definition of a normal topological space.

Definition 9.17 If (S, τS) is both T1 and T4 then we will say that S is a normal space.1

Before we proceed with various examples, we provide a few important characteriza-

tions of those spaces we refer to as being “normal”.

1Readers may find that, in the literature, many authors invert the definitions of “T4” and “normal” aswell as for “T3” and “regular”. So when consulting other texts we caution the reader to verify carefullywhich version the particular author has a preference for. Some writers may feel strongly about their chosenversion. But as long as one is aware of such possible discrepancies, most readers will easily adapt to aparticular version, as long as the author is consistent with its use throughout the body of the text.


Theorem 9.18 Let (S, τS) be a topological space. The following statements about S are

equivalent.

a) The space S is normal.

b) For every proper closed subset, F in S, and proper open set, U containing F , there

exists an open set, V containing F , such that F ∈ clSV ⊆ U .

c) For every pair of disjoint closed sets, F and W in S, there exists and open subset Usuch that F ⊆ U and clSU ∩W = ∅.

d) For every pair of disjoint closed sets, F and W in S, there exists disjoint open subsets

U and V containing F and W , respectively, whose closures, clSU and clSV , do notintersect.

Proof : We are given that (S, τS) be a topological space.

( a ⇒ b ) Suppose S is normal. Let F be a proper closed subset of S and U be aproper open neighbourhood of F . Then S\U is a closed subset of S which does notmeet F . Since S is normal there exists disjoint open subsets V and W containing F

and S\U , respectively. Then clSV ⊆ S\W , where clSV ∩ S\U = ∅.Hence F ⊆ clSV ⊆ U .

( b ⇒ c ) Suppose that, for every proper closed subset, F in S, and proper open set,

U containing F , there exists an open set, V containing F , such that F ∈ clSV ⊆ U .Let F and W be a pair of disjoint closed sets in S. Then S \W is an open subset

which contains F . By hypothesis, there exists an open set, V containing F , such thatF ∈ clSV ⊆ S\W . Then clSV ∩W = ∅.

( c ⇒ d ) Suppose that for every pair of disjoint closed sets, F and W in S, there

exists and open subset U such that F ⊆ U and clSU ∩W = ∅. Let F and W be apair of disjoint closed sets in S. By hypothesis, there exists and open subset U such

that F ⊆ U and clSU ∩W = ∅. Again by hypothesis, there exists and open subset Vsuch that W ⊆ V and clSV ∩ clSU = ∅.

( d⇒ a ) Suppose that for every pair of disjoint closed sets, F andW in S, there exists

disjoint open subsets U and V containing F and W , respectively, whose closures, clSUand clSV , do not intersect. Then there exists disjoint open subsets U and V containing

F and W , respectively, such that U and V , do not intersect. So S is normal.

Some readers will find it useful to know that the class of all normal spaces containsthe class of all spaces metrizable spaces.


Theorem 9.19 Metrizable topological spaces are normal spaces.

Proof : Metrizable spaces are Hausdorff, hence T1. Let ρ be a metric on S.

We now show that S is T4. Let F and W be disjoint closed subsets of S. If x ∈ F

and y ∈ W there exists αx and βy such that Bαx(x) ∩W = ∅ and Bβy(y) ∩ F = ∅ .Then

U = ∪Bαx/3(x) : x ∈ FV = ∪Bβy/3(y) : y ∈W

are open sets containing F and W , respectively.

We claim that U ∩ V = ∅. Suppose not. Suppose q ∈ U ∩ V . Then there exists somea ∈ F and b ∈W such that q ∈ Bαa/3(a)∩Bβb/3(b). Suppose, WLOG, αa ≥ βb. Then

ρ(a, b) ≤ ρ(a, q) + ρ(b, q)

< αa/3 + βb/3

≤ 2αa/3

< αa

But αa was chosen so that Bαa(a)∩W = ∅. Contradiction. So U∩V = ∅, as claimed.So S is T4. We conclude that S is normal, as required.

Since, for every n ≥ 1, Rn equipped with the usual topology is metrizable then...

For every natural number n, Rn is a normal topological space

In example 5 on page 131 we showed that the countably infinite product of a metric

space is metrizable. Then S =∏

n∈N R is metrizable; so S is normal.

We will see that normal spaces are not hereditary, which is unlike the regular spaces.

But they are closed-hereditary (that is, closed subsets inherit the normal property).Also, unlike regular spaces, the normal property does not generally carry over from

factors to their product space, even when the product is finite. However, if the prod-uct space is normal then so will be each of its factors.

Theorem 9.20 Let (S, τS) be a normal topological space.

a) If the function f : S → f [S] = T is a closed continuous function onto the space (T, τT)then T is normal.


b) If T is a closed subspace of S then T is normal.

c) If S =∏

i∈I Si is a normal product space of topological spaces Si : i ∈ I, then eachof its factors, Si, is normal.

Proof : Let (S, τS) be a normal topological space.

a) Suppose f : S → f [S] = T is a closed continuous function. Since S is T1 and f isclosed then T is T1.

Let F and W be two disjoint closed subsets of T . Continuity of f guarantees thatf←[F ] and f←[W ] are disjoint closed subsets of S. Then there exists disjoint open sets

U and V containing F and W , respectively. Let A = T \f [S\U ] and B = T \f [S\V ].Since f is closed A and B are disjoint open neighbourhoods of F and W , respectively.It quickly follows that T is T4. So T is normal.

b) Let T be a closed subspace of S. Then T is T1. If F and W are disjoint closed

subsets of the closed subspace T , then they are disjoint closed subsets of S. It quicklyfollows that T is T4. So T is normal.

c) We are given that S =∏

i∈I Si is normal. There exists a homeomorphism,

h : Si → S, which embeds each Si in S (theorem 7.14). Then S \h[Si] is easilyverified to be open in S and so h[Si] is a closed subspace of S. Since closed subspaces

of normal spaces are normal, then h[Si] is normal and so, Si is normal.

Ordinal spaces of the form [0, ωα] or [0, ωα) are normal.

We now investigate whether ordinal spaces such as [0, ωα] or [0, ωα) satisfy the normalproperty. The ordinal space has a topology, τω, which has an open base whose ele-

ments are of the form (α, β]. Also, since the ordinal space is well-ordered, non-emptysets have an ordinal which is a supremum (infimum) of that set.

Example 16. Let ωα be any ordinal number and let S = [0, ωα]. Show that theordinal space (S, τω) is normal, independent of the value ωα. The proof that shows[0, ωα) that is normal is practically the same. (See the reference to “ordinal space” at

definition 5.14 on page 89.)

Solution : Verifying that S is T1 is left to the reader. Let F and W be disjoint

non-empty closed subsets of S. If µ ∈ F and µ 6= 0 let

Sµ = x ∈W : x < µ = [0, µ) ∩W

If Sµ is empty, let xµ = 0. Then (xµ, µ] is an open neighbourhood of µ ∈ F .Suppose Sµ is not empty. In this case, let xµ = supSµ.1 Since W is closed, then

1The existence of the supremum can be justified by the fact that the class of all ordinals is well-ordered.See the related statements from set theory in the appendix.


xµ ∈ W . Then the open set, (xµ, µ] is a basic open neighbourhood of µ ∈ F whichdoes not intersect W .

This is repeated for each element, µ, of F , to obtain, the open neighbourhood

UF = ∪(xµ, µ] : µ ∈ F

of F . The set UF clearly contains no elements of W .

By applying precisely the same procedure we obtain an open neighbourhood

UW = ∪(xα, α] : α ∈W

of W , which contains no elements of F .

We need now only verify that UW ∩ UF = ∅. If so, then S is T4 and hence is normal.

The standard approach is to suppose UF ∩ UW 6= ∅.

Then there exists some µ ∈ F and some α ∈ W , such that (xµ, µ] ∩ (xα, α] is non-empty. Without loss of generality, suppose α < µ. But, this can’t be, since this would

force α ∈ (xµ, µ], when we know that (xµ, µ] ∩W = ∅.

So UW ∩ UF = ∅. We conclude that S is T4 hence is normal. Proceed similarly for

S = [0, ωα)

Example 17. Show that the deleted Tychonoff plank

T = [0, ω1] × [0, ω0] \ (ω1, ω0)

is not normal. (See page 135 for a description of this space.)

Solution : Recall that T is topologized as a product space. Consider U = [0, ω1)

viewed as a closed interval subset of [0, ω1) itself and consider V = ω0, a singletonset. Then the subset F = U ×V is a closed subset (the top edge) of T with respect to

the subspace topology (since its complement is easily seen to be open). We will alsoconsider the disjoint closed subset,

K = ω1 × [0, ω0) ⊆ T

If T is normal then we should be able to separate F from K with a pair of non-

intersecting open sets.

Suppose M is an open set in T which contains K.

A basic neighbourhood of u in K which would be entirely contained in M would beof the form

Wu = (αu, γu] × (βu, µu]


If Wu = (αu, γu]× (βu, µu] ⊆M for each u ∈ K

K ⊆M = ∪Wu : u ∈ K

We claim that M ∩ F 6= ∅. Since there can be at most countably many u’s in K,

sup αu : u ∈ K = ρ < ω1

since the sup of a countable set cannot reach the uncountable ordinal ω1. So (ρ, ω1)∩M 6= ∅. So no open neighbourhood M of K can miss F , as claimed.

So T is not normal.

Example 18. Show that the subspace of a normal space need not be normal.

Solution : We showed in the examples above that the Tychonoff plank is normal but

that its subspace, the Deleted Tychonoff plank, is not.

Concepts review:

1. What is the formal definition of a T0-space?

2. Is there a topological space which is not T0? If so provide an example.

3. Show that an infinite second countable T0-space cannot have a cardinality larger than2ℵ0.

4. What is the formal definition of a T1-space, emphasizing in the process how it is

different from a T0-space.

5. Give a characterization of a T1-space.

6. Does a subspace of a T1-space necessarily inherit the T1 property?

7. Does a closed function f carry the T1 property of its domain, S, to its range f [S].

8. Does the T1 property of all its factors, Si, carry over to the product space,∏

i∈I Si?

9. If∏

i∈I Si possesses the T1 property does each of the factors necessarily possess theT1 property?

10. Define the Hausdorff property (T2 property) on a topological space.


11. Provide three different characterizations of the Hausdorff topological property.

12. Is the Hausdorff property carried over from the domain of a continuous function,f : S → T , to its range f [S]? What if the function f was just closed? How about if

f was both one-to-one and closed?

13. If all factors of a product space are Hausdorff, is the product space in question neces-sarily Hausdorff?

14. If f : S → T is a function and G = (u, f(u) : u ∈ S is the graph of f in S × T , givea condition on S or T that will guarantee that the graph, G, is closed in S × T .

15. Define both a T3-space and a regular space.

16. Does T3 imply Hausdorff? Does “regular” imply Hausdorff?

17. Give two characterizations of “regular space”.

18. Are metrizable spaces necessarily “regular”? Are regular spaces always metrizable?

19. Is “regular” a hereditary property?

20. Does the “regular” property always carry over from a product to its factors? What

about, from the factors to the product?

21. Define both a T4-space and a normal space.

22. Does T4 imply “regular”? Does “normal” imply “regular”?

23. Give three characterizations of “normal space”.

24. Are metrizable spaces necessarily “normal”?

25. What kind of function will always carry a normal property from its domain to itsrange?

26. Is the normal property hereditary? If not, what kind of subspace will inherit thenormal property from its superset?

27. Is the product of normal spaces necessarily normal?

28. Are open ordinal spaces [0, ωα) normal spaces?

29. Is the deleted Tychonoff plank normal?


EXERCISES

1. Let (S, τS) and (T, τT) be two topological spaces and f : S → T and g : T → S be twocontinuous functions satisfying the property: (g f)(x) = x for all x in S.

a) Show that if T is Hausdorff, then S must also be Hausdorff.

b) Show that f [S] is closed subset of T .

2. Suppose S = x0, x1, . . . , xn is a finite set equipped with a topology, τS, which makesof (S, τS) a Hausdorff space. Describe all such topologies on S.

3. Let (S, τS) and (T, τT) be topological spaces where S is Hausdorff and contains a dense

subset D. Suppose f : S → T is a continuous function such that f |S : D → f [D] is ahomeomorphism. Show that f [S\D] ⊆ T \f [D].

4. Suppose (S, τS) is regular and (T, τT) is a topological space. Suppose the functionf : S → T is a continuous function mapping S onto T . Show that if f is both an open

and closed function as well, then T is Hausdorff.

5. Suppose (S, τS) is a topological space and K is a non-empty subset of S. SupposeτS = U ⊆ S : K ⊆ U ∪ ∅, S. Verify that this is indeed a valid topology on S. If

so, are there any other properties that must be satisfied if we want S to be regular?

6. Suppose (S, τS) is a finite regular topological space. Is (S, τS) necessarily normal?

Why?

7. Suppose (S, τS) is a regular topological space and K is an infinite subset of S. Showthat there exists an infinite family of open subsets, Ui ⊆ S : i ∈ N, Ui ∩ K 6= ∅,such that, if i 6= j, then clSUi ∩ clSUj = ∅.

8. Suppose (S, τS) is a normal topological space and that R is an equivalence relation on

S. Let (S, τθ) be the quotient space induced by the natural map θ : S → S/R. Showthat, if θ is both open and closed, then (S, τθ) is a normal topological space.

9. Let S = [0, 1] and τ denote the usual topology on S. If T = Q ∩ [0, 1], let τS be the

topology on S which is generated by the subbase, S = τ ∪ P(T ). Show that (S, τS)is a normal topological space.

194 Section 10: Separation with continuous functions

10 / Separation with continuous functions.

Summary. In this section we will investigate another method for separating

closed sets. A function, f : S → [0, 1], which continuously maps a topologicalspace, S, onto the closed interval [0, 1] in such a way that f has different con-

stant values on a pair of disjoint closed subsets can be used to separate sets.

10.1 Normal spaces revisited.

Suppose we are given a space (S, τS) and a continuous function f : S → [0, 1]. If, for

a pair of disjoint closed subsets A and B of S, A ⊆ f←[0] and B ⊆ f←[1] thenwe will say that

“f separates A and B′′

Suppose that, for a pair of disjoint closed subsets, A and B, there exists a continuousfunction, f : S → [0, 1] which separates A and B, then we can easily generate a pairof disjoint open sets, U = f← [ [0, 1/2) ] and V = f← [ (1/2, 1] ], which “separate” A

and B. If this holds true for every pair of disjoint closed sets we could conclude thatsuch a space, S, is a normal topological space. This provides us with another way of

recognizing a normal space.

On the other hand, suppose a space, S, is known to be normal. If A and B are

disjoint closed sets in S, are we guaranteed the existence of a “separating function”,f : S → [0, 1], which will separate A and B? The answer to this question is not

obvious. Proving the existence of a separating real-valued continuous function on Swhich serves a particular purpose is not a trivial task. In fact, it can be quite hard.

The statement,

“For each pair of disjoint closed subsets, A and B, of S there exists aseparating function for A and B if and only if S is normal”

is titled Urysohn’s lemma.1

Its proof invokes the previously proven characterization of the normal property:

The topological space S is normal if and only if, for every non-empty closedset F and open neighbourhood U of F , there exists another open neigh-

bourhood V of F such that clSV ⊆ U .

Constructing the required continuous function on a normal topological space just fromthis characterization of normality requires quite a bit of mathematical prowess. In

this chapter we will carefully work our way through this proof.

1Named after the Russian mathematician Pavel Urysohn (1898-1924). Urysohn was a gifted mathemati-cian destined to make great contributions to mathematics. He drowned while on holiday at the young age of26 years old. The proof of Urysohn’s lemma is an illustration of his deep insight in a mathematical problem.


10.2 Urysohn’s lemma

The proof of Urysohn’s lemma will be presented in two parts. We will first discuss amethod for constructing a function on a topological space and then prove continuityof this function. The proof of Urysohn’s lemma will follow.

We will begin by highlighting a few properties of a countable union of subsets, Vn, of

[0, 1]. For each n ∈ N\0 we define a subset Vn as

Vn =m

2n: m = 1, 2, 3, . . . , 2n − 1

For example,

V1 =

1/21

V2 =

1

22,

2

22,

3

22

V3 =

1

23,

2

23,

3

23,

4

23,

5

23,

6

23,

7

23

...

V100 =

1

2100,

2

2100,

3

2100, . . . ,

2100 − 1

2100

...

Note that, for each n, Vn ⊂ Vn+1. So every Vn contains the elements of its predecessorsand some. Let

J = ∪Vn : n = 1, 2, 3, . . .The set, J, is referred to as the dyadic rationals in [0, 1]. Then J is countably infinite,dense1 and linearly ordered in [0, 1]. That is, [0, 1]\J contains no intervals and, eventhough J contains neither 0 nor 1, we have 0 = inf (J) and 1 = sup (J). So J can be

used as a countably infinite indexing set.

With this definition in mind, we are now set to formally state and prove Urysohn’s

lemma. Even though we presented it as a theorem it is generally referred to, in theliterature, as a lemma.

Theorem 10.1 Urysohn’s lemma1. Let (S, τS) be a T1 topological space. The topologicalspace (S, τS) is normal if and only if given a pair of disjoint non-empty closed sets, F andW , in S there exists a continuous function f : S → [0, 1] such that, F ⊆ f←[0] and

W ⊆ f←[1].1To see “dense”: For any interval (a, b), 1

2n < b−a, for some n. Then, 1 < b2n−a2n implies a2n+1 < b2n,hence there must exist an integer, m, such that a2n < m < b2n ⇒ a < m/2n < b.

1This should not be confused with Urysohn’s Extension theorem


Proof : Let (S, τS) be a T1 topological space.

( ⇐ ) We are given a pair of disjoint closed sets, F and W . Also, we are given

that there exists a continuous function f : S → [0, 1] such that F ⊆ f←[0] andW ⊆ f←[1]. Then f←[ [0, 1/3) ] and f←[ (2/3, 1] ] are disjoint open sets containing

F and W , respectively. It follows that S is normal.

( ⇒ ) Suppose S is normal and F and W are disjoint non-empty closed sets in S.

Since S \W is an open neighbourhood of F there exists and open subset U1/2 suchthat F ⊆ U1/2 ⊆ clSU1/2 ⊆ S\W . Normality of S allows us to repeat this step with

dyadic rationals, 122 ,

222 ,

322 as index set, to obtain the chain

F ⊆ U1/4 ⊆ clSU1/4 ⊆ U2/4 ⊆ clSU2/4 ⊆ U3/4 ⊆ clSU3/4 ⊆ S\W

More generally, for each n, let Vn = U i2n

: i = 1 to 2n − 1 where

F ⊂ U 12n

⊆ clSU 12n

⊂ · · · ⊂ U i−12n

⊆ clSU i−12n

⊂ U i2n

⊂ · · · ⊂ U 2n−12n

⊂ S\W

From Vn we inductively construct Vn+1 with the same properties and where Vn ⊂ Vn+1.Let V = ∪Vn : n = 1, 2, 3, . . .. We obtain the countably infinite set, Ui : i ∈ Jwhere, for all j < i in J,

F ⊂ Uj ⊆ clSUj ⊆ Ui ⊆ clSUi ⊂ S\W

W ⊆ S\clSUi ⊆ S\Ui ⊂ S\Uj ⊆ S\clSUj ⊂ S\F

Also, for all i, W ⊂ S\clSUi.

Define

U0 = ∩Ui : i ∈ J = ∩clSUi : i ∈ JU1 = ∩S \ clSUi : i ∈ J = ∩S \Ui : i ∈ J

where neither U0 nor U1 is empty. Then

S\U1 = S\∩S \ clSUi : i ∈ J = ∪clSUi : i ∈ J

So we have the strictly increasing chains of inclusions,

F ⊆ U0 ⊂ · · · ⊂ Ui ⊆ clSUi ⊂ · · · ⊆ ∪Uii∈J ⊆ ∪clSUii∈J = S\U1 ⊂ S\W

W ⊆ U1 = ∩S\clSUii∈J ⊂ · · · ⊂ S\clSUi ⊆ S\Ui ⊂ · · · ⊆ S\U0 ⊆ S\FNote that for a point x in S

x ∈ U0 ⇔ x ∈ Ui for all i ⇔ x ∈ clSUi for all i

x ∈ U1 ⇔ x ∈ S\clSUi for all i ∈ J ⇔ x ∈ S\Ui for all i


So,

x 6∈ U0 ∪ U1 ⇔ x 6∈ U0 and x 6∈ U1

⇔ x 6∈ clSUt and x 6∈ S\Uk for some k, t ∈ J

⇔ x ∈ S\clSUt and x ∈ Uk, for some k, t ∈ J (∗)

⇔ x ∈ S\clSUt ∩ Uk, for some k, t ∈ J (∗∗)

So if x 6∈ U0 ∪ U1 x ∈ S\clSUt ∩ Uk, an open neighbourhood of x in S.

For any given x ∈ S, let

Jx = i ∈ J : x ∈ UiJ∗x = i ∈ J : x ∈ S\Ui

a proper subset of [0, 1]. See that glbJx = lubJ∗x .

We will define a function, f : S → [0, 1], as follows:

f(x) =

0 If x ∈ Ui for all i ∈ JglbJx = lubJ∗x otherwise.

See that

x ∈ U0 ⇔ x ∈ ∩Ui : i ∈ J⇔ x ∈ Ui for all i

⇔ glbi : x ∈ Ui = glb(0, 1] = 0 = f(x)

So f←(0) = U0

Also see that

x ∈ U1 ⇔ x ∈ ∩S\Ui : i ∈ J⇔ x ∈ S\Ui for all i

⇔ lubi : x ∈ S\Ui = lub[0, 1) = 1 = lubJ∗x = f(x)

So f←(1) = U1. We then have,

F ⊆ U0 = f←[0]W ⊆ U1 = f←[1]

Claim #1: For each k ∈ J,

f [clSUk] ⊆ [0, k]

Proof of claim: Let x ∈ clSUk, then for any i > k, x ∈ Ui (since clSUk ⊂ Ui). So Jx

contains (k, 1]. So glbJx ≤ k. That is, f(x) ≤ k. So f [clSUk] ⊆ [0, k], as claimed.


So, if f(x) > k then x 6∈ clSUk.

Claim #2: For each k ∈ J,

f [S\Uk] ⊆ [k, 1]

Proof of claim: Let x ∈ S\Uk, then for any i < k, x ∈ S\Ui (since S\Ui ⊂ S\Uk).So So J∗x contains [0, k). Then lubJ∗x ≥ k. That is f(x) ≥ k. So, f [S\Uk] ⊆ [k, 1], as

claimed. So if x 6∈ Uk then f(x) > k.

We now prove that f : S → [0, 1] is a continuous function on S.

Let q ∈ S and a, b ∈ [0, 1] such that f(q) ∈ (a, b). Then there exist r, t ∈ J such that

a < r < f(q) < t < b

Let W = Ut\clSUr.

Claim #3 : W is an open neighbourhood of q.Proof of claim: We are given r < f(q) < t. Since, by claim #2, if q 6∈ Ut then f(q) ≥ t

we must then haveq ∈ Ut

Since, by claim #1, if f(q) ≥ r then q cannot belong to clSUr. Then q ∈ Ut\clSUr.So W is an open neighbourhood of q in S, as claimed.

Claim #4 : That f [W ] ⊆ (a, b) .

Proof of claim: Suppose x ∈ Ut\clSUr. Then x ∈ clSUt. So, by claim #1, f(x) ∈ [0, t].Since x 6∈ clSUr, by claim #2, f(x) ∈ [r, 1]. So f(x) ∈ [r, t] ⊆ (a, b).

So f [W ] ⊂ (a, b), as claimed.

Sof(q) ∈ f [W ] ⊆ [r, t] ⊂ (a, b)

We conclude f is continuous on S.

Since f [F ] ⊆ f [U0] = 0 and f [W ] ⊆ f [U1] = 1, we are done.

Urysohn’s lemma provides another way of recognizing normal topological spaces.

Definition Given a pair of disjoint closed sets, F and W , in a space S, we will referto the continuous function, f : S → [0, 1], introduced in Urysohn’s proof, as a

“Urysohn separating function for F and W”

We also say that this function is one which separates the disjoint closed sets F and

W . For future reference, we define these formally.


Definition 10.2 Suppose A and B are disjoint non-empty subsets of a space S. We willsay that A and B are completely separated in S if and only if there is a continuous functionf : S → [0, 1] such that A ⊆ f←(0) and B ⊆ f←(1). In such a case, we say the function, f ,

(completely) separates A and B. 1

We now include the following useful fact.

Fact : Given a real-valued continuous function, f : S → R, we define a zero-set, Z(f),

of f as,Z(f) = x ∈ S : f(x) = 0

Two sets in S are completely separated if and only if they are contained in disjointzero-sets in S.

Proof of this fact: For( ⇐ ), suppose A ⊆ Z(f) and B ⊆ Z(g) where Z(f) and Z(g)

are disjoint zero-sets in S. Then

h(x) =|f(x)|

|f(x)|+ |g(x)|

is a well-defined continuous function on S. See that h[A] = 0 and h[B] = 1. SoA and B are (completely) separated. For ( ⇒ ), suppose A and B are completely

separated. Then there exists a continuous real-valued function, f : S → [0, 1], suchthat f [A] = 0 and f [B] = 1. Then A ⊆ Z(f) and B ⊆ Z(f − 1), Z(f) and

Z(f − 1) are disjoint zero-sets of the functions, f and f − 1, respectively.

We will say more on zero-sets further down in this chapter.

10.3 Completely regular spaces.

Note that, if there exists a Urysohn separating function for each pair of disjoint closed

subsets in a normal space, then we can find a Urysohn function that can separate apoint from a closed subset or separate any pair of points in any T1 space. But there

is nothing in what we have seen that allows us to conclude that a non-normal regularor Hausdorff space can generate its own Urysohn separating function like a normal

space can. In fact, as we shall soon see, Hausdorff and regular spaces cannot, ingeneral, produce, by themselves, such a function. If we are presented with a regular

space in which it is shown that, for any point and disjoint closed set, there exists a

1Compare this definition with a similar one provided in definition 7.16, in another context, one where wespeak of a family of continuous functions, not necessarily real-valued. Here we speak of a single real-valuedcontinuous function. The reader is however encouraged to verify that when the family is one of continuousreal-valued functions then the two definitions are equivalent (since R is completely regular).


Urysohn separating function, then this topological space belongs to a separate classof topological spaces called “completely regular spaces”. We formally define this class.

Definition 10.3 Suppose (S, τS) is a T1-space. If, for any given point x and a non-empty

closed subset F disjoint from x, there exists a continuous function, f : S → [0, 1], suchthat x ⊆ f←0 and F ⊆ f←1, then the space, S, is called a completely regular space,

or a Tychonoff space 1 A completely regular space is often referred to as a T3 12-space.

We add a few remarks concerning this definition. In this text, our definition em-phasizes that only T1 spaces are completely regular. (Some authors may not require

the T1 condition.) Obviously, since a Urysohn separating function separates disjointclosed sets of a normal space, and since a regular space is T1, then

normal ⇒ completely regular

Also, if S is completely regular, there exists a Urysohn separating function, f , suchthat x ⊆ f←[0] and F ⊆ f←[1]. It follows that x ⊆ f←[ [0, 1/3) ] and

F ⊆ f←[ (2/3, 1] ] where [0, 1/3) and (2/3, 1] are open in [0, 1]. Then, (see implica-tion chart on page 184),

completely regular ⇒ regular ⇒ semiregular ⇒ Hausdorff

regular ⇒ completely Hausdorff ⇒ Hausdorff

There are, however, regular spaces which are not completely regular. The standard

example of a regular non-completely regular space is called the Tychonoff corkscrewspace1. This space is rather involved and lengthy to describe. So we will not describeit in this text. So, the Tychonoff corksrew, justifies,

regular 6⇒ completely regular

Also, there are completely regular spaces which are not normal. The Moore plane(Niemytzki’s plane) is such an example. At the end of this section (on page 214) we

prove that the “Moore plane is completely regular but not normal”. From which wededuce

completely regular 6⇒ normal

1Named after the Soviet and Russian mathematician Andreı Tychonoff, (1906-1993).1Interested readers will easily find a description of the Tychonoff corkscrew online.


This means, a T312-space is strictly in between T3 and T4 (which partially explains the

tongue-in-cheek terminology, “ T312-space”, used by some authors).

The description of the completely regular property does not simply refer to the separa-tion of closed sets by open sets. It is defined in terms of the existence of a continuous

function. If we want to refer to the“completely regular property” as a “topologicalproperty” (or topological invariant) we should prove that it is one first. Fortunately,the proof is fairly straightforward.

“Completely regular ” is a topological property : Suppose S is completely regular andh : S → T is a homeomorphism mapping S onto the space T . Suppose the singleton

set, x, and the closed subset, F , are disjoint in T . Then there exists a functionf : S → [0, 1] such that h←[x] ⊆ f←[0] and h←[F ] ⊆ f←[1]. Let g = f h←.

Then g[x] = 0 and g[F ] = 1. Then T is also completely regular. So “completelyregular” is indeed a topological property, as expected.

Readers who deal mostly with metrizable topological spaces will be glad to know that

all such spaces are completely regular. We prove this below.

Theorem 10.4 If (S, τ) is a metrizable space then it is completely regular.

Proof : We are given that (S, τ) is metrizable. Then there is a metric, ρ, such that (S, ρ)and (S, τ) are equivalent topological spaces. Let F be a non-empty closed subset of

S and x be a point in S \F . We are required to construct a continuous real-valuedfunction which separates F from x.

Consider the function, dF : S → R defined as,

dF (u) = ρ(u, F ) = inf ρ(u, y) : y ∈ F

The real-valued function, dF (u), numerically provides a reasonable measure of the

smallest distance which separates u from F . Claim : That the function, dF , is a con-tinuous function on S.

Proof of claim : Let u ∈ S. Let ε > 0 and δ = ε/3. It suffices to show that ρ(x, u)< δimplies |dF (x)− dF (u)| < ε.

Then with the variables, x, u and y, we write the two possible triangle inequalities


with y on one side.

ρ(x, y) ≤ ρ(x, u) + ρ(u, y)

ρ(u, y) ≤ ρ(x, u) + ρ(x, y)

implies

inf ρ(x, y) : y ∈ F ≤ ρ(x, u) + inf ρ(u, y) : y ∈ Finf ρ(u, y) : y ∈ F ≤ ρ(x, u) + inf ρ(x, y) : y ∈ F

implies

dF (x) ≤ δ + dF (u) = ε/3 + dF (u)

dF (u) ≤ δ + dF (x) = ε/3 + dF (x)

implies

−ε < −ε/3 ≤ dF (x) − dF (u) ≤ ε/3 < ε

So |dF (x) − dF (u)| < ε. Then dF is continuous on S, as claimed.

Clearly F ⊆ d←F (0). We have a fixed point, x in S \F . Since F is closed then x iscontained in an open ball Bε(x) disjoint from F . So dF (x) equals some k ≥ ε > 0.Since x ∈ Z(dF (x)− k) and F ⊆ Z(dF [F ]) then F and x are completely separated

(see the fact proven on page 199). So S is completely regular.

We conclude that every metrizable space is completely regular.

The following two theorems confirm that completely regular spaces are hereditary andalso carry over from a family of factors to their products.1

Theorem 10.5 Let S be a completely regular space and T be a subspace of S. Then T is

completely regular.

Proof : We are given that S is a completely regular space and T is a subspace of S.

First note that T is clearly T1. Let F ∩ T be a closed subset of T , where F is closedin S, and let u ∈ T \F . Since S is completely regular there exists a continuous func-

tion f : S → [0, 1] such that x ∈ f←[0] and F ⊆ f←[1]. Then the function,f |T : T → [0, 1] separates x and F ∩ T and so T is completely regular.

1Continuous images of completely regular spaces are not always completely regular. A standard exampleof this fact is to produce a continuous function on the completely regular Moore plane which maps it ontoa Hausdorff, but non-regular space.


Theorem 10.6 Let Si : i ∈ I be a family of topological spaces and S =∏

i∈I Si be aproduct space. Then S is completely regular if and only if each factor, Si, is completely

regular.

Proof : We are given that Si : i ∈ I is a family of topological spaces and S =∏

i∈I Si isa product space.

( ⇒ ) Suppose the product space, S =∏

i∈I Si, is completely regular. By theorem7.14, each Si is homeomorphically embedded in the product space, S. By the pre-

ceding theorem, every subspace is completely regular, so each Si is completely regular.

( ⇐ ) We are given that each Si is completely regular. Let K be a closed subset ofS =

∏

i∈I Si and <xi>i∈I be a point in S\K.

Then

<xi>i∈I ∈ ∩π←i [Ui] : i ∈ F ⊆ S\Kfor some finite F ⊆ I and open Ui’s in Si, where xi ∈ Ui for i ∈ F . Then, for eachi ∈ F , there exists a continuous function, fi : Si → [0, 1], such that xi ∈ f←i [1] andSi\Ui ⊆ f←i [0]. We define h : S → [0, 1] as

h(<xi>i∈I) = inf fi(xi) : i ∈ F

Then h = inf fiπi : i ∈ F is a continuous function. Note that h(<xi>i∈I) = 1 and

S\K ⊆ h←(0). So S is completely regular.

10.4 Topic: On zero-sets in normal spaces.

Recall that, in the proof of Urysohn’s lemma (in theorem 10.1), we showed that, for

disjoint closed subsets F and W of the normal space, S, there is a continuous function,f : S → [0, 1], such that

F ⊆ U0 = ∩Uii∈J = f←(0)

W ⊆ U1 = ∩S \ clSUii∈J = f←(1)

The continuous real-valued function, f , on S was referred to as a Urysohn separating

function for the closed subsets F and W .

The set, J, is a countably infinite indexing set so both U0 and U1 are Gδ’s (countable

intersection of open sets). Note that we did not prove that F and W are equal to U0

and U1. Usually, they are not. Although they appear in the proof, neither U0 nor U1


are referred to in the Urysohn lemma statement itself.

We will briefly continue our discussion of normal spaces (in relation to U0 and U1).First we need the following definition.

Definition 10.7 A subset of a topological space, S, of the form, f←[0], for some contin-uous real-valued function, f , on S is called a zero-set. Such a set is denoted by, Z(f).

A cozero-set of S is a set of the form S\Z(f) for some continuous real-valued function f .

The cozero-set of f is denoted by coz(f).

We note a few interesting facts about zero-sets.

When we speak of a zero-set in a topological space, S, we are referring to a specific

subset, A, of S associated to some function f : S → R such that A = f←(0) = Z(f).A zero-set is sometimes described as the “fibre” of a continuous real-valued function

f : S → R. To see this, if f(u) = r ∈ R, then u ∈ f←(r); if g : S → R is the functiondefined as g(x) = f(x) − r then

g(u) = f(u) − r

= r − r

= 0

so g is a continuous real-valued function.

Then the fibre, f←(r), is the zero set, Z(g), which contains the element u. So thefibre of a continuous real-valued function, f , on S is a zero-set.

Also see that, for any zero-set, Z(f) of the continuous function, f ,

Z(f) = ∩ f←[ (−1/n, 1/n) ] n>0

Since f is continuous,

. . . any zero-set, Z(f), is a Gδ

Countable intersections of zero-sets. We know that any zero-set is a countable inter-section of open sets. What can we say about countable intersections of zero-sets? We

claim: The countable intersection of zero-sets is a zero-set .


Proof of the claim : Let Z(fn) : n ∈ N be a countable family of zero-sets withnon-empty intersection. We are required to show that ∩Z(fn) : n ∈ N is a zero-set.

For each fn, let1

hn = |fn| ∧1

2n

We see that 0 ≤ hn(x) ≤ 12n on S. Since the series,

∑

n∈N12n , converges,

∑

n∈N hn(x)converges uniformly to a continuous function h(x) on S.2 That is,

∑

n∈N

hn(x) = h(x)

Since convergence is uniform then h is continuous.

x ∈ ∩Z(fn) : n ∈ N ⇔ fn(x) = 0 for all n

⇔(

|fn| ∧1

2n

)

(x) = 0 for all n

⇔ hn(x) = 0 for all n

⇔ h(x) = 0

⇔ x ∈ Z(h)

SoZ(h) = ∩Z(fn) : n ∈ N

as required. So countable intersections of zero-sets are zero-sets, as claimed.

In the proof of Urysohn’s lemma we showed that F ⊆ f←(0) and W ⊆ f←(1). This

means F ⊆ Z(f) and W ⊆ Z(f − 1). We can then confidently reformulate onedirection of the Urysohn lemma statement as follows:

If (S, τS) is a normal topological space and F and W are disjoint non-empty

closed subsets of S, then F and W are contained in disjoint zero-sets.

The leads to an interesting question: Is the converse of this statement true? Thatis, does it hold true that “if F and W are respectively contained in disjoint zero-sets

then S is normal”? We show that it is true, in general.

Theorem 10.8 Let S be a T1 topological space. The following are equivalent.

a) The space S is a normal space.

b) If A and B are disjoint closed subsets of S then they are contained in disjoint zero-sets.

1(f ∧ g)(x) = minf(x), g(x)2By the Weierstrauss M-test.


c) If A and B are disjoint closed subsets of S then they are completely separated.

Proof : We are given that (S, τS) is normal.

( a ⇒ b ) This direction follows immediately from Urysohn’s lemma stated in theparagraph above.

( b ⇐ a ) We are given that disjoint closed subsets F and W are contained in disjointzero-sets, Z(f) and Z(g), respectively. Since the two zero-sets are disjoint 0 6∈ f [Z(g)]

and 0 6∈ g[Z(f)].1

We are required to show that S is normal. To do this we seek a Urysohn separatingfunction, h : S → R, for F and W .

Let the function h : S → R, be defined as

h(x) =|f(x)|

|f(x)|+ |g(x)|

Then h is continuous and real-valued on S with range in [0, 1].

If x ∈ F ⊆ Z(f) then h(x) = 00+|g(x)| = 0, so F ⊆ Z(f) ⊆ h←(0).

If x ∈W ⊆ Z(g) then h(x) = |f(x)||f(x)|+0 = 1, so W ⊆ Z(g) ⊆ h←(1).

So h separates F and W . By theorem 10.1 S is normal.

( a ⇔ c ) Is Urysohn’s Lemma 10.1.( b ⇔ c ) Since( a ⇔ b ) and ( a ⇔ c ), then ( b ⇒ c ) and ( c ⇒ b ).

From the above theorem we can then say that...

. . . if A and B are completely separated and r 6= k there exist h ∈ C∗(S)such that if h[A] = r and h[B] = k.

by composing the function h with t(x) = (k − r)x+ r.

On normal spaces and C∗-embedded subsets.

The following statement refers to the notions of C∗-embedded subsets and C-embedded

subsets.

A subset A of a topological space S is said to be a C∗-embedded subset

(C-embedded subset) of S if for each g ∈ C∗(A) (respectively, g ∈ C(A))there exists f ∈ C∗(S) (respectively, C(S)) such that f |A = g.

1Since, if 0 ∈ f [Z(g)], there is some u ∈ Z(g) such that f(u) = 0 hence u ∈ Z(f) ∩ Z(g); so Z(f) ∩ Z(g)is not empty, a contradiction.


One of the most important theorem in basic point-set topology is the Urysohn ex-tension theorem. It is often referred to as being a “deep theorem” since the proof

involves innovative mathematical techniques introduce by Pavel Urysohn.1 We invokethe Urysohn extension theorem in the proof of the following result. But its proof is

deferred to theorem 21.8 on page 400.

Urysohn extension theorem: Let T be a subset of the completely regularspace S. Then T is C∗-embedded in S if and only if pairs of sets which can

be completely separated by some function in C∗(T ) can also be separated bysome function in C∗(S).

The proof also in invokes a statement proven in theorem 23.2 whose proof is deferred

at that point.

Theorem 23.2: Suppose U be C∗-embedded in S. The set U is also C-

embedded in S if and only if U is completely separated from any disjointzero-set in S.

Theorem 10.9 Let S be a T1 topological space and A and B be any pair non-emptydisjoint closed subsets of S. The following are equivalent.

a) The space S is a normal space.

b) Each closed subset of S is C∗-embedded in S.

c) Each closed subset of S is C-embedded in S.


( a ⇒ b ) Suppose S is normal.

Let A be a closed subset of S. We are required to show that A is C∗-embedded in S.

Let B and C be completely separated subsets of A. Then, by 10.8 b) ⇔ c), B andC are contained in disjoint zero sets, ZB and ZC in A. Since A is closed both ZB

and ZC are closed in S. Since S is normal, by theorem 10.8 a) ⇔ c), ZB and ZC arecompletely separated in S. Then B and C are completely separated subsets of S. By

Urysohn’s extension theorem stated above, A is C∗-embedded in S.

( b ⇒ c ) Suppose each closed subset of S is C∗-embedded in S. Let A be a closedsubset of S. We are required to show that A is C-embedded in S.

By hypothesis, since A is closed in S it is C∗-embedded in S.

1Pavel Urysohn was a brilliant Russian mathematician who drowned at the young age of 26 years oldwhile on vacation in France in 1924.


We claim that A is completely separated from any zero-set disjoint from A. SupposeZ is a zero-set in Z[S] which is disjoint from A. Then the function, h : A∪Z :→ 0, 1,defined as

h[A] = 1h[Z] = 0

is continuous on A∪Z. Then the function h in C(A∪Z) completely separates Z andA. So A is completely separated from any zero-set disjoint from A, as claimed.

By theorem 23.2 A is C-embedded in S.

We mentioned above that zero-sets of a continuous function are closed Gδ’s. Can weturn this phrase around and say that “closed Gδ’s are zero-sets”? The answer is no.

The statement “Closed Gδ’s are zero-sets” is however true in, at least, one class oftopological spaces, namely, normal spaces.

Theorem 10.10 Suppose F is a non-empty closed subset of the normal topological space,(S, τ). Then F is a Gδ if and only if F is a zero-set produced by some continuous function

f : S → [0, 1].


( ⇐ ) This direction is obvious since, if F = Z(f) ⊆ S where f : S → [0, 1] is contin-uous then F = ∩f← [ [0, 1/i) ] : i ∈ N\0 , a Gδ in S.

( ⇒ ) Suppose F is a closed Gδ in the space S. If F is also open then it is clopen andso is easily seen to be a zero-set. Suppose the closed Gδ, F , is not open. Then, by def-

inition, there exists a countably infinite family, Vi : i = 1, 2, 3, . . . , Vi is open in S such that F = ∩Vi. Suppose

U1 = V1

U2 = V1 ∩ V2

...

Un = V1 ∩ V2 ∩ . . .∩ Vn

...

Then Ui : i = 1, 2, 3, . . . ⊂ τS, F ⊆ Ui+1 ⊆ Ui, for all i > 0, and F = ∩Ui.


Normality of S implies that, for each i > 0, there is a continuous function, fi : S → [0, 1],such that F ⊆ f←i 0 and S\Ui ⊆ f←i [1].So we have a countable family, fi : i > 0, of functions each mapping S into [0, 1].

Consider the function, f : S → R, defined as,

f(x) =

∞∑

i=1

fi(x)

2i(∗)

See that, since 0 ≤ fi(x)2i ≤ 1

2i on S, and∑∞

i=112i = 1, then, by the WeierstrassM -test,

the series (∗) converges uniformly to the function f(x) on S. Uniform convergence

guarantees continuity of f on S.

Also see that, since F = ∩Vi, F = f←[0]. So the Gδ, F , is the zero-set, Z(f),

produced by the continuous function, f : S → [0, 1], described in (∗).

Notation. If C(S) denotes the set of all continuous functions on S, we let the subsetsof S,

Z[S] = Z(f) : f ∈ C(S)coz[S] = S\Z(f) : f ∈ C(S)

denote the family of all zero-sets and cozero-sets of a space S, respectively. The familyof all zero-sets (cozero-sets) in a completely regular space provides us with anotherway of describing a base for closed (open) sets of S.

Theorem 10.11 If (S, τ) is completely regular then,

a) The family Z[S] = Z(f) : f ∈ C(S) of all zero-sets forms a base for the closed setsin its topology.

b) The family coz[S] = S\Z(f) : f ∈ C(S) of all cozero-sets forms a base for the open

sets in its topology.

Proof : a) Let F be a non-empty proper closed subset in S. Given that S is completelyregular, then for every p ∈ S\F , there exists a function f ∈ C(S) such that F ⊆ Z(f)

and p ∈ Z(f − 1). Then F is the intersection of zero-sets. So, by definition of “basefor closed sets” (see definition 5.2) Z[S] forms a base for the closed sets in S.

The proof of part b) is left as an exercise.


Example 1. Show that if S is a topological space in which the family of all zero-sets,

Z[S], forms a base for closed sets then S must be completely regular.

Solution: Suppose S is a space such that Z[S] forms a base for closed sets. Suppose

F is a closed subset of S and p is a point not in F .

By hypothesis, the closed set F is the intersection of zero-sets. Then, for every p ∈ S\F ,there exists a zero-set Z(f) such that F ⊆ Z(f) ⊆ S\p. Let h : S → R be defined

as

h(x) =

(

1

f(p)

)

f(x)

Then h is a continuous real-valued function which maps F to 0 and p to 1. Bydefinition, S is completely regular.

Example 2. Suppose that F is a closed subset of a metric space S. Show that F is a

zero-set of continuous real-valued functions. Conclude that F is a Gδ.

Solution: Suppose F is a closed subset of a metric space, (S, ρ). Define a real-valuedfunction fF : S → R as fF (u) = ρ(u, F ) = inf ρ(u, y) : y ∈ F. Then fF [F ] = 0and fF (u) 6= 0 for u 6∈ F . To show that F is a zero-set it now suffices to show that

fF is a continuous function.

Claim : That the function, fF , is a continuous function on S.

Proof of claim : Let ε > 0. It suffices to show that there is a δ > 0 such that

f [Bδ(x)] ⊆ Bε(f(x)). Let ρ(x, u) < δ = ε/3.

Since,

ρ(x, y) ≤ ρ(x, u) + ρ(u, y)

ρ(u, y) ≤ ρ(x, u) + ρ(x, y)

then,

inf ρ(x, y) : y ∈ F ≤ ρ(x, u) + inf ρ(u, y) : y ∈ Finf ρ(u, y) : y ∈ F ≤ ρ(x, u) + inf ρ(x, y) : y ∈ F

Hence,

fF (x) ≤ ε/3 + fF (u)

fF (u) ≤ ε/3 + fF (x)

Then

−ε < −ε/3 ≤ fF (x) − fF (u) ≤ ε/3 < ε


So |fF (x)− fF (u)| < ε. Then fF is continuous on S, as claimed.

So the closed set F is a zero-set, Z[fF ], of fF .

Since a zero-set is a G− δ, then F is a Gδ.

10.5 Topic: Perfectly normal topological spaces.

In theorem 10.10 we showed that in normal spaces closed Gδ’s and zero-sets are one

and the same. We should, however, be cautious and not deduce from this that closedsubsets of normal spaces are necessarily Gδ’s. We will soon exhibit a normal space

which contains a non-Gδ singleton set. Those normal spaces in which all non-emptyclosed subsets are Gδ’s form a special class of topological spaces which we will now

define.

Definition 10.12 A normal topological space, (S, τS), is said to be perfectly normal if andonly if every closed subset of S is a Gδ.

To say that every closed subset is a Gδ is equivalent to saying that every open set is

an Fσ (a countable union of closed sets) as we shall now see.Suppose F is closed in S and Ui : i ∈ N is a family of open sets in S.

Let U = S\F and Fi = S\Ui for each i.

F = ∩Ui : i ∈ N, is a closed Gδ ⇔ S\F = S\∩Ui : i ∈ N, is open

⇔ U = ∪S\Ui : i ∈ N, is open

⇔ U = ∪Fi : i ∈ N, is an open Fσ

Example 3. Show that all second countable normal spaces are perfectly normal

spaces.

Solution : Suppose S is a second countable normal space. If x ∈ S, let Ux denote an

open base element of S which contains x. Since S is second countable, we can chooseB = Ux : x ∈ S to be countable. Let F be a closed subset of S and V = S \F .

Since S is normal, if x ∈ V , there exist an open base element, Ux, of x such that

x ∈ clSUx ⊆ V

Then we can find clSUx : x ∈ V , a countable set of closed subsets each of which is

contained in V . ThenV = ∪clSUx : x ∈ V


an Fσ in S. This implies that F = S\V is a Gδ. So every closed set is a Gδ. Then Sis perfectly normal.

Theorem 10.13 Suppose (S, τS) is a normal topological space. Then S is perfectly normalif and only if every non-empty closed subset of S is a zero-set.


( ⇒ ) We are given that S is perfectly normal. Suppose F is a non-empty closedsubset. By hypothesis, F is a Gδ. By theorem 10.10, since F is a Gδ in a normal

space then F is a zero-set, Z(f), for some continuous f : S → [0, 1].

( ⇐ ) Suppose S is normal and every non-empty closed subset F is a zero-set. Zero-sets have been shown to be closed Gδ’s. So S is perfectly normal.

We know from theorem 9.19 that metric spaces are normal. We raise the bar slightlyhigher to show that they are even perfectly normal.

Example 4. Show that a metric space is perfectly normal.

Solution : By theorem 9.19, a metric space is normal. In an example found on page

210 it is shown that closed subsets of a metric space are zero-sets. By the abovetheorem a metric space is perfectly normal.

Normal but not perfect

Are there normal spaces which are not perfectly normal? We have shown in an ex-ample on page 211 that second countable normal spaces are perfect. If we seek a

“non-perfect” normal space we investigate normal spaces which are not second count-able.

Example 5. A normal space which is not perfectly normal. Let ω1 denote the first

uncountable ordinal. Show that the ordinal space, S = [0, ω1], (previously discussedon page 89) shown previously to be normal (on page 189) is not perfectly normal.

Solution : Let (S, τω) denote the ordinal space [0, ω1] where ω1 is the first uncountable

ordinal. It will suffice to show that the closed singleton subset, ω1, of S is not a Gδ

and hence S is not perfect.

Recall that (µ, ω1] is an open neighbourhood base element of ω1. Suppose

ω1 ⊆ ∩Ui : i ∈ N


where Ui : i ∈ N is a countable family of open neighbourhoods each containing ω1.We will show that ω1 6= ∩Ui : i ∈ N.For each Ui, there exists µi such that ω1 ∈ (µi, ω1] ⊆ Ui. Note that each µi is a

countable ordinal (since µi < ω1). Let U = µi : i ∈ N and

γ = supU ≤ ω1

We consider two cases based on the nature of γ.

Case 1. The ordinal γ is a limit ordinal : Then γ = ∪µi : i ∈ N, a countable union

of countable ordinals. Countable unions of countable sets are countable. Hence γ is acountable ordinal. Since ω1 is uncountable, γ < ω1. Then (γ, ω1] 6= ω1.

Case 2. The ordinal γ is not a limit ordinal : Then γ ∈ U = ∪µi : i ∈ N. So againγ is a countable ordinal. That is γ < ω1. Then (γ, ω1] 6= ω1.

So in both cases, (γ, ω1] 6= ω1. Also,

(γ, ω1] ⊆ ∩(µi, ω1] : i ∈ N ⊆ ∩Ui : i ∈ N

So ω1 6= ∩Ui : i ∈ N. This shows that ω1 cannot be a Gδ. So S is not perfect,

as required.

We provide another example.

Example 6. A normal space which is not perfectly normal. Let S = R2 and p = (0, 0)and suppose S has a topology defined as follows:

τ = T ⊆ S : S\T is finite or p 6∈ T

Show that (S, τ) is normal but not perfectly normal.

Solution : We claim that S is T1. A set which does not contain p is declared to beopen so the set p is closed. Complements of finite sets are open so, if x 6= p, since

S \ x is open then x is closed. So S is T1, as claimed.

Claim : That S is T4. Suppose F and K are disjoint closed subsets of S. If p 6∈ K ∪Fthen K and F are both open. So F and K are clopen and so separated by open sets.Suppose p ∈ F . Then K is clopen and finite. Now S\K is open since K is finite. So F

and K are contained in the disjoint open sets S\K and K. So S is normal, as claimed.

Claim : That S is not perfect. We know that R2 is uncountable. Let Ui : i ∈ N bea countable family of sets in S each containing the point p. Then each Ui is closedin S and p ∈ ∩Ui : i ∈ N. If each Ui is also open S \Ui if finite. So each Ui is

uncountable. Then S\∩Ui : i ∈ N = ∪S\Ui : i ∈ N a countable subset of S. So∩Ui : i ∈ N is uncountable. So p 6= ∩Ui : i ∈ N. So p is not a Gδ. So S is not

perfectly normal, as claimed. The completes the problem.


So we have, the implications,

normal 6⇒ perfectly normal

perfectly normal ⇒ normal ⇒ completely regular ⇒ regular ⇒ semiregular ⇒ Hausdorff

⇓completely Hausdorff ⇒ Hausdorff

We have a particular characterization of the perfectly normal property.

Theorem 10.14 Suppose S is a normal topological space. The space S is perfectly normalif and only if every closed subset of S is a continuous pre-image of some closed subset of aperfectly normal space.

Proof : ( ⇐ ) Let F be a closed subset of S. Suppose that there exists a continuous func-tion, f : S → T , mapping S into the perfectly normal space, T , such that F = f←[K]

for some closed subset K of T . We are required to show that S is perfectly normal.It will suffice to show that F is a zero-set. Since K is a closed subset of a perfectly

normal T , then by theorem 10.13 K is a zero-set in T . Then K = h←(0), for somecontinuous function h : T → R. Then (hf)←(0) = f←(h←(0)) = F . So F is a zero-set in S. Since F is an arbitrary closed subset, by theorem 10.13, S is perfectly normal.

( ⇒ ) Suppose S is perfectly normal and F is closed subset of S. Then F is a zero-setand so F = f←(0), for some f ∈ C(S). Since R is a metrizable it is perfectlynormal. Let K = 0 a closed subset of R. Then F = f←[K], as required

10.6 Topic: Example of a “Completely regular non-normal” space.

Up to now, we have not yet witnessed a completely regular space which is not normal.The Moore plane is such a space, as we now show.

Example 8. Show that the Moore plane is completely regular.

Solution : Let (S, τM) denote the Moore plane. We have already shown in a previous

example (on page 185) that (S, τM) is regular. We have also seen on page 74, that if τis the usual topology on S, τ ⊂ τM . We know that metrizable spaces are completely

regular so (S, τ) is completely regular.

We proceed to show that (S, τM) is completely regular.


For what follows we let F = (x, 0) ∈ S : x ∈ R and W = S \ F .

Suppose K is a non-empty closed subset of S and (a, b) is a point which does notbelong to K. We seek a continuous function which completely separates K and (a, b).

We consider two cases.

Case 1: Suppose b 6= 0.

Now F is closed with respect to both τ and τM .

Since (S, τ) is completely regular there exists an open neighbourhood V (in τ) of Fsuch that (a, b) 6∈ clSV . Then K ∪ clSV is closed with respect to both topologies.Then there exists a continuous function g : S → [0, 1] (with respect to τ) such that,

K ∪ clSV ⊆ g←(0) and (a, b) ∈ g←(1). Since g is continuous with respect to τ it iscontinuous with respect to τM . Then the continuous function g with respect to τMcompletely separates K and (a, b).

Case 2: We consider the case where (a, b) ∈ F . That is, (a, b) = (a, 0).

Without loss of generality suppose K ∩W 6= ∅. If Bδ = Bδ(a, δ) ∪ (a, 0) is a basicopen neighbourhood of (a, 0) such that Bδ ∩K = ∅; then

F ∩ S\Bδ = F \(a, 0)

is a closed subset of S. So K ∪ (F \(a, 0)) is a closed subset of S which misses Bδ

and intersects W .

Since (S, τM) is regular, there exists a neighbourhood, Bε = Bε(a, ε)∪ (a, 0) of thepoint (a, 0) such that,

Bε ⊆ clSBε ⊆ Bδ

Then clSBε is a closed neighbourhood of (a, 0) in S contained in the set Bδ, and

K ∪ F \(a, 0) ⊆ S \Bδ . Then

clSBε ∩W and W ∩ (S \Bδ)

are disjoint closed subsets of W .

Since W is metrizable it is normal (by theorem 9.19) and so there exists a Urysohn

separating function, g : W → [0, 1], such that

clSBε ∩W ⊆ g←[0] and W ∩ (S \Bδ) ⊆ g←[1]

Given g[clSBε ∩W ] = 0, then we can extend g to a continuous function, gS : S →[0, 1], on S where

gS[clSBε] = 0 and gS[S \Bδ] = 1

Since K ⊆ S \Bδ and (a, 0) ∈ clSBε then gS completely separates K and (a, 0).


So the Moore plane is completely regular.

Example 9. If (S, τM) denotes the Moore plane and F = (x, 0) : x ∈ R show that

every subset of F is closed in (S, τM).

Solution : We are given that (S, τM) denotes the Moore plane and F = (x, 0) : x ∈R. Since S\F is easily seen to be open in S, then F is closed.

Suppose T ⊂ F and (a, 0) ∈ F\T . Then (a, 0) belongs to some basic open neighbour-hood, Ba = Bε(a, ε) ∪ (a, 0). If

UF\T = ∪Ba : (a, 0) ∈ F \T

then F \T ⊆ UF\T where UF\T is an open subset of S, and so T = (S\UF\T ) ∩ F is aclosed subset of S. So T is closed in S.

In the following example, we show that the Moore plane is not normal. To follow thesolution presented, we need to remember that, if T is a set and P(T ) is its power

set (the set of all subsets of T ), then the cardinality of P(T ) is 2|T | where |T | is thecardinality of T . Also it is important to know that if c = |R| and ℵ0 = |N|, then

2c > 2ℵ0 = c.

Example 10. Show that the Moore plane is not a normal space.

Solution : Let (S, τM) denote the Moore plane. We claim S is not normal. To showthis we will assume S is normal.

Let F = (x, 0) : x ∈ R and W = S \ F .

For each (a, 0) ∈ F , a basic open neighbourhood of (a, 0) is of the form

Ba = Bε(a, ε) ∪ (a, 0) for some ε

If T is a non-empty subset of F , then, from the previous example, both T and F \Tare closed in S.

Let

D = (Q × Q) ∩Wa countable dense subset of S.

The cardinality of P(D) is 2|D| = 2ℵ0 = c and the cardinality of P(F ) is 2|F | = 2c.

The contradiction that we seek is that |P(D)| ≥ |P(F )|.We supposed that S is normal, so for each non-empty set T of F , there exists in τMa pair of disjoint open neighbourhoods,

UT = ∪Ba : (a, 0) ∈ TUF\T = ∪Ba : (a, 0) ∈ F \T


of T and F \T , respectively.

For each T ∈ P(F ), let

DT = ∪Ba ∩D : Ba ⊆ UT = UT ∩D ∈ P(D)

DF\T = ∪Ba ∩D : Ba ⊆ UF\T = UF\T ∩D ∈ P(D)

where UF\T ∩ UT = ∅

Let f : P(F ) → P(D) be defined as

f(T ) = DT , f(F \T ) = DF\T , DF\T ∩DT = ∅

Claim. That f maps P(F ) one-to-one into P(D):

Suppose T1 and T2 are distinct elements in P(F ). Say, there exists (b, 0), such that(b, 0) ∈ T2\T1 ⊆ UF\T1

. It suffices to show that DT2\DT1 6= ∅.

Then Bb ∩D ⊆ Bb ∩DT2 ⊆ Bb ∩ UF\T1. Since DT1 ⊆ UT1 and UT1 ∩ UF\T1

= ∅, then

DT2\DT1 6= ∅. So DT2 6= DT1.

We conclude that f(T1) 6= f(T2), which means that f : P(F ) → P(D) is one-to-one

as claimed.

Then the cardinality of the domain of f , P(F ), is less than or equal to the cardinality

of its codomain, P(D). The cardinality of P(F ) is 2|F | = 2c while the cardinality ofP(D) is 2|D| = 2ℵ0 = c. Then from the claim we obtain 2c ≤ c. This contradicts the

fact that c < 2c. The source of our contradiction is our supposition that S is normal.

So the Moore plane is not normal.

10.7 Topic: The embedding theorem revisited.

We remind the reader of the statement in theorem 7.17, titled, The embedding theorem

I, presented earlier as an application of Cartesian products. Assuming that the set,F = fi : i ∈ J, of functions separates points and closed sets of a space, S, and

e : S → ∏

i∈J Xi is an evaluation function with respect to F on S, the Embeddingtheorem I states that . . . ,

“. . . the evaluation map e, with respect to F , embeds a homeomorphic copy,

e[S], of S in the product space,∏

i∈J Xi where each of the factors, Xi, ofthe product is the codomain of the function, fi, in F”.

We now have tools which will allow us to enhance the Embedding theorem I further.

But first, we must do a bit of groundwork with a preliminary lemma.


In the following lemma we prove an interesting characterization of a completely regu-lar space. It shows that the topology of any completely regular space is precisely the

“weak topology” induced by the family, C∗(S), of all real-valued continuous boundedfunctions.

Lemma 10.15 Suppose (S, τ) is a T1 topological space and S = C∗(S), represents the

set of all continuous bounded real-valued functions on S. The topological space, (S, τ), iscompletely regular with respect to τ if and only if its topology, τ , is the “weak topology”

induced by S .

Proof : We are given that (S, τ) is a T1 topological space. Recall that

τS = f←[U ] : f ∈ S , U an open subset of R

is the weak topology generated by S .

( ⇒ ) Suppose (S, τ) is a completely regular space. To show that τ is the weak topol-

ogy induced by S it suffices to show that τS = τ .

Claim #1: τ ⊆ τS : Let x ∈ V ∈ τ . We claim that V ∈ τS . Since S is completely

regular, there exists f : S → [0, 1] such that x ∈ f←[0] and S \V ⊆ f←[1].Then f←[ [0, 1/3) ] ∈ S , an open neighbourhood of x entirely contained in V . Then,V ∈ S ⊆ τS . This means τ ⊆ τS . As claimed.

Claim #2: τS ⊆ τ : See that τS is the weakest topology induced by S = C∗(S).Let V ∈ τS . Then there exists and open set, U ∈ R, such that f←[U ] = V for some

f ∈ C∗(S). But for every function f ∈ C∗(S), f←[U ] is open in S with respect to τ .Then f is continuous on S with respect to τ . So V ∈ τ . So τS ⊆ τ , as claimed.

Combining both claims we obtain τS = τ .

So τ is the smallest topology on S such that every function in S is continuous on S,as required. We are done with ⇒.

( ⇐ ) Suppose τ is the weak topology on S, induced by the family of functions, S .

That is,

τ = V ⊆ S : V = f←[U ], for some open U in R, f ∈ S

We are required to show that (S, τ) is a completely regular space.

Let W be a non-empty open subset of S and x ∈W . To show complete regularity we

must construct a continuous function, h : S → [0, 1], which separates x from S\W .


The following family of open sets

M = f←[Z] : some f ∈ S , Z is a subbase element of R

forms a subbase for τ . (Note that subbase elements for R are of the form (ai,∞) or(−∞, ai).)

Since τ is generated by the subbase M , there exists finitely many subbase elements,Vi : i ∈ F = 1, . . . , n ⊂ M , such that

x ∈ ∩Vi : i ∈ F = 1, . . . , n ⊆W (∗)

where Vi = f←[Ui], then Ui is of the form (ai,∞) or (−∞, ai). To simplify our

expression, note that, if Uj = (−∞, aj) then

f←j [Uj] = f←j [(−1)(−aj,∞)] = (−f)←j [(−aj,∞)]

so we can assume that all Ui’s are of the form Ui = (ai,∞) after having made therequired adjustments on the functions, fi. So (∗) can be expressed as

x ∈ ∩f←i [(ai,∞)] : i ∈ F ⊆W (∗∗)

For each i in F = 1, . . . , n, let ti : S → R be defined as: ti = fi − ai.

See that ti > 0 on f←i [(ai,∞)]. So x ∈ t←i [(0,∞].Also see that ti(x) ≤ 0 everywhere else, including on S\W .

For each i in F = 1, . . . , n, let

t∗i = ti ∨ 01

So t∗i ≥ 0, and

x ∈ ∩t∗i←[(0,∞)] : i ∈ F ⊆W

We define t : S → R, as

t(x) = t∗1(x)t∗2(x)t

∗3(x) · · · t∗n(x)

We then have a real-valued function such that x ∈ t←[(0,∞)] ⊆W and S\W ⊆ t←(0).We restrict the range of the function: Suppose t(x) = δ. Then h = (t/δ)∧ 1 2 maps Sinto [0, 1] and x ∈ h←(1). So h separates x and S\W .

This allows us to conclude that S is completely regular.

1(f ∨ 0)(y) = max f(y), 0(y)2(t/δ) ∧ 1)(y) = min(t/δ)(y), 1(y)


In the Embedding theorem II statement, we will speak of a “cube”. When one casuallyspeaks of a “cube” a three dimensional space, [a, b]3, comes to mind. In topology,

unless it is otherwise specified, the word “cube” refers more generally to any finite orinfinite Cartesian product,

∏

i∈J [ai, bi]

of closed intervals, [ai, bi] in R. Since any closed and bounded interval [a, b] is homeo-morphic to the closed unit interval, [0, 1], then, by theorem 7.13, any cube,

∏

i∈J [ai, bi],

is homeomorphic to a product space,∏

i∈J [0, 1], of the unit interval [0, 1].

Theorem 10.16 The embedding theorem II. Suppose S is a T1 topological space. A space,

S, is completely regular if and only if S can be homeomorphically embedded in some cube,∏

i∈J [ai, bi].

Proof : We are given that (S, τ) is a T1 topological space.

( ⇐ ) Suppose S is embedded in a cube,∏

i∈J [ai, bi]. We are required to show that S

is completely regular.

Since each [ai, bi] is a metric space, then, by theorem 10.4, each [ai, bi] is completelyregular. By theorem 10.6, the cube,

∏

i∈J [ai, bi], is completely regular. Since S can

be viewed as a subspace of∏

i∈J [ai, bi], by theorem 10.5, S is completely regular.

( ⇒ ) Suppose S is completely regular. We are required to show that S is embeddedin a cube.

Two facts follow from S being completely regular.

- By lemma 10.15, the topology on S, is the weak topology induced by C∗(S).

- Since S is completely regular, C∗(S) separates points and closed sets of S.

So, by the Embedding theorem I, the evaluation map,

e : S →∏

i∈J

[ai, bi]

induced by C∗(S), homeomorphically embeds S into the product space,∏

i∈J Xi where

each of the factors, Xi, of the product is the codomain of the function, fi, in F .

Each function, fi, in C∗(S) is bounded and so maps S into some closed interval, [ai, bi],of R, so Xi ⊆ [ai, bi].

So, the evaluation map, e : S → ∏

i∈J [ai, bi], induced by C∗(S), homeomorphically

embeds S into the cube,∏

i∈J [ai, bi].


Since∏

i∈J [ai, bi] and∏

i∈J [0, 1] are homeomorphic spaces (by theorem 7.13) then thecompact space,

∏

i∈J [0, 1], contains a homeomorphic copy of S.

Notation: If S is a space and f : S → [0, 1] then f ∈ [0, 1]S =∏

x∈S[0, 1].

10.9 Topic: On P -points and P -spaces.

Suppose S is a completely regular space. We know that infinite intersections of opensubsets need not be open. We have referred to such sets as Gδ’s. In fact some Gδ sets

are closed (but not open). Some Gδ-sets are points while others may have non-emptyinterior in S. We will be investigating completely regular spaces in which the follow-

ing property is satisfied: If G is a Gδ which contains a point p then p ∈ intSG. Forexample, R would not be such a space since 1 = ∩(1− 1/n, 1 + 1/n) : n ∈ N\0and intS1 = ∅. In to this concept we introduce the following definitions.

Definition 10.17 Let S be a completely regular topological space. We will say that a

point p in S is a P -point in S if, for any Gδ-set B containing p, p ∈ intSB. If every pointin S is a P -point then we will say that S is a P -space.

We list a few properties by P -spaces.

Theorem 10.18 Completely regular spaces P -spaces are zero-dimensional spaces.

Proof : Suppose S is a completely regular P -space.

Let x ∈ S and U0 be some open neighbourhood of x. Since S is regular, there exists

an open subset, U1, of S such that

x ∈ U1 ⊆ clSU1 ⊆ U0

We now inductively construct the decreasing family, Ui : i ∈ N, of open neighbour-

hoods of x where x ∈ Ui ⊆ clSUi ⊆ Ui−1. Let

B = ∩Ui : i ∈ N = ∩clSUi : i ∈ N


a closed Gδ-set containing x. Since S is a P -space then x belongs to the interior of B.Then, for any x ∈ B, x ∈ intSB. This means B ⊆ intSB. So the Gδ, B, is a clopen

set containing x which is itself contained in U0. Then the set of all clopen subsets ofS then forms a base for the open subsets.

Then S has a base of clopen sets. So S is zero-dimensional.

Recall that any zero-set of S is a Gδ. So in a completely regular P -space space every

point is in the interior of every zero-set which contains it. So, has shown in the proofabove, every zero-set in such spaces are clopen. This is the case for all cozero-sets of

S, also. As shown in theorem 10.11, in a completely regulars space, the family of allcozero-sets form a base for open sets. It follows that in completely regular P -space

all zero-sets form a base for open sets.

On the other hand every clopen set is a Gδ is a set equal to its interior. So S would

be a P -space.

So,

. . . , at least for completely regular spaces, P -spaces are precisely those spacesfor which Z[S] forms a base for open sets.

Theorem 10.19 A pseudocompact P -spaces is finite.

Proof : The proof is differed to theorem 23.7 where we will have at our disposal extra tools.

Concepts review:

1. Define “dyadic rationals” used in the proof of Urysohn’s lemma.

2. State Urysohn’s lemma.

3. What are we referring to when we speak of a Urysohn separating function for twoclosed sets. Describe such a function.

4. Define a completely regular space (Tychonoff space).

5. How does a completely regular space differ from a T3 12-space?

6. Is a normal space necessarily completely regular? Is a completely regular space nec-essarily regular? Why?


7. Define a perfectly normal space.

8. What are we referring to when we speak of a zero-set in a topological space?

9. It was proven that, in a certain type of space, Gδ’s and zero-sets mean the same thing.

What type of space did we refer to?

10. Provide an example of a completely regular space that is not normal.

11. Provide an example of a perfectly normal topological space.

12. Provide an example of a normal space which is not perfectly normal.

13. Describe the Moore plane and its topology.

14. Describe an ordinal space and its base for open sets.

15. Is the completely regular property hereditary?

16. Suppose S =∏

i∈J Si. If S is completely regular does it follow that each Si is com-pletely regular?

17. Suppose S =∏

i∈J Si. If each Si is completely regular does it follow that S is com-

pletely regular?

18. The completely regular topology is the weak topology induced by which family of

functions?

19. The evaluation map, e, which is used embed S into a cube is with respect to whichfamily of functions?

20. Are continuous images of a completely regular space always completely regular?

EXERCISES

1. Let (S, τS) be a normal topological space which contains a subset T which is an Fσ.

Verify whether T is a normal subspace.

2. Are there any normal spaces which are not separable? If so, show one.

3. Is “perfectly normal” a hereditary property?

4. Is it true that a topological space, (S, τS), is normal if and only if any two non-emptydisjoint closed subsets have disjoint closed neighbourhoods? Prove it.


5. Let (S, τS) be a normal topological space which contains the non-empty closed subsetF . If U is an open neighbourhood of F , show that F ⊂ V ⊂ U for some Fσ-set, V .

6. Let (S, τS) be a countable completely regular space. Show that S is normal.

7. Let (S, τS) be a completely regular topological space. Let V be an open neighbourhood

of a point x. Show that there exists a Urysohn separating function, f : S → [0, 1], forx and S\V if and only if x is a Gδ.

8. We showed that closed subspaces of normal spaces are normal. A topological space(S, τS) is said to be completely normal if and only if every one of its subspaces is a

normal space. Show that S is completely normal if and only if, whenever clSU ∩V =∅ = U ∩ clSV , there exists disjoint open neighbourhoods for U and V , respectively.

Note: To prove “⇒” consider S\(clSU ∩ clSV ).

Part IV

Limit points in topological spaces

225

Part IV: Limit points in topological spaces 227

11 / Limit points in first countable spaces.

Summary. In this section we will investigate whether the notion of limit points

and there sequences can be applied to topological spaces. We know that, in met-ric spaces, closed subsets can be characterized by the limit points they possess.

Also, continuous functions can be characterized in terms of how they act onsequences and their limits. We will show that first countable topological spaces

have sequences and limits points which can be handled in the usual way.

11.1 Introduction.

Limit points of sets play an important role in the study of normed and metric spaces.

In particular, they are involved in characterizations of closed sets and the sequentialdefinition of continuity. Up to this point limit points are assumed to be the limit of

sets called sequences, all indexed by countably infinite linearly ordered sets. We willinvestigate limit points in a more general context of topological spaces.

11.2 Sequences and limit points in topological spaces.

We will first start with the formal definition of a sequence of elements in a topologi-cal space. It is analogous to the definitions introduced in normed vector spaces and

metric spaces.

Definition 11.1 Let S be any topological space.

a) Sequence. A sequence in S is a function, f : N → S, mapping N into S. It can bedenoted as, f(i) : i = 0, 1, 2, 3, ..., x0, x1, x2, x3, . . . , xn, . . . where xi = f(i) for

each i ∈ N, or, xi : i = 0, 1, 2, 3, . . . or, simply, xi.For Tj = i ∈ N : i ≥ j, we will say that f [Tj] = xi : i ≥ j is a “tail end” of thesequence xi.

b) Convergence. A sequence, xi, is said to converge to a point p in S if and only if,for every open neighbourhood U of p, there exists, N , such that, when n > N , xn

belongs to U . We also say “...if and only if every open neighbourhood of p containsa tail end of the sequence”.1 The expression, xi → p, is used to indicate that the

sequence converges to p.

c) Limit point of a sequence. If xi → p, then p is a limit point of the sequence xi.d) Limit point of a set. If F is a non-empty set, we say that p is a limit point of the

set, F , if and only if p is the limit point of some sequence, xi, which is entirelycontained in F .

1The expressions “xi converges to p if the sequence eventually belongs to every neighbourhood, U ofp” is also used.

228 Section 11: Limit points in first countable spaces

e) Accumulation point of a sequence. We say that k is an accumulation point of thesequence, A = f(i), if every open neighbourhood of k has a non-empty intersection

with every tail end, f [Tj] = xi : i ≥ j, of A.

The reader is no doubt familiar with the following characterization of “closed set”

given in the study of metric spaces:

“A non-empty subset, F , of the metric space, (M, ρ), is closed in M , if and

only if, F contains the limit point of each and every sequence contained inF .”

Also, we have another definition which involves sequences, namely the sequential def-

inition of continuity for functions mapping a metric space to a metric space.

“A function, f , mapping a metric space, M1, into a metric space, M2, is

said to be continuous at a point p if and only if, whenever p is the limitpoint of a sequence, xi, in M1 f(p) is the limit point of the corresponding

sequence, f(xi), in M2.”

Can we easily generalize the statements from “metric spaces” to non-metrizable topo-logical spaces? We will show that the transition from sequences in metric spaces to

sequences in topological spaces flows fairly smoothly whenever the topological spacesare first countable; that is, if every point has a countable neighbourhood base.

We begin by providing the following definition.

Definition 11.2 Let S and T be topological spaces.

We say that a function f : S → T is sequentially continuous at p if and only if, wheneverp is limit point of a sequence, xi, in S, then f(p) is the limit point of the corresponding

sequence, f(xi) in T .

Theorem 11.3 Let (S, τS) and (T, τT) both be first countable topological spaces.

a) A non-empty subset, F of S, is closed in S if and only if every limit point of F belongsto F .

b) A function mapping S into T is continuous on S if and only if it is sequentiallycontinuous at each point, p, in S.


Proof : Let (S, τS) and (T, τT) both be first countable topological spaces.

a) ( ⇒ ) Suppose F is a closed subset of S and let p be a limit point of F .

Then, by definition, there is a sequence, xi ⊆ F , which converges to p. Supposep ∈ S\F . Then, if U is any open neighbourhood of p, U ∩ (S\F ) contains a tail end

of the sequence xi. Since this is impossible, then p ∈ F .

( ⇐ ) Suppose F contains all its limit points.

Suppose there exists a point p ∈ clSF \F . By hypothesis, there exists, Ui : i ∈ N,a countable open neighbourhood base of p. If, for n ∈ N, Vn = U1 ∩ U2 ∩ · · · ∩ Un,

then Vn : n ∈ N is a countable neighbourhood base of open sets of p, such thatVn+1 ⊆ Vn. Then p = ∩Vn : n ∈ N.

Since p is a boundary point of F , we can choose, for each i ∈ N, xi ∈ Vi ∩ F (Choice

function). Since every Vi contains a tail end of the sequence xi, then p is a limitof the sequence in F ; so it is a limit point of F . By hypothesis, p, being a limit of

F , should belong to F . This contradicts, p ∈ clSF \F . To avoid the contradiction wemust have, clSF \F = ∅, which implies F = clSF , and so F is closed.

b) ( ⇒ ) Suppose f : S → T is continuous on S and let p ∈ S. Also, suppose that p isthe limit point of a sequence, xi, in S.

If V is an open neighbourhood of f(p), then there exists an open neighbourhood Uof p such that f [U ] ⊆ V .

Then U contains some tail end, say xi : i > N, of the sequence xi. Thenf(xi) : i > N ⊆ f [U ] ⊆ V . So V contains a tail end of the sequence f(xi). Since

V is an arbitrary open neighbourhood of f(p), then f(xi) → f(p). This establishesthe proof of ( ⇒ ).

( ⇐ ) Suppose f : S → T is such that xi → p in S implies f(xi) → f(p).

We are required to prove that f is continuous at p. By hypothesis, there exists,

Ui : i ∈ N, a countable open neighbourhood base of p, where Ui+1 ⊆ Ui. Then∩Ui : i ∈ N = p.We will proceed by contradiction. That is, suppose f is not continuous at p. Thenthere exists an open neighbourhood, V , of f(p) such that, for each i, f(xi)\V 6= ∅.

Then, for each j, there exists yj ∈ Uj such that f(yj) 6∈ V . Then yj → p; but thetail end of the sequence, f(yj), is excluded from V and so can’t converge to f(p),

a contradiction. So f is continuous at p.

The above theorem allows us to say that, in a first countable space S, if F is a subset

of S, then the closure, clSF , of F is the set of all the limit points of F . Also, in firstcountable spaces, to say that a function, f : S → T , is continuous on S is to say that


f is “sequentially continuous” at each point p in S.

11.3 Subsequences of a sequence.

Sequences in a first countable topological space don’t always converge to a point. But

infinite subsets of a sequence are themselves sequences and might still “converge”.

Definition 11.4 Let S be a first countable topological space. Let

A = f(i) : i = 1, 2, 3, . . . , = xi

be a sequence in S. Suppose g : N\0 → N\0 is a strictly increasing function. Then the

sequence,

B = f(g(i)) : i = 1, 2, 3, . . . , = xg(i)is a called a subsequence of the sequence A. By strictly increasing we mean i > j ⇒ g(i) >

g(j).

Note that the notation, xg(i), means that xg(i) is the element of the sequence xiwith the index number g(i). The terms in a subsequence always respect the order in

which they appear in the sequence.

Example 1. If A = 11 ,

12 ,

13 ,

14 , . . . ,

1n , . . . , , where an = 1/n, and g(n) = 2n, then

B = ag(n) = a2n is the subsequence

B =

1

2,1

4,1

6, . . . ,

1

2n, . . . ,

Every element of a subsequence must always be an element of the sequence from which

it is derived. The key is that the index function must be strictly increasing. Also,remember that a sequence always has infinitely many terms (but not necessarily have

infinitely many distinct elements). In this example, the sequence 12 ,

12 ,

13 ,

13 ,

14 ,

14 , . . . ,

is not a subsequence of the sequence A since the indexing function used is not strictlyincreasing.

Example 2. Let

A = 1, 1, 2, 1, 3, 1, 4, 1, . . . ,

Then B = 1, 2, 3, 4, 5, . . . , and C = 1, 1, 1, 1, 1, 1, 1, . . . , are both subsequencesof A. The subsequence B does not converge and does not have a subsequence which


converges. On the other hand, the subsequence C of A converges to 1.

So the element, 1, is, by definition, an accumulation point of the sequence A.

Note that, if we view the sequence A as simply the “set of all non-zero natural num-bers” then 1 is not, by definition, a “cluster point”1 of the set A even though 1 is an

accumulation point of A. It will be understood in this textbook that “cluster point”refers to a point in relation to some set. While accumulation point is used in relation

to sequences (or later, to nets).Note. Some authors use “cluster point” or “accumulation point” interchangeably,

which may cause some confusion in some theorem proofs. So readers should verifycarefully how authors define both words.

Theorem 11.5 Let A = f(i) : i ∈ N = xi be a sequence in a first countable topologicalspace S. Then p is an accumulation point of the sequence, A, if and only if it is a limit

point of some subsequence, f(g(i)) : i ∈ N = xg(i), of A.

Proof : Let A = f(i) : i ∈ N = xi be a sequence in a first countable topological space S.

( ⇐ ) Suppose A has a subsequence, T , which converges to the point p. Then every

open neighbourhood of p contains a tail end of T , and so intersects a tail end of A.So p is an accumulation of A.

( ⇒ ) Suppose p is an accumulation point of A. We are required to find a subsequenceof A which converges to p.

Since S is first countable, p has a countable open neighbourhood base,

Bp = Bi : i = 1, 2, 3, . . .

such that Bi+1 ⊆ Bi for all i. If Tj = i : i ≥ j is a tail end of N, then for all i andj, f [Tj] ∩ Bi 6= ∅.

We inductively construct a subsequence, B, of A which converges to p. Let g(1) bethe smallest number in T1, such that f(g(1)) ∈ B1. Suppose that, for g(1) < g(2) <

· · · < g(j), f(g(i)) ∈ Bi. Let g(j + 1) be the smallest number in Tg(j+1) such thatf(g(j + 1)) ∈ Bj+1. We thus obtain a subsequence

B = f(g(i)) : i = 1, 2, 3, . . . ,

of A, which converges to p, as required.

1Reminder: In the chapter on closures on page 49 we defined a cluster point of a set A as a point p suchthat any open neighbourhood of p contains some other point x in A.


We will later see that, if S is not first countable, the statement above will fail to be

true, in general. So we will have to redefine our notions about “sequence”.

Concepts review:

1. If S is a first countable topological space what does it mean to say that xi is asequence in S?

2. If S is a first countable topological space what does it mean to say that p is limit

point of xi, a sequence in S?

3. If S is a first countable topological space what does it mean to say that p is a limit ofa subset, U , in S?

4. If S is a first countable topological space and the subset, U of S, contains all its limitpoints state a topological property possessed by U .

5. If S is a first countable topological space and F is a closed subset of S what can we

say about the set of all limit points of F?

6. If S is a first countable topological space define “the closure of F” in terms of limit

points.

7. Define a subsequence of a sequence in a first countable topological space.

8. Define an accumulation point of a sequence in a first countable topological space.

9. If p is and accumulation point of a sequence is p a limit point of the sequence? Explain.

EXERCISES

1. Let (S, τS) be a first countable topological space. Define the “interior of the subsetF” in terms of limit points.


12 / Limit points of nets.

Summary. In this section, we will expand the definition of “limit point” so

that it applies to arbitrary topological spaces, including those which are not firstcountable. The indexing set of all natural numbers used for sequences will be

substituted by a more general indexing set called “directed set”. The sets whoseelements are ordered by a “directed set” will be called “nets”. Nets will serve as

a tool which will assist in recognizing closed subsets and continuous functions inarbitrary topological spaces. They will also, eventually, be used to identify setswhich are “compact” once this concept has been properly defined.

12.1 Directed sets.

Suppose “≤” is a partial order relation on a set, S. That is,

– for all x ∈ S, x ≤ x, (reflexive property)

– if x ≤ y and y ≤ x then x = y, (anti-symmetric property)

– if x ≤ y and y ≤ z then x ≤ z (transitive property)

The second condition (referred to as the “anti-symmetric property”) is not absolutely

required in what follows, but it is easier to set the stage in this way to introducethe notion of a “directed set” whose definition requires some sort of predefined order

relation on a given set.

Definition 12.1 Let D be a non-empty infinite set on which is defined a partial orderingrelation “≤”. Suppose the partially ordered set, (D,≤), also satisfies the property:

If x, y ∈ D, there exists z ∈ D such that x ≤ z and y ≤ z (The directed set property)

Then we refer to (D,≤) as a directed set.

We already know of a directed set, namely, (N,≤). Simply see that, if m, n ∈ N, thereexists k ∈ N such that m ≤ k and n ≤ k. So a directed set is a generalization of the

linearly ordered countable sets we normally use as indexing sets for sequences. As wecan see, the definition of directed set does not require that its order relation, “≤”, be

linear. It is usually appears in the form of a partial ordering.

Example 1. Given a set, S, its power set, (P(S),⊆), is partially ordered by the inclu-sion relation, ⊆. This order relation can be seen to satisfy the directed set property.

234 Section 12 : Limit points of nets

Then (P(S),⊆) is a directed set where ∅ ⊆ T for all T ∈ P(S) and T ⊆ S for allT ∈ P(S).

Example 2. Consider the set of all finite partitions, Part, of the closed interval [0, 10]of real numbers. For example, M = [0, 4]∪ [4, 7]∪ [7, 10] ∈ Part. We define a partialorder on Part as follows: M ≤ K if and only if K is a finer partition then M . That

is, each closed interval of M is a union of closed intervals in K. For example,

M = [0, 4]∪ [4, 7]∪ [7, 10] ≤ [0, 2]∪ [2, 4]∪ [4, 7]∪ [7, 9]∪ [9, 10] = K

It is easy to verify that “≤” respects the directed set property on Part. For example,suppose M = [0, 5]∪ [5, 10] and J = [0, 2]∪ [2, 10]. If

K = [0, 2]∪ [2, 5]∪ [5, 7]∪ [7, 10]∪ [9, 10]

then M ≤ K and J ≤ K. See that [0, 10] ≤ W for all W ∈ Part. Then (Part,≤) is adirected set.

We will now show how directed sets can be used to establish a direction in arbitrarysubsets of a topological space. The definition of a net, in terms of a function, is similar

to the definition of a sequence except that the index set is not necessarily linear andso can be “larger” than N.

Definition 12.2 Let S be a set.

a) A net in S is a function, f : D → S, mapping a directed set, (D,≤), into S. It is

usually denoted as

f(i) : i ∈ D = f(i)i∈D = xii∈D

or, simply, by xi whenever it is obvious that the index set is the directed set, D.

b) Tail-end of N : Let N = f(i)i∈D be a net in S, k ∈ D and Tk = i ∈ D : i ≥ k.We will refer to a subset of the form, f [Tk] = f(i) : i ≥ k as a tail end of the net,

N . We can obtain many tail ends of N by varying the value of k in D.

c) Convergence : Suppose (S, τ) is a topological space and N = xii∈D is a net in S,ordered by a directed set, D. The net, N , is said to converge to a point p in S (with

respect to τ and the given ordering) if and only if every open neighbourhood, U ofp, contains some tail end, f [Tk], of the net N . That is, there exists k ∈ D such that,

whenever j ≥ k, then xj ∈ U . The expression,

xi → p

is used to indicate that this net converges to p.


d) Limit point : If a net, xii∈D in a topological space, S, converges to the point p, wesay that p is a limit point of the net xii∈D.

e) Limit point of a set :If F is a non-empty set in a topological space, S, we say that p

is a limit point of the set, F , if and only if p is the limit point of some net, xii∈D,which is entirely contained in F .

f) Accumulation point : Suppose (S, τ) is a topological space. We will say that y is anaccumulation point of the net, N = f(i)i∈D in S, or that A accumulates at y, if for

every tail end f [Tk] = f(i) : i ≥ k, of N and every open neighbourhood By of y,

f [Tk] ∩ By 6= ∅

Since (N,≤) is a directed set, then a sequence, xi : i ∈ N , is a particular kind of net.

Example 3. Consider the net (sequence),

N = n+ (−1)nn : n ∈ N\0 = 0, 4, 0, 8, 0, 12, . . . ,

with directed set N\0 in R (with the usual topology). This net does not have a

limit point, since a sufficiently small open neighbourhood, (p − ε, p + ε), of any pcannot contain a tail end of N . However, any neighbourhood, (−ε, ε), of the number,

0, intersects every tail end of the net (since all tail ends of N contain zeroes). So 0 isan accumulation point.2

It is important to remember that limit points or convergence of a net can only be

discussed with respect to a particular topology in mind. Given a net, N = f(i)i∈D,in a topological space, if we change the topology on the set, then we may be alteringits convergence properties. However, the directed set, (D,≤), which indexes the net

elements is not topologized.

By definition, nets are indexed subsets of a topological space. If N is a net in S we

stated that N may have a limit point, p, and may have an accumulation point, y. Ifwe say that the net, N , “has” a limit point, p, we do not imply that p ∈ N . Also, by

definition, every neighbourhood of a limit point p contains a tail end of N so everyneighbourhood of p intersects a tail end of N and so, by definition, every limit point,

p, of N is an accumulation point of N . Of course, an accumulation point of N neednot be a limit point, as the example above shows. So accumulation points of a net

2In some texts, authors use the word “cluster of a net” instead of “accumulation point of a net”. Thismay lead to confusion since the word “cluster” is already defined in the context, “cluster point of a set”(see page 49). Note that the net N in this example, when viewed as a set, (according to definitions) has nocluster points but does have an accumulation point.


are of interest only when the net does not converge to a point in S.

Example 4. Consider the product space, S = [0, ω1)× [0, ω1) and S∗ = [0, ω1]× [0, ω1]where ω1 is the first uncountable ordinal. Let p = (ω1, ω1). The set,

B = (α, ω1]× (β, ω1] : α < ω1, β < ω1

forms a neighbourhood base of open sets for the point p ∈ S∗. Construct a net in S

whose limit point is p.

Solution : We are given B = (α, ω1] × (β, ω1] : α < ω1, β < ω1 forms a neighbour-

hood base of open sets for the point p ∈ S∗. For U, V ∈ B we define U ≤ V if andonly if V ⊆ U . Then “≤”, thus defined on B, is a partial ordering that satisfies the

directed set property, and so forms a directed set, (B,≤).

Let f : B → S be a choice function which maps each U ∈ B to one point of ourchoice, say f(U) = xU in U ∩ S. We thus obtain, by definition, a net

f(U) : U ∈ B = xU : U ∈ B

an uncountable subset of the product space, S = [0, ω1)× [0, ω1), with directed set B.

We claim that, p = (ω1, ω1) is a limit point of the set, S. It suffices to show that,xU → p.

Let K = (α, ω1]× (β, ω1] be an arbitrary open neighbourhood of p in S∗. There existsγ, µ such that V = (γ, ω1] × (µ, ω1] ⊂ K. Then V ≥ K and so xV ≥ xK . That is,

a tail end, f [TK ] = xU : U ≥ K is contained in the open neighbourhood, K of p.Hence the tail end, f [TK ], of the net, xU : U ∈ B, in S is a subset of the open

neighbourhood, K, of p. Hence xU → p.

Since S contains a net, xU : U ∈ B, which converges to p then p is a limit point ofthe set, S, as claimed.

12.2 Closed sets and continuity in terms of nets.

We now present the results which describe the closure of a set and the continuity ofa function in terms net convergence.

Theorem 12.3 Let (S, τS) and (T, τT) both be topological spaces.

a) Let E be a subset of S. A point p belongs to clSE if and only if there is a net in Ewhich converges to p.


b) A function, f : S → T , mapping S into T is continuous at p in S if and only ifwhenever p is limit point of a net, xi, in S then f(p) is the limit point of the corre-

sponding net, f(xi) in T .

Proof : Let (S, τS) and (T, τT) both be topological spaces.

a) ( ⇒ ) Suppose E ⊆ S and p belongs to clSE. We can index all open neighbourhoodsof p and partially order them by reverse inclusion to form the directed set D = U : U

is an open neighbourhood of p. Then, for each open neighbourhood, U , of p, we can,by a choice function, choose xU in U ∩ E. We obtain a net, xU : U ∈ D, in E,

which converges to p.

( ⇐ ) Suppose xi is a net in E which converges to p.

Then, by definition of “convergence”, each neighbourhood of p intersects E. Then pcannot belong to the open set S\clSE. So p ∈ clSE.

b) ( ⇒ ) Suppose f : S → T is continuous at p ∈ S and p is a limit point of a net,A = xi : i ∈ I. Let U be an open neighbourhood of f(p) in T .By definition of continuity, f←[U ] is an open neighbourhood of p in S. Since the net

A converges to p then, by definition, f←[U ] contains a tail end, xi : i ≥ k, of A.

Then f [xi : i ≥ k] = f(xi) : i ≥ k is a tail end of f [A] which belongs to U .

So f [A] converges to f(p).

( ⇐ ) Left as an exercise for the reader.

The above result shows that, given a subset, F , of a topological space, S, we can use

nets as a tool to determine the closure, clSF , of the set F .

a) A set F in a topological space, S, is closed in S if and only if F contains the limitpoint of every net which is contained in F .

b) Equivalently, p is a limit point of a closed subset of S if and only if every openneighbourhood of p intersects F .

We choose sequences or nets depending on the nature of the topological space consid-

ered.

12.3 Subnet of a net

We know that sequences have special subsets called “subsequences”. Nets can have

proper subsets which are analogous to subsequences. We formally define this conceptin such a way that it generalizes the concept of a subsequence.


Definition 12.4 Let S be a set and N = f(i) : i ∈ D be a net in S with directed set D.

Suppose Tj = i : i ≥ j denotes a tail end of D (with “least” element j).

a) Cofinal : Suppose U is a subset of D such that for each d ∈ D there is an elementu ∈ U such that u ≥ d. Then we say that

“U is cofinal in D”

Verify that, if U is cofinal in D, then U is also a directed set. By extension, we willsay that M = f(i) : i ∈ U is a cofinal net in N . In this case, M intersects every tail

end of N . We can then say that, “if every neighbourhood U of y, U ∩N is cofinal inN then y is an accumulation point of N”.

b) Subnet of a net : Let D be a directed set (possibly different from D).

Suppose g : D → D is an increasing function on D (i.e., i ≥ j ⇒ g(i) ≥ g(j) ) so thatg[D ] is cofinal in D; then (fg)[D ] is cofinal in N .

LetB = f(g(i)) : i ∈ D = xg(i) : i ∈ D

be a cofinal subset of N . Then B is a called a subnet of the net N .

A few remarks on the notion of a subnet. In the above definition the function g pointsto a cofinal subset of the directed set D while the function f associates this subsetto a cofinal subset (the subnet) of the net N . We alert the reader to the fact that,

in the usual definition of a subsequence we require that the function g be “strictly”increasing. So g chooses not only a cofinal subset of N but a ”subsequence” of N. The

definition of “subnet” requires that g be only “increasing”. So g may be pointing toa cofinal subset ofD, but some of the elements of this cofinal subset may be repeating.

We verify how this may make a difference. If f(n) = 1/n, so that the sequence,

A = f(n) = 1, 1/2, 1/3, 1/4, . . . , , and g(n) = 1, 1, 2, 2, 3, 3, . . . , , then we endup with a subnet B = f(g(n)) = 1, 1, 1/2, 1/2, 1/3, 1/3, . . ., a subset of A, but

clearly not a subsequence of A (at least not as we viewed it previously). So B is asubnet of A, but B does not satisfy the definition of a subsequence of A.

Given our experience with subsequences, we expect that, if a net, N , in the space S,

has an accumulation point x in S, the net will have a subnet B which will convergeto it. We confirm that this holds true in the next theorem.

Theorem 12.5 Let S be a topological space containing a net, N = f(i) : i ∈ D andu ∈ S. Then u is an accumulation point of N if and only if N has a subnet converging tou.


Proof : Let S be a topological space containing a net N = f(i) : i ∈ D with directed setD.

( ⇐ ) SupposeN = f(i) : i ∈ D = xi has a subnet B = f(g(i)) : i ∈ D = xg(i)which converges to u (where D is a directed set). We must show that for any open

neighbourhood Bu of u Bu is cofinal in N . Let Bu be any open neighbourhood of u.Since the subnet, B, converges to u, then Bu contains a tail end,

f [g[Tr]] = f(g(i)) : i ≥ r = xg(i) : i ≥ r

of B, where r ∈ D . Letf [Tk] = f(i) : i ≥ k

be some tail end of N . By definition, g[D ] is cofinal in D. Then there exist q ∈ D suchg(q) > max k, g(r). Then f(g(q)) ∈ f [Tk] ∩ f [g[Tr]] ⊆ Bu. So every tail end of N

intersectsBu. So Bu is cofinal inN . So, by definition, u is an accumulation point of N .

( ⇒ ) Suppose u is an accumulation point of the net N = f(i) : i ∈ D. We are

required to construct a subnet, B, of N which converges to u.

For every j ∈ D, let Tj = i ∈ D : i ≥ j. Let Bu be an open neighbourhood base ofu in S. By definition of accumulation point of N , if U ∈ Bu, U is cofinal in N . (That

is, U ∩ f [Tj] is non-empty, for all j ∈ D.)

Step 1 : Defining the set D. Consider the set

D = (i, U) ∈ D × Bu : f(i) ∈ U

Step 2 : Ordering the elements of D. We will order the elements of D with, “≤”, asfollows:

(i, U) ≤ (j, V ) ⇔ [ i ≤ j and V ⊆ U ]

Then (D ,≤) is a partially ordered set.

Step 3 : We claim that (D ,≤) is a directed set. Suppose (i, V ), (j, U) ∈ D and

W = V ∩ U . Since D is directed, there exists k ∈ D such that i ≤ k and j ≤ k andu ∈ W . Since u is an accumulation point of N , the tail end, f [Tk], intersects W and

so W contains some f(q) for q ∈ Tk. Then (q,W ) ∈ D where (i, V ) ≤ (q, W ) and(j, U) ≤ (q, W ). We then have a directed set, (D ,≤) .

Step 4 : Constructing the subnet B. Define g : D → D as, g(j, U) = j, for each

(j, U) ∈ D . Then g assigns each (i, U) to i in D such that f(i) ∈ U .

To show that (fg)[D ] is a subnet of N it suffices to show that it is cofinal in N . LetTk be a tail end of D. It suffices to show that g[D ] ∩ Tk is non-empty.

Let V be an open neighbourhood of u. Since u is an accumulation point of N , there ex-ists q ∈ Tk such that f(q) ∈ f [Tk]∩V . Then (q, V ) ∈ D . Then q = g(q, V ) ∈ g[D ]∩Tk.


So g[D ] is cofinal in D, as claimed.

Then f [ g[D ] ] is cofinal in N , so

B = f(g(i, Uu)) : (i, Uu) ∈ D

satisfies the definition of a subnet of N .

Step 5 : Claim that the subnet, B, converges to u. Let U ∈ Bu. To show that thesubnet B converges to u we are required to find a tail end of B which is entirely

contained in U . Let Tz be a tail end of D.

Since u is an accumulation point of N , U ∩ f [Tz] 6= ∅ so we can choose f(j) ∈U ∩ f [Tz] 6= ∅. Then (j, U) ∈ D .

Suppose (j, U) ≤ (k, V ). Then j ≤ k and f(k) ∈ V ⊆ U .So (fg)(k, V ) = f(k) ∈ U . So the tail end of all elements in B passed the point

(fg)(j, Uu)) belongs to Uu. So the subnet B converges to u, as claimed.

Then the subnet B converges to the accumulation point, u, as required.

Remark . In the above theorem, the net N may be either a sequence or an uncountablenet. The cardinality of N does not play a role in the proof, so the statement holdstrue for countable and uncountable nets.

12.3 A few examples.

A comment on a few infinite countable ordinals. Ordinals can also serve as an indexing

set for a net. Recall that there are two types of ordinals: Those ordinals, α, whichhave an immediate predecessor, say β, where β + 1 = α and limit ordinals, γ, where

γ = supα : α < γ = ∪α : α ∈ γ or, equivalently, γ is an ordinal which does notcontain a maximal ordinal. The following represent countable limit ordinals.

ω0 = 0, 1, 2, 3, . . . = [0, ω0)

2ω0 = 0, 1, 2, 3, . . . , ω0, ω0 + 1, ω0 + 2, . . . , ω0 + n, . . . , = [0, 2ω0)

3ω0 = 0, 1, 2, . . . , ω0, ω0 + 1, ω0 + 2, . . . , 2ω0, 2ω0 + 1, 2ω0 + 2, . . . , = [0, 3ω0)

...

nω0 = 0, 1, 2, . . . , ω0, ω0 + 1, ω0 + 2, . . . , (n− 1)ω0, (n− 1)ω0 + 1, . . . , = [0, nω0)

...

ω0ω0 = [0, ω0ω0) = 0, . . . , ω0, . . . , 2ω0, . . . , 99ω0 + 100, . . . , . . . ,


Those ordinals which are smaller than ω0 are simply the natural numbers. Ordinalssuch as ω0, 2ω0, 3ω0, . . . , nω0 . . . , , are all countably infinite limit ordinals.

With these facts in mind we present a solution for the question in the following ex-ample.

Example 5. Consider the set n + 1m : n ∈ N, m ∈ N\0. We will use the set of

ordinals, D = [ω0, ω0ω0), as an index to a subset of Q ∩ [0,∞). The set, D, need notbe topologized.

Define f : [ω0, ω0ω0) → Q ∩ [0,∞) as follows where, for n ≥ 1, m ≥ 0:

f(nω0 +m) = m+1

n+ 1

Then N = f(i) : i ∈ [ω0, ω0ω0) is a net in Q∩ [0,∞) with [ω0, ω0ω0) as directed set.

With this definition, the elements of the net are described as follows:

12 , 1

12 , 2

12, 3

12 , . . . ,

13 , 1

13, 2

13 , 3

13, . . . ,

1n , 1

1n , 2

1n, 3

1n, . . . ,

1n+1 , 1

1n+1 , 2

1n+1 , 3

1n+1 , . . . ,

Show that each natural number is an accumulation point of the net N . Then con-struct a subnet which converges to each natural number.

Solution: A tail end of [ω0, ω0ω0) is of the form

Taω0+b = [aω0 + b, ω0ω0) = nω0 +m : nω0 +m ≥ aω0 + bSo a tail end of N is of the form

f [Taω0+b] = f(nω0 +m) : nω0 +m ≥ aω0 + b = m+1

n + 1: nω0 +m ≥ aω0 + b

Let q be a natural number. We claim that q is an accumulation point of N .

Proof of claim: To see this, let (q − ε, q + ε) be a small open neighbourhood of q.

It suffices to show that f [Taω0+b] ∩ (q − ε, q + ε) 6= ∅.

Then there exists k > a such that q + 1/k ∈ (q − ε, q + ε). Then kω0 + q > aω0 + b,so f(kω0 + q) ∈ f [Taω0+b]. Since f(kω0 + q) = q + 1/k ∈ (q − ε, q + ε), then

f [Taω0+b] ∩ (q − ε, q + ε) 6= ∅. So q is an accumulation point of N , as claimed.

Since q is an accumulation point of N then, by the preceding theorem, N must contain

a subnet which converges to q. Define the function g : [ω0, ω0ω0) → [ω0, ω0ω0) as

g(nω0 +m) = nω0 + q (For all m.)

We see that g is increasing (but not strictly increasing) on [ω0, ω0ω0), since for any a

and b such that a < b,

g(aω0 +m) = aω0 + q < bω0 + q = g(bω0 +m) (So incrsg)

g←(aω0 + q) = aω0 +m : m ∈ N (So not strictly incrsg)


We also have that f(g(nω0 +m)) = f(nω0 + q) = q + 1n+1 .

Then

(fg)[D] = q +1

n + 1: n ∈ N

mapping D to a subnet q + 1n+1 : n ∈ N of N which converges to q as n→ ω0. As

required.

In the following example we witness how in certain circumstances the theory govern-

ing sequences and their subsequences can have shortcomings.

Example 6. Let S = [0, 1)× [0, 1) = (a, b) : a, b ∈ [0, 1). We will equip S with theusual topology inherited from R2. Let

D = (c, d) : c, d ∈ (0, 1), both c and d are irrationals

lexicographically ordered, untopologized. By “lexicographically ordered” we mean that

(e, k) ≤ (c, d) if e < c, and, when e = c, then k ≤ d.2 Then, with this ordering, (D,≤)will serve as a directed set for a net in S. The net,

N = f(c, d) : (c, d) ∈ D

in the space, S, is defined as follows: f(c, d) = (c, d). Then N linearly orders the

ordered pairs, (c, d), in the space S, where (0, 0) < (c, d)< (1, 1).

a) Is the point, p = (1, 1) in S, a limit point of the net, N?

b) Is the point, p = (1, 1) in S, an accumulation point of the net, N?

Solution :

a) Let Bε(1, 1) be an open neighbourhood of the point p = (1, 1) in S. SupposeT(a,b) is a tail end of D. Then f [T(a,b)] = T(a,b) is a tail end of N . Suppose (a, b) ∈Bε(1, 1). See that (a + 1−a

2 , b − 2ε) > (a, b) with respect to the lexicographic

ordering since b 6= 1 ⇒ 1−a2 6= 0. But (a + 1−a

2 , b − 2ε) does not belong toBε(1, 1). So no tail end of the net N = f(c, d) : (c, d) ∈ D belongs entirely in

Bε(1, 1). So p is not a limit point of the net, N .

b) We claim that p = (1, 1) is an accumulation point of the net, N . By the theoremabove it suffices to show that N has a subnet converging to (1, 1).

Let g : N → N be defined as: g(a, b) = (a, a) if a ≥ b and g(a, b) = (b, b) ifa < b. We see that g is an increasing function with respect to the lexicographic

ordering of the ordered pairs. We claim that g[D] is cofinal in D. For (a, b) ∈ D,

2This means, we order the ordered pairs by using the same principle as the one used to order words inthe dictionary.


if a < b, (b, b) > (a, b), if b < a, (a, a) > (a, b). Then for any (a, b) in D, thereexists an element in g[D] which is larger than (a, b). So g[D] is cofinal in D.

So fg : N → (a, a) : a ∈ [0, 1) defines a subnet of N . If Bε(1, 1) is an openneighbourhood of (1, 1) then (fg)[T(1− ε

2,1− ε

2)] ⊆ Bε(1, 1). So (fg)[D] converges

to (1, 1).

Example 7. Consider the directed set, (Part,≤), of all finite partitions of the closed

interval [0, 10] introduced in an example on page 234. Given two partitions P and Qin Part, P ≤ Q if Q is a finer partition than P . That is, each closed interval in P is a

union of closed intervals in Q.

Let f : [0, 10] → R be any function on [0, 10]. Let Pαn = Pi : i = 1..n represent a

partition of [0, 10]. Let pi be the width of the ith interval, Pi, of the partition Pαn .Let

Part = Pαn : Pαn is a partition of [1, 10]

and (Part,≤) be its representation as a directed set of partitions.

For a given partition, Pαn , let mi, li denote the two points of the ith interval, Pi, suchthat f(mi) is maximal at mi ∈ Pi, and f(li) is minimal at li ∈ Pi, respectively. Then

U(Pαn) =

n∑

i=1

f(mi)pi = xPαn

L(Pαn) =

n∑

i=1

f(li)pi = yPαn

both approximate the area beneath the curve of f on [0, 10]. Then U : Part → R and

L : Part → R define two nets,

NU = U(Pαn) : Pαn ∈ Part = xPαn: Pαn ∈ Part

NL = L(Pαn) : Pαn ∈ Part = yPαn: Pαn ∈ Part

If the two nets NU and NL have a common limit, say q ∈ R, the theory of integration

confirms that∫ 10

0f(x) dx = q

Example 8. Let the space, R2, be equipped with the radial topology τr. Recall that(R2, τr) is called the radial plane. This topology is described in example 2 on page 75.In example 11 on page 84, we show that the radial plane is not first countable. Consider

the set

S = R2\(0, 0)


equipped with the subspace topology inherited from τr. Construct a net which con-verges to p = (0, 0).

Solution : In what follows, (r, θ), represents the radial coordinates of a point in R2.Let D = (r, θ) : r > 0, θ > 0 be an uncountably infinite linearly ordered set where

(r1, θ1) = (r2, θ2) if r1 = r2 and θ2 = θ1 + 2nπ

and

(r1, θ1) < (r2, θ2) if

r2 < r1or

r1 = r2 and θ1 + 2nπ < θ2 + 2nπ

Then D is a directed set. Suppose f : D → S is defined as f(r, θ) = ( rθ , θ). Then

N = f(r, θ) : (r, θ) ∈ D = ( rθ , θ) : (r, θ) ∈ D

describes an uncountable net in S.

Claim: That the net, N , accumulates at p = (0, 0).

Proof of claim: Suppose B is an open neighbourhood of p, with respect to τr.

Let (rq, θq) ∈ D andT(rq,θq) = (r, θ) : (r, θ) ≥ (rq, θq)

be a tail end of D. Then

f [T(rq,θq)] = f(r, θ) : (r, θ) ≥ (rq, θq) = ( rθ , θ) : (r, θ) ≥ (rq, θq)

is a tail end of N .

By definition of accumulation point it suffices to show that f [T(rq,θq)] intersects B. If

so, then N contains a subnet which converges to p.

Let (rk, θk) be any point in B. We can choose m such that

rs =rk

θk + 2mπ< rk

Let θs = θk +2mπ. See that (rs, θs) is a point on the ray passing through (rk, θk) andso belongs to B. Also (rs, θs) > (rk, θk), so (rs, θs) ∈ T(rk,θk). Then

f(rs, θs) =

(

rsθs, θs

)

∈ f [T(rk,θk)] ∩B

So every neighbourhood of p intersects a tail end of N . So the net N accumulates at

p = (0, 0), as claimed.


Then by the above theorem, the net N has a subnet which converges to p.

Example 9. Let C denote the Cantor set. Recall from its formal definition 7.18 and its

graphic representation given on page 141 that C = ∩Ci : i ∈ N where Ci+1 ⊂ Ci forall i. Also every element of C is the set of all elements in [0, 1] of the form

∑∞i=1

mn3n

where mn ∈ 0, 2. Let W = [0, 1]× P([0, 1]) be partially ordered by ≤ defined as

(x, A) ≥ (y, B) ⇔ x ∈ A ⊆ B

Then D = (x, Ci) : i = 0, 1, 2, . . . , x ∈ Ci is a partially ordered set. If m ≥ n and

(x, Cm), (y, Cn) ∈ D where x ∈ Cm, then (x, Cm) ≥ (y, Cn). So D is a directed set.

Define the function f : D → S as f(x, Ci) = x. Let

N = f(x, Ci) : (x, Ci) ∈ D

be a corresponding net in [0, 1]. Show that every element of C is an accumulation

point of the net N .

Solution : Let k ∈ C. Then k ∈ Ci for all i ∈ N. Suppose B is a basic open neigh-

bourhood of k and j ∈ N.

If B ∩ f(x, Ci) = ∅ for all i ≥ j then

z ∈ B ⇒ z 6∈ f(x, Ci) : i ≥ j⇒ z 6∈ Ci for all i ≥ j

⇒ k 6∈ Ci for all i ≥ j

⇒ k 6∈ C

a contradiction. We can only conclude that the open neighbourhood B of k is cofinalin the net N = f(x, Ci) : Ci ∈ C. So k is an accumulation point of N .

Remark . In the example immediately above, we have shown that k is an accumulation

point of the net,

N = f(x, Ci) : (x, Ci) ∈ DThen N must have a subnet, say

M = (fg)(x, Ci) : (x, Ci) ∈ D

which converges to k. We will produce such a subnet by defining an appropriate in-

creasing function g : D → D.

Suppose k =∑∞

i=1tn3n where tn ∈ 0, 2 and let km =

∑mi=1

tn3n .

Define

g(x, Cm) = (km, Cm)


for m ∈ N. Then (fg)(x, Cm) = km. So

(fg)[D] = [km : m ∈ N]

is a subnet which clearly converges to k.

12.5 Convergence of functions

We will now consider the notion of “convergence to a limit point” for the particular

case of a set, S, whose elements are functions. This will be considered from a topolog-ical point of view. Since convergence to a limit point depends on the topology defined

on the set, our first step is to decide which topology on S is the most suitable for ourpurposes.

Consider, for example, the set,

S = R[0,1]

representing the family of all functions, f : [0, 1] → R, mapping [0, 1] into R (wherethe functions are not necessarily continuous). The standard topology on Cartesian

products,∏

i∈I Si, can be described as being the set of all functions mapping the indexset, I = [0, 1], into ∪i∈ISi. We will examine convergence in

∏

i∈[0,1] Ri equipped with

the product topology where each factor, Ri is R. An element of the product space∏

i∈[0,1] R can be viewed as an element of R[0,1].

The product topology on S = R[0,1].

We examine what the product topology on S = R[0,1] when viewed as∏

i∈[0,1] R looks

like.

Suppose f = yi : i ∈ [0, 1] = f(i) : i ∈ [0, 1] is an element in S. Then,for f ∈ S and j ∈ [0, 1], πj(f) = f(j) and, for k ∈ R,

π←j (k) = f ∈ S : f(j) = k

With the product topology on S, a basic open neighbourhood, Bf , of f is of the form

Bf = ∩π←i [Ui] : i ∈ Ffinite ⊆ [0, 1], Ui ⊆ R

where F is a finite subset of [0, 1] and Ui is an open neighbourhood of yi = f(i) in R.

The product topology declares that πi[Bf ] = R for all i 6∈ F and yi ∈ πi[B] = Ui, forall i ∈ F . If g = xi : i ∈ [0, 1] ∈ Bf (where g(i) = xi) this means that xi ∈ Ui for

each i ∈ F and xi is anything else in R for i 6∈ F . This means the element g is “close”to f if g(i) and f(i) both belong to Ui for i ∈ Ffinite.

The basic open neighbourhood, Bf , of f = yi establishes restrictions on conver-gence of a net only on finitely many index elements at a time, not on the whole set


yi : i ∈ [0, 1] simultaneously.

So a sequence or a net of functions, gα : α ∈ J, will converge to a limit point,f ∈ S = R[0,1] =

∏

i∈[0,1] R provided gα(i) → f(i) at finitely may points i ∈ [0, 1] at

a time. We will refer to this type of function convergence as “point-wise convergence”.

The box topology on S = R[0,1].

On the other hand, if we equip S = R[0,1] =∏

i∈[0,1] R with the (stronger) box topologyan open neighbourhood of f = yi : i ∈ [0, 1] = f(i) : i ∈ [0, 1] is of the form

Bf = ∩π←i [Ui] : i ∈ [0, 1], Ui ⊆ R

where Ui is an open neighbourhood of yi in R. In this case, g = xi : i ∈ [0, 1]belongs to Bf provided xi = g(i) ∈ Ui for each i ∈ [0, 1].

When equipped with the box topology, convergence in S = R[0,1] =∏

i∈[0,1] R isreferred to as “uniform convergence”.

Notice that a topology on [0, 1] is not relevant in the choice of topology on S. We

formally define these in a more general context, S = TA.

Definition 12.6 Let A be a set and T be a topological space. Let S = TA represent theset of all functions which map A into the topological space, T .

a) Point-wise convergence : Suppose S is expressed as S =∏

i∈A T and is equipped with

the product topology.In this case a basic open neighbourhood of f ∈ S, is of the form

Bf = ∩π←i [Ui] : i ∈ Ffinite ⊆ A, Ui ⊆ T

Then convergence of a sequence or net, gα : α ∈ J to f will occur if gα(i) converges tof(i) at finitely many values of i ∈ A at a time. Convergence in S = TA, equipped with theproduct topology, is called point-wise convergence.

b) Uniform convergence : Suppose S = TA is equipped with the box topology.In this case a basic open neighbourhood of f ∈ S, is of the form

Bf = ∩π←i [Ui] : i ∈ A, Ui ⊆ T

In this case, g is considered to be near f if g(i) ∈ Bf for all i ∈ A. Then convergence of asequence or net, gα : α ∈ J to f will occur if gα(i) converges to f(i) for all values of

i ∈ A simultaneously. In this case, we say that this sequence or net converges to f uniformly.

c) Uniform norm topology on C(S) : If S is a topological space it is the custom for analyststo represent the set of all continuous real-valued functions on S by C(S) (or C(S,R)). The


uniform norm topology is equivalent to topology generated by the sup-norm ‖ ‖∞ : C(S) →R, defined as

‖f‖∞ = supx∈S

|f(x)|

It is equivalent to the topology whose open sets are the basic open sets that are associatedto the ones in the box topology.

In the following example, if f is a real-valued function on [0, 1] (not necessarily con-

tinuous), let Z(f) = f←(0) and

κ(f) = |Z(f)| (the cardinality of Z(f))

with κ(f) = c if Z(f) is uncountably large, κ(f) = ℵ0 = |N| if Z(f) is countably

infinite and κ(f) ∈ N otherwise.

Example 10. Let S = [0, 1][0,1] be the topological space of all functions mapping

[0, 1] into [0, 1] equipped with the product topology (that is, the topology in whichconvergence is pointwise). Let

H = f ∈ S : f [0, 1] ⊆ 0, 1

Let z = 0(x) denote the zero-function on [0, 1].

a) Show that there is a net, N , in the subset, H of S, which accumulates at z = 0(x).

b) Let D∗ = f ∈ H : κ(f) ∈ N, κ(f) > 0 (all uncountable strings of zeroesand ones which contain finitely many zeroes). Show that no sequence in the

uncountable set D∗ can converge to z.

Solution : We are given that S = [0, 1][0,1] is equipped with the product topology and

H = f ∈ S : f [0, 1] ⊆ 0, 1

a) If f ∈ H , κ(f) represents the“cardinality of Z(f) = f←(0)”. Let

D = f ∈ H : κ(f) ∈ N

So κ(f) is a natural number for all f ∈ D. Since [0, 1] is uncountable, the set ofall finite subsets of [0, 1] is uncountable, so D is an uncountable set of functions

which map [0, 1] to finitely many zeroes. We will define “≤” on D as follows:

f ≤ p if and only if κ(f) ≤ κ(p)

f = p if and only if κ(p) = κ(f)

If s, t ∈ D are such that κ(s) = m < k = κ(t) then there exists r ∈ D such thatκ(r) = k + 1, so r > s and r > t. So D is a directed set. A tail end in D is as


follows: If g ∈ D, Tg = f : κ(f) ≥ κ(g), κ(f) < ℵ0.Let h : D → H be defined as h(f) = f .

LetN = h(f) : f ∈ D

be a net in H .

Claim#1: That the net N accumulates at 0(x). Consider an arbitrary basic open

neighbourhood B of 0(x). Then, there is Ffinite ⊂ [0, 1], where |Ffinite| = k,

B =∏

i∈[0,1]Ai

where, for i ∈ Ffinite, Ai = 0, Ai = 0, 1, otherwise.Let g ∈ N such that κ(g) = m. Then

h[Tg] = f ∈ N : κ(f) ≥ κ(g), κ(f)< ℵ0

is a tail end of N .

There exists h ∈ N such that κ(h) > max m, k, such that h(i) = 0 for i ∈ Ffinite.Then h ∈ B. So h[Tg] ∩B 6= ∅. So 0(x) is an accumulation point of the net N ,

as required for claim#1.

Claim#2: That the net N does not converge to 0(x). Consider an arbitrary basicopen neighbourhood B of 0(x). Then, there is Ffinite ⊂ [0, 1], where |Ffinite| = k,

B =∏

i∈[0,1]Ai

where, for i ∈ Ffinite, Ai = 0, Ai = 0, 1, otherwise.

It suffices to show that B does not contain a tail end of N . If g ∈ B, leth[Tg] = f ∈ N : f ≥ g be a tail end in N . Then for i ∈ F , g(i) = 0. Suppose

F ∗ ⊂ [0, 1] such that |F ∗| = k + 1 and F \F ∗ 6= ∅.

Suppose h ∈ N such that, for i ∈ F ∗, h(i) = 0 for i ∈ F ∗ and h(i) = 1 otherwise.Then, since κ(h) > κ(g), h > g, so h ∈ h[Tg]. But since h(i) = 1 for some

i ∈ F \F ∗ then h 6∈ B. So B cannot contain a tail end of N . So the net, N inH , does not converge to 0(x), as required. This proves claim#2.

b) Let A = fk : k ∈ N be a sequence in D. For each k ∈ N, let Z(fk) = f←k (0).

Since each Z(fk) is a finite subset of [0, 1] then

U = ∪Z(fk) : k ∈ N

is a countable proper subset of [0, 1]. Let F be a finite non-empty subset of [0, 1]such that F ∩ U = ∅. Let B =

∏

i∈[0,1]Ai where, for i ∈ F , Ai = 0 and

Ai = 0, 1, otherwise.

Then fk(i) = 1 for all i ∈ F and k ∈ N. Such fk’s cannot belong to the basicopen neighbourhood, B, of 0(x). So no tail end of A is contained is contained in

B. So the sequence A cannot converge to 0(x).


Example 11. Let the set, S = [0, 1][0,1], of all functions mapping [0, 1] to [0, 1] beequipped with the product topology. Here we view S as

∏

i∈[0,1][0, 1]i. Show that S

is not first countable.

Solution : We are given that the space S = [0, 1][0,1] viewed as∏

i∈[0,1][0, 1]i.

Suppose S is first countable and let

f =<f(i)>i∈[0,1] =<xi>i∈[0,1]

be a point in S. Then S has a countable neighbourhood base,

Bf = Bn : n ∈ N

at f . Then each open base neighbourhood, Bn, of f in∏

i∈[0,1][0, 1]i, can be expressedas,

<xi>i∈[0,1]∈ Bn = ∩π←i [Ui] : i ∈ FBn ⊆ [0, 1], Ui ⊆ [0, 1]where FBn is finite and Ui is open. Then ∪FBn : N is countable union of finite subsets

and so is a countable proper subset of [0, 1]. There must exist k ∈ [0, 1]\∪FBn : Nsuch that

πk

[

∏

i∈[0,1][0, 1]i

]

= πk [Bn ] = [0, 1]

for all n ∈ N.

Now, let V be a proper open neighbourhood of πk(f) = xk ∈ πk[S] = [0, 1]. Since V

is proper in [0, 1] thenxk ∈ πk[Bn] = [0, 1] 6⊆ V

for all Bn ∈ Bf . Then, for any Bn,

f ∈ π←k (xk)

⊆ Bn

6⊆ π←k [V ]

This contradicts that Bf is a countable neighbourhood base of f . So S is not firstcountable.

Concepts review:

1. Define a directed set.

2. Are the natural numbers a directed set?


3. Show that the set of all finite partitions of a closed interval of real numbers is adirected set.

4. Define a net of elements in a set.

5. Define a tail end of a net in a set.

6. What does it mean to say that a net converges to a point p in a topological space?

7. What is a limit point of a net.

8. What does it mean to say that a set is cofinal in a net?

9. What does it mean to say that a set B is a subnet of a net A?

10. If B is a subnet of the sequence A is B necessarily a subsequence of A?

11. Is it true that a point p belongs to the closure of a set B if and only if B contains a

net which converges to p?

12. Define an accumulation point of a net.

13. If the sequence A has an accumulation p is there necessarily a subnet which convergesto p?

14. If the sequence A has an accumulation p is there necessarily a subsequence whichconverges to p?

15. Complete the statement in terms of nets: A function f : S → T is continuous at p if

and only if...

16. Complete the statement in terms of subnets: The point u is an accumulation point of

the net A if and only if...

17. Describe two topologies used on sets of function TA and describe the convergence

properties with respect to these particular topologies.

EXERCISES

1. Show that the subnet of a subnet of a net A is also a subnet of A.

2. Show that a net, xi : i ∈ D, in a product space, S =∏

α∈A Sα, converges to p ifand only if, for any µ ∈ A, the net πµ(xi) : i ∈ D converges to πµ(p) in Sµ.


3. Let J denote the set of all irrationals in R and let ~p = (π, π). If S = J2\~p we willdefine a partial ordering, “≤”, on S as:

~a ≤ ~b whenever ||~b− ~p || ≤ ||~a− ~p ||

where “|| ||” refers to the distance between points in J2.

a) Verify that the pair, (S,≤), forms a directed set in S.

b) Suppose f : J2 → R is a real-valued function on J2. Then, when f is restrictedto S, the set, A = f(~x), ~x ∈ S , defines a net in R. Verify that the net, A,converges to a limit L if and only if lim~x→~p f(~x) = L (in the usual sense).


13 / Limit points of filters.

Summary. In this section we will introduce a form of convergence by setsrather than the convergence by points. These families of sets are called filters.

The construction of a filter is modeled closely on the structure of a neighbour-hood base at a point. We will formally define a “filter of sets” and explain how

to determine whether it has a limit point or not. We will also define an “accu-mulation point” which belongs to the closure of all sets in a filter. Finally, we

will introduce “ultrafilters” and a few of their characterizations; these will helpus distinguish them from regular filters.

13.1 Filters and some of their properties.

Recall that a net is essentially a set whose elements are indexed by either a linearly or

partially ordered set. The ordered elements may or may not converge to a particularelement we called a “limit point”. The type of convergence we will consider in thissection is more topological in nature. Essentially, neighbourhood bases serve as the

main inspiration for this type of convergence.

Definition 13.1 Let S be a set. Let F ⊆ P(S). If F satisfies the two properties:

1) ∅ 6∈ F ,2) For every U, V ∈ F , there exists F ∈ F such that F ⊆ U ∩ V .

then we will say that F is a filter base of sets.

If F is a filter base which also satisfies the condition:

3) Whenever U ∈ F and there is V ∈ P(S)\∅ such that U ⊆ V , then V ∈ F .

Then F is called a filter of sets.1

A quick way of describing a filter of sets is to say: “A filter is a subset of P(S)\∅which is closed under finite intersections and supersets.”2 Note that the notion of

a filter is a set-theoretic concept, not a topological one. Once we want to discuss“convergence” we will be in the context of a filter in some topological space.

1Not that there can be many types of filters: filters of sets, filters of closed sets, filters of zero-sets. Thedefinition of “filters of sets” is independent of a topology on S.

2If A ⊆ B then we say that B is a “superset” of A. It is worth noting that the set, P(S), is not a filter.Why?

254 Section 13 : Limit points of filters

A filter is, itself, a filter base, while a filter base, if not a filter, can be completed tobecome a filter. Given a filter base, F , we define the larger set, F ∗, as follows:

F ∗ = U ∈ P(S) : F ⊆ U , for some F ∈ F

We have simply adjoined to F all its supersets. It is easily verified that F ∗ is a filter.

We will say that the filter base, F , generates the filter F ∗.

Example 1. For a topological space, (S, τ), and a point x ∈ S, a neighbourhood base

of open sets, Bx, is clearly a filter base of sets in S since it satisfies the filter baseproperty,

“ For any two open neighbourhoods U and V of Bx there exists W ∈ Bx

such that x ∈W ⊆ U ∩ V ”.

The following proposition states that a filter which is generated by a filter base is the

intersection of all filters which contain the filter base in question.

Proposition 13.2 If F ∗ is a filter generated by the filter base, F then,

F ∗ = ∩H : H is a filter and F ⊆ H

Proof : Let H be a filter which contains the filter base F . It suffices to show that F ∗ ⊆ H .Let A ∈ F ∗. Then, by definition, there is a non-empty set U ∈ F such that U ⊆ A.

Since U ∈ H and H is a filter, A ∈ H . Then F ∗ ⊆ H .

Definition 13.3 Let (S, τ) be a topological space and F be a filter base of sets in P(S).

a) If ∩clSF : F ∈ F 6= ∅ then we will say that F is fixed in S.

b) Otherwise, we will say that F is free in S.

Example 2. Let (S, τ) be a topological space and K be a non-empty subset of S. Let

F ∗ be the set of all subsets, U , of S such that K ⊆ intSU .

a) Show that F ∗ is a filter of sets in S.

b) If K = x, describe a filter base, F , of F ∗.

c) Is the filter base F free or fixed in S?


Solution : We are given F ∗ = U ∈ P(S) : K ⊆ intSU for a fixed K.

Let U, V ∈ F ∗ and M ∈ P(S) such that U ⊆M .

a) Since K 6⊆ intS∅, then ∅ 6∈ F ∗.

Since U, V ∈ F ∗ then K ⊆ intSU and K ⊆ intSV . Then

K ⊆ intSU∩ intSV = intS(U ∩ V )

So U ∩V ∈ F ∗. Then F ∗ is closed under finite intersections, which implies that

F ∗ is a filter base of sets in P(S).

Since U ∈ F ∗ and U ⊆ M , then K ⊆ intSU ⊆ intSM then M ∈ F ∗. So F ∗ is

closed under supersets.

So F ∗ is a filter of sets.

b) Suppose K = x. Then F is the set, Bx, of all neighbourhoods of x. The set,F , satisfies the property: “If A and B are two neighbourhoods of x in F then

there exists a neighbourhood C in F such that x ∈ C ⊆ A ∩ B”. So the filterbase, F = Bx, is a neighbourhood base of x.

c) Since K ⊆ intSU and intSU ⊆ clSU for all U ∈ F then K ⊆ ∩clSU : U ∈ Fso F is fixed.

13.2 Limit points and accumulation points of filters.

As we have done for sequences and nets, we define the limit point and accumulation

point of a filter base and of a filter. In what follows, filters are family of subsets ofsome topological space.1

Definition 13.4 Let F be a filter base of sets on the topological space S, and x ∈ S. LetBx be a neighbourhood base at the point x.

a) We will say that the filter base of sets, F , converges to x if and only if, for every open

neighbourhood, B of x, there is some U ∈ F such that U ⊆ B. This implies that, ifBx is an open neighbourhood base of x, then Bx ⊆ F ∗, the filter generated by F .So the filter, F ∗, is “finer” than Bx. We abbreviate the “convergence of F to x” by

the expression, F → x. In this case, we will say that

“x is a limit point of the filter base F”

1Later in the text we will discuss filters whose elements are zero-sets.


b) We will say that x is an accumulation point for F if and only if every open neigh-bourhood of x intersects each F ∈ F . That is,

x ∈ clSF for all F ∈ F

c) The set of all accumulation points of F is called that the adherence of F . The

adherence of F is denoted by a(F ). That is, a(F ) = ∩clSF : F ∈ F.

It is worth verifying that, if F is a filter base, then a(F ) = a(F ∗). Also note that,by definition, a limit point of a filter base, F , is an accumulation point of F .

Observe that, for x to be a limit point of a filter base F , x need not belong to of anyof its elements. Witness the following example.

Example 3. Consider the set, F = (4, 4 + δ] : δ > 0. See that F doesn’t contain ∅

and satisfies the filter base property and so qualifies as a filter base. The element 4

does not belong to any of its elements. But if we take any open neighbourhood of 4,say B = (4− ε, 4 + µ) there exists, say U = (4, 4+ µ/4) in B, such that U ⊆ B. So 4

is a limit point of the filter base F .1

Furthermore, distinct filter bases can converge to the same point. Witness such a casein the following example.

Example 4. Consider the sets, F1 = (4, 4 + δ] : δ > 0 and F2 = [4− δ, 4) : δ > 0.See that both F1 and F2 are filter bases which converge to the limit point 4.

By definition, any filter base that has an accumulation point is fixed. Filter bases canhave multiple accumulation points.

Example 5. Consider the set, A = 2 ∪ (3, 4] in R, equipped with the usual topology.

Let F = U ∈ P(R) : A ⊆ U. Every point, x ∈ A, belongs to every F ∈ F . Soevery point in A is an accumulation point of F . But also, every open neighbourhood

of 3 will intersect every set which contains A, so 3 is also an accumulation point. Thefilter base, F , has no limit point since any open interval of a point in A with radius

less than 1 cannot be contained in a set which contains all of A.

However, verify that if, F1 = U ∈ P(R) : 2 ∈ U, is a fixed filter where F ⊆ F1.

13.3 Filters derived from nets.

Suppose we are given a net in a topological space S. We will show how a filter canbe constructed from this net, simply by gathering together all tail-ends of the net.

Eventually, we will show they have the same limit points.

1However, by definition, the set, B = (4 − ε, 4 + µ), will belong to F∗, the filter generated by F .


Theorem 13.5 Let S be a set and D be a directed set. Suppose the function, f : D → S,defines a net, A = f(i) : i ∈ D, in S. If u ∈ D, let Tu = i ∈ D : i ≥ u, a tail end of D.

Then,

f [Tu] = f(i) : i ≥ u in D

represents a tail end of the net, N , starting with f(u). Let

Ff = f [Tu] : u ∈ D

be the set of all tail ends of the net, N . Then Ff is a filter base of sets in S.

Proof : We are given Ff = f [Tu] : u ∈ D where

N = f(i) : i ∈ D

is a net in the set S. First note that ∅ 6∈ Ff .

We claim that Ff is closed under finite intersection: If j, k ∈ D, then there exist,h ∈ D, such that j ≤ h and k ≤ h. Then Th ⊆ Tj ∩ Tk and so,

f [Th] ⊆ f [Tj] ∩ f [Tk]

So Ff is closed under finite intersections, as claimed. So the set, Ff , of all tail ends

of the net, N , is a filter base of sets in S.

Definition 13.6 Let S be a set and N = f(i) : i ∈ D be a net in S. Let Tu = i ∈ D :

i ≥ u. The filter base, Ff = f [Tu] : u ∈ D, forming the set of tail ends, of the net, N ,in S, is referred to as the filter base in S which is determined by the net N .

13.4 A net derived from a filter.

Having seen how we can construct a filter from a net, we now show how a net can be

constructed given any filter.


Definition 13.7 Let S be a set and F be a filter of sets. We define the relation, ≤, onthe set,

DF = (u, F ) ∈ S × F : u ∈ Fas follows: (s, F ) ≤ (u,H) if and only if u ∈ H ⊆ F . Then (DF ,≤) forms a directed set.

With this directed set in hand, and the function, f : DF → S, defined as, f(u, F ) = u, wedefine the net

NF = f(u, F ) : (u, F ) ∈ DFin the set, S. The net NF is called a net generated by the filter F .

13.5 Corresponding limit points of a filter and its net.

We will now confirm that the limit point or accumulation point of a net will always

be a limit point or accumulation point of the filter corresponding to this net, andvice-versa.

Theorem 13.8 Let S be a topological space. Suppose F is a filter base of sets in S andA = h(i) : i ∈ D is a net.

a) Suppose the net, A, is a net generated by F ∗. The point, u, in S is a limit point(accumulation point) of F if and only if u is a limit point (accumulation point) of A.

b) Suppose the filter base, F , is determined by the net, A. The point, u, in S is a limit

point (accumulation point) of F if and only if u is a limit point (accumulation point)of A.

Proof : We are given that S is a topological space.

a) We are given that F is a filter base and the net,

A = h(u, F ) : (u, F ) ∈ DF , u ∈ F ∈ F

is one which is generated by filter F ∗.

( ⇒ ) Case: Limit point. Suppose u is a limit point of the filter base F . Then

u is a limit point of F ∗. We are required to show that u is a limit point of thenet A. Let U be an open neighbourhood of u. To show that A converges to u

we must show that U contains a tail end of A.

Since u is a limit point of F and U is a neighbourhood u, there exists F ∈ Fsuch that F ⊆ U . Then U belongs to the filter, F ∗. We can choose some point


v ∈ U , so that the pair, (v, U) belongs to DF .

We claim that the tail end, h[T(v,U )] in A, is contained in U : Let (x, V ) ∈ DF

such that (v, U) ≤ (x, V ). Then T(x,V ) ∈ T(v,U ) and x ∈ V ⊂ U . Thenh(x, V ) = x ∈ U . So neighbourhood of u, U , contains the tail end of h[T(v,U )],

as claimed. So the net, A, converges to u.

( ⇒ ) Case: Accumulation point. Suppose u is an accumulation point of thefilter base F and A = h(u, F ) : (u, F ) ∈ DF, the net generated by F . Weare required to show that u is an accumulation point of A. Let U be an open

neighbourhood of u. By hypothesis, U ∩ F 6= ∅ for all F ∈ F (equivalently,u ∈ clSF for each F ∈ F ).

To show that u is an accumulation point of the net, A, we must show that Uintersects each tail end of A. Let h[T(v,K)] be a tail end in A and M ∈ F . Then

V = M ∩ K is an element of F which is a subset of K. Then, by hypothesis,there exists

t ∈ U ∩ V ⊆ V ⊆ K

The expression, t ∈ V ⊆ K, translates to, (v,K) ≤ (t, V ). Hence

t = h(t, V ) ∈ h[T(v,K)] ∩ U

So U intersects the tail-end h[T(v,K)]. We conclude that u is an accumulation

point of the net corresponding to F .

( ⇐ ) Case: Limit point. Suppose u is a limit point of the net A. We are

required to show that the filter, F , generated by A converges to u. Let U be anopen neighbourhood of u. To show that F converges to u we must show that U

contains an element of F .

By hypothesis, U contains a tail end of A. That is, h[T(s,F )] ⊆ U for some (s, F ).

Then h(r, F ) ∈ U for all r ∈ F . But h(r, F ) = r. So F ⊆ U . Since F belongs toF then F converges to u.

( ⇐ ) Case: Accumulation point. The proof is left as an exercise for the reader.

b) We are given that the filter base, Fh, is determined by the net A = h(i) : i ∈ D.Recall that Fh = h[Tu] : i ≥ u where Tu = i ∈ D : i ≥ u. That is, it is the

set of the tail ends of the net, h(i) : i ∈ D.( ⇒ ) Case: Limit point. Suppose x is a limit point of the filter base, Fh. We

must show that the net A also converges to x. Suppose U is any open neigh-bourhood of x. Then there is a ∈ D such that h[Ta] ⊆ U . Then every open

neighbourhood contains a tail end of the net, h(i) : i ∈ D. So the net con-verges to x.

( ⇐ ) Proving the converse of the “Case, limit point”, is left as an exercise.


( ⇒ ) Case: Accumulation point. Suppose y is an accumulation point of thefilter base, Fh. We must show that y is an accumulation point of the net A. Let

U be any open neighbourhood of y. Then U ∩ h[Tu] 6= ∅ for all u ∈ D. That is,U contains an element of all tail ends of the net, h(i) : i ∈ D. Then y is an

accumulation point of this net.

( ⇐ ) Proving the converse of the “Case, accumulation point” is left as an exercise.

13.6 Other properties of filters.

In the following theorem, we see that “uniqueness of limits of filters (nets)” charac-

terizes the Hausdorff property on a topological space.

Theorem 13.9 Let S be a topological space. The space S is Hausdorff if and only if a

convergent filter, F ∗, has only one limit point.

Proof : We are given that S is a space.

( ⇒ ) Suppose S is Hausdorff. Suppose the points x and p are limits of the filter base

of sets, F , in S. Then neighbourhood bases, Bx and Bp, are subsets of the filter,F ∗, generated by F . If x 6= p, then, by hypothesis, we can choose disjoint open

neighbourhoods Ux and Vp which separate p from x. Since F ∗ converges to p and xthere exists, W and M , in F ∗ which are contained in Ux and Vp, respectively. But

then W ∩M = ∅, contradicting the fact that F ∗ is a filter. So x = p.

( ⇐ ) We are given that limits of any filter, F ∗, of sets in S are unique. Suppose Sis not Hausdorff. Then there is a pair of points, x and p, with open neighbourhoodbases, Bx and Bp such that, U ∩ V 6= ∅ for all U and V in Bx and Bp. It easily

follows that F = U ∩ V : U ∈ Bx and V ∈ Bp is a filter base of sets in S.

We claim that F ∗ converges to both x and p. Let Wx and Dp be elements of Bx andBp, respectively. It suffices to show that there are elements, F1 and F2 in F ∗ such

that F1 ⊆ Wx and F2 ⊆ Dp, respectively. This would contradict convergence to asingle element.

There exists Ux ∈ Bx and Vp ∈ Bp such that

x ∈ F1 = Ux ∩ Vp ⊆Wx

and Tp ∈ Bp and Cx ∈ Bx such that,

p ∈ F2 = Tp ∩Cx ⊆ Dp


By definition of F ∗ both F1 and F2 belong to F ∗. By definition of convergence, F ∗

converges to both x and p. This contradicts our hypothesis. So S must be Hausdorff.

The statement in the above theorem translates to nets. That is, limit points of nets

in a topological space, S, are unique if and only if S is Hausdorff.

In the following theorem we show that, for any accumulation point, p, of a filter, F ,it is possible to increase the size of F to a larger filter which will converge to p. So

every accumulation point of a filter is the limit point of a larger filter which contains it.

Theorem 13.10 Suppose S is a topological space and p is an accumulation point for afilter base, F , of sets in S. Then p is the limit point of some filter, H ∗, which contains

F ∗.

Proof : We are given that S is a space and that p is an accumulation point of a filter, F .

Let Bp denote the set of all open neighbourhoods of p. Since p is an accumulation

point of F , every open neighbourhood of p intersects every element of F . We canthen consider the family

H = U ∩ F : U ∈ Bp, F ∈ F

itself a filter base. Let H ∗ be the filter generated by H .

We claim that H converges to p. If V is an element of Bp, for F ∈ F , V ∩ F isnon-empty so is an element of H . Since V ∩ F is a subset of V , by definition of

convergence, H converges to p, as claimed.

Suppose F ∈ F . Since U ∩ F ⊆ F then F ∈ H ∗ so F ⊆ H ∗. In fact, F ∗ ⊆ H ∗.

We conclude that, if p is accumulation point of F , then it is the limit point of H ∗,

a filter which contains F ∗.

Theorem 13.11 Suppose S is a topological space. Suppose F1 and F2 are two fixed filter

bases in S. If F1 and F2 have the same adherence set then every element of the filter, F1,intersects every element of the filter, F2.


Proof : We are given that F1 and F2 are two fixed filter bases in S.

Suppose a(F1) = a(F2). Then, if p ∈ a(F1), p is an accumulation point of both F1

and of F2. There exists a filter, H ∗, which contains both F1 and of F2 and whichconverges to p. (Note: There are some straightforward details involving theorem 13.10

to work out for proving this; these are left as an exercise.) Since H ∗ is closed underfinite intersections, every element of F1 intersects every element of F2.

13.7 Filter bases describe closures of sets and continuity.

Not surprisingly, given our experience with sequences and nets, the closure of a setcan be described in terms of the limit points of its filter bases.

Theorem 13.12 Let F be a non-empty subset of a topological space, S. Then a point ubelongs to the closure, clSF , of F if and only if u is the limit point of some filter base in F .

Proof : We are given that S is a space and F is a non-empty subset of S.

( ⇒ ) Suppose u ∈ clSF . Let Bu be a neighbourhood base of open sets for u. Then

F = U ∩ F : U ∈ Buforms a filter base of sets in F which converges to u.

( ⇐ ) Suppose F is a filter base in F converging to some u ∈ S. Let U be any open

neighbourhood of u. Then there exists some V ∈ F such that V ⊆ U . Since V ⊆ F ,U intersects F and so u ∈ clSF .

Just as for nets, continuity of a function f : S → T can be expressed in terms offilters. But first we must determine whether the continuous image of a filter can itself

be seen as a filter.

About the image, f [F ∗], of a filter, F ∗. Given a function f : S → T and given that

F ∗ is a filter of sets in S we define,

f [F ∗] = f [F ] : F ∈ F ∗

We claim f [F ∗] is a filter base in the range of the function, f . Let f [U ] and f [V ] be

two elements of f [F ∗] where U and V are elements of F ∗. Then there exists non-empty W ∈ F ∗ such that W ⊆ U ∩ V . Then f [W ] ⊆ f [U ∩ V ] ⊆ f [U ] ∩ f [V ] 6= ∅.So...

f [F ∗]is a filter base which generates the filter f [F ∗]∗.


Theorem 13.13 Let S and T be two topological spaces. Let f : S → T be a function andu ∈ S. Let F ∗ be a filter. The function f is continuous at u if and only if whenever u is alimit point of the filter, F ∗, then f(u) is a limit point of the filter, f [F ∗]∗.

Proof : We are given that S and T are two topological spaces and f : S → T is a function.

( ⇒ ) Suppose f is continuous at u ∈ S and F ∗ converges to u. We are required to

show that f [F ∗]∗ converges to f(u).

Suppose V is an open neighbourhood of f(u). It suffices to show that an element

of f [F ∗]∗ is contained in V . Since f is continuous at u, then there exists an openneighbourhood U of u such that f [U ] ⊆ V . Since F ∗ is a filter then U belongs to the

filter F ∗. Since f [U ] ∈ f [F ∗] and is contained in V , then V ∈ f [F ∗]∗. So f [F ∗]∗

converges to f(u).

( ⇐ ) Suppose that whenever F ∗ converges to u then f [F ∗]∗ converges to f(u). Let

F ∗ be the neighbourhood filter of u. Since f [F ∗]∗ converges to f(u), each neighbour-hood V of f(u) contains some element of f [F ∗]∗. Then for some neighbourhood U of

u, f [U ] ⊆ V . This shows that f is continuous at u.

The statement in the following example involves a subspace of S = [0, 1][0,1] similarto the one found in the example discussed in the chapter on nets (on page 248).

Example 6. Let S = [0, 1][0,1], the set of all functions mapping [0, 1] into [0, 1], be

equipped with the product topology. Let H be a subspace of the subspace, 0, 1[0,1]

of S which is defined as follows:

H = h ∈ S : h(x) = 0 for finitely many values, otherwise h(x) = 1

Construct a filter in H , which accumulates at the zero element, z.

Solution : Given S = [0, 1][0,1], the set of all functions mapping [0, 1] into [0, 1],equipped with the product topology and

H = h ∈ S : h←(0) is finite, otherwise h(x) = 1

a subspace of 0, 1[0,1]. Let z represent the zero function, 0(x).

Let Tn = f ∈ H : n ≤ | f←(0) | < ℵ0. Let

F = Tn : n = 1, 2, 3, . . . ,

where Tn ⊆ H for each n. If f ∈ Tm, f(i) = 0 for m or more values of i in [0, 1].So Tm ⊆ Tm−1. By finite induction, if n > m then Tn ⊆ Tm. Hence, if m > n then


Tm ∩ Tn = Tm 6= ∅. Since Tn is never empty, F is then a filter base of sets in S.

Let BF be a basic open neighbourhood of z in S where |F | = q, for some q ∈ N\0and f(x) = 0 for the q values of x in F ⊂ [0, 1].

To show z is an accumulation point of F , we must show that, given Tk ∈ F ,

BF ∩ Tk 6= ∅. If k > q, then, for some f ∈ Tk, f ∈ BF ∩ Tk 6= ∅. If k < q,since Tq ⊆ Tk then, again, for some f ∈ Tk, f ∈ BF ∩Tk 6= ∅. So BF ∩Tk 6= ∅. Then

z is an accumulation point of the filter base, F .

13.8 Partial ordering the family, F , of all filters in P(S).

We will now discuss some properties possessed by families of filters of sets. The reader

can assume that all filters discussed here are subsets of P(S) for some topologicalspace, S. Let

F = F ⊆ P(S) : F is a filter of sets of S

denote the family of all filters on a space S. We will define an ordering of the elements

in F by inclusion, “⊆”.

Given two filters, F and H , in F , if F ⊆ H , we will say that,

“H is finer than F” or “F is coarser than H ”

For example, the sets F ∗ = (2−1/n, 2+1/n) : n ∈ N\0∗ and H ∗ = (2−ε, 2+ε) :ε > 0∗ are easily verified to be two filters both converging to 2. We see that sinceF ∗ ⊆ H ∗, then H ∗ is finer than F ∗. This ordering is a partial ordering.1

13.9 Ultrafilters.

We have seen that a filter base can contain an other smaller filter base and can oftenbe itself contained in some other larger filter of sets. Filter bases with large adherence

sets can be built up so that they eventually will converge to at most one of thosepoints. But, as we shall soon see, at some point, a filter of sets will have attained its

maximum size. We often call such filters, “maximal filters”. Although the use of thewords “maximal filter” are by themselves sufficiently descriptive, most writers have

become accustomed to using the single term, “ultrafilter”. We will formally definean “ultrafilter” and prove some of its characterizations. These characterizations will

make it easier to recognize an ultrafilter when we see one. We will then discuss someultrafilter properties. Ultrafilters play an important role in certain branches of general

topology.

1Of course, “⊆” is not a “linear ordering” of F since it is not always the case that one filter is a subsetof another.


Definition 13.14 Let S be a topological space and (F ,⊆) denote the set of all filters in

P(S), partially ordered by “⊆”. Suppose F is an element of F .

We will say that the filter F is an ultrafilter of sets if and only if, for H ∈ F ,

F ⊆ H ⇒ F = H

That is, F cannot be properly contained in some other distinct filter.

The most commonly used characterizations of ultrafilters of sets in P(S) are stated

and proved below. These are extremely useful. It is a good exercise to try to provethem before consulting the proposed proofs.

Theorem 13.15 Let S a topological space and F be a filter of sets in P(S). The following

statements are equivalent.

a) The set F is an ultrafilter in P(S).

b) If U ∈ P(S) is such that it intersects every element of F , then U ∈ F .

c) For any U ∈ P(S), if U 6∈ F then S\U ∈ F


( a ⇒ b ) We are given that F is an ultrafilter. Suppose U ∈ P(S) is such thatU ∩ F 6= ∅ for all F ∈ F . We are required to show that U ∈ F .

Let

H = U ∩ F : F ∈ FThe set, H , is easily seen to be a filter base of sets in S. Then the filter, H ∗, whichis generated by H , contains both U and the set F . Then, F ⊆ H ∗. But, since Fis a maximal filter, then F = H ∗. Then U ∈ F , as required.

( b ⇒ c ) Suppose that, whenever a set U meets every element of F , then U ∈ F .Suppose U 6∈ F . It suffices to show that S \U ∈ F .

Since U 6∈ F , then U ∩ F = ∅ for some F ∈ F . Then F ⊆ S \ U . If F1 ∈ F andF1 6= F , then F1 ∩F ⊆ S\F . Since F1 ∩F is non-empty and is a subset of S\U , then

S\U intersects every element of F . By hypothesis, S\U ∈ F .


( c ⇒ a ) We are given that, if U 6∈ F then S \ U ∈ F . We claim that F is anultrafilter.

Let H ∗ be any filter such that F ⊆ H ∗. If V ∈ H ∗ such that V 6∈ F thenS \ V ∈ F . This means that H ∗ contains both V and its complement. This wouldimply that H ∗ is not a filter. Then there can be no such V . So H ∗ ⊆ F . This

means that F is an ultrafilter.

Example 7. Let u be an element of the space S. Let F = U ∈ P(S) : u ∈ U. We

claim that F is an ultrafilter. To see this, let U ∈ P(S). If u ∈ U then U ∈ F ; ifu 6∈ U , then u ∈ S\U ∈ F . So, by the above theorem, F is an ultrafilter.

We have seen that (F ,⊆) is partially ordered by “⊆”. For every filter, H , in F does

there exist an ultrafilter which contains it? Many readers see this as an “existencestatement”. It suggests that, to prove it we may have to invoke Zorn’s lemma1. We

remind ourselves what Zorn’s lemma formally states:

“If every chain in a partially ordered set, (S,≤), has a maximal element

then S has a maximal element.”

Theorem 13.16 Let S be a topological space and (F ,⊆), denote the set of all filters inP(S), partially ordered by inclusion. If F ∈ F , then there exists an ultrafilter, U , in Fwhich contains F .

Proof : We are given that F ∈ F . We wish to prove the existence of an ultrafilter whichcontains F . We set up the problem so that Zorn’s lemma applies.

For a given F ∈ F , consider the set,

SF = H ∈ F : F ⊆ H

of all filters which contain F . We are required to show that some filter in SF is anultrafilter (i.e., SF has a maximal element). If SF = F then F is an ultrafilter.

Suppose SF contains filters other than F . Now F can be the filter base element ofmany chains in SF . In fact, SF can be viewed as the union of a family of filter-chains,

Cα : α ∈ J1Zorn’s lemma is proven to be equivalent to the Axiom of choice. A proof appears in R. Andre, Axioms

and set theory


each with base element, F . Pick an arbitrary filter-chain, Cγ = Fi : i ∈ I in SF .It looks like,

F ⊆ F1 ⊆ F2 ⊆ F3 ⊆ · · ·

Let T = ∪Fi : Fi ∈ Cγ. Quite clearly, Fi ⊆ T for each i. We claim that thefamily of sets, T , is itself a filter in the chain Cγ . If U and V are elements of T , thenU and V are elements of some Fj in the chain, Cγ , so U ∩ V is an element of Fj; so

U ∩ V is also an element of T . Then T is a filter base of sets which is a maximalelement in the chain Cγ .

So every chain, Cα, in SF has a maximal element. We can invoke Zorn’s lemma, toconclude from this that SF , must also have a maximal filter element, U . Since, SF

contains every filter which contains F , then U is the maximal filter which contains

F . Then, by definition, U is the unique ultrafilter which contains F . This provesthe statement.

Example 8. Show that an ultrafilter of sets in a Hausdorff space can have, at most,one accumulation point.

Solution : If U is a free ultrafilter then, then ∩clSF : F ∈ U = ∅, so U has noaccumulation point. Suppose, on the other hand that U is fixed and u and p are two

accumulation points. Then, by theorem 13.10, there exists a filter, H ∗, such thatU ⊆ H ∗, H ∗ → p and H ∗ → u. Since U is an ultrafilter then U = H ∗. So Uconverges to both u and p. By theorem 13.9, since S is a Hausdorff space the limitof a filter is unique. So u = p. We conclude that an ultrafilter can have at most one

accumulation point.

13.11 Convergence of filters on product spaces.

Suppose Si : i ∈ I is a family of topological spaces and S =∏

i∈I Si is a product

space.1 Suppose there is a filter, F , in S which converges to a point, <xi>i∈I . Then,for each i ∈ I ,

πi[F ] = πi[F ] : F ∈ F ∗ = Fi

is a filter base in Si.2

1We are assuming that the product space is non-empty. This is a consequence of Axiom of choice.2To see this simply note that if πi[F1] and πi[F2] belong to Fi then πi[F1 ∩ F2] ⊆ πi[F1] ∩ πi[F2] where

F1 ∩ F2 6= ∅ and πi[F1] is never empty.


Since F converges to < xi >i∈I , by continuity of πi, πi[F ] = Fi, will converge toπi(<xi>i∈I) = xi (by theorem 13.13). So if a filter converges to a particular point in

a product space,∏

i∈I Si, then the corresponding filter in each factor converges to thecorresponding point under πi. The following theorem shows that the converse holds

true.

Theorem 13.17 Let Si : i ∈ I be a family of topological spaces and p =<xi>i∈I be a

point in the product space, S =∏

i∈I Si. Suppose that F ∗ is a filter of sets in S such that,for each i ∈ I , the filter,

πi[F∗] = f [F ] : F ∈ F ∗ = Fi

converges to xi. Then F converges to the point, p =<xi>i∈I .3

Proof : Given: A filter, F ∗, in the product space, S =∏

i∈I Si, and that p =<xi >i∈I isa point in S. For each i, the filter, πi[F ∗] = Fi, converges to xi. We are required to

show that F ∗ converges to p.

Let F be a finite subset of I . For each i ∈ F , let Ui be an open neighbourhood of

xi ∈ Si. Then

B = ∩π←i [Ui] : i ∈ F

forms a basic open neighbourhood of p =<xi >i∈I∈ S. To show that F ∗ converges

to p we must find an element, A of F ∗ which is contained in B.

Since Fi converges to xi, Ui ∈ Fi (since Fi is a filter), so π←i [Ui] ∈ F . For each

i ∈ F , since πi is continuous there exists an open neighbourhood, Ai, of p, in F suchthat xi ∈ πi[Ai] ⊆ Ui. By the filter base property, there exists A ∈ F such that

p ∈ A ⊆ ∩Ai : i ∈ F. Now, for each i ∈ F ,

πi[A] ⊆ πi[∩Ai]⊆ ∩πi[Ai]⊆ πi[Ai]

⊆ Ui

So we have found A ∈ F such that p ∈ A ⊆ ∩π←i [Ui] : i ∈ F. So F converges top =<xi>i∈I , as required.

3This result will be useful to show that the product of compact spaces is compact in the next section.


Corollary 13.18 Let Sj : j ∈ J be a family of topological spaces and

S =∏

j∈JSj

be a corresponding product space. Let B = β(i) : i ∈ N be a sequence in S, where

β(i) =< βj(i) >j∈J . For each j ∈ J, let

πj(β(i)) = πj(< βj(i) >j∈J ) = βj(i)

Then, for each j ∈ J,

πj[B] = βj(i) : i ∈ Nthe corresponding sequence in Sj.

Suppose that, for each j ∈ J, the sequence βj(i) : i ∈ N converges to pj in Sj.

Then the sequence β(i) : i ∈ N converges to <pj>j∈J in S.

Proof : We are given the product space, S =∏

j∈J Sj, and a sequence, B = β(i) : i ∈ Nin S. Also, for each j ∈ J,

πj [β(i) : i ∈ N] = βj(i) : i ∈ N

Suppose that, for each j ∈ J, βj(i) : i ∈ N → pj ∈ Sj.

Let F be the filter base generated by the sequence B and, for each j ∈ J, Fj = πj[F ].

Then, by theorem 13.8, the filter, Fj, in Sj generated by the sequence βj(i) : i ∈ Nconverges to pj.

By the theorem 13.17, the filter F converges to p =<pj >j∈J∈∏

j∈J Sj. Then the

sequence, β(i) : i ∈ N, in∏

j∈J Sj, converges to p =<pj>j∈J , as required.

13.12 Ultranets: A close relative of ultrafilters.

We will now formally define what is also known as maximal nets.

Definition 13.19 Let S be a topological space and A = f(i)i∈D be a net in S. We willsay that A is an ultranet if, for any subset, B, in S, a tail end of A either belongs to B and,

if not, belongs to its complement, S\B.


A net which is constant on its tail end, is an example of an ultranet. It is not imme-diately obvious from this definition that non-constant ultranets exist. We will be able

to address this question below.

Theorem 13.20 Let A = f(i) : i ∈ D be an ultranet in a space S. If E is a subnet of Athen E is a also an ultranet.

Proof : Let S be a topological space which contains an ultranet, A = f(i) : i ∈ D.Suppose E is a subnet of A. Let B be a non-empty subset of S. Then B or S \Bcontains a tail end of A. Suppose, without loss of generality, that B contains a tail

end of A. Then B must contain the tail end of any subnet of A. Then every tail endof E belongs to B. By definition of “ultranet” E is an ultranet.

For what follows recall, from the definition 13.19, that “ultranets are those nets, xi,such that, for each B ∈ P(S), either B or S\B contains a tail end of xi”.

Example 9. Let S be a topological space. Suppose A = f(i) : i ∈ D is a net in Sand F is a filter of sets in S.

a) Show that, if A is an ultranet and F is the filter generated by A, then F is an

ultrafilter.

b) Show that, if x ∈ S is the limit point of the ultranet A, then a tail end of A is

constant. So a convergent ultranet must have a constant tail end.

c) Show that, if F is an ultrafilter and A is a net determined by F , then A is anultranet.

d) Show that a non-constant ultranet exists.

Solution : We are given that S is a topological space and A = f(i) : i ∈ D is a netin S. Also, F is a filter of sets in S.

a) Suppose A is an ultranet and F = f [Tu] : u ∈ D, where Tu = i ∈ D : i ≥ u,is the filter generated by the net A. We are required to show that F is an ultra-filter.

Let B ∈ P(S)\∅. By theorem 13.15, either B or S\B belongs to an ultrafilterof sets. Suppose S \B does not belong to F . Then S \B does not contain a

tail end, say f [Tj], of A (for, if it did, then S\B would belong to F ). Then, bydefinition of “ultranet”, f [Tj] ⊆ B. Since F is a filter, then B ∈ F . So F is an

ultrafilter.


b) We are given that the ultranet, A, converges to x and that F = f [Tu] : u ∈ Dis the ultrafilter generated by A. Then F must also converge to x.

We are required to show that some tail end of A is constant. Suppose no tail end

of A is constant. Then, for k ∈ D, there exists two non-empty non-intersectingcofinal sets, C and E of D, such that f [Dk] = C ∪E and both C and E convergeto the net limit point x. Since F is an ultrafilter, either C or S\C belongs to

F . Suppose C ∈ F . Then S\C does not belong to F . Since F is an ultrafilter,S \C does not intersect some element F of F . Then, since S \C contains E,

the cofinal set, E, does not intersect some F ∈ F , that is some tail end of theultranet, A. But this can’t be since E is cofinal in A. So A must be constant on

a tail end. Since x is a limit point of A then the constant is x.

c) We are given that F is an ultrafilter and A = f(y, F ) : F ∈ F and y ∈ F isthe net determined by F . Recall that (u, F ) ≤ (v, H) if and only if v ∈ H ⊆ F

and f(u, F ) = u. We are required to show that A is an ultranet.

Let B ∈ P(S) \ 0. Then B ∈ F or S \B ∈ F . Suppose s ∈ B ∈ F .

Suppose (u,H) ∈ DF such that (u,H) ≥ (s, B). Then u ∈ H ⊆ B. Thenf(u,H) = u ∈ B. Then the tail end, f(y, F ) : (y, F ) ≥ (s, B), of A belongs to

B. By definition, A is an ultranet.

d) Since convergent ultranets have a constant tail end, we will only investigate non-

converging ultranets. We begin with a free ultrafilter. Let F be a free ultrafilterof sets in S. Let κ : F → S be a choice function defined as κ(F ) = xF , where

xF ∈ F . Let A = f(xF , F ) : F ∈ F be the free ultranet determined by F .Since A is free it cannot be constant on any tail end, otherwise it would converge

to that constant element. So A is a free ultranet with no constant tail end. Asrequired.

Theorem 13.21 Let A = f(i) : i ∈ D be a net in a topological space, S. Then A has a

subnet U which is an ultranet.

Proof : We are given that A = f(i) : i ∈ D is a net in a space, S. Let F = f [Tu] :u ∈ D be the filter generated by the net A, where Tu = i ∈ D : i ≥ u and

f [Tu] = f(i) : i ∈ Tu. Then there exists an ultrafilter, U , which contains the filterF .

We now construct a net derived from the ultrafilter, U . Let

(DU ,≤) = (i, F ) ∈ D × U : xi ∈ F


where (i, F ) ≤ (j, H) if and only if j ≥ i and xj ∈ H ⊆ F . So DU is a set directed by≤.

We define the function, g : DU → D, as g(j, F ) = j if xj ∈ F .

Then (f g)(j, F ) = f(j) ∈ A where i ≥ j implies (f g)(i, F ) = f(i) ≥ f(j). So

(f g) maps DU in A, respecting the order, and AU = (f g)[DU ] is cofinal. We haveshown above that the net,

AU = (f g)(u, F ) : (u, F ) ∈ DU

generated by an ultrafilter is an ultranet. So AU is an ultranet which is a subnet of A.

Concepts review:

1. Define filter base and filter.

2. How does a filter base generate a filter?

3. What does it mean to say that a filter base is fixed? When is it free?

4. If A is a net describe a filter base of sets determined by the net, A.

5. What does it mean to say that a filter base of sets converges to a point x?

6. How do we define an accumulation point of a filter base of sets?

7. Given a filter base, F , how do we define the adherence of F?

8. In what kind of topological space do convergent filter bases have precisely one limit

point?

9. If a filter base, F , has a non-limit accumulation point, p, describe a filter whichcontains F and converges to p.

10. How do we recognize the points in the closure, clSF , of a set, F , in terms of filters?

11. Give a characterization of a continuous function, f : S → T , at a point in terms of

filters.

12. What does it mean to say that the filter, F , is finer than the filter, H ?

13. Define an ultrafilter of sets, in P(S).

14. Give two characterizations of an ultrafilter of sets.


15. If S is a Hausdorff space, how many accumulation points can an ultrafilter in P(S)have?

16. Define an ultranet.

EXERCISES

1. Suppose H is a filter base of sets which is finer than the filter base F . If F convergesto p does H necessarily converge to p?

2. Suppose p is an accumulation point of the ultrafilter of sets, U . Does U necessarilyconverge to p?

3. Let f : S → T be a function mapping the topological space S onto the topological

space T . If F is an ultrafilter of sets in P(S) and f [F ] = f [F ] : F ∈ F is f [F ]necessarily a filter?

4. If V is a non-empty open subset of the topological space S and the filter of sets, F ,

converges to some point p ∈ V , does V necessarily belong to F?

5. Suppose U is a non-empty subset of the space S. Suppose U belongs to every filter,F , which converges to a point p of U , is U necessarily open in S?

6. Let u be a point in a topological space S. Let F = F ∈ P(S) : u ∈ F and S\F is

finite.

a) Show that F is a filter of sets in P(S).

b) Show that F is a neighbourhood system of u.

7. Let S be a topological space and F and H be two filters in P(S). Let

F × H = F ×H : F ∈ F and H ∈ H

Is F × H a filter base?

8. Let S be a topological space an F is a filter of sets in P(S). Let a(F ) be the

adherence set of F . Show that a(F ) is a closed subset of S.

9. Suppose U is an ultrafilter in P(S). If U and V are disjoint non-empty subsets of Ssuch that U ∪ V belongs to U show that either U or V belongs to U .

Part V

Compact spaces and relatives

275

Part V: Compact spaces and relatives 277

14 / Compactness: Definition and basic properties.

Summary. In this section we define and describe properties of compactness

in a topological space. The concept of compactness is a generalization of the“closed and bounded” property in the Euclidean space R. It was introduced byPavel Urysohn in 1929. We give the most general definition of compactness in

terms of covers and subcovers. The few characterizations given for this propertywill simplify the task of determining whether a space is compact. These char-

acterizations also provide a deeper understanding of what it means for a set tobe “compact”. We also show that the compact property is carried over from one

topological space to another by continuous functions. Furthermore, the compactproperty carries over from the factors of a product space to the product space

itself. Subsets of a compact space, S, will be seen to be closed only if S is Haus-dorff. But closed subsets of any compact space are always compact.

14.1 Introduction.

Many readers may already be familiar with the notion of “compactness”. It is oftenencountered early in a course of real analysis, in a form that is somewhat different

from the one we will encounter here. Initially, some readers will wonder whether weare talking about the same property. If viewed in this context a possible definition

could have been: “A subset, F , of Rn is compact if and only if every sequence in Fhas a convergent subsequence with its limit point inside F”. Or, in the context of

normed vector spaces, the reader may have encountered a theorem which states: “If Vis a finite dimensional vector space, the compact subsets of V are precisely the closed

and bounded subsets”. One of many reasons the notion of compactness would havebeen introduced in a previous course is to access a statement commonly called the

“Extreme value theorem” (EVT) which says: “Given that f : V → R is a continuousfunction mapping a normed vector into R, if F is a compact subset of V , then fattains a maximum value and a minimum value on F”. We will be studying precisely

the same notion of compactness in the more general context of a topological space.Compact spaces are encountered in numerous other fields of mathematics. In general,

most students will feel that spaces which are both compact and Hausdorff are moreintuitive; they seem to be “well-behaved” since they come with many useful tools that

can be used to solve various problems.

In what follows, we will be referring to an “open cover” of a subset F of a space, S.

We say that U = Ui : i ∈ I is an open cover of F if each Ui is an open subset of Sand F = ∪Ui : i ∈ I. A “finite subcover”, Ui : i ∈ K, is a finite subset of an open

cover, U , of F , which also covers F .

278 Section 14: Compactness: Definition and basic properties.

Definition 14.1 Let F be any non-empty subspace of the topological space, S. We saythat F is compact if and only if every S-open cover of F has a finite subcover.

Since the reader may have encountered other definitions of “compact” we will refer to

14.1 above as the topological definition of compactness. We provide a few examples ofcompact and non-compact spaces to help the reader develop a better feeling for what

“every open cover has an a finite subcover” means.

Example 1. Note that the cover, V = (i − 2/3, i + 2/3) : i ∈ Z, forms an open

cover of each of the four spaces, R, Q, J, (the irrationals) Z with the usual topol-ogy. But removing just one set from V will leave a point which is not covered by the

other sets in V . So V has no finite subcover. So none of these four spaces are compact.

Example 2. Consider the set, F = 1/n : n = 1, 2, 3, . . . ∪ 0 with the subspace

topology inherited from R. Let V be an R-open cover of F . There exists U ∈ V suchthat 0 ∈ U . Then there are at most finitely many points in F \U . For each of these

points we can choose precisely one set from V which contains it. The set U alongwith each one of these sets covers F . So F is compact. If we remove the point, 0,

from F then the resulting space is not compact. (Verify this!)

It will be useful to remember the particular technique exhibited in the following exam-

ple, since it will be called upon to help solve various questions in the next few chapters.

Example 3. Consider the bounded closed interval, S = [3, 7], viewed as a subspace of

R. Let V = Vi : i ∈ I be an open covering of S. Let

U = u ∈ [3, 7] : where [3, u] is covered by finitely many sets from V

Let k = sup u : u ∈ U . Then 3 ≤ k ≤ 7 and [3, 3] ⊆ [3, k] ⊆ [3, 7].There exists Vj ∈ V such that k ∈ Vj.

Suppose k < 7. Then there exists ε > 0 such that k ∈ (k − ε, k + ε) ⊆ Vj ⊆ [3, 7].By definition of k, [3, k− ε

2 ] has a finite subcover, say VF . Then VF ∪ Vj is a finite

subcover of [3, k]. Also, [3, k + ε2 ] has a finite subcover, VF ∪ Vj. This contradicts

the definition of k. The source of our contradiction is our supposition that k < 7. Sok = 7. Hence S has a finite subcover so S is compact.

14.2 Characterizations of the compact property.

We will be needing a few efficient ways of determining whether a subspace is compact(or not). The following characterizations will be useful.

But first we should develop some familiarity with the following concept. It is oftenabbreviated by the acronym, FIP.


A family, F , of sets is said to satisfy the finite intersection property if“every finite subfamily of F has non-empty intersection”.

A “filter of sets” is a prime example of a family of sets which satisfies the finite inter-

section property.

We present the following various ways of recognizing a compact set.1

Theorem 14.2 Let S be a topological space. The following are equivalent.

a) The space S is compact.

b) Any family, F , of closed subsets of S which satisfies the finite intersection propertyhas non-empty intersection.

c) Every filter of sets in S has an accumulation point.

d) Every ultrafilter of sets in S has a limit point.

e) Every net in S has an accumulation point.

f) Every ultranet in S has a limit point.


( a ⇒ b ) Let S be a compact topological space. Suppose the family, F , of closed

subsets of S satisfies the finite intersection property. That is, for every finite subset,W , of F , ∩F : F ∈ W 6= ∅.

We are required to show that ∩F : F ∈ F 6= ∅.

Suppose ∩F : F ∈ F = ∅. Then, S\F : F ∈ F is an open cover of S. Then, byhypothesis, there exists a finite subset, W , of F , such that S = ∪S\F : F ∈ W .Equivalently, ∩F : F ∈ W = ∅, contradicting the given fact that F satisfies thefinite intersection property. So, ∩F : F ∈ F 6= ∅, as required.

( b ⇒ a ) Suppose that, whenever a family, F , of closed sets satisfies the finite inter-

section property, then ∩F : F ∈ F 6= ∅. Let U be an open cover of S. If, for anyfinite W of U , ∪F : F ∈ W 6= S, then the set S\F : F ∈ U satisfies the finiteintersection property, and so ∩S\F : F ∈ U 6= ∅. This can only mean that, U is

not an open cover of S, a contradiction. So U has a finite subcover of S.

( b ⇒ c ) We are given that if a family of closed sets satisfies the FIP then it hasnon-empty intersection. Let F be a filter base of subsets in S. We are required

1Those readers who have skipped studying the chapters on nets, will find properties b), d) the mostuseful. These are the characterizations that are the most often called upon, in future proofs.


to show that F has an accumulation point. That is, there is a point x such thatx ∈ clSF for all F ∈ F . Then, by definition of filter base, F satisfies the finite

intersection property. Then, if G is a finite subset, of F , ∩F : F ∈ G 6= ∅. Then∩clSF : F ∈ G 6= ∅. By hypothesis, there exists x ∈ ∩clSF : F ∈ F 6= ∅. Then

F has an accumulation point.

( c ⇒ d ) Suppose every filter of sets has an accumulation point. Let U be an ultra-

filter of sets in P(S). We must show that U has a limit point. By hypothesis, Umust have an accumulation point, say p. By theorem 13.10, p is a limit point of some

filter, H ∗ containing U . Since U is a maximal filter, then U = H ∗. So p is a limitpoint of U .

( d ⇒ c ) We are given that every ultrafilter has a limit point. We must show that ev-

ery filter of sets has an accumulation point. Let F be a filter base. Then, by theorem13.16, F is contained in some ultrafilter, U . Let p be the limit point of U . Then, for

any open neighbourhood, U ∈ Bp, of p, there is F ∈ U such that F ⊆ U . This meansevery open neighbourhood of p intersects every F ∈ U . Then p ∈ ∩clSF : F ∈ F.So p is an accumulation pont of F

( c ⇒ b ) We are given that every filter base in P(S) has an accumulation point.

Suppose the family, F , of closed subsets of S satisfies the finite intersection property.

By hypothesis, F has an accumulation point, say p.

We are required to show that ∩F : F ∈ F 6= ∅.

Note that, F , satisfies the main filter base property and so F a “filter base of closedsets” which has an accumulation point, p. Then, by definition, p ∈ ∩clSF : F ∈F 6= ∅. Since each F is closed in S, p ∈ ∩F : F ∈ F 6= ∅, as required.

( f ⇔ d ) By theorem 13.8 and the example on page 270, an ultrafilter has limit pointp if and only if its corresponding ultranet has limit p.

( e ⇔ c ) By theorem 13.8 a filter has accumulation point p if and only if its corre-sponding net has accumulation limit p. By theorem 12.5, if a net has an accumulation

point p then it has a subnet which converges to p.

The reader will soon discover that, compact spaces are more intuitive and richer in

interesting properties when equipped with the Hausdorff separation axiom. It is of-ten assumed that “compact sets are always closed” since, in many books, to removeclutter from the main body of the text, a blanket assumption is initially made that all

spaces are hypothesized to be Hausdorff. But a compact space may not be closed ifthe space is not Hausdorff. When referring to various texts, readers should keep this

in mind, and check at the begin of the text to see if any blanket assumptions are made.


14.3 Properties of compact subsets.

We will now verify whether the compactness property is carried over by continuousfunctions from its domain to its codomain. Also, we will show that closed subspacesof compact spaces are compact. These results will be familiar to readers with some

experience in real analysis. The proofs confirm that these properties generalize nicely,from Rn to topological spaces.

Theorem 14.3 Let S and T be topological spaces.

a) Suppose f : S → T is a continuous function mapping S into T . If F is a compactsubset of S then f [F ] is compact in T . So continuous images of compact sets are

compact.

b) Suppose S is compact. If H is a closed subset of S, then H is compact.

c) The union of a finite collection of compact sets is compact.

d) Suppose S is Hausdorff.

i) If H is a compact subset of S then H is closed in S.

ii) If S is also compact then S is regular.

iii) If S is also compact then S is normal.

iv) The intersection of any collection of compact subsets is compact. (Note: TheHausdorff property is required in the proof of this statement.)


a) We are given that the function, f : S → T , is continuous and that F is a compact

subset of S.

Let U be an open cover of f [F ]. Then f←[U ] ∩ F : U ∈ U is an open cover

of F . Since F is compact, there is a finite subcover, f←[U ] ∩ F : U ∈ F, ofopen subsets of F . Since,

F ⊆ ∪f←[U ] ∩ F : U ∈ F ⇒ f [F ] ⊆ f [∪f←[U ] ∩ F : U ∈ F]

then,

f [F ] ⊆ ∪f [f←[U ] ∩ F ] : U ∈ F⊆ ∪f [f←[U ]] ∩ f [F ] : U ∈ F= ∪U ∩ f [F ] : U ∈ F

So U ∩ f [F ] : U ∈ F is a finite subcover of f [F ]. So continuous images of

compact sets are compact.


b) Suppose S is a compact space. We are given that H is a closed subset of S. LetF = Fi : i ∈ I be a family of closed subsets in the subspace, H , such that

F satisfies the finite intersection property. Note that, since H is closed, thenclSFi = Fi for all i ∈ I . So that F is a family of closed subsets of S such that Fsatisfies the FIP. Since S is compact, there is a point p ∈ ∩F : F ∈ F. Sincep ∈ F ⊆ H , ∩F : F ∈ F is a non-empty subset of H . So H is compact.1

c) Let K1, K2, . . . , Kn represent a finite collection of compact subsets of a space,

S. Let K = ∪i=1 to nKi. Let U = Uα : α ∈ A be an open cover of K. Thenfor each i, Ki has a finite subcover Uαi : αi ∈ Fi. Then ∪i=1to n[∪αi∈Fi

Uαi ]forms a finite subcover of K. So K is compact.

d) For what follows, suppose S is Hausdorff.

i) Let H be a compact subset of S (where S need not be compact). We are

required to show that H is closed in S. It will suffice to show that S \H isopen in S. Let u ∈ S \H . Since S is Hausdorff, then for each x ∈ H , there

exists disjoint open neighbourhoods, Ux and V xu , of x and u respectively. Since

H is compact, the open cover, Ux : x ∈ H, will have a finite subcover, say

Uxi : i ∈ F and finite open neighbourhoods of u, say V uxi

: i ∈ F, whereUxi ∩ V u

xi= ∅, for i ∈ F . Then, for each i ∈ F ,

Uxi ∩ [∩V uxi

: i ∈ F] = ∅

Let Wu = ∪Uxi : i ∈ F and Mu = ∩V uxi

: i ∈ F.Then Wu ∩Mu = ∅. So Mu ∩H = ∅. Then Mu ∩H = ∅ holds true for anychoice of u ∈ S \H . We have then shown that every point, u, in S \H has an

open neighbourhood, Mu, which has empty intersection with H . So S \H isopen; hence H is closed. (Note how the Hausdorff property plays a role in theproof.)

ii) We are given that S is compact Hausdorff. We are required to show that Sis regular. Let H be closed and u ∈ S \H . Then, since S is compact, H is

compact. In part ii) we constructed disjoint open neighbourhoods, Wu andMu, of H and u, respectively. So S is regular.

iii) We are given that S is compact Hausdorff. Let H and K be disjoint closed

subsets of S. Then, since S is compact, H and K are compact. Let u ∈ K.We showed in part ii) that S is regular and so there exists disjoint open neigh-

bourhoods, Wu and Mu of H and u, respectively. Then Mu : u ∈ K formsan open cover of K which has a finite subcover, say Mui : i ∈ F. Then

∩Wui : i ∈ F and ∩Mui : i ∈ F form disjoint open neighbourhoods of H

1Note that the Hausdorff property for S is not required for this to hold true. That is, “Closed subsets of

compact spaces are always compact”.


and K. So S is normal.

iv) Let Kα : α ∈ A be a collection of compact subsets of a Hausdorff space, S.Let K = ∩α∈AKα. Since S is Hausdorff and each Kα is closed; then so is K.

Let αi ∈ A. Since, by hypothesis, Kαi is compact and K ⊆ Kαi then K iscompact.

Some readers may find it useful to “think” of (or manipulate) compact subsets ofHausdorff spaces as if they are “points”.

Example 4. Construct a topological space, S, with two compact subspaces, A and B,

such that A ∩B is non-compact.

Solution : Given theorem 14.3 part d) we need only consider spaces which are not

Hausdorff. Consider the set, N, equipped with topology generated by the base,

B = ∅, N, N\0, N\1 ∪ n : n ≥ 2

Let A = N\0 and B = N\1. Let UA be an open cover of A. Then the only

element of B which will cover 1 is A = N\0. So A is compact. Similarly, B iscompact. But any cover of A ∩ B contains infinitely many isolated points. So eventhough both A and B are compact, A ∩ B is not compact.

Example 5. Let F be a compact subset of a topological space S. Suppose f : S → Tis a continuous one-to-one function mapping F into a Hausdorff space, T . Show that

f [F ] is a homeomorphic copy of F .

Solution : By theorem 6.9, it will suffice to show that f : F → T is a closed functionon F . Let K be a closed subset of F . Then, by the above theorem, K is compact.

Then f [K] is compact in T . Since T is Hausdorff, then f [F ] is Hausdorff. Since f [K]is a compact subset of f [F ], then it is closed. Then f is a closed mapping on F . So

f : F → f [F ] is homeomorphism.

The following very important theorem titled The Tychonoff theorem will often be

referred to in the proofs that will follow. It is surprising how often this theorem isinvoked in proofs of topology and real analysis. It shows that the Cartesian product

of any number of compact spaces is compact.

Theorem 14.4 The Tychonoff theorem . Let Si : i ∈ I be a family of topological spaces.Then the product space, S =

∏

i∈I Si, is compact if and only if Si is compact for each i ∈ I .


Proof : Let Si : i ∈ I be a family of topological spaces.

( ⇒ ) We are given that S =∏

i∈I Si is compact. Then, since the projection map,πi : S → Si, is continuous for each i ∈ I , and continuous images of compact sets are

compact, then πi[S] = Si is compact, for each i ∈ I .

( ⇐ ) We are given that Si is compact for each i ∈ I .

Let

U = π←i [U ] : U open in Si, i ∈ I

be a cover of open subbase sets of S. Let

Ui = U : U ⊆ Si, π←i [U ] ∈ U

be an open cover of Si corresponding to U .

Let

UF = π←i [U ] : finitely many open U ’s in each Si, i ∈ F

be a finite subset of U . For each i ∈ F , let U ∗i be a finite subset of Ui which covers Si.

Then

U ∗F = π←i [U ] : U ∈ U ∗

i , i ∈ F

is a finite subset of U .

If U doesn’t not have a finite subcover for S then there must be a point p =<xi>i∈I

∈ S such that p 6∈ π←j [U ] for all j ∈ F and all U ∈ U ∗j . This means xj 6∈ U ∈ U ∗

j

for all j ∈ F . This contradicts that U ∗i covers Si for each i ∈ F . So U has a finite

subcover for S.

The proof for the case where U is a cover of basic open sets mimics the one provided

above for the case of the cover of open subbase sets of S.

Alternate approach for the proof of ( ⇐ ). We are given that Si is compact for eachi ∈ I . Let U be an ultrafilter in S =

∏

i∈I Si. By theorem 14.2, compact spaces are

precisely those spaces in which every ultrafilter of sets converges to a point in thatset. So it will suffice to show that U has a limit point in S. Then

Ui = πi[U ] = πi[U ] : U ∈ U

is a filter base in Si.1

1Let πi[U ] and πi[V ] be two elements of πi[U ] where U and V are elements of U . Then there existsnon-empty W ∈ U such that W ⊆ U ∩ V . Then πi[W ] ⊆ f [U ∩ V ] ⊆ f [U ]∩ f [V ] 6= ∅. So, πi[U ] is a filterbase.


Since U is a maximal filter then Ui = πi[U ] must be a maximal filter for each i ∈ I .2

Since Si is compact each Ui converges to a point xi. By theorem 13.17, U convergesto < xi >i∈I . Then every ultrafilter of sets in S converges to a point of S. So the

product space S is compact.

We will now see that, in Euclidean spaces, Rn, the compact subsets are precisely the

“closed and bounded” subsets of Rn.

In the following theorem we consider the Euclidean topological space Rn. We will saythat a subset, U , is bounded in Rn if there exists k ∈ R+ such that

U ⊆ [−k, k]n ⊆ Rn

The following theorem states that. . . ,

“. . . in Rn, compactness is equivalent to the closed and bounded properties combined.”

Theorem 14.5 Let Rn be the topological space equipped with the product topology. Thenon-empty subset, T of Rn, is a compact subset if and only if T is both closed and bounded

in Rn.

Proof : When we invoke the Tychonoff theorem, the proof is straightforward and so isassigned as an exercise question.

Theorem 14.6 Let F be a compact subset of a topological space, S. If f is a continuous

real-valued function on F then f attains its maximum and minimum values inside F .

Proof : We are given that F is compact in S and f [F ] is the continuous image of F inR. Then f [F ] is compact in R (by 14.3) and so is closed and bounded (by 14.5).

Then boundedness of f [F ] guarantees that f [F ] ⊆ [−k, k], for some k. Since f [F ] isclosed in R then sup f [F ] and inf f [F ] must both belong to f [F ]. So f attains bothits maximum and minimum values on F .

2To see this, suppose Ui is a subset of Si which intersects every element of πi[U ]. Then πi←[Ui] intersects

every element of U . Since U is an ultrafilter, πi←[Ui] ∈ U . Then πi[πi

←[Ui]] = Ui ∈ πi[U ].


Example 6. Suppose S and T are topological spaces. We know that projection mapson product spaces are open maps. (See theorem on page 121)

Show that, in the case where the space S is Hausdorff and the space T is compact

then the projection map, π1 : S × T → S, is a closed map.

Solution : Let K be a closed subset of S × T . We are required to show that π1[K] isclosed. It then suffices to show that S\π1[K] is open in S.

Let u ∈ S\π1[K]. Then (u × T ) ∩K = ∅. Since T is compact then we easily seethat u × T is compact. Since S is Hausdorff, for each (u, x) ∈ u × T there is an

open neighbourhood, V xu × Ux, which does not meet K. In this way we obtain an

open cover

V xu × Ux : x ∈ T

of u × T which then has a finite subcover

V xiu × Uxi : xi ∈ F ⊆ T

Then ∩V xiu × Uxi : xi ∈ F ⊆ T forms an open neighbourhood of u which does not

intersect π1[K]. So S\π1[K] is open in S, as claimed. We conclude that π1 : S×T → Sis a closed projection map.

Example 7. For the Hausdorff topological space, S, and the compact space, T , let

f : S → T be a function mapping S into T . The set, G, will represent the graph of f ,defined as

G = (x, f(x)) : x ∈ S ⊆ S × T

Show that, if G is a closed subspace of the product space, S × T , then f : S → Tmust be a continuous function.

Solution : Given: S is Hausdorff and T is compact. Also, we are given that G =(x, f(x)) : x ∈ S is a closed subset of S × T . Let K be a closed subset of T . To

show continuity of f , it will suffice to show that f←[K] is a closed subset of S.

By the result stated in the example immediately above, π1 : S × T → S is a closedmap. See that π←2 [K] ∩ G is the intersection of two closed sets and so is a closedsubset of S × T . Since

π←2 [K] ∩G = (x, f(x)) : f(x) ∈ K

then π1[π←2 [K] ∩G] = f←[K].

Since π1 is a closed map, f←[K] is closed. We have shown that f pulls back closed

sets to closed sets; so f is continuous.


14.4 Topic : The Embedding theorem III.

Recall the statement of the Embedding theorem II (theorem 10.16) which says that aspace S is completely regular if and only if the evaluation map with respect to C∗(S)embeds S inside a cube, T =

∏

i∈I[ai, bi].

Now we know something else about cubes: Since they are the product of compact

sets, the Tychonoff theorem guarantees that they are compact. Since∏

i∈I [ai, bi]and

∏

i∈I[0, 1] are homeomorphic then S can be embedded in the compact space,∏

i∈I[0, 1]. With this in mind, we promote the Embedding theorem II to Embeddingtheorem III, in the form of the following theorem.

Theorem 14.7 Embedding theorem III. If a topological space S is completely regular

then S can be densely embedded in a compact Hausdorff space.

Proof : We are given a completely regular topological space S.

Then, by the theorem 10.16, the evaluation map with respect to C∗(S) embeds S intoa cube, T =

∏

i∈I [ai, bi]. Each factor of the product T is Hausdorff. Since a product

of Hausdorff spaces was shown to be Hausdorff, T is Hausdorff. By the Tychonofftheorem, T is a compact Hausdorff space.

Since clT e[S] is a closed subset of a Hausdorff compact space T , it is Hausdorff com-pact. Then the homeomorphic image, e[S], of S is dense in clT e[S], a Hausdorff

compact set, as required.

Theorem 14.8 Let S be a topological space. Then S is completely regular if and only if

S is a subspace of a Hausdorff compact space.

Proof : We are given that S is any topological space.

( ⇒ ) Suppose S is a subspace of a compact Hausdorff space, T . By Urysohn’s lemma,there is a continuous bounded function which separates any two closed subsets (so one

that separates a point and a closed set). So T is completely regular. By theorem 10.5,since S is a subspace of a completely regular space, it is completely regular, as required.

( ⇐ ) Suppose S is completely regular. Then, by theorem 14.7, S is densely embeddedin a compact Hausdorff space T . That is, a topological copy of S is a subspace of T .

So S is a subspace T .


14.5 Topic : Completely normal spaces.

With the few results involving compact spaces behind us we can now delve a bitdeeper on the topic of complete separation of sets. We first remind ourselves of somepreviously proven results.

– The Cartesian product (equipped with the product topology) of compact spaces

is compact. (Tychonoff theorem 14.4)

– Any compact Hausdorff space is normal. (Theorem 14.3 c) )

– Metrizable spaces are normal. (Theorem 9.19)

– Since subspaces of metrizable spaces are metrizable (Example on page 87), thosenormal spaces which are metrizable have normal subspaces.

We have never proven that subspaces of normal spaces are always normal. The mainreason is that it is impossible to do so. The Embedding theorem III (Theorem 14.7)

confirms that every completely regular space can be (densely) embedded in a compactHausdorff space. We know, however, that there are completely regular spaces which

are not normal. The Moore plane is such an example. (See the Moore plane onpage 214). Then...

... the Moore plane is a non-normal subspace of the compact (normal) space

in which it is embedded.

Those normal spaces whose subspaces are normal form a particular class of topologicalspaces.

Definition 14.9 A completely normal space is a space in which every subspace is normal.

Then every metrizable space is completely normal.

By the given definition, since a topological space is a subspace of itself,

. . . completely normal spaces are normal spaces

Recall that a perfectly normal space is a normal space whose closed subsets are Gδ’s.We investigate how “completely normal” spaces compare with “perfectly normal”

spaces.


Suppose S is perfectly normal. We claim that S must then be completely normal.Then, by definition, S is normal. Let T be a subspace of S. To show that S is

completely normal it suffices to show that T is a normal subspace. Let F be a closedsubset of T . Then F = F ∗ ∩ T for some closed subset, F ∗, of S. Since S is perfectly

normal, F ∗ is a Gδ in S, then F is a Gδ of T . So T is perfectly normal and so is anormal subspace of S. We can then say that a perfectly normal space, S, is completely

normal. So we have the chain,

perfectly normal ⇒ completely normal ⇒ normal ⇒ completely regular

In the next example we have a characterization of a completely normal space.

Example 8. Show that S is completely normal if and only if for any pair of subsets Aand B such that A ∩ clSB = ∅ = clSA ∩ B there exists disjoint open subsets U and

V such that A ⊆ U and B ⊆ V .1

Solution: ( ⇐ ) Let S be a topological space. Suppose that for any pair of subsets A

and B such that A ∩ clSB = ∅ = clSA ∩ B there exists disjoint open subsets U andV such that A ⊆ U and B ⊆ V . If C and D are disjoint closed subsets of S then

C ∩ clSD = ∅ = clSC ∩D. Then, by hypothesis, there exists disjoint open subsets Uand V such that C ⊆ U and D ⊆ V . So S is normal.

We now claim that S is completely normal. Let T be a subspace of S. It suffices toshow that T is normal. Suppose F and K are disjoint non-empty closed subsets of

T . Then it suffices to show that F and K are contained in disjoint open sets. LetF ∗ and K∗ are closed subsets of S such that F = F ∗ ∩ T and K = K∗ ∩ T . ThenF ∗ ∩ clSK

∗ = ∅ = clSF∗ ∩K∗. By hypothesis, there exists disjoint open subsets of

S, U∗ and V ∗, such that F ∗ ⊆ U∗ and K∗ ⊆ V ∗. Then F ⊆ U∗ ∩ T and K ⊆ V ∗ ∩ T .Then T must be normal. We can conclude that, by definition, S is completely normal.

The direction ( ⇒ ) is left as an exercise.

A normal space which is not perfectly normal. The following example illustrates a

completely normal space which is not perfectly normal.

Example 9. Suppose S is an uncountable set and and p ∈ S. We equip S with atopology, τ , defined as follows:

τ = T ⊆ S : S\T is finite or p 6∈ T

Show that S is completely normal but not perfectly normal.

Solution : We claim that S is T1. The set p is closed and if x 6= p then S\x isopen so x is closed. So S is T1, as claimed.

1This is an exercise question which appears in 15B of S. Willard’s Topology.


We claim that S is completely normal. Suppose F and K are disjoint closed subsetsof S. If p 6∈ K ∪ F then K and F are both open. So F and K are clopen and so

F ∩ clSK = ∅ = clSF ∩ K. Suppose p ∈ F . Then K is clopen and finite. That isK = clSK is open. Now S\K is open since K is finite. So clSF and K are contained

disjoint open sets, S\K and K as well as F and clSK. So, by the statement provenin example immediately above, S is completely normal, as claimed.

We claim that S is not perfectly normal. Let Ui : i ∈ N be a countable fam-

ily of sets in S each containing the point p. Then each Ui is closed in S andp ∈ ∩Ui : i ∈ N. If each Ui is also open S \Ui if finite. So each Ui is uncount-

able. Then S \∩Ui : i ∈ N = ∪S \Ui : i ∈ N, a countable subset of S. So∩Ui : i ∈ N is uncountable. So p 6= ∩Ui : i ∈ N. So p is not a Gδ. So S is notperfectly normal, as claimed.

So we can confidently state that . . . ,

completely normal 6⇒ perfectly normal

14.6 Topic : Compactness in terms of filters of zero-sets.

For a given topological space, (S, τ), the symbols C(S) and C∗(S) denote the set of

all real-valued continuous functions and the set of real-valued continuous boundedfunctions, respectively, on S. We recall a few concepts presented in an earlier chapter

on separation of closed subsets with functions.

– If S is a topological space and f ∈ C(S) then Z(f) = f←(0) is referred to as a

“zero-set” in S. We represent the family of all zero-sets in S by

Z[S] = Z(f) : f ∈ C(S)

– Recall that zero-sets are closed Gδ’s.

– On the other hand, given a closed Gδ, F , of S, F need not necessarily be azero-set.

– However, we have shown that, in a normal space, S, every closed Gδ is a zero-setin S.

Even though Z[S] is just a particular type of subset of P(S), the expression Z[S]makes sense only if S is a known topological space. After all, by definition, Z(f) is

the zero-set associated to a continuous real-valued function on S.

We would like to discuss subfamilies of Z[S] which are “filters”. To distinguish a filter

in Z[S] from a filter of sets in P(S) we refer to filters restricted to Z[S] as

z-filters


Even though z-filters are, in most ways, analogous to filters of sets, for the sake ofcompleteness, we include the following definitions.

Definition 14.10 Let S be a topological space. Let F ⊆ Z[S].

If F satisfies the two properties:

1) ∅ 6∈ F ,

2) For every U, V ∈ F , there exists F ∈ F such that F ⊆ U ∩ V .

then we will say that F is a z-filter base.

If F is a z-filter base which also satisfies the condition:

3) Whenever U ∈ F and there is V ∈ Z[S]\∅ such that U ⊆ V , then V ∈ F .

Then F is called a z-filter.

4) If F is a z-filter which is not contained in any other strictly larger z-filter then wesay that F is a maximal z-filter or, more commonly, a z-ultrafilter.

A quick way of describing a z-filter is to say:

“A z-filter is a subset of Z[S]\∅ which is closed under finite intersections

and supersets both in Z[S].”

Suppose F = Z(f) : f ∈ U is known to be a z-filter where f and g both belong to

U . Then, by definition of z-filter, Z(f)∩Z(g) and Z(f)∪Z(g) must both be zero-setswhich belong to F . We must then assume that both Z(f) ∩ Z(g) and Z(f) ∪ Z(g)

are zero-sets. This should nevertheless be verified. Verify that

Z(f2 + g2) = Z(f) ∩ Z(g)

Z(fg) = Z(f) ∪ Z(g)

A z-filter is, itself, a z-filter base, while a z-filter base, if not a z-filter, can be completedto become a z-filter. Given a z-filter base, F , we define the larger set, F ∗, as follows:

F ∗ = U ∈ Z[S] : F ⊆ U , for some F ∈ F

We have simply adjoined to F all its supersets which belong to Z[S]. It is easilyverified that F ∗ is a z-filter. We will say that the z-filter base, F , generates the


z-filter F ∗.

Every z-filter is usually expected to be a proper subfamily of a filter of sets. In the

event that every subset is a zero-set (such as in a discrete space, for example) a z-filteris simply a filter of sets.

The use of the notation, Z(f), for the zero-set generated by f suggests that Z can be

viewed as a function

Z : C(S) → Z[S]

where C(S) is the algebraic ring, (C(S), +, ·), of continuous functions on S, andZ[S] is a subset of P(S). So Z maps f to Z(f) = f←(0).

A subset (U, +, ·) of C(S) which is also a ring is called a subring.

Those subrings, (U, +, ·), of C(S) satisfying the property,

“For every f ∈ U , gf ∈ U , for all g ∈ C(S)”

are called

“ideals”

For obvious reasons we tend to represent an ideal in P(C(S)) by I .1 If J is an idealin C(S) such that for any ideal I , J ⊆ I implies I = J we say that J is a maximal

ideal in C(S). Ideals in C(S) are relevant in our study since the image Z[I ] of an idealI under the map Z : C(S) → Z[S], will be seen to be z-filters. Also the function Z

will pull back a z-filter, U , in Z[S] to a corresponding ideal, I = Z←[U ].2

The following theorem formally describes the association between ideals in the ringof continuous functions, C(S), and the z-filters in the family of all zero-sets, Z[S], of

the topological space, S.

Theorem 14.11 Suppose S is a topological space. Let I be an ideal in C(S) and F be az-filter in Z[S] (where Z : C(S) → Z[S] is viewed as a function mapping f in C(S) to Z(f)

in Z[S]).

a) Then the subset, Z[I ], of Z[S] is a z-filter.

b) Then the subset, Z←[F ], of C(S) is an ideal in C(S).

c) If J is a maximal ideal then Z[J] is a z-ultrafilter. Conversely, if F is a z-ultrafilterthen Z←[F ] is maximal ideal.

1When we will say that I is an ideal in C(S), we will always mean a “proper ideal”.2Note that (C∗(S), +, ·) also constitutes a ring, and so we can speak of ideals and maximal ideals in

C∗(S) also.


d) If M is the set of all maximal ideals in P(C(S)) and Z is the set of all z-ultrafiltersin P(Z[S]), then Z : M → Z is one-to-one and onto.

e) If Z = Z[M ] is a z-ultrafilter in P(Z[S]) and Z(f) is a zero-set which meets every

member of Z then Z(f) ∈ Z .

Proof : We are given that I is a (proper) ideal in C(S) and F is a z-filter in Z[S].

a) Given: I is a proper ideal. We are required to show that Z[I ] is a z-filter.

Since I is proper, then the unit function 1 is not an element of I . Then theempty-set, in the form of Z(1), is not in Z[I ]. Suppose Z(f) and Z(g) belong toZ[I ]. Since f, g ∈ I and I is an ideal, f2 + g2 ∈ I and so

Z(f2 + g2) = Z(f) ∩ Z(g) ∈ Z[I ]

Suppose f ∈ I and Z(g) ∈ Z[S] such that Z(f) ⊆ Z(g). It suffices to show thatZ(g) ∈ Z[I ]. Then gf ∈ I (since I is an ideal). See that

Z(g) = Z(f) ∪ Z(g) = Z(fg) ∈ Z[I ]

So Z[I ] is a z-filter, as claimed.

b) Given: F is a z-filter. We consider Z←[F ]. We claim that Z←[F ] is an ideal.

Since ∅ 6∈ F , the unit function 1 is not in Z←[F ]. Suppose f and g belong

to Z←[F ]. To show that Z←[F ] is an ideal we must show it is closed underaddition and fk ∈ Z←[F ] for all k ∈ C(S).

Then Z(f), Z(g) and Z(−g) belong to F . So Z(f) ∩ Z(−g) ∈ F . Now, if

u ∈ Z(f)∩Z(−g), then f(u) = 0 = −g(u) then (f + g)(u) = 0, so Z(f)∩Z(g) ⊆Z(f + g). Since F is a z-filter, then Z(f + g) ∈ F . This implies f + g ∈ Z←[F ].

So Z←[F ] is closed under addition.

Let k ∈ C(S). Since Z(f) ⊆ Z(fk), then Z(fk) ∈ F so fk ∈ Z←[F ]. SoZ←[F ] is an ideal, as claimed.

c) Suppose J is a maximal ideal. Then Z[J] is a z-filter (by part a).

We are required to show that Z[J] is a z-ultrafilter. Suppose F is a z-filter suchthat Z[J] ⊆ F . It suffices to show that F = Z[J].

Z[J] ⊆ F ⇒ J ⊆ Z←[Z[J]] ⊆ Z←[F ]

⇒ J ⊆ Z←[F ] (Where J is maximal ideal.)

⇒ J = Z←[F ] (Since Z←[F ] is an ideal (by part b).)

⇒ Z[J] = Z[Z←[F ]] = F

So Z[J] is a z-ultrafilter, as claimed.


On the other hand, if F is a z-ultrafilter, then Z←[F ] is an ideal (by part b).We are required to show that Z←[F ] is a maximal ideal. Suppose K is an ideal

such that Z←[F ] ⊆ K. It will suffice to show that K ⊆ Z←[F ].

See that F = Z[Z←[F ]] ⊆ Z[K]. Then Z[K] = F (since Z[K] is a z-filter andF is a maximal z-filter). So K ⊆ Z←[Z[K]] = Z←[F ]. Then K ⊆ Z←[F ]. So

Z←[F ] is a maximal ideal.

d) The result follows immediately from the definition of “maximal”.

e) The proof is analogous to the similar statement referring to an ultrafilter of sets

proven theorem 13.15.

In the following theorem we guarantee that, given any z-filter, F , in Z[S] there exists

in Z[S] a z-ultrafilter which contains it.

Theorem 14.12 Let S be a topological space and (Z ,⊆), denote the set of all z-filters inP(S), partially ordered by inclusion. If F is any z-filter belonging to Z then there exists

a z-ultrafilter, U , in Z which contains F .

Proof : We are given that F ∈ Z . We wish to prove the existence of a z-ultrafilter whichcontains F . We set up the problem so that Zorn’s lemma applies.

For a given F ∈ Z , consider the set SF = H ∈ Z : F ⊆ H of all z-filters whichcontain F . We are required to show that some z-filter in SF is a z-ultrafilter (i.e.,

SF has a maximal element). If SF = F then F is a z-ultrafilter. Suppose SF

contains other filters. Now F can be the base element of many chains in SF . In fact,SF can be viewed as the union of a family of filter-chains, Cα : α ∈ J, each with

base element, F . Pick an arbitrary filter-chain, Cγ = Fi : i ∈ I in SF . It lookslike,

F ⊆ F1 ⊆ F2 ⊆ F3 ⊆ · · ·

Let T = ∪Fi : Fi ∈ Cγ. Quite clearly, Fi ⊆ T for each i. We claim that the

family of sets, T , is itself a z-filter in the chain Cγ . If U and V are elements of T ,then U and V are elements of some Fj in the chain, Cγ , so U ∩ V is an element of

Fj; so U ∩ V is also an element of T . Then T is a z-filter base of sets which is amaximal element in the chain Cγ .

So every chain, Cα, in SF has a maximal element. We can invoke Zorn’s lemma, to

conclude from this that SF , must also have a maximal z-filter element, U . Since,SF contains every z-filter which contains F , then U is the maximal z-filter which

contains F . Then, by definition, U is the unique z-ultrafilter which contains F .


Definition 14.13 Let S be a topological space. Let I be an ideal in C(S) or C∗(S) and

F = Z[I ] be the corresponding z-filter.

a) If ∩Z(f) : f ∈ I 6= ∅ then we will say that F is a fixed z-filter and that I is a fixedideal.

b) Otherwise, we will say that F is a free z-filter and I is a free ideal.

Theorem 14.14 Suppose S is a completely regular topological space. Then the following

are equivalent.

a) The space S is a compact space.

b) Every z-filter in Z[S] is fixed. Every ideal in C(S) is fixed. Every ideal in C∗(S) is

fixed.

c) Every z-ultrafilter is fixed. Every maximal ideal in C(S) is fixed. Every maximal

ideal in C∗(S) is fixed.

Proof : We are given that S is a completely regular topological space.

(a ⇒ b) Suppose S is compact. Since S is compact then every function in C(S) is bounded;so C(S) = C∗(S). Suppose F = Z(f) : f ∈ I is a z-filter in Z[S] corresponding

to an ideal I in C(S).

Since S is compact and F satisfies the finite intersection property then ∩Z(f) :

f ∈ I is non-empty. So, by definition, F is fixed. Equivalently I is a fixed idealin C(S).

(b ⇒ a) Suppose every z-filter in Z[S] is fixed. We are required to show that S is compact.

Suppose F is a family of closed subsets with the FIP. It suffices to show that Fhas non-empty intersection.

If F ∈ F and p ∈ S\F then, since S is completely regular, there exists f ∈ C(S)such that F ⊆ Z(f) and p ∈ f←(1). Then the set F = ∩Z(f) : f ∈ U, the

intersection of zero-sets. Then

∩F : F ∈ F = ∩Z : Z ∈ G ⊆ Z[S]

the intersection of zero-sets in Z[S]. Since Z : Z ∈ G ⊆ Z[S] is a fixed z-filter,

then ∩F : F ∈ F is non-empty so S is compact.

(b ⇒ c) Suppose every z-filter in Z[S] is fixed. Then every z-ultrafilter in Z[S] is fixed. Aswell every ideal is fixed and so every maximal ideal is fixed.


(c ⇒ b) Suppose every z-ultrafilter in Z[S] is fixed. Then every maximal ideal is fixed.Suppose F is a z-filter in Z[S]. Then, by theorem 14.12, F is contained some

z-ultrafilter, G . By hypothesis, G is fixed, so there exists p ∈ ∩F : F ∈ G . Thenp ∈ ∩F : F ∈ F. So F is fixed. Also every corresponding ideal in C(S) is fixed.

Concepts review:

1. Define an open cover of a space. What does it mean to say that it has a finite subcover?

2. State the formal topological definition of compact subspace.

3. What does it mean to say that a family of sets satisfies the “finite intersection prop-erty”?

4. State a characterization of compact space involving the “finite intersection property”.

5. What kind of properties do nets in a compact space satisfy?

6. What kind of properties do filters in a compact space satisfy?

7. What can we say about ultrafilters of sets in a compact set?

8. What can we say about continuous images of compact sets?

9. What can we say about closed subsets of compact sets?

10. Under what conditions on S does the statement “Compact subsets of S are closed”hold true, in general?

11. Complete the statement: A one-to-one continuous function, f : F → T , mapping acompact space, F , into a Hausdorff space, T , is....

12. Suppose S =∏

i∈I Si is a product space with compact factors, Si. Is this statementtrue or false? “The product space S maybe compact but it depends on the size of the

index I”.

13. What separation axioms will compact Hausdorff spaces satisfy?

14. Define the separation property referred to as completely normal.

15. Give a characterization of “completely normal”.


16. How does a completely normal space compare with a normal space and a perfectlynormal space.

17. Define a z-filter base and a z-filter.

18. Define a z-ultrafilter.

19. Define an ideal and a maximal ideal in C∗(S).

20. Describe a relationship between ideals in C∗(S) and z-filters.

21. Define a fixed filter and a free filter. How are they recognized?

22. Give a characterization of a compact space in terms of z-filters.


EXERCISES

1. Show that a closed and bounded subset of R is compact.

2. Show that disjoint compact sets in a Hausdorff space are respectively contained indisjoint open subsets.

3. Let F and H be be disjoint closed subsets of [0, 5]n. Show that they are contained indisjoint open neighbourhoods.

4. Suppose g : S → T is a function mapping the topological space S into the compact

Hausdorff space T . Show that g is continuous on S if and only if the graph of g,(x, g(x)) : x ∈ S is a closed subset of S × T .

5. Show that the finite union of compact subsets of a topological space is compact.

6. The set T is a compact subset of Rn if and only if T is both closed and bounded inRn.

7. Suppose Fi : i ∈ I is a family of compact sets in a Hausdorff space. Show that∩Fi : i ∈ I is compact.


15 / Countably compact spaces.

Summary. In this section we discuss a slightly weaker version of the compact-ness property called “countable compactness”. We formally define it and discussits properties. We will see that there are countably compact spaces which are not

compact spaces. We also show that, in some topological spaces (such as metricspaces) the countably compact and compact properties are equivalent.

15.1 The countably compact property.

The compact property imposes strong restrictions on a topological space. Sometimes,a weaker version of, “Every open cover has a finite subcover” will suit our purposes. A

slightly weaker version, can be expressed as “Every countable open cover has a finitesubcover”. We will investigate properties which are specific to countable compactness.

Definition 15.1 Let F be any non-empty subset of the topological space, S. We say thatF is countably compact if and only if every countable open cover of F has a finite subcover.

Just as for compact subsets, it will be useful to obtain various ways of recognizingthose spaces which are countably compact. It will also allow us to better see how

countably compact spaces distinguish themselves from those which satisfy the com-pact property. Recall that p is said to be a “cluster point of the set A if every openneighbourhood of p contains some other point of A”.

Theorem 15.2 Let S be a topological space and T be a non-empty subset of S. Then thefollowing are equivalent.

a) The space S is a countably compact set.

b) If T is a countably infinite subset of S, then T has at least one cluster point.

c) If A = xi : i ∈ N is a sequence with an infinite number of distinct points, then Ahas at least one accumulation point.

d) A countable family, F = Fi : i ∈ N, of closed sets in S satisfying the finite intersec-tion property has non-empty intersection in S.

300 Section 15 : Countably compact spaces

Proof : Given: S is a topological space.

( a⇒ b ) Given: S is countably compact and T is countably infinite subset of S.

We are required to show that T has a cluster point. Suppose T has no cluster point.

Claim #1: That T = clST . If not, T ⊂ clST . Then there is a point, p, in clST∩(S\T ).So p is a boundary point of T . This would mean that every open neighbourhood of p

intersects T \p; this would make of p a cluster point,a contradiction. So T = clST ,as claimed.

By claim #1, we can say that S\T is open in S. Since T has no cluster point, thenevery point x in T is contained in an open neighbourhood Vx of S which does not

intersect T\x. Then Vx∩T = x. Then Vx : x ∈ T is a countably infinite familyof open subsets of S which covers T . Clearly, it has no finite subset which covers T .

Then,

(S\T ) ∪ Vx : x ∈ T

is a countably infinite open cover of S with no finite subcover, contradicting the factthat S is countably compact. So T must have a cluster point.

( b ⇒ c ) We are given that, a countably infinite subset of S must have a clusterpoint. Let A = f(i) : i ∈ N be an infinite sequence. We are required to show that

the sequence, A, has an accumulation point.

By hypothesis, A has a cluster point, say z, in T . Then, for any neighbourhood, Bz, ofz, Bz ∩A contains infinitely many distinct points. So Bz ∩ f(i) : i > q is non-emptyfor all q ∈ N. This means that any neighbourhood of z intersects a tail end of A. So,

by definition, z is an accumulation point of the sequence, A.

( c ⇒ a ) Suppose that any sequence in any subset, T , of S has an accumulation point.Recall that p is an accumulation point of a sequence if every open neighbourhood of

p intersects the tail end of the sequence. We are required to show that S is countablycompact.

Let U = Ui : i ∈ N be a countable family of open sets which covers S. We arerequired to show that U has a finite subcover. Suppose U has no finite subcover.

Then, for every n, it is possible to choose,

f(n) ∈ T \∪Ui : i = 1 to n

Then f(n) : n ∈ N\0 forms a sequence in T .

Then, for each k,

f [ [k,∞) ] ⊆ T \∪Ui : i = 1 to k

So no element of U can intersect a tail end of the sequenc, f(n), in T . This con-tradicts our hypothesis stating that every sequence in T has an accumulation point.


So U must have a finite subcover.

( a⇒ d ) We are given that S is countably compact. Let F = Fi : i ∈ N be count-able family of closed sets in S with the finite intersection property. We are required

to show that ∩Fi : i ∈ N = ∅.

Suppose ∩Fi : i ∈ N is empty. Then S \Fi : i ∈ N forms an open cover of S withno finite subcover, contradicting the property of countable compactness on S.

( d ⇒ a ) We are given that, if F is a countable family of closed sets in S with the

finite intersection property, then ∩F is not empty. We are required to show that Sis countably compact.

Suppose U = Ui : i ∈ N is an open cover of S. We are required to show that U con-

tains a finite subcover of S. Suppose U has no finite subcover. Then S \Ui : i ∈ Nforms a family of closed sets satisfying the finite intersection property which has empty

intersection. This contradicts our hypothesis. So S is countably compact.

15.2 Bolzano-Weierstrass property

We will examine a bit more closely the countable compactness characterization given

in part c) of the characterization theorem stated above. It states that, in countablycompact spaces,...

“The space S is countably compact if and only if every sequence,

A = xi : i ∈ N

in S has an accumulation point, p, in S.”

The statement is quite similar to the one found in the compactness characterizationtheorem 14.2. . . ,

“A space S is compact if and only if every net in S has an accumulation

point, p, in S.”

By theorem 12.5, “p is an accumulation point of a net N if and only if N has a subnetconverging to p”. We can then say:

“The space S is a compact space if and only if every net N in S has a subnet

which converges to some accumulation point, p, of N .”

and


“The space S is a countably compact space if and only if every sequence Ain S has a subnet which converges to some accumulation point, p, of A.”

To some readers, this may be reminiscent of a real analysis statement called the

Bolzano-Weierstrass theorem. It states that

“Every bounded sequence in R has a convergent subsequence.”

There is, however, a fundamental difference between the statement of the Bolzano-

Weierstrass theorem (BWT) and this particular characterization of the countablecompactness. We see that the BWT refers only to R, not an arbitrary topological

space, and only to those sequences which are “bounded”. (Note that the property of“boundedness” has not yet been defined for arbitrary topological spaces.)

For future reference we will formally provide a name for the property described in theabove countably compact characterization theorem.

Definition 15.3 A topological space, S, is said to satisfy the

Bolzano-Weierstrass property

(BWP) if and only if every sequence, A = xi : i ∈ N, in S has an accumulation point in S.

By the characterization theorem above, countably compact topological spaces are pre-

cisely those spaces which satisfy the Bolzano-Weierstrass property.

15.3 Properties of countably compact spaces.

Before presenting some examples, it will help to list a few properties of countably

compact spaces. The first statement says that a countably compact space is “closed-hereditary”.

Theorem 15.4 Closed subspaces of countably compact spaces are countably compact.

Proof : Given: The space, S, is a countably compact topological space and F is a non-

empty closed subset of S.

Suppose F is a countable family of closed subsets of F which satisfies the finite inter-

section property. It suffices to show that F has non-empty intersection. Then everyset in F is also a closed subset in S. Since S is countably compact the family of sets

in F has non-empty intersection inside F . So F is also countably compact.


We know that continuous images of compact sets are compact. The following theoremshows that an analogous statement for countably compact spaces holds true.

Theorem 15.5 Suppose S is countably compact and the continuous function, f : S → T ,maps S into the topological space, T . Then f [S] is a countably compact subspace of T .

Proof : Given: The space S is countably compact and T is a topological spaces. Supposef : S → T is continuous.

Let U = Ui : i ∈ N be a countable open cover of f [S]. Then V = f←[Ui] : i ∈ Nis a countable open cover of S. Since S is countably compact, V has a finite subcoverVF = f←[Ui] : i ∈ F of S. Then

f [S] ⊆ f [∪f←[Ui] : i ∈ F]= ∪f [f←[Ui]] : i ∈ F ]= Ui : i ∈ F

So U has a finite subcover UF = Ui : i ∈ F of f [S]. So f [S] is also countablycompact.

We now consider how the countable compactness property carries over products.

Theorem 15.6 Let Sj : j ∈ N be a family of topological spaces. We consider the

corresponding product space, S =∏

j∈N Sj, with countably many factors.

a) If the product space, S, is countably compact then so is each one of its Sj factors.

b) If each Sj factor of the product space, S, is both countably compact and first countablethen the infinite product space, S, is countably compact.

Proof : Given: The set, S =∏

j∈N Sj, is a product space.

a) Suppose the product space, S, is countably compact. Since each projection map,πj, is continuous, then, for each factor Sj, Sj = πj[S], the continuous image of

a countably compact space. So, by theorem 15.5 each factor, Sj, is countablycompact. We are done with part a).


b) We are given that Sj : j ∈ N is a family of first countable and countably com-pact topological spaces. We are required to show that S =

∏

j∈N Sj is countably

compact.

Let B = β(i) : i ∈ N be a sequence in S. To show that S is countably compactit suffices to show that B has an accumulation point.

Step 0 : Since π0(β(i)) = π0(< βj(i) >j∈N) = β0(i), then

π0[B] = β0(i) : i ∈ N

is a sequence in S0. Since S0 is countably compact, then, by theorem 15.2, π0[B]

has an accumulation point, say p0. By theorem 11.5, since S0 is first countable,π0[B] has a subsequence,

β0(g0(i)) : i ∈ N

in S0 converging to p0.

Let

B0 = β(g0(i)) : i ∈ N

Then B0 is a subsequence of B in S =∏

j∈N Sj.

Step 1 : We consider the sequence π1[B0] in S1 where

π1(β(g0(i))) = π1(< βj(g0(i)) >j∈N) = β1(g0(i))

Then

π1[B0] = β1(i) : i ∈ N

is a sequence in S1.

Since S1 is first countable, π1[B0] has a subsequence,

β1(g1(g0(i))) : i ∈ N

in S1 converging to p1.

Let B1 = β(g1(g0(i))) : i ∈ N. Then B1 is a subsequence of B in S =∏

j∈N Sj.

Step n : Eventually we obtain a sequence βn(gn(· · · (g0(i))) · · ·)) : i ∈ N, in Sn,converging to pn ∈ Sn.From this, we obtain the subsequence

Bn = β(gn(gn−1(· · · (g0(i))) · · · ) : i ∈ Nof B.


Let γ : N → B be defined as,

γ(n) = β(gn(gn−1(· · · (g0(n))) · · ·)That is,

γ(0) = β(g0(0))

γ(1) = β(g1(g0(1)))

γ(2) = β(g2(g1(g0(2))))

...

γ(n) = β(gn(gn−1(· · · (g0(n))) · · · )...

Then A = γ(i) : i ∈ N is a subsequence of B.

We claim that the subsequence, A, converges to p =< pj >j∈N.

See that, for each j ∈ N, πj(γ(i)) : i ∈ N converges to pj. By corollary 13.18,

γ(i) : i ∈ N ⊂ B ⊂ S =∏

j∈NSj

converges to p =< pj >j∈N, as claimed.

Since B has a converging subsequence, A = γ(i) : i ∈ N, then B has an accumu-

lation point and so the product space, S, is countably compact. As required.

15.4 Examples.

Example 1. Consider the set S = [0, ω1], of all countable ordinals union the singleton

set, ω1, where ω1 is the first uncountable ordinal. The Hausdorff topology on S isgenerated by the base, B, for open sets defined as,

B = (α, β] : α < β

a) Show that S = [0, ω1] is a compact space.

b) Show that the subspace, T = [0, ω1), of S is not compact but is, nevertheless,

countably compact.

Solution : Given: S = [0, ω1] and T = [0, ω1).


a) We will use the formal definition of compactness. The technique mimics the oneused to show that closed and bounded intervals of R are compact. Let U = Bi :

i ∈ I be an open cover of S. It suffices to show that S has a finite subcover.

Let V = u : [0, u] has a finite subcover.1 Suppose ω1 6∈ V . Since the ordinalspace, S, is well-ordered every non-empty subset has a least element. Let k be theleast element in S\V . Then [0, k] does not have a finite subcover. There is some

element, Bj ∈ U , which contains k. Then there exists some m ∈ S such that thebasic open neighbourhood of k, (m, k] ⊆ Bj . But [0, m] has a finite subcover, say

UF ⊂ U . Then UF ∪ Bj is a finite subcover of [0, k]. This is a contradiction. SoV = S. Hence S = [0, ω1] has a finite subcover and so is compact.

b) Consider T = [0, ω1), a subspace of compact S. It is not closed seeing that it doesnot contain the boundary point, ω1 (since every basic open neighbourhood of ω1 in-

tersects T ). Since compact subsets of Hausdorff spaces are closed, then T = [0, ω1)is not compact.

We claim that T is countably compact. Suppose A = xi : i ∈ N is a sequence

in T . By a characterization of “countably compact” it suffices to show that thesequence A has an accumulation point.

As explained on page 88, if supA = ω1, then ω1 = ∪xi : i ∈ N. Then ω1 wouldbe the countable union of countable ordinals and so would be countable, a contra-

diction. So supA < ω1. Then there must exist k ∈ T such that A ⊆ [0, k]. Since[0, k] is a closed subset of the compact space, [0, ω1], then [0, k] is itself compact.We know that, since [0, k] is compact, every net in [0, k] has an accumulation point

in [0, k] (this is a characterization of the compact property). Then the sequence Amust have an accumulation point, say p in [0, k]. Since p ∈ [0, k] ⊂ [0, ω1), then

every sequence in T has an accumulation point in T . Then T is countably compact.

15.5 On countably compact spaces which are compact

In the following results we see how the countably compact property compares with

the compact property in certain spaces. Recall that second countable spaces are thosespaces which have a countable base for open sets.

Theorem 15.7 Suppose S is a second countable topological space. Then S is compact if

and only if S is countably compact.

Proof : Given: The space S is a second countable topological space.

( ⇒ ) It is always true that, if S is compact then S is countably compact.

1The set S is well-ordered by ≤.


( ⇐ ) Suppose S is countably compact. Let U be an open cover of S. It suffices toshow that U has a finite subcover.

Since S is second countable then S has a countable open base, B.

Let x be a point in S. Since U is an open cover, there exist Ux ∈ U which contains

x. Since B is a base for open sets, there exists in B, at least one Bx such thatx ∈ Bx ⊆ Ux. For each p ∈ S, let

Ap = Bp ∈ B : p ∈ Bp ⊆ Up

Let A = ∪Ap : p ∈ S. Then A ⊆ B.

Since S is second countable, then A is a countable set of open neighbourhoods, say

Bi : i ∈ N\0. For each i ∈ N\0, p ∈ Bi ⊆ Up, for some p. For each i, chooseexactly one Upi such that Bi ⊆ Upi . Then V = Upi : i ∈ N\0 forms a countable

subcover of S.

Since S is countably compact, then V contains a finite subcover W = Upi : i ∈ Fof S. We have shown that U contains a finite subcover, W , of S. So the secondcountable property combined with the countably compact property on S implies S is

compact. As required.

Theorem 15.8 Those metric spaces which are countably compact are compact metricspaces.

Proof : Suppose (S, ρ) is a countably compact metric space. It suffices to show that S is

separable since, by theorem 5.11, this implies S is second countable and, by theorem15.7, second countable countably compact spaces are compact.

Let ε > 0. Recall that since S is countably compact every infinite subset V has acluster point.

Claim 1: There exists a finite set xi : i ∈ F ⊆ S such that S ⊆ ∪Bε(xi) : i ∈ F.Proof of claim: Suppose not. Then there is some δ > 0 such that S 6⊆ ∪Bδ(xi) : i ∈F for all finite sets F . Suppose y0 ∈ S. If Bδ(y0) does not cover S, then there is

y1 ∈ S such that ρ(y0, y1) ≥ δ. Inductively, suppose yi : i = 1, 2, 3, . . . , k are suchthat S 6⊆ ∪Bδ(yi) : i = 0, 1, 2, 3, . . . , k. then there exists

yk+1 ∈ S\∪Bδ(yi) : i = 0, 1, 2, 3, . . . , k

Then, we can construct in this way, the infinite subset, V = yi : i ∈ N, of S. ThenV has no cluster point, contradicting the countable compactness property. This es-

tablishes the claim.


Claim 2: The metric space (S, ρ) is separable.

Proof of claim: Let n ∈ N\0. By claim 1, for this value, n, we can construct,

y(i,n) : i = 1, 2, 3, . . . , kn such that, if S ⊆ Bn = ∪B1/n(y(i,n)) : i = 1, 2, 3, . . . , kn.By constructing in this way the sets, Bn : n ∈ N\0, we extract the subset,

D = y(i,n) : i = 1, 2, 3, . . .kn, n = 1, 2, 3, . . .

where S ⊆ Bn for each n = 1, 2, 3, . . .. Then D is easily seen to be dense in S. So Sis separable, as claimed.

By theorems 15.7 and 5.10, S is compact.

Example 2. Show that the space S = [0, ω1) is non-metrizable.

Solution : We have shown in the previous example that [0, ω1) is countably compact.By theorem 15.8, if S was metrizable it would be compact. But we showed in the

example that S is not compact. So S = [0, ω1) is a non-metrizable topological space.

Many of the results above allow us to confirm that the real line, equipped with theusual topology, is not countably compact. One of the characterization states that any

sequence will have an accumulation point. We already mentioned that the sequenceconstructed from the natural numbers has no accumulation point. One could also

raise the argument that, since R is known to be second countable, then, if R wascountably compact, it would have to be compact. In the theorem below we show that

continuous real-valued functions on countably compact spaces are bounded functions.If R was countably compact, then every real-valued continuous function on R wouldhave to be bounded, which is not the case.

Theorem 15.9 Let S be a countably compact topological space. If f is a real-valuedcontinuous function on S, then f [S] is compact.1

Proof : Given: The space S is a countably compact topological space and f [S] is the con-tinuous image of S in R, equipped with the usual topology.

Then, by theorem 15.5, f [S] is countably compact in R. In the example on page 84, it

is shown that the real line, R, equipped with the usual topology is second countable.In theorem 5.13, it is shown that the second countable property is hereditary. So f [S]

is second countable in R. Then by theorem 15.7, f [S] is compact.

1Later in the text we will refer to those spaces, S, in which all continuous real-valued functions arebounded on S as being pseudocompact spaces. This theorem can then correctly be paraphrased as “Allcountably compact Hausdorff spaces are pseudocompact”.


Concepts review:

1. Define a countably compact space.

2. Give three characterizations of the countable compactness property.

3. State the Bolzano-Weierstrass property.

4. Are closed subsets of countably compact spaces necessarily countably compact?

5. What can we say about continuous images of countably compact spaces?

6. Suppose S is the product of countably compact spaces. If we want some guaranteethat S be countably compact, what conditions must be satisfied?

7. Give an example of a space which is countably compact but not compact.

8. Identify topological spaces in which the compact property is equivalent to the count-able compactness property.

9. What can we say about continuous real-valued functions on a countably compactspace.

EXERCISES

1. Show that a space S is countably compact if and only if whenever a family, F =Fi : i ∈ N of closed non-empty sets in S, is such that Fi+1 ⊆ Fi for all i, then

∩Fi : i ∈ N is non-empty.

2. Suppose f : S → T is a one-to-one continuous function mapping a countably compactspace, S, onto a first countable space, T . Show that T is a homeomorphic image of

S.

310 Section 16 : Lindelof spaces

16 / Lindelof spaces.

Summary. We will now investigate a property which can be seen as a weak

relative of the compact property. It is called the Lindelof property. We will for-mally define it and study its characterizations and basic properties. We compare

its characteristics to those of the countable compactness property. Finally, wewill provide a few examples.

16.1 Introduction.

There are various ways to weaken the “open cover - finite subcover” property used

to described compactness of a space. The “countably compact” property, for exam-ple, refers to those spaces for which every countable open cover has a finite subcover.

There is another approach we can use to weaken the compactness property. We canrequire that arbitrarily large open covers have a countable subcover (rather than the

stricter “finite subcover”). This property is referred to as the Lindelof property. Al-though, some readers may not necessarily view the Lindelof property as a relative of

compactness, we included it in this part of the text since it often appears in the lit-erature combined with other weaker compact properties. Furthermore, it shares withthe “compact relatives” a property which allows for the reduction of a “large” open

cover to one of a “smaller” size. We will see that this property may be of interestwhen combined with the properties of other non-compact topological spaces.

16.2 The definition.

The following definition applies to arbitrary topological spaces. Some text define theLindelof property only for regular spaces, possibly because it is in such spaces that

the Lindelof property is most commonly studied.

Definition 16.1 A topological space, S, is said to be Lindelof, or satisfies the Lindelofproperty, if every open cover of S has a countable subcover.1

1Named after the Finnish mathematician Ernst Lindelof (1870 - 1946). He made contributions to thefields of real analysis, complex analysis and general topology.


Obviously, since finite sets are by definition countable, if an open cover has a finitesubcover then this cover has a countable subcover; so we can say that, . . .

Every compact space is Lindelof.

Also, by combining the Lindelof property with the countably compact property, we

see that the Lindelof property reduces an open cover to a countable one, and thecountable compact property reduces the countable open cover to a finite one. So, . . .

Every countably compact Lindelof space is a compact space.

There are other properties which guarantee that an open cover will have a countablesubcover.

Theorem 16.2 If S is second countable then S must be Lindelof.

Proof : Given: We are given that S is a second countable space and that U is an open

cover on S.

If p ∈ S, there exist Vp ∈ U which contains p. By hypothesis, S has a countable openbase, B, for S. Then, we can choose, from B, at least one Bp such that p ∈ Bp ⊆ Vp.

LetAp = Bp ∈ B : p ∈ Bp ⊆ Vp

Let A = ∪Ap : p ∈ S.Since S is second countable, then A can be reduced to a countable subfamily, A ∗,say A ∗ = Bi : i ∈ N\0, without sacrificing its open base property. For each

i ∈ N\0, p ∈ Bi ⊆ Vp, for some p. For each i, choose exactly one Vpi such thatBi ⊆ Vpi . Then V = Vpi : i ∈ N\0 forms a countable subcover of S. So, if S is

second countable, then S must be Lindelof.

16.3 Characterizations.

Those who only work with metrizable topological spaces will soon notice that the

Lindelof property provides nothing new in their topological region of the universe.This is due to the fact that, in metric spaces, the Lindelof property is equivalent to

the second countable property. We will say that the family of subsets, F , of a spaceS satisfies the countable intersection property (CIP) if, for any countable subfamily,FN, of F ,

∩F : F ∈ FN 6= ∅


Theorem 16.3 Let S be a topological space.

a) The space, S, is Lindelof if and only if each filter of closed sets with the countableintersection property has non-empty intersection.

b) Suppose (S, ρ) is a metric space. Then the following are equivalent.

i. The space S is Lindelof.

ii. The space S is second countable.

iii. The space S is separable.

c) Suppose S is a regular space. Then the following are equivalent.

i. The space S is Lindelof.

ii. If each open cover, U = Ui : i ∈ I, of S has a subfamily, Uij : j ∈ N, suchthat clSUij : j ∈ N covers S, then Uij : j ∈ N covers S.

iii. Every filter of open sets in S with the countable intersection property has anaccumulation point.


a) We are given that F = Fi : i ∈ I represents a filter of closed sets in the space S.

( ⇒ ) Suppose S is Lindelof and F satisfies the CIP. That is, countable subfamiliesof F have non-empty intersection. We are required to show that ∩Fi : i ∈ I 6= ∅.

Suppose not. That is, suppose ∩Fi : i ∈ I = ∅. Then S \Fi : i ∈ I is an opencover of S. Since S is Lindelof, S\Fij : j ∈ N forms a countable subcover. Thismeans, ∩Fij : j ∈ N = ∅. This contradicts the fact that F satisfies the CIP.

So ∩Fi : i ∈ I 6= ∅, as required.

( ⇐ ) Given: Every filter of closed sets with the countable intersection property has

non-empty intersection. Suppose U = Ui : i ∈ I is an open cover of S.

We are required to show that U has a countable subcover.

Suppose not. That is, suppose that, for any countable subfamily, Uij : j ∈ N,∩S \Uij : j ∈ N 6= ∅. Then F satisfies the CIP. Then ∩S \Ui : i ∈ I 6= ∅.

This contradicts that U is an open cover. So U has a countable subcover. Thismeans that S is Lindelof.

b) We are given that S a metric space.


( i ⇐ ii ) Given that S is second countable. By theorem 16.2, second countablespaces are Lindelof spaces, always. So S is Lindelof. We are done with ( i ⇐ ii ).

( i ⇒ ii ) Suppose the metric space, S, is a Lindelof space. We are required to showthat S has a countable base for open sets.

For each i ∈ N\0, we construct an open cover, Ui = B1/i(u) : u ∈ S, of S. Byhypothesis, each Ui has a countable subcover, say

Bi = B1/i(uij) : uij ∈ S, j ∈ N

Claim. That the countable family of open sets, B = ∪Bi : i ∈ N\0, is a basefor open sets of S. That is, every open subset is the union of elements from B.

Proof of claim : Suppose U is an open neighbourhood of p ∈ S. We are required toshow that p belongs to some element, B, of B such that B ⊆ U . Then there exists

some natural number, k, such that B1/k(p) ⊆ U . Now, B3k has been hypothesizedto be an open cover of S. So there is some v ∈ S, such that p ∈ B1/3k(v). Then

x ∈ B1/3k(v) ⇒ ρ(x, p) <2

3k<

1

k

So x ∈ B1/k(p) ⊆ U . Then p ∈ B1/3k(v) ⊆ U where B1/3k(v) is an element of B.Then B is a countable base for open sets, as claimed.

So S is second countable, as required. We are done with ( i ⇒ ii ).

( ii ⇐ iii ) If S a separable metric space, by theorem 5.11, S is second countable.

( ii ⇒ iii ) Suppose S is second countable. Then, by theorem 5.10, S is separable,always.

c) We are given that S is a regular topological space.

( i ⇒ ii ) Suppose S is Lindelof. Then S has a countable subcover, Uij : j ∈ Nand so clSUij : j ∈ N is a countable subcover. So ii) holds true.

( i ⇐ ii ) We are given that, if V = Ui : i ∈ I is an open cover, thenclSUij : j ∈ N covers S for some countable subfamily, Uij : j ∈ N, of V .

We claim that Uij : j ∈ N must also be a subcover.

Proof of claim: Let y ∈ S. Then there exist Uy in the open cover, V , which con-tains y. By regularity, there exists an open neighbourhood, By , of y such that

y ∈ By ⊆ clSBy ⊆ Uy. Then By : y ∈ S forms an open cover of S. By hy-pothesis, we have clSByj : j ∈ N covers S. Since clSByj ⊆ Uyj for each j, then

Uyj : j ∈ N covers S, as claimed.

We conclude that S is Lindelof. We are done for ⇐.

( i ⇒ iii ) Suppose S is Lindelof.


Let U = Ui : i ∈ I be a filter base of open sets in S which satisfies the countableintersection property.

We are required to show that ∩clSUi : i ∈ I is non-empty.

Consider the family V = clSUi : i ∈ I. Since U satisfies the countable intersec-tion property then so does V . Then V is a filter of closed sets in S. Since S is

Lindelof, by part a), ∩clSUi : i ∈ I is non-empty. If p ∈ ∩clSUi : i ∈ I, then bydefinition, p is an accumulation point of the filter of open sets, U .

( i ⇐ iii ) We are given that filters of open sets in S which satisfy the countableintersection property have an accumulation point.

Suppose U = Ui : i ∈ I is an open cover of S. We are required to show that Uhas a countable subcover, Uij : j ∈ N.To obtain a contradiction, let us suppose U does not have a countable subcover.

Let x ∈ S. Then there exists Ux in U such that x ∈ Ux. By regularity, there exists

Wx such that x ∈Wx ⊆ clSWx ⊆ Ux. Then W = Wx : x ∈ S as well as

W ∗ = clSWx : x ∈ S

cover S. Then, for any countable subfamily, clSWxj : j ∈ N, of W ∗,

∩S\clSWxj : j ∈ N

is non-empty. Then, S\clSWx : x ∈ S is a family of open sets which satisfies thecountable intersection property. By hypothesis, it must have an accumulation point

p. This means p ∈ ∩clS [S \clSWx] : x ∈ S. But clS[S \clSWx] = S \Wx. Thenp ∈ ∩S\Wx] : x ∈ S. Then p is not in any Wx, contradicting that Wx : x ∈ Sis an open cover of S. The source of the contradiction is the assumption that Udoes not have a countable subcover.

So U has a countable subcover. We conclude that S is Lindelof.

Example 1. We have shown in chapter five (page 83) that the metrizable set of all real

numbers, equipped with the usual topology, is second countable and so is separable.Then, by part b) of the above characterization theorem, R is a Lindelof space.

Example 2. We have seen that the real numbers, (R, τS), equipped with the upper

limit topology (Sorgenfrey line) is not second countable (see page 83). Show that itis Lindelof.

Solution : Let U be an open cover of R. We will fix n ∈ N. Let Sn = [u, n] : [u, n]has a countable subcover.Let kn = inf u : [u, n] ∈ Sn. We claim that kn 6∈ R. Suppose kn ∈ R. Then [kn, n]has a countable subcover, Un = Ui ∈ U : i ∈ N. This means that kn ∈ Uj , for


some j ∈ N. Since Uj is open, there is some ε > 0 such that kn ∈ (kn − ε, kn] ⊆ Uj.Then [kn − ε/2, n] has a countable subcover contradicting the definition of kn. So,

(−∞, n] has a countable subcover, Un, as claimed. This is independent of the valueof n. Then ∪Un : n ∈ N is a countable subcover of ∪(−∞, n] : n ∈ N = R. So R

is Lindelof.

Notice that (R, τS) has been found to be Lindelof in spite of it not being second count-

able. Does this contradict theorem 16.3, part b)? The answer is “No it doesn’t”. Itsimply guarantees that (R, τS) is not metrizable.

Example 3. In the example on page 305, we showed that the ordinal space, S = [0, ω1),

is countably compact but not compact. Then it cannot be Lindelof, for, if it was bothcountably compact and Lindelof, it would have to be compact.

16.4 Two invariance theorems for the Lindelof property.

Arbitrary subspaces generally do not inherit the Lindelof property from its superset.

Also, products of Lindelof spaces need not be Lindelof. But closed subspaces do in-herit the Lindelof property from its superset.

We will also show that the Lindelof property is carried over by continuous functions.

Theorem 16.4 If S is Lindelof and F is a closed subset of S then F is Lindelof.

Proof : Given: That S is Lindelof and F is closed in S.

Let U = Ui : i ∈ I be a family of open sets in S which covers F . ThenU ∗ = U ∪ S \F covers S. Since S is Lindelof then U ∗ has a countable sub-cover Uij : j ∈ N ∪ S\F. Then Uij : j ∈ N is a countable subcover of F . So F

is Lindelof.

Theorem 16.5 If S is Lindelof and f : S → T is continuous and onto T , then T is Lindelof.

Proof : Given: That S is Lindelof and f : S → T is continuous and onto T .

Let U = Ui : i ∈ I be a family of open sets which covers T . Then V = f←[Ui] :

i ∈ I covers S. Then V has a countable subcover, f←[Uij ] : j ∈ N, of S. Thenf [f←[Uij ]] : j ∈ N = Uij : j ∈ N is a countable subcover of T . So T is Lindelof.


16.5 Regular Lindelof spaces are normal.

We now show that regular spaces that are equipped with the Lindelof property are in

fact normal spaces.

Theorem 16.6 If S is both regular and Lindelof then it is normal.

Proof : Given: That S is regular and Lindelof.

Let F and K be disjoint closed sets. If x ∈ F and y ∈ K, then, by regularity, there

exists an open Bx such that clSBx ∩K = ∅ and open Dy such that clSDy ∩ F = ∅.Then

BF = Bx : x ∈ FDK = Dy : y ∈ K

form open covers of F and K, respectively. Since S is Lindelof then so are the closed

subsets F and K. So we can obtain

BF∗ = Bxi : i ∈ N

DK∗ = Dyi : i ∈ N

as countable subcovers of F and K, respectively.

We now inductively construct disjoint open neighbourhoods of F and K:

For F1 = Bx1 let K1 = Dy1\clSF1

For i ∈ N if Fi = Bxi \clS(K1 ∪ · · · ∪Ki−1) let Ki = Dyi \clS(F1 ∪ · · · ∪ Fi)

Then ∪Fi : i ∈ N and ∪Ki : i ∈ N form disjoint open neighbourhoods of F andK, respectively. So S is normal.

Concepts review:

1. Define the Lindelof property on a topological space.

2. What kind of space is obtained if we combine countable compactness with the Lindelofproperty?


3. How does a space with the second countable property compare with a Lindelof space?

4. Give a characterization of the Lindelof property in terms of a filter of closed sets.

5. In a metric space, state two properties each of which is equivalent to the Lindelofproperty.

6. In a regular space, state two properties which are equivalent to the Lindelof property.

7. Is there a quick argument you can use to conclude that R is Lindelof, based on thetheory developed in the chapter.

8. What can you say about subspaces of Lindelof spaces?

9. What about continuous images of Lindelof spaces.

10. In what way does the Lindelof property enhance a regular space?

11. Provide examples of Lindelof and non-Lindelof spaces.

EXERCISES

1. Suppose that the space S is such that it can be expressed as a countable union of

compact spaces. Show that S is Lindelof.

2. Suppose S is a topological space. We say that the open cover, U = Ui : i ∈ I,refines (or is a refinement of) the open cover, V = Vi : i ∈ I, if for every Vi there is

some Uj such that Uj ⊆ Vi. Show that S is Lindelof if and only if every open cover ofS has a countable refinement.

3. Recall that a T1-space, S, is perfectly normal if and only if, for any pair of non-emptydisjoint close subsets, F and K, there exists a continuous function f : S → [0, 1] such

that F = f←(0) and K = f←(1). Suppose S is a regular space such that each of itsopen subsets is Lindelof. Show that S is perfectly normal.

318 Section 17 : Sequentially and feebly compact spaces

17 / Sequentially and feebly compact spaces.

Summary. In this section we define the sequentially compact property and de-

termine in which precise circumstances it is just another way of referring to thecompactness property. At the same time we outline conditions under which a

compact space is not sequentially compact. To illustrate this we provide variousexamples. We also introduce two closely related classes of spaces: feebly compact

and pseudocompact spaces.

17.1 Sequentially compact spaces

For many readers sequentially compact is just another way of saying “compact” intwo words instead of one. In many cases, this is fine. As long as we keep in mind

that the two properties are, in general, not equivalent. We will begin with a formaldefinition and then examine specific cases where the sequential compact spaces are

equivalent to compact ones. Based on our experience with countably compact spaceswe expect that countability properties will, somehow, be relevant in many arguments.

Definition 17.1 Let S be a topological space. We say that a subspace, T , of S is sequen-

tially compact if and only if every sequence in T has a subsequence which converges to apoint in T .

Theorem 17.2 Let S be a topological space.

a) If S is second countable, then the following are equivalent.

i. The space S is compact.

ii. The space S is countably compact.

iii. The space S is sequentially compact

b) If S is metric, then the following are equivalent.

i. The space S is compact.

ii. The space S is countably compact.

iii. The space S is sequentially compact

c) If T ⊆ S = Rn the following are equivalent.


i. The subspace T is compact.

ii. The subspace T is closed and bounded.

iii. The subspace T is sequentially compact


a) We are given that S is second countable.

( i ⇔ ii ) In second countable spaces, the equivalence of compactness and countablecompactness is proved in theorem 15.7.

( ii ⇔ iii ) We must show “countably compact if and only if sequentially compact”.

Recall that second countable spaces are first countable, always.

S ctbly cpct ⇔ every seq in S has an accum’tion pt. (Thm 15.2)

In a first countable space S,

p = accum’tion point of a seq ⇔ p = limit point of a subseq. (Thm 11.5)

So, in a second countable space,

S → sequentially compact ⇔ S → countably compact

b) We are given that S is a metric space.

( i ⇔ ii ) In theorem 15.8, it is shown that in a metric space, compactness and

countable compactness are equivalent.

( ii ⇔ iii ) Recall that metric spaces are first countable. As shown in ( ii ⇔ iii ) ofpart a), in first countable spaces, countable compactness and sequential compact-ness are equivalent.

c) We are given that T ⊆ S = Rn with the usual topology.

( i ⇔ iii ) Since S is a metric space, S is compact if and only if S is sequentiallycompact. (By part b))

( i ⇔ ii ) For ⇒. We are given that T is compact. Since S is Hausdorff, compactsubspaces are closed in S, so T is closed in S.

We now show boundedness of T . For u ∈ T , let f : T → R be defined as, f(x) =

distance(u, x) (known to be continuous, see page 210). Since f is continuous onthe compact set T , f attains a maximum value, k, at some element v ∈ T , so k =dist(u, v) ≥ dist(u, x) independently of the choice of x in T . So, for any a, b ∈ T ,

dist(a, b) ≤ dist(a, u) + dist(u, b) ≤ k + k = 2k. So T is bounded, as required.

For ⇐. We are given that T is closed and bounded in Rn. Since T is bounded in

Rn there exists k ∈ R+ such that T ⊆ [−k, k]n. Since [−k, k] is compact then so isits product,

∏ni=1[−k, k]i. Then, since T is a closed subset of a compact space, T is

compact.


Many readers may already be familiar with the following result about compact subsets

of metric spaces.

Theorem 17.3 If S is a metric space and T is compact in S, then T is closed and bounded.

Proof : The proof is left as an exercise for the reader.

Note: The converse fails. Consider an infinite discrete space with metric ρ(a, b) = 1

if a 6= b.

Theorem 17.4 If S is a sequentially compact space then it must necessarily be countablycompact.

Proof : We are given that S is a sequentially compact space. Then every sequence in S

has a subsequence which converges to a point in S. We are required to show that itis countably compact.

Suppose S is not a countably compact space. Then S has a countable open cover,U = Ui : i ∈ N, with no finite subcover. For each n ∈ N, we can then choose

pn ∈ S\∪Ui : i = 0, 1, 2, 3, . . . , n

Let p ∈ S. We claim that p cannot be the limit of a subsequence of A = pi : i ∈ N.Since U covers S then p ∈ Uk for some k ∈ N. If m > k then

pm 6∈ ∪Ui : i = 0, 1, 2, 3, . . . , m

Then the tail end, pi : i ≥ k, of A, does not intersect the open neighbourhood, Uk,

of p. Then p cannot be the limit of a subsequence of A, as claimed.

This contradicts the fact that S is sequentially compact. So spaces which are sequen-

tially compact are countably compact.

We now verify that sequential compactness is preserved by continuous functions.


Theorem 17.5 Let S be a topological space and T be a Hausdorff topological space. Iff : S → T is a continuous function and S is sequentially compact then f [S] is sequentially

compact in T .

Proof : We are given that S is a sequentially compact space, T is Hausdorff and f : S → T

is continuous. We are required to show that f [T ] is sequentially compact.

The proof follows directly from the definition of sequentially compact. The details areleft as an exercise.

17.2 Example of a compact space which is not sequentially compact.

We have seen that, if S is either second countable or is metrizable then

S sequential compactness ⇔ S countable compactness

We have seen in 15.2 that “countably compact spaces are precisely those spaces inwhich every sequence has an accumulation point”. We can then also say:

If S is either second countable or is metrizable then

Every sequence in S has a convergent subsequence in S⇔

Every sequence in S has a convergent subnet in S

This leads us to wonder whether there are any spaces in which sequences have a con-vergent subnet but not a convergent subsequence.

We studied convergence properties of S = [0, 1][0,1] on page 248. In that example, we

saw that S is a compact space in which a sequence has an accumulation point but nosubsequence converging to that point. We will revisit the example, S = [0, 1][0,1], to

show that a compact space need not be sequentially compact.

Example 1. Let S = [0, 1][0,1] be equipped with the product topology. That is, we view

S as∏

i∈[0,1][0, 1]i. Show that S is compact, hence countably compact.

a) Show that S is compact, hence countably compact.

b) Show that, in spite of its compactness, S is not sequentially compact.

Solution : We are given that the space S = [0, 1][0,1] viewed as∏

i∈[0,1][0, 1]i is equippedwith the product topology.


a) Since [0, 1] is compact and the any product space of compact sets is compact (byTychonoff theorem) then S =

∏

i∈[0,1][0, 1]i is compact. Since any compact set is

countably compact, then S is also countably compact.

b) We are given that S = [0, 1][0,1] is equipped with the product topology. Thatis, we view S as

∏

i∈[0,1][0, 1]i. We will construct a sequence in S which has no

converging subsequence. Suppose each element, x, of [0, 1] is expressed in itsbinary expansion form. For each n ∈ N\0 we define the one-to-one function

fn : [0, 1] → 0, 1 as

fn(x) = “the nth digit in the binary expansion of x”

For example, f3(0.10101010 . . .) = 1.1 This allows us to view S as an uncountable

family,T = fn(x) : n = 1, 2, 3, . . . : x ∈ [0, 1]

of sequences of 0’s and 1’s.

We define a sequence xk : k = 1, 2, 3, . . . in S, such that

N = fn(xk) : n = 1, 2, 3, . . . : xk ∈ S

where fk(xk) = 1 for all values of k, and fn(xk) = 0 if n 6= k. That is, the

diagonal of N in T is made only of 1’s and all other entries are 0’s.

We claim that N cannot have a convergent subsequence. To see why, suppose N

has a convergent subsequence, say

M = fn(xkn) : n = 1, 2, 3, . . . : xkn ∈ xk : k = 1, 2, 3, . . .

Then every column of M is constant on a tail end and so must converge (to zero).

If so, what can we say about the point (1, 0)? This particular point becomes alimit point of a sequence of both a string of 0’s (above it) and a string of 1’s (on

the diagonal). A contradiction. The source of the contradiction is our supposi-tion that M converges.

Since M was an arbitrary subsequence of N , N cannot have a convergent subse-quence. So S cannot be sequentially compact.

The example does not contradict theorem 17.2. It simply shows that S is notmetrizable.

17.3 Topic : A compact characterization for metric spaces.

Completeness is a concept which is defined only in the context of metric spaces.

A sequence xn : n ∈ N in a metric space (M, ρ) is said to be a Cauchy sequenceif, for any ε > 0, there exists N such that, whenever n,m > N , ρ(xn, xm) < ε. If

every Cauchy sequence in (M, ρ) converges to a point in M , then (M, ρ) it is saidto be a complete metric space.

1One-to-one if we avoid infinite tail ends of zeroes. For example, 0.1111111 . . . = 1.000000 . . ..


For example the metric space, Rn, is known to be complete. But we can still associatecompleteness in a metric space to the topological notion of compactness. For example,

it is not difficult to show that all compact metric spaces are complete metric spaces,as shown below.

Example 10. Verify that a compact metric space is a complete metric space.

Solution : Suppose (M, ρ) is a compact metric space and S = xn : n ∈ N is a

Cauchy sequence in M . Then, for any k ∈ N \ 0 there exists Nk such that, ifxn : n > Nk ⊆ B1/k(xn) 6= ∅. So for every 1/k, there is some n such that B1/k(xn)contains a tail-end xn, xn + 1, . . . , of S. Then the set of closed subsets,

F = clM[

B1/k(xn) ∩ xn : n > Nk]

: k ∈ N\0

of M satisfies the FIP. Since M is compact, ∩F is non-empty and so contains a point,

say z ∈ M . Since every open neighbourhood of z will intersect every element of F .Then every open neighbourhood of z contains a tail-end of S. It follows that z is the

limit point of the sequence, S. So the Cauchy sequence, S, converges. Then M iscomplete.

Since the metric space, R, is known to be complete and non-compact, the converseobviously does not hold true. A compact metric space would also have to satisfy someother property. We introduce the following notion for metric spaces:

Definition. A metric space, (M, ρ), is said to be totally bounded if for any ε > 0 there

exists a finite subset F of M such that Bε(x) : x ∈ F which covers M .

Theorem 17.6 A metric space is compact if and only if it is both complete and totally

bounded.

Proof : Suppose (M, ρ) is a metric space.

( ⇒ ) We have already shown that compact metric spaces are complete. To show to-

tally bounded, let U = Bε(x) : x ∈ I be an open cover of M . Since M is compact,U has a finite subcover and so M is totally bounded.

( ⇐ ) Suppose (M, ρ) is both a complete and totally bounded metric space. We are

required to show that M is compact.

To show compactness of M it suffices to show sequential compactness of M . That is,

that any sequence must have a convergent subsequence.

Let κ0 = xi : i ∈ J0 be a sequence in M . Since M is complete, it will suffice toshow that κ0 has a subsequence which is Cauchy.

Suppose M1 is an open cover of M made only of balls of radius 1. Since M is totallybounded M1 has a finite covering, F1. Let B1 be an element of F1 which contains


infinitely many points, say κ1 = xi : i ∈ J1 ⊆ κ0, for some countable J1 ⊆ J0.

Suppose M2 is an open cover of M made only of balls of radius 1/2. Since M is totally

bounded M2 has a finite covering, F2. Let B2 be an element of F2 which containsinfinitely many points, say κ2 = xi : i ∈ J2 ⊆ κ1, for some countable J2 ⊆ J1.

More generally, suppose we are given κn = xi : i ∈ Jn ⊆ κn−1 and suppose Mn+1

is an open cover of M made only of balls of radius 1/(n+ 1). Then Mn+1 has a finitesubcover, Fn+1, of M . Let Bn+1 be an element of Fn+1 which contains infinitely

many points κn+1 = xi : i ∈ Jn+1 ⊆ κn of M where Jn+1 ⊆ Jn.

Applying this process, we have a nested family of sequences κn : n = 1, 2, 3, . . .where κn+1 ⊆ κn. Then we can choose xj ∈ κj for each j ∈ N to form a subsequenceκ = xj : j ∈ N of κ0 where for any ε > 0, κ has a tail-end contained in a ball ofradius ε. So κ is Cauchy.

By hypothesis, M is complete so κ converges to, say, a point y ∈ M . Then everysequence has a convergent subsequence. That is, M is sequentially compact. So M is

compact.

17.5 Feebly compact spaces

We introduce a new class of spaces called feebly compact. Feebly compact spaces havea few very interesting characterizations. We will see that hey form a proper subclass

of the all pseudocompact spaces (those spaces for which C∗(S) = C(S)). We will alsosee that, for Tychonoff spaces, pseudocompact spaces are always feebly compact. The

definition refers to “locally finite collections of open subsets”.

Definition. A locally finite collection, C , of subsets of a topological space, S, is one

for which every point in S has at least one neighbourhood which meets at mostfinitely many elements of C .

Definition 17.7 Let S be a topological space. A apace S for which there can be no infinite

locally finite collection of open sets is called a feebly compact space.

The formal definition is not particularly enlightening. The following characterizations

will provide different perspectives on this property.


Theorem 17.8 Let S be any topological space. Then the following are equivalent.

1. Any locally finite collection of pairwise disjoint open subsets of S is finite.

2. The space S is feebly compact.

3. For any decreasing collection, Ui : i ∈ N of open subsets of S, ∩clSUi : i ∈ N 6= ∅.

4. Any countable open cover, Ui : i ∈ N, of open subsets of S has a finite subcollection,

Ui : i ∈ F, such that ∪Ui : i ∈ F is dense in S.


(1 ⇒ 2) Suppose any locally finite collection of pairwise disjoint open subsets of S is

finite. We are required to show that S is feebly compact.

Suppose S is not feebly compact. Then there exists an infinite locally finite collectionof open sets. Let U = Ui : i ∈ N be such a collection. Choose x0 ∈ U0. There existsan open neighbourhood, V0 of x0 which intersects at most finitely many elements of

U . Then there is a n1 ∈ N that if n ≥ n1, Un ∩ V0 = ∅.

We proceed inductively. Let ni : i = 0...m be a strictly increasing set of naturalnumbers, and let Vi : i = 0...m−1 be a sequence of open sets such that Uni ∩Vi 6= ∅

but, for i < j implies Vi ∩ Unj = ∅. We can then choose xm ∈ Unm for which we can

find an open set Vm and a natural number nm+1 > nm such that xm ∈ Unm ∩ Vm andVm ∩ Uj = ∅ whenever j ≥ nm+1.

Then Vi ∩ Uni : i ∈ N forms and infinite pairwise disjoint locally finite collection ofopen sets in S. This is contrary to our hypothesis. So S must be feebly compact.

(2 ⇒ 3) Suppose S is feebly compact. Then any locally finite collection of open sub-

sets of S is finite. Suppose U = Ui : i ∈ N is a collection of non-empty open subsetsof S such that Ui+1 ⊆ Ui for all i.We are required to show that ∩clSUi : i ∈ N 6= ∅.

Suppose ∩clSUi : i ∈ N = ∅. Then, if x ∈ S there exists some kx such that

x 6∈ clSVi for all i ≥ kx. Then, if i ≥ kx, S\clSVi is an open neighbourhood of x whichintersects only finitely many elements of Ui : i ∈ N. So U is an infinite locallyfinite collection of open sets contrary to our hypothesis about the properties of U . So

∩clSUi : i ∈ N 6= ∅.

(3 ⇒ 4) Suppose S is a space which satisfies the property: Whenever U = Ui : i ∈ Nis a collection of non-empty open subsets of S such that Ui+1 ⊆ Ui for all i, then∩clSUi : i ∈ N 6= ∅.

Let V = Vi : i ∈ N be any countable open cover of S. We are required to show thatV has a subfamily whose union is dense in S.

For each n ∈ N we define

Bn = S\clS[∪Vi : i ∈ 0, n]


Suppose Bn is non-empty for all n ∈ N. Then Bn : n ∈ N is a collection of non-empty open sets such that Bn+1 ⊆ Bn.

By hypothesis, ∩clSBn : n ∈ N 6= ∅. But we also have,

∩clSBn : n ∈ N ⊆ S\∪clSVi : n ∈ N = ∅

We have a situation where ∩clSBn : n ∈ N is both empty and non-empty. The factthat ∩clSBn : n ∈ N 6= ∅ followed from our assumption that Bn is non-empty for all

n ∈ N. So this assumption is not valid. There must be some k ∈ N such that Bk = ∅.So ∪clSVi : i ∈ 0, k = clS ∪ Vi : i ∈ i ∈ 0, k = S. Then ∪Vi : i ∈ i ∈ 0, kis dense in S.

(4 ⇒ 1) Suppose that, if V = Vi : i ∈ N be a countable open cover of S, then Vhas a subfamily whose union is dense in S. We are required to show that all locally

finite families of pairwise disjoint open subsets are finite.

We will suppose that an infinite locally finite family, U = Ui : i ∈ N, of pairwisedisjoint open subsets exists. For each x ∈ S, let Mx be an open neighbourhood of xand Fx be a finite subset of N such that Mx ∩ Ui 6= ∅ for i ∈ Fx. Then Mx : x ∈ Sforms an open cover of S.

Let F = F : F is a finite subset of N. Note that F is a countably infinite subsetof P(N). 1 Let W : F → S be defined as

W (F ) = ∪Mx : x ∈ S, F = Fx

See that, for each F , W (F ) is a non-empty open neighbourhood of x in S. SinceS ⊆ ∪W (F ) : F ∈ F,

W = W (F ) : F ∈ F is an open cover of S

Since F is countable then so is the open cover, W .

We will show that no finite subfamily of W can be dense in S. To see this, suppose

W (Fi) : Fi ∈ F1, F2, . . . , Fn is any finite subfamily of W .

Choose any Fk ∈ W \Fi : i ∈ 0, n.We claim that Uk ∩ ∪W (Fi) : Fi ∈ F1, F2, . . . , Fn = ∅.

For, if there is a y in ∪W (Fi) : i ∈ 0, n ∩ Uk, then y ∈ My and Fy = Fk. Since

Fk 6∈ Fi : i ∈ 0, n, we have a contradiction. The source of the contradiction is oursupposition that ∪W (Fi) : i ∈ 0, n ∩ Uk contains an element y.

1The number of finite subsets of N is countable: For each n ∈ N, let Un = A ∈ P(N) : maxA ≤ n,the set of all subsets of 0, 1, 2, . . . , n. Then, for each n, |Un| = 2n+1. The countable union of countablesets is countable (See 19.3 of Appendix B.), so ∪n∈NUn is countable. If F is a finite subset of N, F ∈ Um

for some m. Then ∪n∈NUn contains all finite subsets of N. Then the number of finite subsets of N iscountable.


So Uk ∩ ∪W (Fi) : Fi ∈ F1, F2, . . . , Fn = ∅, as claimed.

Then we can conclude that for any finite collection W (Fi) : Fi ∈ F1, F2, . . . , Fn ⊆W , S\∪W (Fi) : Fi ∈ F1, F2, . . . , Fn contains a non-empty open set. So no finitesubfamily of W can be dense in S. This contradicts our hypothesis.

The assumption that there is an infinite locally finite family, U = Ui : i ∈ N, of

pairwise disjoint open subsets cannot possibly hold true. So U must be finite. Weare done with (4 ⇒ 1).

Feebly compact and countably compact spaces share a common property. It is that

neither topological space admits unbounded continuous real-valued function.

Theorem 17.9 Let S be a feebly compact topological space. Then every continuousreal-valued function on S is bounded.

Proof : We are given that S is a feebly compact topological space. We are required to showthat any continuous real-valued function on S is bounded. Suppose f ∈ C(S)\C∗(S).

Note that we can assume, without loss of generality, that f ≥ 0. Then f [S] containsa sequence, xn : n ∈ N, of real numbers where xn+1 ≥ xn + 1. See that

U = f←[(xn − 13 , xn + 1

3 )] : n ∈ N

is an infinite locally finite collection of open subsets of S. So S cannot be feebly

compact, a contradiction of our hypothesis.

17.5 Pseudocompact spaces

We know that the continuous image of a compact space is always compact. Thismeans that continuous real-valued functions on a compact space are always bounded.However, a space S in which the property “all real-valued continuous functions are

bounded on S” need not be compact. A countably compact space is an example. Aspace which satisfies this property is said to be pseudocompact. We formally define

this concept.


Definition 17.10 Let S be a topological space. If every continuous real-valued functionon S is bounded in R then S is pseudocompact.

Using the newly introduced terminology above, we can say that all compact spaces

are pseudocompact spaces. The set of all real numbers, R, is, of course, not pseudo-compact (as witnessed by f(x) = x3 on R). We have seen in theorem 15.9, that

. . . every continuous real-valued function on a countably compact space, S,is bounded on S.

So countably compact spaces are pseudocompact.

What about sequentially compact spaces? Consider a continuous real-valued func-

tion, f : S → R, on a sequentially compact space S. By theorem 17.4, S must also becountably compact. We have seen in theorem 15.9 that f must be bounded on S. Sosequentially compact spaces are pseudocompact.

We have already encountered an example of a space which is pseudocompact but not

compact. We saw that the ordinal space,

S = [0, ω1)

is countably compact and not compact. So S = [0, ω1) is pseudocompact but notcompact.

The next theorem illustrates how closely related the feebly compact and pseudocom-pact properties are. We see that, in the particular case where a space is completely

regular, the pseudocompact property and the feebly compact property simply differ-ent ways of representing the same topological property.

Various characterizations of the pseudocompact property will be discussed in chap-ters to come. But we can present immediately characterizations of the pseudocompact

property for completely regular spaces.

Theorem 17.11 Let S be a completely regular space. Then the following are equivalent.

1. The space S is pseudocompact.

2. If Ui : i ∈ N is a nested collection of non-empty open sets in S (i.e., Ui+1 ⊂ Ui forall i ∈ N) then

∩clSUi : i ∈ N 6= ∅


3. For every open cover, Ui : i ∈ A, of the space S, there is a finite subset, F , of Asuch that clSUi : i ∈ F covers S.

4. The space S is feebly compact.

Proof : We are given that S is a completely regular space.

( 1 ⇒ 2 ) Suppose S is pseudocompact. Let Ui : i ∈ N be a collection of non-empty

open subsets in S such that U0 ⊃ U1 ⊃ U2 ⊃ · · · . We are required to show that∩clSUi : i ∈ N 6= ∅.

Suppose ∩clSUn : n ∈ N = ∅.

For each n, choose a point xn ∈ Un to form a closed infinite sequence xn : n ∈ N.For each xn we can inductively choose a closed neighbourhood, Vn, of xn such thatVn ⊆ Un and Vm ∩ Vn = ∅ whenever m 6= n.

Since S is completely regular there is, for each n, a continuous function gn : S → [0, 1]

such that gn(xn) = 1 and gn[S\Vn] = 0.Let fn = ngn. Then fn maps S into [0, n]. We define the function,

h(x) =∑

n∈N

fn(x)

on S.

We investigate properties of this function carefully. First see that h is an unbounded

function.

Also see that, for any u ∈ S, then fn(u) 6= 0 for no more than one n. Then

h(u) = fm(u), for some m.

We now proceed to show that h is a continuous function. Let u ∈ S. We claim that

u has an open neighbourhood Wu which intersects at most a single Vn.

Proof of claim. Suppose every open neighbourhood in h[S] of u intersects infinitelymany Vn’s. Then u would belong to ∩clSUn : n ∈ N which has been hypothesizedto be empty. So u has an open neighbourhood, say W , which intersects finitely many

Vn’s, say Vn1, Vn2, . . . , Vnk. Since each Vni is closed u has an open neighbourhood

Wu which intersects only Vn1, as claimed.

Then, for some m, h agrees with fm on a neighbourhood Wu of u.

Let h(y) ∈ h[S] and U be an open neighbourhood of h(y) in R. Then h(y) = fm(y) forprecisely one m ∈ N. We have shown that h = fm on some open neighbourhood, Wy,

of y in S. Since fm is continuous on Wy, then there exists an open neighbourhood Bof y such that h(x) = fm(x) on B and fm[B] ⊆ U So h[B] ⊆ U . We conclude that h

is continuous at each point y ∈ S. Then h is continuous on S, as desired.

Since h is unbounded, we have a contradiction. So ∩clSUi : i ∈ N 6= ∅.

( 4 ⇒ 1 ) This follows from theorem 17.9. This does not require complete regularity.


Concepts review:

1. Define a sequentially compact space.

2. If S is a second countable or metrizable topological spaces which satisfies the sequen-

tially compact property, name two other properties which are equivalent properties insuch spaces.

3. Describe a topological space in which sequentially compact subsets are precisely theclosed and bounded ones.

4. Is it true that, in all metric spaces, the compact subsets are alway closed and bounded?

5. Is it true that, in all metric spaces, the the closed and bounded subsets are always

compact?

6. Given a sequentially compact property and the countably compact property, one al-

ways implies the other. Which one? Can you provide a counter example for the onethat doesn’t.

7. Define a “pseudocompact space”?

8. Is a sequentially compact space pseudocompact?

9. Is a countably compact space necessarily pseudocompact?

10. Are there any spaces which are pseudocompact and not compact? If so, give anexample.

11. Are there any spaces which are compact and not pseudocompact? If so, give anexample.

EXERCISES

1. Show that, if F is a non-empty closed subset of a sequentially compact space, then Fis sequentially compact.


2. Let f : S → T denote a function mapping S into a Hausdorff space T . Show that, iff is continuous on S and S is a sequentially compact space, then f [S] is sequentially

compact.

3. Suppose f : S → R is a continuous function mapping the sequentially compact spaceinto R, equipped with the usual topology. Can the function f be “onto” R?

332 Section 18 : Locally compact spaces

18 / Locally compact spaces.

Summary. We will discuss another way of weakening the compact property

without sacrificing many of the properties we find desirable in a topological space.It is called the “locally compact” property. Rather than have the compact prop-

erty assigned to the whole space, we assign it on the elements of a certain neigh-bourhood base of each point. After defining this property formally we examine

what are its main characteristics. Even though local compactness is usually seencombined with the Hausdorff property our formal definition will be in its most

general form. We have seen that all Hausdorff compact spaces are normal. Butlocally compact Hausdorff spaces will be proven to be only completely regular. All

metrizable spaces will also be seen to be locally compact and Hausdorff. Theseconcepts will set the stage for the future study of “compactifications of locallycompact Hausdorff spaces”.

18.1 The locally compact property

We now present a formal definition of the locally compact property. We define this

property for arbitrary topological spaces. We alert the reader to the fact that manyauthors define local compactness for Hausdorff spaces only, probably because it sim-plifies its definition, and, more often then not, it is mostly referred to in the context

of Hausdorff spaces. We prefer to define it in its most general form.

Definition 18.1 Let S be a topological space. We say that S is locally compact if, for

every point, x, in S there is a compact set, K, such that x ∈ intSK ⊆ K.

When a particular topological property is applied only to some unspecified neighbour-hood of each point of a space rather than on the whole space, we traditionally precede

the name of this property with the adverb “locally”. In this particular case, we arereferring to a compact neighbourhood.

18.2 A characterization of local compactness.

Note that, in the above definition, in the particular case where the space S is not

Hausdorff, it may occur that the compact neighbourhood is not a closed subset of S.The following characterization referring specifically to locally compact spaces which


are also Hausdorff allows us to say, “there is an open set U with compact closure suchthat x ∈ U ⊆ clSU”.

Theorem 18.2 Let S be a Hausdorff topological space. The space, S, is locally compactif and only if each point, x, in S has an open neighbourhood base, Bx = Bi : i ∈ I, such

that clSBi is compact.


( ⇐ ) If B is an open neighbourhood of a point, x in S, such that clSB is compact

then, by definition, x has a compact neighbourhood. So S is locally compact.

( ⇒ ) We are given that S is locally compact and Hausdorff. Let p ∈ S and U be

an open neighbourhood of p and K be any compact neighbourhood of p. Then p ∈(intSK) ∩ U ⊆ K. Since K is compact and S is Hausdorff then, by theorem 14.3 b)

iii), K is a regular subspace. This means that there exists an open set V such that

p ∈ V ⊆ clKV ⊆ (intSK) ∩ U ⊆ K

Since clKV is closed in the compact set K, it must be compact. So p ∈ intSclKV ⊆clKV ⊆ U . We can conclude that each point of S has an open neighbourhood base

whose elements have a compact closure.

In the proof of the above theorem, note why the Hausdorff property on S is required.

The above characterization of local compactness on Hausdorff spaces allows us toquickly see that every compact Hausdorff space, S, is locally compact: Since compact

Hausdorff spaces are regular we can construct a compact neighbourhood of a point,p, inside any open neighbourhood of p.

Example 1. Consider the space R with its usual topology. Show that R is locally

compact.

Solution : Let Bp = (a, b) : a < p < b be an open neighbourhood base for the pointp. For each (a, b) ∈ Bp, there exists ε > 0, such that,

p ∈(

p− ε

3, p+

ε

3

)

⊆[

p− ε

3, p+

ε

3

]

⊆ (p− ε, p+ ε) ⊆ (a, b)

Since the closed interval in this chain of containments is the closure of the first interval

and is compact, then R, with the usual topology, is locally compact.


The space R2 is locally compact. Note that R2 is also locally compact since, for(a, b) ∈ R2, [a+ ε, a− ε] × [b+ ε, b− ε] is compact and

(a, b) ∈ intR2[a+ ε, a− ε] × [b+ ε, b− ε]

= (a+ ε, a− ε) × (b+ ε, b− ε)

Example 2. Consider the space, Z, of all integers with the discrete topology. Showthat Z is locally compact.

Solution : If n ∈ Z, and U is an open set containing n, clZn is a compact neigh-

bourhood of n, so,

n ∈ n = intZn = clZn = n ⊆ U

Then Z, with the discrete topology, is locally compact. In fact, every discrete spaceis locally compact.

18.3 Some basic properties of local compactness.

We now examine some invariance properties. A subset of a locally compact spacedoes not always inherit the locally compact property from its superset, as we shall

soon see. The locally compact property is carried over from the domain of a contin-uous function to its range provided the function is an open mapping. For a productspace, the locally compact property is carried over from the factors to the product,

provided all factors are locally compact and at most finitely many of those factors arenot compact. We now immediately prove these facts.

Theorem 18.3 Let S be a locally compact Hausdorff topological space.

a) If F is a closed subspace of S, then F is locally compact. So a closed subspace inheritsthe locally compact property from its Hausdorff superset.

b) If U is an open subspace of S, then U is locally compact. So an open subspace inherits

the locally compact property from its Hausdorff superset.

c) In any Hausdorff topological space, the intersection of two locally compact subspaces

of S is locally compact. In particular, if S is locally compact and Hausdorff, any subsetwhich is the intersection of an open set and a closed one is a locally compact subspace.

d) Let S be any Hausdorff topological space which contains a non-empty subspace, W .If W is locally compact, then W is the intersection of an open subset with a closed

subset of S.



a) We are given that F is a closed subset of the locally compact Hausdorff space S. Letx ∈ F and V be an open neighbourhood of x in S. Since S is locally compact and

Hausdorff, there exist an open neighbourhood U of x such that x ∈ U ⊆ clSU ⊆ V .Then x ∈ U ∩F ⊆ clSU ∩F ⊆ V ∩F , where clSU ∩F is a closed subset of the compact

set clSU , hence is a compact neighbourhood of x in F . So F is locally compact.

b) Let V be an open subset of the locally compact Hausdorff space S and U be an open

neighbourhood of x ∈ V . Then U ∩ V is an open neighbourhood of x in S. Since Sis locally compact and Hausdorff, there exist an open neighbourhood, W , such that

x ∈ W ⊆ clSW ⊆ U ∩ V , where clSW is compact in V . So V is locally compact, asrequired.

c) We are given that S is a Hausdorff space containing the two locally compact subspacesU and V with non-empty intersection. Since S is Hausdorff so are U , V and U ∩ V .

Let x ∈ U ∩ V and Z be an open subset of S such that x ∈ Z. Then there exists anopen subset, A, in U such that

x ∈ A ⊆ clUA ⊆ Z ∩ U ⊆ U

and an open subset, B, in V such that

x ∈ B ⊆ clVB ⊆ Z ∩ V ⊆ V

where clUA and clVB are compact; because of the Hausdorff property, they are closed.Then U ∩ V is locally compact, as required.

Combining parts a, b, and c the second part of the statement quickly follows.

d) Suppose W is a subset of a Hausdorff topological space. Suppose W is locally compact.

We are required to show that W is the intersection of an open set and a closed set inS. If we can show that W is open in S then W = W ∩ clSW , an intersection of an

open and closed subset of S, and we will be done. To show W is open we will provethat every point in W belongs to an open subset of S, contained in W .

Let x ∈W . Since W is locally compact and Hausdorff, S contains an open neighbour-hood, A, of x such that

x ∈ A ∩W ⊆ clW (A ∩W ) = clS(A ∩W ) ∩W

where clW (A ∩W ) is a closed and compact set in W and so clS(A ∩W ) ∩W is also

closed and compact in S. Since A∩W is contained in the closed set, clS(A∩W ) ∩W ,it must follow that clS(A ∩W ) ⊆ clS(A ∩W ) ∩W , and so, clS(A ∩W ) ⊆ W . So x ∈intS(clS(A∩W )) ⊆W . Then each point of W is contained in an open neighbourhoodof S and W is open. So W = W ∩ clSW , as required.

Corollary 18.4 Let S be a compact Hausdorff topological space and D be a dense subsetof S. Then the subset, D, of S is locally compact if and only if D is an open subset of S.


Proof : We are given that S is a compact Hausdorff topological space and D is a densesubset of S.

( ⇒) Suppose the dense subset, D, is a locally compact subspace of S. Then, by theabove corollary 18.4, D is an open subset of clSD = S. So D is open in S.

( ⇐ ) Suppose D is a an open dense subset of the compact space, S. Since S is com-pact it is locally compact. As shown in theorem 18.3, open subsets of locally compact

sets are locally compact. So D is locally compact. We are done.

Example 3. Determine whether the set, Q, of all rationals is locally compact or not.

Solution : The set Q is dense in the locally compact R. If Q was locally compact inR it would have to be open. Knowing that every interval containing an element of Q

contains an irrational, Q cannot be locally compact.

We know that the compact property is carried over from the domain to the codomain

of a continuous function. A similar result holds for the locally compact property butonly if the function is both continuous and open.

Theorem 18.5 Let S be a locally compact topological space and suppose T is any topo-logical space. If f : S → T is a continuous open function mapping S into T then f [S] islocally compact.

Proof : We are given that S is a locally compact topological space and f : S → T is acontinuous open function.

We are required to show that f [S] is locally compact. Let u ∈ f [S] and U be any

open neighbourhood of u. Suppose v ∈ f←[u] ⊆ f←[U ]. Then, by hypothesis, vhas a compact neighbourhood, say K, such that v ∈ intSK ⊆ K ⊆ f←[U ]. Since f is

open and continuous and

u = f(v) ∈ f [intSK] ⊆ f [K] ⊆ f [f←[U ]] = U

where f [intSK] is open and f [K] is compact, then every point in f [S] has a compact

neighbourhood. So f [S] is locally compact.

Note that the Hausdorff property is not required in the above statement.

Example 4. Show that the closed continuous image of a locally compact space need

not be locally compact.


Solution : Recall that in the usual topology, R2 is locally compact. Let A = (x, 0) :x ∈ R. Let D denote the decomposition of R2,

D = A ∪ (x, y) : y 6= 0

Given the quotient map, q : R2 → D , D is equipped with the quotient topology in-

duced by q.

Claim #1. We claim that function q is a closed map. Suppose B is a closed subsetof R2. Case 1: Suppose B ∩ A = ∅. Then q←[q[B]] = B. So q←[D \ q[B]] =R2 \ q←[q[B]] = R2 \B which is open in R2. Thus D \q[B] is open in D with respect

to the quotient topology. So q[B] is closed. Case 2: Suppose B ∩ A 6= ∅. Thenq←[q[B]] = A ∪B. So

q←[D\q[B]] = R2\q←[q[B]] = R2\(A∪ B)

an open subset of R2. Then D\q←[B] is open in D (by definition of quotient topology);

so q[B] is closed. Claim #1 is proved.

Claim #2. We claim that D is not locally compact. Suppose S is an open neigh-bourhood of A. It suffices to show that clDS is not compact. Then V = q←[S ] is anopen subset of R2.

q[clR2V ] = q[clR2q←[S ]]

= clD(q[q←[S ]])

= clDS

Consider the open neighbourhood, V ∪ (x, y) ∈ R2 : −n < x < n\A, of A in R2.

Then q[

V ∪ (x, y) ∈ R2 : −n < x < n\A]

is an open neighbourhood of A in D , foreach n ∈ N.

Then the collection

q[

V ∪ (x, y) ∈ R2 : −n < x < n\A]

: n ∈ N

forms an open cover of clDS , with no finite subcover. So clDS , is not compact. Weconclude that D is not locally compact.

In the following statement we show that finite products of locally compact spacesare locally compact, but one has to be cautious when considering infinite products of

locally compact spaces.



i∈I Si be aproduct space. Then S is locally compact if and only if every factor, Si, is locally compact

and at most, finitely many of the factors are not compact.


i∈I Si is

a product space.

( ⇒ ) Suppose S =∏

i∈I Si is locally compact. Since every projection map is open(see page 121) then, by theorem 18.5, every factor is locally compact. We now claim

that, at most, finitely many of the factors are not compact spaces.

If u ∈ S, then, since S is locally compact, there exists a compact set K in S, suchthat u ∈ intSK ⊆ K. Then there exists a finite subset, F , of I such that

u ∈ U = ∩π←i [Ui] : i ∈ F ⊆ intSK

where Ui is an open subset of Si. We claim that, if i 6∈ F then Si is compact. Consider

i 6∈ F : Then πi[U ] = Si ⊆ πi[K] (since U ⊆ K). So Si = πi[K]. Since πi is continuousand K is compact then Si is compact, for all i 6∈ F , as claimed. Done.

( ⇐ ) Suppose every factor, Si, is locally compact and at most finitely many are not

compact. We are required to show that the product space, S, is locally compact.

Let <xi>i∈I ∈ S =∏

i∈I Si and U = ∩π←i [Ui] : i ∈ F be a basic open neighbour-hood of <xi>i∈I . Suppose Fc is the largest subset of I , such that, if i ∈ Fc, Si is not

compact. Then, by hypothesis, Fc is finite. Let F1 = F ∪ Fc. Then F1 is finite. LetU1 = ∩π←i [Ui] : i ∈ F1 ⊆ U .

We claim that <xi>i∈I has a compact neighbourhood, K, which is contained in U1.

Given that, for each i ∈ F1, Si is locally compact, there must be a compact set, Ki,

such that xi ∈ intSiKi ⊆ Ki ⊆ Ui.

Then

<xi>i∈I ∈ ∩π←i [intSKi] : i ∈ F1 ⊆ ∩π←i [Ki] : i ∈ F1 = K ⊆ U1

where K is a neighbourhood of <xi>i∈I . See that, for each i 6∈ F1, Xi is a compact

factor of K, and, for each i ∈ F , Ki, is a compact factor of K. So K is a compactneighbourhood of <xi>i∈I , as claimed. So S is locally compact.


18.4 More on the Hausdorff locally compact property.

In the Embedding theorem part III, we arrived at the conclusion that “. . . any com-

pletely regular space can be densely embedded in a Hausdorff compact space”. Wewould like to state a similar result involving Hausdorff locally compact spaces. That

is, we will show that “. . . any locally compact Hausdorff space, S, can be embeddedin a Hausdorff compact space”.

But in this particular context the compact space in question is not the product ofclosed intervals of R, as one might expect. This compact space, denoted by, Sω, is

constructed by adding a single point to S and then defining an appropriate topologyon it so that it is compact.

Theorem 18.7 Let (S, τ) be a Hausdorff topological space and let ω be a point which doesnot belong to S. The set ωS is defined as

ωS = S ∪ ω

The topology, τω, on ωS is defined as follows: Firstly, τ ⊆ τω. Secondly, if ω ∈ U ⊆ ωS and

ωS\U is a compact subset of S, then U ∈ τω.

a) Then the family of sets, τω, thus defined, is a valid topology on ωS.

b) The topological space, (ωS, τω), is a compact space.

c) The compact space, (ωS, τω), densely contains a homeomorphic copy of S.

d) The compact space, (ωS, τω), is Hausdorff if and only if S is locally compact.

Proof : We are given that (S, τ) is a Hausdorff topological space and that ω is a point

which does not belong to S. Also, ωS = S ∪ ω.a) The proof is routine and so is left as an exercise.

b) We are given (ωS, τω). Let U = Ui : i ∈ I be an open cover of ωS.

By hypothesis, there exists, Uk, such that ω ∈ Uk and S\Uk is compact. Then S\Uk

has a finite subcover, Ui : i ∈ F. Then Ui : i ∈ F ∪ Uk is a finite subcover of ωS

and so ωS is compact.

c) The identity map, i : S → S ∪ ω, is easily seen to be a continuous one-to-one openmap onto i[S] = S.

d) We are given that ωS is compact.

( ⇒ ) Suppose ωS is Hausdorff. Then S is a Hausdorff subspace (since the Hausdorff

property is hereditary). Also, for any x ∈ S, there exists disjoint open neighbourhoodsU and V such that ω ∈ U and x ∈ V . Then S is open and dense in ωS. By corollary


18.4, S is locally compact.

( ⇐ ) We are given that S is a Hausdorff locally compact dense subspace of the com-pact space, ωS. We are required to show that ωS is Hausdorff. It follows immediately

that distinct pairs of points in S are contained in disjoint open neighbourhoods of ωS.Suppose p ∈ S. Since S is Hausdorff and locally compact there exists an open neigh-bourhood U of p such that p ∈ U ⊆ clSU ⊆ S, where clSU is a compact neighbourhood

of p. By hypothesis, S\clSU is an open neighbourhood of ω disjoint from U . So ωS isHausdorff, as required.

We now show that the local compactness property on a space S, when combined withthe Hausdorff property satisfies the completely regular separation property.

Theorem 18.8 A locally compact Hausdorff space is completely regular.

Proof : We are given that S is a Hausdorff locally compact topological space. Then S isembedded in the compact Hausdorff topological space ωS = S ∪ ω. Then by theo-

rem 14.3, ωS is normal and so is completely regular. The completely regular propertyis hereditary, so S is completely regular.

From the above result we again argue that if S is locally compact and Hausdorff it

can be densely embedded in at least two essentially different Hausdorff compact spaces.

We now know that S is embedded in ωS = S ∪ ω; secondly, from the Embedding

theorem 14.7, we know that...

A locally compact Hausdorff space, S, can be densely embedded in a compact

Hausdorff subspace of a cube.

18.5 On spaces which are σ-compact.

Some spaces can be described as being the union of a countably infinite number ofcompact sets. Such spaces need not be compact but still have properties that are

worth discussing. In particular, we will see that a locally compact Hausdorff spacewhich is a countable union of compact sets is a Lindelof Hausdorff space.

Definition 18.9 Let S be a topological space. We say that S is σ-compact if S is the

union of countably many compact sets.


Theorem 18.10 Suppose S is a locally compact Hausdorff space. Then S is σ-compact if

and only if S is Lindelof.

Proof : We are given that S is a locally compact Hausdorff topological space.

( ⇐ ) Suppose S is Lindelof. We are required to show that S is σ-compact. Since Sis locally compact and Hausdorff, we can construct an open covering of

B = Ux : x ∈ S

of sets with compact closures,

B∗ = clSUx : x ∈ S

Let Uxi : i ∈ N be a countable subcover of B. Then

S = ∪Uxi : i ∈ N ⊆ ∪clSUxi : i ∈ N

So S is σ-compact, as required.

( ⇒ ) Suppose S is σ-compact. We are required to show that S is Lindelof.

Then S = ∪Ci : i ∈ N, Ci is compact , a countable union of compact sets.

Consider the compact subset, C0.

Claim : That there is an open neighbourhood, U0, with compact closure, clSU0, such

that C0 ⊆ U0 ⊆ clSU0.

Proof of claim : Since S is locally compact and Hausdorff and C0 is compact any opencover, U , of C0 has a finite cover Vi : i = 1...k, such that each clSVi is compact for

i = 1 to k. Let U0 = ∪Vi : i = 1...k. Then

C0 ⊆ U0 ⊆ clSU0 = ∪clSVi : i = 1...k

is compact in S, as claimed.

We can inductively construct Ui : i ∈ N, such that each closure, clSUi, is compact

whereC0 ⊆ U0 ⊆ clSU0

C1 ∪ clSU0 ⊆ U1 ⊆ clSU1

C2 ∪ clSU1 ⊆ U2 ⊆ clSU2...

...Ci ∪ clSUi−1 ⊆ Ui ⊆ clSUi

......

SetKi = clSUi ⊆ Ui+1 ⊆ clSUi+1 = Ki+1


we obtain a family of compact sets, Ki : i ∈ N where Ki−1 ⊆ Ki and S = ∪Ki :i ∈ N.Suppose V = Vj : j ∈ J is an open cover of S. To show Lindelof it suffices to showthat V has a countable subcover.

Since each Ki is compact, there exist a finite subset, Fi ⊆ J such that

Ki ⊆ ∪Vjk: k ∈ Fi

Then V ∗ = Vjk: k ∈ Fi, i ∈ N is a countable subfamily of V .

Since S = ∪i,kVjk: k ∈ Fi, i ∈ N, V ∗ is a countable subcover of S. So S is Lindelof,

as required.

Example 5. Verify if the Moore plan (Niemytzki’s plane), Γ, is locally compact or not.

Solution : The Moore plane, Γ, has (x, y) ∈ R2 : y ≥ 0 as underlying set. Basicneighbourhoods of points in R2, and, if z = (x, 0), then z ∪A is a basic neighbour-

hood of z in R2 where A is an open disc tangent to the x-axis at z.

We claim that the Moore plane is not locally compact. If it was, then (0, 0) wouldhave a compact basic open neighbourhood with compact closure. If

V = (0, 0)∪ (x, y) : x2 + (y − δ2) < δ2

is such a basic neighbourhood then

clΓV = (x, y) : x2 + (y − δ2) ≤ δ2

We claim that clΓV is not compact. To see this consider the open cover

V ∪ (x, y) ∈ Γ : y > 1/n : n ∈ N, n > 0

of clΓV . It has no finite subcollection which covers clΓV .

Example 6. Is the set R line equipped with the lower limit topology, (R, τS), locally

compact?

Solution : The answer is no. Recall that an open base for τS, is B = [a, b) : a < b.Let r ∈ R. Suppose there is a δ > 0, such that clR[r, r+δ) is a compact neighbourhood

of r. See thatR\[r, r+ δ) = ∪∞r=1[r − n, r) ∪ [r+ δ, r+ δ + n]

is open, so clR[r, r+ δ) = [r, r+ δ). But [r, r+ δ) is not compact since

[r, r+ δ − 1/n) : n ∈ N, n > 0

is an open cover of [r, r+ δ) with no finite subcover.


Concepts review:

1. Define the locally compact property on a topological space.

2. State a characterization of Hausdorff locally compact spaces.

3. If S is locally compact and Hausdorff what kind of subsets are guaranteed to inheritthe locally compact property?

4. If T is a locally compact subspace of the Hausdorff space, S, what property does T

satisfy?

5. If D is a dense subset of a compact Hausdorff space, S, what can we say about D?

6. Provide an easy example of a non-locally compact space.

7. Are continuous images of locally compact spaces necessarily locally compact? Explain.

8. If S is a product space which is locally compact. What can say about its factors?

9. What conditions must be satisfied if we want the locally compact factors to be carriedover to their product.

10. Describe the Hausdorff compact space, ωS, which contains a dense homeomorphiccopy of a locally compact Hausdorff space, S.

11. If S is a Hausdorff locally compact space, what other separation axiom does it satisfy?

12. If S is a Hausdorff locally compact space, is there another compact space, other thanωS, which contains a dense copy of S?

13. Define a σ-compact space.

14. If S is locally compact and Hausdorff. In such a case provide a characterization ofthe σ-compact property.

EXERCISES

1. Suppose f : S → T is a continuous open function mapping a locally compact space,

S, onto a space T . Show that any compact subspace, K, of T is the image under f ofsome compact subspace, F , of S.


2. Show that the real numbers equipped with the upper limit topology (Sorgenfrey line)is not locally compact.

3. A perfect function, f : S → T , is a continuous closed and onto function which pulls

back each point in T to a compact set in S. Suppose S is Hausdorff and T is locallycompact. Show that f : S → T is a perfect function if and only if f pulls back

compact sets in T to compact sets in S.

4. Let S be a locally compact regular space and K be a closed and compact subset of

S. Suppose K is a subset of an open set, U , in S. Show that there is some compactset V , such that F ⊆ intSV ⊆ V ⊆ U .

5. Show that the Moore plane is not locally compact.

6. Suppose S is a Hausdorff space which contains a dense subset, D. Suppose x ∈ D.Show that, if V is compact neighbourhood of x in D, then V is a neighbourhood of x

in S.


19 / Paracompact topological spaces.

Summary. We will develop in this section some familiarity with the paracom-

pact property. After giving a formal definition we provide a few examples anddiscuss its invariance properties. Finally we show that all metrizable spaces areparacompact.

19.1 Paracompact topological spaces.

We now introduce a last important class of topological spaces closely related to the

family of compact spaces. Its importance was discovered in the role it played in char-acterizations of metrizable spaces. We begin by introducing some new terminology.

Note that some of what we present here may appear somewhat familiar since we havealready (informally) introduced the concept of a “locally finite” family of sets before

(on page 110). For convenience we formally reintroduce the concept here along withassociated terms.

Definition 19.1 Let S be a topological space.

a) Let U = Ui : i ∈ I be a family of subsets of S. We say that U is a locally finitefamily of sets if each point in S has a neighbourhood which intersects only finitely

elements of U . 1

b) We say that V = Vj : j ∈ J is a refinement of, (or refines) the family, U = Ui :

i ∈ I, if every element of V is contained in some element of U . If the elements of Vare open, then we will more specifically say that V is an open refinement of U .

c) Let U be a family of subsets of the topological space S. We say that U has a locallyfinite open refinement if there is a family, V , of open subsets in S which both refines

U and satisfies the locally finite property in S.

Whether a family of sets is locally finite or not in the space, S, depends on the topol-ogy of S (since the definition refers to neighbourhoods of S). Whether V refines Uor not refers to a set-theoretic property which characterizes a particular relationshipbetween between the elements of V and those of U . Open refinement adds a partic-

ular topological characteristic to the elements of the refinement.

1The word “neighbourhood-finite” is sometimes used instead of “locally finite”.

346 Section 19 : Paracompact topological spaces

Note that if, for any open cover U of a space S, there is a finite subfamily, F , of Uwhich covers the space S, then by definition, F refines U and, again by definition,

F is locally finite in S.

We can then say that . . . ,

. . . if S is compact and U is an open cover then U has a locally finite open

refinement V .

In the particular case described above, V simply turns out to be a finite subfamily ofU . We now define the main subject of this section.

Definition 19.2 Let S be a topological space. The space S is said to be a paracompact

space if, for any open cover U of S, there exists a locally finite open cover, V , of S whichrefines U .1

We alert the reader to the fact, in some books, the Hausdorff property is incorporatedinto the formal definition of the paracompact property, in the sense that, for these

authors, all paracompact spaces are hypothesized to be Hausdorff. In this book, aparacompact space is Hausdorff only when we explicitly state it as such.

Theorem 19.3 Any compact space is a paracompact space.

Proof : This statement has been proven in the paragraph immediately preceding the defi-

nition above.

Since all compact spaces are paracompact we then know of a large family of topological

spaces which are paracompact. We consider the following example of a non-compactparacompact space.

Example 1. Let S be an infinite space equipped with the discrete topology. SinceV = x : x ∈ S is an open cover with no subcover, the space, S, is not compact.

Show that S is, nevertheless, paracompact.

1Note that V satisfies four conditions: 1) V ’s elements must be open sets, 2) V must refine U , 3) V

must be locally finite, 4) V covers S.


Solution : Let U = Ui : i ∈ I be an arbitrary open cover of S. To show that Sis paracompact it suffices to show that U has a locally finite open refinement. From

the definition, the family V = x : x ∈ S is an open refinement of U (since everyone of its elements, x, is a subset of some set, Ui, in U ). It then suffices to show

that V is a locally finite family of sets. Let x ∈ S. Consider the open neighbourhood,B = x, of x. The neighbourhood B intersects only one element, x ∈ V . So V is

a locally finite family of sets in S.

So V is a locally finite open refinement of U . We can conclude that S is paracompact.

Example 2. Consider the space S = R equipped with the usual topology. We knowthat this metrizable space is unbounded and so is not compact. Show that S is para-

compact.

Solution : Suppose we have an open cover, U = Ui : i ∈ I, of S. We are requiredto construct an open refinement of U which is locally finite.

Firstly, we will first construct an open refinement, V , of U which covers S.For each n ∈ N\0, let (−n, n) denote an open interval centered at 0 of radius

n. Since, for each n, [−n, n] is closed and bounded it is a compact subset of S. So,for each n, [−n, n] has a finite open cover, say, Un = Ui : i ∈ Fn ⊆ U . That is,[−n, n] ⊆ ∪Un = ∪Ui : i ∈ Fn. For each i ∈ Fn, let

Vi = Ui\[−n+ 1, n− 1] ⊆ Ui

and let

Vn = Vi : i ∈ Fn

So each Vi in Vn is open in S and is contained in some Uj. Then, for each n, Vn is an

open cover of [−n, n] \[−n+ 1, n− 1]. So,

V = ∪Vn : n ∈ N\0

is an open refinement of U which covers all of S, as required.

Secondly, we will show that V is locally finite in S. If p ∈ S, then p ∈ [−m,m] \[−m+ 1, m− 1], for some m, and so p has a neighbourhood which intersects at mostfinitely elements of Vm ⊆ V . So V is locally finite.

We have constructed the family, V , which is both an open refinement of the open

cover, U , and is locally finite. So S = R is paracompact. We are done.

Example 3. Consider the space S = Rn equipped with the usual topology. We knowthat, for any n, this metrizable space is not compact. Show that S is paracompact.

Solution : To solve this we mimic the procedure in the previous example, replacing

the interval (−n, n) with the open ball, Bn(0), centered at 0.


We know that Hausdorff compact spaces are normal. We have a similar result forHausdorff paracompact spaces. We prove that Hausdorff paracompact spaces are

guaranteed to be “normal” topological spaces.

In the proof of the following theorem we invoke a lemma (6.16 on page 110) which werestate here, for convenience.

Lemma 6.16 : If U = Vi : i ∈ I is a locally finite collection of sets in S thenU ∗ = clSVi : i ∈ I is also locally finite. Furthermore,

clS [∪Vi : i ∈ I] = ∪clSVi : i ∈ I

Theorem 19.4 Let S be a Hausdorff paracompact topological space.

a) The space, S, is a regular space.

b) The space, S, is a normal space.

Proof : We are given that S is a Hausdorff paracompact topological space.

a) We are required to show that S is regular.

Suppose H is a closed subset of S and u ∈ S\H . It suffices to show that there is an open

set, W , such that H ⊆W ⊆ clSW ⊆ S\u.Since S is Hausdorff and H is closed, then, for each x ∈ H , there exists an open neighbour-

hood, Ux, of x such that u 6∈ clSUx. Now,

U = Ux : x ∈ H ∪ S\H

forms an open cover of S. Since S is paracompact and U is an open cover of S, there existsa locally finite collection of open sets

V = Vi : i ∈ I ∪ V

which covers S and refines the open cover, U (where each Vi ⊆ Uyi , for some yi ∈ H , and

V ⊆ S\H).

Then, for each i,

Vi ⊆ clSVi ⊆ clSUyi ⊆ S\u (For some yi ∈ H)

Let W = ∪Vi : i ∈ I. Then by the lemma 6.16,

clSW = ∪clSVi : i ∈ IIt follows that

H ⊆W = ∪Vi : i ∈ I ⊆ clSW = ∪clSVi : i ∈ I ⊆ S\u

So S is regular. As required.


b) We are now required to prove that S is a normal space. Let H be a closed set in S. Toprove the normal property holds we replace the point u in the proof of part a) with a closed

set, F , and invoke the regular property and mimic the steps in the proof above.

Suppose H is a closed subset of S and F is a closed subset such that F ∈ S\H .

For each x ∈ H , there exists an open neighbourhood, Ux such that F ∩ clSUx = ∅ (sincewe have shown that S satisfies the regular property).

Now,

U = Ux : x ∈ H ∪ S\H

forms an open cover of S (with F ⊆ S \H). By hypothesis, there exists a locally finite

collection of open sets

V = Vi : i ∈ I ∪ V

which refines U , where each Vi is a subset of some Ux and V ⊆ S\H .

Then, for each i,

Vi ⊆ clSVi ⊆ clSUx ⊆ S\F (For some x ∈ H.)

Again, by lemma 6.16, ∪clSVi : i ∈ I = clS(∪Vi : i ∈ I).Since clS(∪Vi : i ∈ I) = ∪clSVi : i ∈ I ⊆ (∪clSUx : x ∈ H) ⊆ S\F , S is normal, thedesired property.

We know that the compact property is carried over from the domain to the codomain

by arbitrary continuous functions. This is not the case for the paracompact property.But, if the function is one which is closed and continuous, it does, as the following

result shows.

Theorem 19.5 Let S be a paracompact topological space and T be any space. If f : S → T

is a closed and continuous function then f [S] is a paracompact subspace of T .

Proof : The proof is lengthy and involved. A reasonably complete proof is found in [En-gelking].

Theorem 19.6 Let S be a paracompact topological space and F be a closed subset of S.Then F inherits the paracompact property from S.


Proof : We are given that S is a paracompact topological space and F is closed in S. Weare required to show that F is also paracompact.

Let U = Ui : i ∈ I be an open cover of F . Then, for each i, Ui = Vi ∩ F , for some

open subset Vi of S. We then obtain an open cover, V = Vi : i ∈ I ∪ S\F of S.Since S is paracompact, then there is a locally finite open refinement, W = Wi : i ∈I, of V .

The family Wi ∩F : Wi ∩F 6= ∅ forms a locally finite open refinement of U . ThenF is paracompact.

19.2 A non-paracompact space.

We now present an example of a space which is not paracompact.

Example 4. A standard, non-trivial, example of a non-paracompact space is the deletedTychonoff plank, S = [0, ω1] × [0, ω0)\(ω1, ω0). On page 190, we showed that the

space S is not normal. Since paracompact spaces have been shown to be normal, thenS cannot be paracompact.

Also the Moore plane was shown to be completely regular but not normal. So this is

another example of a non-paracompact space.

19.3 Topic: Metrizability and paracompactness.

Any result which can bring us a step closer to a characterization of metrizable spacesis considered to be important and worth the effort required to study the proofs of

related statements. The following definitions and technical results will lead to thethreshold of a proof of the statement, “Metrizable spaces are paracompact spaces”.

Definition 19.7 Let S be a topological space and U be a subfamily of P(S). If

U = ∪Un : n ∈ N

where each Un is a locally finite subset of P(S) then we say that U is σ-locally finite.


Lemma 19.8 Suppose S is a metrizable space and U is an open cover of S. Then thereis an open cover, E , which both refines U and is σ-locally finite.1

Proof : Let S be a metrizable topological space. Then there is a metric, ρ, which allows us

to express S as a metric space, (S, ρ). Suppose U is an open cover of S.

We will construct a subfamily, E , of P(S) such that

1) E is an open cover of S,

2) E refines U ,

3) E = ∪En : n ∈ N\0 where each En is a locally finite collection of open sets.

We will index the elements of the open cover, U , with ordinals Ω: U = Uα : α ∈ Ω.That is,

U = U0, U1, . . . , Uω0, Uω0+1 . . . , Uω1 , Uω1+1, Uω1+2, . . . , 2

Step 1: Construction of the collection of sets, E .

For each Uγ ∈ U and a fixed n in N\0 let

Sn(Uγ) = x ∈ S : B1/n(x) ⊆ Uγ

Let

Tn(Uγ) = Sn(Uγ)\∪Uα : α < γ

Now see that Tn(Uγ) ⊆ Uγ .3 If we repeat this for each element, Uα, of U , then

Tn = Tn(Uα) : α ∈ Ω

is a refinement of U .

We claim that the elements of Tn are pairwise disjoint.

Proof of claim : Consider Tn(Uβ) and Tn(Uγ) where β < γ. Choose element c ∈Tn(Uγ). Then for any point b ∈ Tn(Uβ), b ∈ Sn(Uβ). So b ∈ B1/n(b) ⊆ Uβ . Since

c ∈ Tn(Uγ) and β < γ, c 6∈ Uβ . So c 6∈ B1/n(b). So c 6∈ Tn(Uβ). We have shown that,

b ∈ Tn(Uβ) and c ∈ Tn(Uγ) ⇒ c 6∈ B1/n(b) (∗)

So the elements of Tn are pairwise disjoint, as claimed.

1Note that E is “σ-locally finite” (not locally finite) so we cannot conclude immediately that metrizablespaces are paracompact. This will come later.

2Here we are invoking the Well-ordering theorem which permits such an ordering. It is a statement whichis equivalent to the Axiom of choice.

3For, if x ∈ Tn(Uγ), then x ∈ B1/n(x) ⊆ Uγ .


Even if Tn is a refinement of U , its elements may not be open, so we have more workto do.

We will construct an open set En(Uγ) such that

Tn(Uγ) ⊆ En(Uγ) ⊆ Uγ

To construct En(Uγ), let

En(Uγ) = ∪B1/3n(x) : x ∈ Tn(Uγ)

We repeat this for each element, Uα, of U , to form the collection of sets,

En = En(Uα) : α ∈ Ω

Step 2: Proof that the family of sets,

E = ∪En : n ∈ N\0

satisfies all the required properties.

1) The collection E is an open refinement of U :

Each En(Uα) is easily seen to be open so, for each n, En is a family of open sets, the

same then holds true for E .

We claim that En refines U .

u ∈ En(Uγ) ⇒ u ∈ B1/3n(c) for some c ∈ Tn(Uγ)

Tn(Uγ) = Sn(Uγ)\∪Uα : α < γ ⇒ c ∈ Sn(Uγ)

Sn(Uγ) = x ∈ S : B1/n(x) ⊆ Uγ ⇒ B1/n(c) ⊆ Uγ

u ∈ B1/3n(c) ⊆ B1/n(c) ⇒ u ∈ Uγ

Then En(Uγ) ⊆ Uγ . Hence En is an open refinement of U , as claimed.

This satisfies one of the required properties of E .

2) The collection E is σ-locally finite. That is, E is the countable union of locallyfinite families of sets:

Claim A. We first claim that the elements of En = En(Uα) : α ∈ Ω are pairwise

disjoint.

Proof of claim. Consider the pair En(Uβ) and En(Uγ) where β < γ. Let v ∈ En(Uγ)and u be any element in En(Uβ). We will show that v 6= u.

u ∈ En(Uβ) = ∪B1/3n(x) : x ∈ Tn(Uβ) ⇒ u ∈ B1/3n(b) for some b ∈ Tn(Uβ)

v ∈ En(Uγ) = ∪B1/3n(x) : x ∈ Tn(Uγ) ⇒ v ∈ B1/3n(c) for some c ∈ Tn(Uγ)

b ∈ Tn(Uβ) and c ∈ Tn(Uγ) ⇒ c 6∈ B1/n(b) (By (∗).)


Then

1/n ≤ ρ(b, c)

≤ ρ(b, v)+ ρ(c, v)

< [ρ(b, u)+ ρ(u, v)]+ 1/3n

< 1/3n+ ρ(u, v) + 1/3n

= ρ(u, v) + 2/3n

implies ρ(u, v) > 1/3n (∗∗)

So v 6∈ B1/3n(u).

We have shown that,

u ∈ En(Uβ) and v ∈ En(Uγ) ⇒ v 6∈ B1/3n(u) (∗ ∗ ∗)

So v 6= u and u ∈ En(Uβ) then v 6∈ En(Uβ). We conclude, if β < γ, En(Uβ)∩En(Uγ) =∅; this proves claim A.

Claim B. We now claim that each En is locally finite.

Proof of claim. Let z ∈ S. By claim A, z belongs to at most one element of En, sayEn(Uγ) = ∪B1/3n(x) : x ∈ Tn(Uγ). Say z ∈ B1/3n(y). By (**) the distance betweenpoints in different sets in En is larger than 1/3n. So the ball B1/6n(z) intersects only

En(Uγ). Then En is locally finite, as claimed.

We are done with the second property.

3) The collection E is an open cover of S : Let y be any point in S. We must showthat y must belong to some member, Em(Uµ) ∈ Em, of E .

Let µ be the least ordinal such that y ∈ Uµ ∈ U .1 Since Uµ is open we can choose a

natural number m such that B1/m(y) ⊆ Uµ. Then y ∈ Sm(Uµ) = x ∈ S : B1/m(x) ⊆Uµ. Since µ is the least ordinal such that y ∈ Uµ, then y ∈ Tm(Uµ) = Sm(Uµ)\∪Uα :

α < µ. By definition, B1/3m(y) ⊆ Em(Uµ) = ∪B1/3m(x) : x ∈ Tn(Uµ), soy ∈ Em(Uµ) ∈ Em ⊆ E . Then each element of S belongs to some element of E .

So E is an open cover of S.

So E is the countable union of locally finite collections of open sets which both covers

S and refines U .

We are done.

1The ordinal µ exists since the ordinals are “well-ordered”.


In the next theorem, takes us a step closer to the statement “Metrizable spaces areparacompact spaces”. The reader will recognize the use of techniques similar to the

ones used in the proof of the lemma 19.8.

Theorem 19.9 Suppose the space S is metrizable and U is an open cover of S. Then Uhas a refinement, C , which both covers S and is locally finite. 2

Proof : Suppose S is a metrizable space and U is an open cover of S. By lemma 19.8 thereis an open cover, E , which both refines U and is σ-locally finite. Let

E = ∪En : n ∈ N\0

be such an open cover where each En = En(i) : i ∈ In is a locally finite collection ofopen sets which refines U . For each m ∈ N\0, let

Vm = ∪Em(i) : i ∈ Im = ∪ Em

For n ∈ N\0, and i ∈ In let

Sn(En(i)) = En(i)\∪Vm : m < n = En(i)\∪

∪Em(i) : i ∈ Im : m < n

We are required to construct a collection of sets, C , which covers S, refines E and islocally finite. (Note that the members of C need not be open.)

LetCn = Sn(En(i)) : i ∈ In

Since Sn(En(i)) ⊆ En(i) ∈ En then Cn refines En.

Let

C = ∪Cn : n ∈ N\0

Since each Cn refines En and each En refines U then C is a refinement of U

Claim A: It is claimed that C covers S.

Proof of claim : Let p ∈ S. We know that E = ∪En : n ∈ N\0 is an opencover of S. Let k = minn : p belongs to some element of En. Let j be such that

p ∈ Ek(j) ∈ Ek. Then p ∈ Sk(Ek(j)) ∈ Ck ⊆ C . Then p belongs to some element of C .So C covers S. This establishes claim A.

Claim B: It is claimed that for p ∈ S, p belongs to a neighbourhood B ⊆ Ek(j) which

meets at most finitely many elements of C . That is, C is a locally finite collection ofsets in S.

2Note that we are not hypothesizing that the members of C are open in S.


Proof of claim : Let p ∈ S and k = minn : p belongs to some element of En (asdefined in the proof of claim 1). Then p ∈ Sk(Ek(i)) ⊆ Ek(i), for some i ∈ Ik.

Recall that, for each n, En is locally finite. So, for each n, there exists a neighbour-hood, Bn, of p which intersects only finitely many elements in En = En(i) : i ∈ In.Because of this, this Bn can only intersect finitely may elements of Cn = S(En(i)) :

i ∈ In for, whenever it intersects S(En(i)) ⊆ En(i), it intersects En(i). Also, if n > k,Ek(j) ∩ Sn(En(i)) = ∅ (by definition of Sn(En(i))).

Let B = ∩Bn : n ∈ 1, 2, . . . , k. Then p ∈ B and so B ∩ Ek(i) is an openneighbourhood of p which meets at most finitely many elements of C (these are in∪C1,C2, . . . ,Ck, ). So C is locally finite.

Then the collection, C , refines U , is locally finite and covers S. As required.

Lemma 19.10 Suppose the space S is metrizable and U is an open cover of S. Then Uhas a refinement of closed sets which both covers S and is locally finite.

Proof : Suppose S is a metrizable space and U is an open cover of S. Since metrizablespaces are regular (by theorem 9.19), then S is regular.

Then, for each U ∈ U and for each x ∈ S, there is some open neighbourhood, Bx, ofx such that clSBx ⊆ U . Then B = Bx : x ∈ S is an open cover of S which refines

U where Bx ⊆ U implies clSBx ⊆ U .

By theorem 19.9, there is a collection of sets, C = C : C ∈ C , which refines B, cov-

ers S and is locally finite. By lemma 6.16, if C locally finite then C ∗ = clSC : C ∈ C is also locally finite. Then the family C ∗ covers S (since C covers S) and is a refine-

ment of U (since C refines U ).

So C ∗ is the desired refinement of U whose members are closed and which covers S.

Example 5. Suppose F is a locally finite collection of closed subsets in P(S). Show

that ∪F : F ∈ F is a closed subset of S.

Solution : We are given that each F ∈ F is closed where F is locally finite. LetM = ∪F : F ∈ F. From lemma 6.16, we know that, if F is locally finite,clSF : F ∈ F is locally finite and clSM = ∪clSF : F ∈ F. Since

∪clSF : F ∈ F = ∪F : F ∈ F

then M = clSM . So M is closed in S.


We are finally set for the main result of this section.

Theorem 19.11 If a space is metrizable then it is paracompact.

Proof : Suppose S is a metrizable space and U is an open cover of S. By theorem 19.9,there exists a family of sets, H = H : H ∈ H , which refines U , covers S and is

locally finite.

Since H is locally finite, for each x ∈ S we can find open a neighbourhood, Vx, of x

which intersects at most finitely many elements of H . Let

V = Vx : x ∈ S

be the collection of all such sets.

Since V is an open cover of S, lemma 19.10 applies: There is a collection of closed

sets,

C = C : C ∈ C which refines V , covers S and is locally finite.

For each C ∈ C ,

C intersects with finitely many elements of H (∗)

(since for each C, C ⊆ Vx, for some x, and Vx itself intersects with finitely manyelements of H ).

Since H refines the open cover, U , for each H ∈ H , we can choose an element , UH ,in U , such that

H ⊆ UH ∈ U

For each H ∈ H . let

EH = S\∪C ∈ C : C ⊆ S\H

See that EH is open, since C is locally finite, ∪C is closed (see the example imme-

diately preceding this theorem). We claim that H ⊆ EH . For, if y ∈ H , we canargue,

y 6∈ EH ⇒ y ∈ ∪C ∈ C : C ⊆ S\H⇒ y ∈ C, for some C ∈ C such that C ∩H = ∅

⇒ y 6∈ H


So H ⊆ EH , as claimed.

Then H is a subset of the open subset, EH ∩ UH . Then we define

D = EH ∩ UH : H ∈ H

Then,

1) D is an open cover of S: Since H covers S, and H ⊆ EH ∩ UH .

2) D refines U : Since, [y ∈ EH0 ∩ UH0 some H0] ⇒ [y = UH0 ∈ U ].

3) D is locally finite. (If so, D is the family of sets we seek.)

Proof of 3): Let p ∈ S. Since C is locally finite, p has a neighbourhood, B, whichintersects finitely many elements, Ci : i = 1, . . . , k ⊆ C . Since C is a cover of S,

thenB ⊆ ∪Ci : i = 1, . . . , k

It suffices to show that each element of C intersects EH ∩ UH for, at most, finitelymany H ; for, if so, B intersects only finitely many elements of D .

Suppose that, for some j, Cj intersects EH1 ∩ UH1 ; say, y ∈ Cj ∩ [EH1 ∩ UH1 ]. Theny ∈ Cj ⊆ UH1 ∈ U . Also,

y ∈ Cj ∩EH1 ⇒ y ∈ Cj ∩ S\∪C ∈ C : C ∩H1 = ∅⇒ y 6∈ ∪C ∈ C : C ∩H1 = ∅⇒ y belongs to Cj such that Cj ∩H1 6= ∅

⇒ Cj intersects some UH1 ∈ U

Recall that C in C can intersect at most finitely many H ’s in H (see (*)). Then each

Cj can intersect at most finitely many elements of D .

We conclude that B intersects at most finitely many elements of D .

We conclude that D is locally finite. This proves statement 3).

We conclude that D is an open refinement of U which both covers S and is locally

finite in S. So the metrizable space, S, is paracompact.

Concepts review:

1. What does it mean to say that U is a locally finite family of subsets of a space S?

2. What does it mean to say that the family of subsets, V , is an open refinement of thefamily U ?


3. What does it mean to say that the family of subsets, V , is a locally finite openrefinement of the family U ?

4. Define the paracompact property of a topological space.

5. What class of topological spaces has elements which are guaranteed to be paracom-

pact?

6. Provide an example of a paracompact space and briefly summarize arguments which

confirms your answer.

7. Provide an example of a non-paracompact space.

8. If a paracompact space is Hausdorff what other separation axioms does it satisfy?

9. If f : S → T is a function mapping the paracompact space S into T , what properties

must be satisfied by f if we want to guarantee that f [S] is paracompact?

10. If S is a paracompact space what kind of subsets of S will share this property?

EXERCISES

1. Show that, if a paracompact space S is countably compact, then S is compact.

2. Suppose S is a Lindelof space and U is a family of subsets of S. Show that U is a

countable family of subsets.

3. Suppose S is a regular space and U = Fi : i ∈ I is a family of paracompact closedsubsets of S. Show that if U is locally finite then ∪Fi : i ∈ I is paracompact.

4. Let S be a perfectly normal paracompact space. Show that any non-empty subspaceis also paracompact.

Part VI

The connected property

359

Part VI: The connected property 361

20 / Connected spaces and properties

Summary. In this section we formally define the topological property of con-nectedness while providing a few examples of connected spaces, and of spaces

which are not. Continuous images of connected spaces are proven to be con-nected. Arbitrary unions of families of connected sets are proven to be connected

provided they have at least one point in common. If the set A is connected andis dense in a set, B, then B is shown to be connected. In particular, closures of

connected sets are connected. The connected property will be seen to be invariantover arbitrary products of connected spaces, without conditions on the number

of factors. A connected component is defined as being the largest connected sub-space containing a given point. Components will be seen to be closed, but notnecessarily open. Properties of spaces that have an open base of connected neigh-

bourhoods will be briefly discussed. We will also define the pathwise connectedproperty; we will see that sets which satisfy this property are connected. But the

converse does not hold true. Finally we will introduce the totally disconnectedproperty.

20.1 Definition.

The property of connectedness is one that is, more or less, independent of the topologi-

cal properties that we have studied until now, in the sense that we cannot characterizethis property with some combination of the other properties. It is fairly easy to con-struct mental images of “connected spaces”. A good way to start is to think of a

single string and ask yourself “Is this string connected?” to which you would answer,“...of course!”. Then take a pair of scissors and cut it, and ask, “...what about now?”

to which you would intuitively answer “... well no, not anymore”. A bystander mightadd “Both represent the same string, but the second one differs topologically from

the the first”. All this is stated based on an intuitive understanding of the word “con-nected”. But intuition, as a guiding tool, precedes the process of providing a rigourous

mathematical definition. That is, producing a definition which can be interpreted inonly one way while simultaneously satisfying our intuitive understanding of this prop-

erty. A rigourous definition should allow us to answer more difficult questions aboutobjects which possess this property but whose complexity defies our imagination.

Definition 20.1 A topological space S is said to be connected if it is not the disjoint unionof two non-empty open sets.

362 Section 20: Connected spaces and properties

Recall that, in this book, those subsets of a space that are both open and closed arereferred to as being clopen. So, if S is the disjoint union of two open sets A and B if

and only if both A and B are complements of an open sets and so are also closed. Sowe can say that a space . . .

“S is not connected if and only if S contains a proper non-empty clopen subset”.

Example 1. The topological space, R, with the usual topology is connected since no

non-empty proper subset in R is clopen. However, the subspace of all rationals, Q, isnot connected since Q the disjoint union of the two open sets,

U = (−∞, π) ∩ Q and Q\U = (π,∞)∩ Q

So the connected property is not hereditary.

It is not difficult to prove that the only proper non-empty subsets of R which are

connected subspaces are those subsets which are either a single element or those whichare intervals (either open, closed or half open).

Example 2. The topological space R with the upper limit topology is not connectedsince R is the disjoint union of U = (−∞, 3] and R\U = (3,∞) both of which are

open in this topology. (Verify this.)

20.2 Properties of connected spaces.

We now investigate some invariance properties for connectedness.

Theorem 20.2 The continuous image of a connected space is connected.

Proof : We are given that S and T are topological spaces, that S is connected and thefunction, f : S → T , is continuous. We are required to show that f [S] is connected.

Suppose U is clopen subset of f [S] where U 6= f [S]. We claim that U must be empty.

To see this suppose,

h : f [S] → 0, 1

is defined as h[U ] = 0 and h[ f [S] \ U ] = 1. Then h[f [S]] ⊆ 0, 1, so h is

continuous on f [S]. Since f is continuous on S,

hf : S → 0, 1


is also continuous on S. Now 0 and 1 are clopen subsets of h[f [S]], hence

(hf)←[0] = f←[h←(0)] = f←[U ] 6= S (Since U 6= f [S].)

where f←[U ] must be a clopen subset of S (but not equal to S). Since S is connected,

the only clopen subset in S other than S itself is ∅. So f←[U ] = ∅ which implies

U = f [f←[U ]] = f [∅] = ∅

So U = ∅, as claimed. So the continuous image, f [S], contains no non-empty properclopen set. So f [S] must be connected.

Example 3. Suppose (R, τ) is the set of reals equipped with the usual topology and

(R, τs) is the set of reals with the upper limit topology. Show that the identity map

i : (R, τ) → (R, τs)

is not continuous.

Solution : We have seen that (R, τ) is connected and that (R, τs) is not. Since thecontinuous image of a connected set is connected then i : (R, τ) → (R, τs) cannot be

continuous, as required.

Does the union of connected sets preserve the connected property? Our instinct statesthat they must, a least, have non-empty intersection. We should check this out.

Suppose

S = ∩Ui : i ∈ Iwhere each Ui is connected in S. Suppose q ∈ ∩Ui : i ∈ I.If there exists a function, h : ∪Ui : i ∈ I = 0, 1, which is continuous on S then,

since each Ui is connected, h must be constant on each Ui. There is no other option.It must then follow that, for each i, h(q) and h[Ui] have the same value. So h must

be constant on the union of these sets. So ∪Ui : i ∈ I cannot be the union of twonon-empty clopen sets and so must be connected.

It is worth recording this as a theorem, for future reference.

Theorem 20.3 The arbitrary union of a family of connected sets which have at least onepoint in common is connected.

Proof : The statement is proved in the paragraph above.


Example 4. Suppose S = R2 equipped with the usual topology. Show that S is aconnected space.

Solution : For each point x ∈ S, let Lx denote the infinite line containing both x and0.

Then S = ∪Lx : x ∈ S. For each x ∈ S, Lx is homeomorphic to R (previously shownto be connected) and contains (0, 0); hence S is the union of a family of connected

spaces which contain one point in common. Then S is connected.

In the following theorem we see that the closure operation preserves connectedness.We then verify that arbitrarily large products preserve connectedness, provided every

factor is connected. Also that connected product spaces must have connected factors.

Theorem 20.4 The closure of a connected set is connected. Furthermore, any space, S,which has a connected dense subset is connected.

Proof : We are given that T is a connected subspace of the topological space S.

We are required to show that clST is connected. Suppose h : clST → 0, 1 is acontinuous function. To show that clST is connected it suffices to show that h is

constant on clST . Since h is continuous, then h must be constant on the connectedset T . Suppose q ∈ clST\T . If h(q) 6= h[T ], then h←[h(q)] is the pull-back of a clopenset in 0, 1 and so must intersect T , contradicting the fact that h is constant on T .

So h must be constant on clST and so is connected.

By applying a similar reasoning, if S has a connected dense subset then S is connected.


i∈I Si be aproduct space. Then S is connected if and only if each Si is connected.


i∈I Si is a product space.

( ⇒ ) If the product space, S, is connected then each factor, Si, is the continuous

image of S under the projection map, πi : S → Si, and so is connected.

( ⇐ ) Suppose Si is connected for each i ∈ I .

Let’s first consider the case where S has only two connected factors, S1 and S2.


We claim that S1 × S2 is connected. Suppose (q1, q2) is a fixed point in S1 × S2 andlet (x, y) be any other point. Then (q1, q2) ∈ S1 × q2 and (x, y) ∈ x × S2. Then

the set,M(x,y) = (S1 × q2) ∪ (x × S2)

is the union of two connected. Hence M(x,y) is connected and contains the fixed point(q1, q2), for any choice of (x, y). More generally, every element of M(x,y) : (x, y) ∈S1 × S2 is a connected set and contains (q1, q2). So ∪M(x,y) : (x, y) ∈ S1 × S2 isconnected. Since S1 × S2 = ∪M(x,y) : (x, y) ∈ S1 × S2, then S1 × S2 is connected,

as claimed.

By finite induction, the product space of finitely many connected factors is connected.

Let F denote the family of all finite subsets of the index set, I . Supposeq =< qi >i∈I is an element of the product space, S, and let F ∈ F . Let

Ti = Si if i ∈ F

Ti = qi otherwise

Let KF =∏

i∈I Ti. By the claim established above, KF is a connected subspace ofthe product space, S. Hence KF : F ∈ F is a family of connected subspaces of S,

whereq = < qi >i∈I = ∩KF : F ∈ F

By theorem 20.3, V = ∪KF : F ∈ F is a connected subset of S.

We claim that V is dense in S. Let u =< ui >i∈I ∈ S. Let U =∏

i∈I Ui be a basic

open neighbourhood of u. Then there is a finite subset, F ∗ of I , such that, Ui = π←i [U ]for i ∈ F ∗ and equals Si, otherwise. It suffices to show that, U ∩ V 6= ∅. See thatKF∗∩U =

∏

i∈I Vi such that, Vi = Ui for i ∈ F ∗ and Vi = qi, otherwise. So KF∗∩Uis non-empty. Hence V ∩ U is non-empty. So V is dense in S. Since V is dense andconnected then, by theorem 20.4, so is S.

Given a function f : T → T , the point x in T is called a fixed point of f if f(x) = x.If T ⊆ S and f : S → T is a continuous function on S such that every point of T is

a fixed point of f , then we say that T is retract of f . The following theorem showsthat connectedness of [a, b] guarantees that a continuous function f : [a, b] → [a, b] on

[a, b] will have at least one fixed point of f . This well-known statement is commonlyreferred to as the Brouwer’s fixed point theorem.

Theorem 20.6 For a closed interval [a, b] in R a continuous function f : [a, b] → [a, b]mapping [a, b] onto [a, b],auchy will have at least one fixed point of f .


Proof : We are given that f : [a, b] → [a, b] is continuous. Let

h(x) =f(x) − x

|f(x) − x|

Suppose f has no fixed points in [a, b]. The function h(x) is then well-defined andcontinuous on [a, b]. Then h(x) maps [a, b] into 1,−1. Since f is continuous on the

connected set [a, b], h[a, b] is connected. So h is constant on it’s domain. If h(x) = 1,then |f(x) − x| = f(x) − x; so f(x) ≥ x on [a, b]. Then f(b) = b. Similarly, ifh(x) = −1 then −|f(x) − x| = f(x) − x so f(x) ≤ x, so f(a) = a. Then f must have

at least one fixed point.

20.3 The connected components of a topological space.

A topological space, S, which is not connected can always be partitioned in a familyof subspaces which are connected. A trivial example could be the partitioning of

S = x : x ∈ S, into singleton sets, since each singleton set is connected. However,a point x may be contained in a larger connected subspace of S which contains it.

One might want to partition S as

S = ∪Cx : x ∈ S

where Cx denotes “the largest connected subspace of S which contains x”. To remove

any ambiguity about what we mean by “. . . Cx is the largest connected set in S, . . .”we might prefer to say “Cx is the union of all the connected subspaces of S which

contain x”. This leads us to the following definition.

Definition 20.7 If S is a topological space and u ∈ S, then we define the connected compo-nent of u in S as being the union of all connected subspaces of S that contain the point u. 1

Clearly, the only component of a connected space is the space itself.

The definition confirms our perception that the component of a point u in S, is the

unique largest connected subset of S which contains the point u. Also, two compo-nents, C1 and C2, in S cannot intersect, for if z ∈ C1 ∩C2, then C1 ∪C2 is connected;

1When there can be (by context) no confusion we often drop the adjective “connected” and only say“component” .


hence neither C1 nor C2 could be components of z. Since every point of a spaceS belongs to precisely one connected component, then the family of all components

partitions the space, S.

If C is a “connected component” in S, we mean that C is a connected as a subspaceof S. That is, C cannot contain a non-empty clopen sets in C. This means that,. . .

– if U and V are both open in S such that

(U ∩ C) ∩ (V ∩C) = ∅ and (U ∩C) ∪ (V ∩ C) = C

then one (U ∩ C) or (V ∩ C) must be empty.

– if x ∈ S\C, and Cx is the component containing x and T = C∪Cx then T cannot

be a connected subspace of S, and so T must contain a non-empty clopen subsetof T , say B. Since C and Cx are connected in T , C ⊆ B or C ⊆ T \B. Then Cand Cx are clopen in T .

Example 5. Consider the subspace, T = B1(0, 0) ∪ B1(0, 2), the union of two openballs of radius one in R2 with center (0, 0) and (0, 2) respectively. Since T is the union

of two open disjoint subsets of T , then, by definition, T is not a connected space. Thespace, T , has precisely two components, B1(0, 0) and B1(0, 2).

Example 6. Let U = B1(0, 0), V = B1(0, 2) in R2. Consider the subspace,

T ∗ = U ∪ V ∪ (0, 1)

of R2. It is clearly the case that T ∗ is the union of three non-intersecting connected

subsets, the two connected sets U and V and the connected set (0, 1). But wecannot conclude from this that T ∗ is not connected. We can view T ∗ as the union

of the two connected sets, U ∪ (0, 1 and V ∪ 0, 1) with non-empty intersection,(0, 1). So T ∗ is a connected set with itself as the only component.

Theorem 20.8 Suppose C is a component of a space, S. Then C is a closed subset of S.

Proof : We are given that C is a component of the topological space S.

Since C is, by definition, connected, then clSC is also connected. Since C is the largest

connected set containing one of its points, then clSC = C. So C is closed.


The definition of the “connected space” may lead one to surmise that components inS must also be open in S. If the reader is tempted by this assertion, the following

example may incite this reader to reconsider.

Example 7. Consider the subspace, Q with the subspace topology. The subset 5 is

a connected closed subset of Q. Suppose point, u, is an element in Q which is lessthan 5. We claim that u is not an element of, C5, the connected component of 5. Let

i be an irrational number in (u, 5).

Then [(u− 1, i) ∪ (i, 5)]∩ Q . Then ( then the set u, 1/2 is the union of two opensubsets of Q and so is not connected in Q. So 1/2 is a component in Q. It is however

not open in Q, since it is the limit point of a sequence of rationals in Q\1/2, henceits complement cannot be closed. So in spite of being a connected component, 1/2is not open in Q.

Example 8. Let Si : i ∈ I be a family of topological spaces and S =∏

i∈I Si

be the corresponding product space. Suppose that u = < ui >i∈I ∈ S. For eachi ∈ I , let Ci be the unique connected component in Si which contains the point ui.

Suppose C =∏

i∈I Ci. Show that C is the component in S which contains the point u.

Solution : If ui ∈ Ci for each i ∈ I , then u ∈ C =∏

i∈I Ci. By theorem 20.5,C is a connected subset of S containing u. We claim that C is a component in

S. Suppose not. Suppose D is the component in S which contains u and supposev =< vi >i∈I ∈ D \C. Then there must be some j ∈ I such that vj 6∈ Cj. Since

πj : S → Sj is continuous, it must be that πj[D] is a connected subset of Sj whereCj ∪vj ⊆ πj [D]. Since Cj is the largest connected set in Sj which contains uj, then

πj[D] cannot be connected. We have a contradiction due to our supposition that D\Cis non-empty. So C is the component in S which contains u, as claimed.

The previous example shows that. . . ,

. . . the product of connected components is a connected component.

Example 9. Let Si : i ∈ I be an infinite family of discrete topological spaces, noneof which contain only one point. Then the connected components of each Si must be

singleton sets. Let S =∏

i∈I Si be the corresponding product space. Show that theconnected components of S are also singleton sets. Show that the product space, S,

is not a discrete space.

Solution : Let u =< ui >i∈I ∈ S. Since Si is discrete, Ci = ui is a connectedcomponent, for each i. Then, by the example above (where it is shown that the

product of connected components is a connected component),

C =∏

i∈I Ci =∏

i∈Iui = u


is the connected component in S containing u. So the connected components of S aresingleton sets, as required.

We now claim that, even though S is the disjoint union of components each of which

is a singleton set, it is not a discrete space. Simply see that the basic open neighbour-hood, ∩π←i (ui) : i ∈ F, (where F is finite in I) cannot be contained in u. So the

set u is not an open subset of S. The product space, S, is then not a discrete space.

Example 10. Show that the components of the Cantor set are all singleton sets.

Solution : Recall that, in example 2 of page 140 and theorem 7.19, the Cantor set wasdefined as being the image of the infinite product,

∏

n∈Z+0, 2, of the discrete space,

0, 2, under the homeomorphism function ϕ :∏

n∈Z+0, 2 → [0, 1]. By the aboveexample, the components of

∏

n∈Z+0, 2 are singleton sets. The homeomorphic im-age of a component must be a component. So the components of the Cantor set are

singleton sets.

Example 11. Decomposition of a compact space into components. Let S be a compactHausdorff topological space which is not connected. We will partition the space S byits connected components. That is,

DS = Cx : x ∈ S, Cx a component containing x

denotes the family of all connected components of S where x ∈ Cx. Let θ : S → DS

be defined such that θ(u) = Cu if u ∈ Cu in DS . Then we obtain a decomposition

space (DS , τ) where U is open in DS if and only if θ←[U ] is the union of connectedcomponents of S which is open in S. Show that the only connected components of

DS are points.1

Solution : We are given that S is compact Hausdorff and the function

θ : S → DS

defined as

θ(u) = Cu if u ∈ Cu

Let p, q ∈ S, contained in distinct components of S. Then θ(p) and θ(q) are distinct

points in DS .

We claim that there exists a clopen subset, M , of DS, such that θ(q) ∈M ⊆ S\θ(p).If so, then no two points in DS belong to the same component and so we will be ableto conclude that the singleton set, θ(p) in DS, is a connected component containing

θ(p).

1We will later refer to such spaces as being totally disconnected.


Proof of claim. Each connected component, θ←(Cx), of S has been shown to be closedin S. Since S is compact then θ←(Cx) is compact. Now “compact and Hausdorff” ⇒“normal”, so, for each x 6= p in S, there exists a pair of disjoint open sets, Vx and Px,such that θ←(Cx) ⊆ Vx and θ←(Cp) ⊆ Px. Then the family open sets,

V = Vx : x ∈ S, x 6= p ∪ Px : x ∈ S, x 6= p

forms an open cover of S. Since S is compact, then V has a finite subcover (of S)

F = Vxi : xi ∈ S, xi 6= p, i ∈ F ∪ Pxi : xi ∈ S, xi 6= p, i ∈ F

where F is a finite indexing set and Vxi ∩ Pxi = ∅. Then S is the disjoint union ofthe two open subsets,

V = ∪Vxi : xi ∈ S, xi 6= p, i ∈ FP = ∩Pxi : xi ∈ S, xi 6= p, i ∈ F

where θ←(Cq) ⊆ V and θ←(Cp) ⊆ P . Since P and V are both open and disjoint such

that S = P ∪ V , then V and S\V = P are clopen in S and so are both compact. SoDS is the disjoint union of the compact sets θ[V ] and θ[P ]. Each of these must thenbe clopen in DS . Given that the set, θ[V ], is clopen, then

θ(q) = Cq ⊆ θ[V ] ⊆ S\θ(p)

as claimed.

It follows that no two points in DS belong to the same connected component and sothe only connected components of DS are its points.

20.4 Locally connected: Spaces with an open base of connected sets.

Even when a space is not connected it may have a base for open sets whose elementsare connected subspaces. For example, any discrete space, S, can easily be seen to

have an open base of connected subspaces, B = x : x ∈ S. So does the con-nected space of real numbers, R, with the usual topology since it has an open base,

B = (a, b) : a, b ∈ R, a < b.We formally define this particular notion.

Definition 20.9 If S is a topological space. We say that the space S is locally connectedif it has an open base, B = Bi : i ∈ I, where each element, Bi, is a connected subspace.


What would a non-locally connected space look like? We construct a connected sub-space of R2 which is not locally connected.

Example 12. For each n ∈ N\0, let

gn(x) =

(

sin ( π4n)

cos ( π4n)

)

x

For each n, let Ln denote the line,

Ln = (x, yn) : yn = gn(x)

LetT = ∪Ln : n ∈ N\0 ∪ (1, 0)

be a subspace of R2 with the usual topology. Show that T is a connected subspacewhich is not locally connected.

Solution : For each n ∈ N\0, Ln is a line in R2 which is homeomorphic to R andso is connected. Since each Ln contains the point, (0, 0), then ∪Ln : n ∈ N\0is connected. The connected set ∪Ln : n ∈ N\0 is dense in T and so T is connected.

Consider the open neighbourhood base,

B = B1/k(1, 0) ∩ T : k ≥ 2

of the point (1, 0) ∈ T . For each k, the open “ball”, B = B1/k(1, 0) ∩ T , in T contains

countably many line segments (belonging to the Ln’s). If j 6= m there is a line L∗

whose slope is strictly in between the slopes of Lj and Lm, so the line of L∗ does notappear in T . Then B is disconnected at the line L∗. So T is not locally connected at

the point (1, 0).

20.5 Which spaces have clopen connected components?

Suppose S is a locally connected space and U is an open subspace of S. Then S has

an open base,B = Bi : i ∈ I

of connected subsets. If u ∈ U , let Cu denote a connected component of the subspace,U , which contains u. We have already proven that Cu must be closed in U .

We claim that Cu must be a clopen subset in U . To see this simply note that thereis some connected Bj in B such that u ∈ Bj ⊆ U . Since u ∈ Cu and Bj is con-

nected and intersects Cu then Bj ∪Cu is connected so u ∈ Bj ⊆ Cu. Since Bj and Uare open in S, Bj is open in U and so Cu is open in U and so is clopen in U , as claimed.

This leads to an interesting characterization of the locally connected property on a

space.


Theorem 20.10 Suppose S is a topological space.

a) The space S is a locally connected space if and only if every open subspace of S(including S itself) has clopen components.

b) If the space S is both locally connected and compact then it has at most finitely many

connected components.

Proof : For part a): The direction ( ⇒ ) of part a) is proven in the paragraph above.

We prove the direction ( ⇐ ): Suppose that every open subspace of S has clopencomponents. We are required to show that S has an open base of connected sets.

Let V be an open neighbourhood of a point x in S. Let Cx be a connected componentof V which contains x.

We claim that Cx is a connected subset of S. If not, then there is a clopen subset, U ,of S which intersects only a part of Cx. Then U ∩ V is a clopen subset of V whichintersects only a part of Cx, contradicting the fact that Cx is a connected subset of

V . So Cx is connected in S, as claimed.

By hypothesis, Cx is an open subset of V . So Cx is an open neighbourhood of x inS. So Cx is an open connected neighbourhood of x in S which is contained in V . So

every open neighbourhood V of x contains an open connected set Cx in S such thatx ∈ Cx ⊆ V . This implies that S is a locally connected space.

For part b): Suppose S is compact and locally connected. Then its components are

clopen. Suppose it has infinitely many components then its components will forman open cover of S with no finite subcover, a contradiction. So a locally connected

compact space can have at most finitely many components.

Example 13. Let gn(x) = [sin (π/4n)/(cos (π/4n)]x and Ln = (x, yn) : yn = gn(x)where n ∈ N\0. Consider the space S defined as

S = ∪Ln : n ∈ N\0 \ (0, 0)

Consider any two distinct sets Sj = Lj \(0, 0) and Sk = Lk \(0, 0) of slope sjand sk, respectively, in S. These can easily seen to be separable by two disjoint opensubsets of R2 and so can be seen as four disjoint open subsets of S. It can also be

shown that each Si is a connected component of S and so is closed in S. (Note thatthe closure of S in R2 adds the line 0(x) = 0 to S. So S is the union of infinitely

many clopen components of S.



i∈I Si be thecorresponding product space. Then S is locally connected if and only if every factor, Si, is

locally connected and, at most, finitely many of the factors are not connected.


i∈I Si isa product space.

( ⇒ ) Suppose S =∏

i∈I Si is locally connected.

Claim #1. That all but finitely many of the factors are connected spaces. Let u ∈ S.

Since S is locally connected, then, by theorem 20.10, u ∈ C for some clopen connectedcomponent, C, of S. Let U be a basic open neighbourhood of u in S such that

u ∈ U ⊆ C. That is, for a finite subset, F , of I ,

u ∈ U = ∩π←i [Ui] : i ∈ F ⊆ C

where Ui is an open subset of Si.

Consider i 6∈ F . Then πi[U ] = Si ⊆ πi[C] (since U ⊆ C). So πi[U ] = Si = πi[C].Since C is a connected component of S and πi is continuous, Si is connected for all

i’s except possibly the ones in F . This establishes the claim #1.

Claim #2. That each Si is locally connected. For i ∈ I let vi ∈ Si. Let W be an open

neighbourhood of vi in Si and v ∈ S such that πi(v) = vi. It suffices to show thatthere is a connected open neighbourhood of vi which is entirely contained in W .

Since S is locally connected, and π←i [W ] is an open neighbourhood of v in S, there ex-ists an open connected subspace, V , in S such that v ∈ V ⊆ π←i [W ]. Since πi is both

continuous and open then πi[V ] is a connected open neighbourhood of vi containedin W . So vi has a connected neighbourhood base. Then Si is locally connected. Thisestablishes claim #2.

We are done with this direction.

( ⇐ ) Suppose every factor, Si, is locally connected and at most finitely many of

these factors are not connected. We are required to show that the product space, S,is locally connected.

Suppose Fc is the largest subset of the indexing set, I , such that, if i ∈ Fc, Si is notconnected. Then, by hypothesis, Fc is finite.

Let

x =< xi >i∈I ∈ S =∏

i∈ISi

and let V = ∩π←i [Ui] : i ∈ F be any open base element containing x. Let F1 = F∪Fc

andU1 = ∩π←i [Ui] : i ∈ F1

Clearly, since F ⊆ F1, then U1 ⊆ V .


We claim that the point, x, has a connected neighbourhood, W , which is containedin U1.

Given that every Si is locally connected, for each i ∈ F1, there must be an openconnected set, Ci in Si, such that xi ∈ Ci ⊆ Ui.

Thenx ∈ ∩π←i [Ci] : i ∈ F1 = W ⊆ U1 ⊆ V

exhibits an open neighbourhood, W , of x in S which is contained in V . If j 6∈ F1,

the jth factor of W is the connected space, Sj . If j ∈ F1, the jth factor of W is theconnected, Cj . By theorem 20.5, W a connected subspace in S.

So x has a connected open neighbourhood, W , contained in U1, as claimed. SinceU1 ⊆ V , S is locally connected, as required.

20.6 Pathwise connected spaces

There are different ways of perceiving the connected poperty on a set. The formaldefinition of the connected space given earlier was “ ... a space that is not the unionof two disjoint open sets.” appeals to its proponents because of its simplicity. It is a

clear and unambiguous definition. However, some critics might have argued that thisdefinitions sounds to much like “... a connected space is one that is not disconnected;

and we all know what disconnected means.” The formal definition, along with a fewexamples, and some discussion of the properties derived from it, allowed users to

develop a better understanding of what a connected space is. But, in spite of this,the space, T , described in the example found on page 371, may not appear to be

connected, by some participants of a random survey. We define below, the

“pathwise connected property”

a slightly stronger definition than that of the connected property in the sense that

pathwise connected spaces are always connected, but not conversely. To some, it is amore appealing form of connectedness even though it is sightly more difficult to work

with.

Definition 20.12 Let S be a topological space. A path in S is a continuous function,

f : [0, 1] → S, mapping the closed unit interval into S. If, for a path, f : [0, 1] → S,a = f(0) and b = f(1), we say that. . .

“f is a path from a to b in S”


where a is referred to as being the initial point of the path f and b is called the terminalpoint of the path f . These terms prescribe a direction for the image, f [0, 1], of f .

The space, S, is said to be a

“pathwise connected space”

if, for any pair of points, a and b in S, there is a path, f : [0, 1] → S, joining a = f(0) tob = f(1).1

Since [0, 1] is connected in R and f is continuous on [0, 1] then the image f [0, 1] isa connected subspace of S. The set of all real numbers, R, is of course pathwise

connected since for any a and b in R

f(x) = (b− a)x+ a

is a path connecting a = f(0) to b = f(1).

This particular perception of “connected” is also quite intuitive. It suggests that if

one can draw a curve joining any two points in S, without lifting the pencil off thepage, then the space is connected. The following theorem guarantees that the path-

wise connected spaces are connected as described by the formal definition.

Theorem 20.13 If the topological space, S, is pathwise connected then it is connected.

Proof : Suppose S is pathwise connected. If a ∈ S, then, for any b ∈ S, there is a path,fb : [0, 1] → S, such that a = fb(0) is its initial point and b = fb(1) is its terminal point.

Since fb is continuous, then fb[0, 1] is connected. Then, the family of subsets of S,

C = fb[0, 1] : b ∈ S, b 6= a

is a family of connected sets which entirely covers S each of which contains the pointa as initial point.

Since S is the union of connected subsets of S each containing a common element,then S is connected.

1If the continuous function, f , from a = f(0) to b = f(1) is a homeomorphism then f is called an arc. Ifevery path on S is a homeomorphism then the space is said to be arcwise connected.


We have then confirmed that

“Pathwise connected space” ⇒ “Connected space”

It is worth noting that, if there is a path, f , from the point a to the point b in a space

S, then there is a path, g : [0, 1] → S, defined as g(x) = f(1 − x) which goes in theopposite direction, from b to a.

A connected space which is not pathwise connected.

Example 14. We reconsider the example presented earlier described as follows: For

each n ∈ N\0, let Ln ⊆ R2 denote the set,

(x, yn) : yn = gn(x) =

(

sin ( π4n)

cos ( π4n)

)

x

Let

T = ∪Ln : n ∈ N\0 ∪ (1, 0)be a subspace of R2 with the usual topology. We have shown (in an example on page371) that T is a connected subspace but is not locally connected. Show that T is notpathwise connected.

Solution : Suppose there is a path joining the point a = (1, 0) in T to a point,b = (u, v) ∈ T\(0, 0). That is, suppose there is a continuous function f : [0, 1] → T ,

such that f(0) = (1, 0) and f(1) = b = (u, v). It suffices to show that continuity of fwill hold true only for f(x) = (1, 0) for all x in [0, 1].

Note that the family Ln\(0, 0) : n > m∪(1, 0) is the union of an infinite familyof connected components of T \(0, 0).Continuity of f : [0, 1] → T guarantees that

0 < ε < 1/10 ⇒ f [0, δ) ⊆ Bε(1, 0) for some δ > 0

Suppose q ∈ Lk ∩ f [0, δ) for some k ∈ N\0. Then f [0, δ) intersects the two compo-

nents, Lk\(0, 0) and (1, 0), of T \(0, 0) and so is not connected, contradictingcontinuity of f . Then f [0, δ) = (1, 0).So continuity of f holds true only for b = (1, 0). So T is not pathwise connected.

Not surprisingly, pathwise connectedness is carried over by continuous functions. Weverify this now.

Theorem 20.14 If g : S → T is a continuous function mapping a pathwise connectedspace S to a space T , then g[S] is a pathwise connected subspace of T .


Proof : Suppose S is a pathwise connected space. Then if u and v are distinct points in Sthere is a continuous function f : [0, 1] → S such that f(0) = u and f(1) = v.

Suppose now that g : S → T is a continuous function mapping S onto g[S] ⊆ T . Weclaim that g[S] is pathwise connected. To see this let a and b be distinct points in

g[S]. Then we can choose distinct points, x ∈ g←(a) and y ∈ g←(b). Since S is path-wise connected there is a continuous function f : [0, 1] → S such that f(0) = x andf(1) = y. Then the continuous function, gf : [0, 1] → g[S], maps 0 to a and 1 to b.

So gf is a path from a to b. We conclude that g[S] is pathwise connected, as required.

We now consider invariance of the pathwise connected property with respect to prod-ucts.

Theorem 20.15 Let Si : i ∈ I be family of topological spaces. Then the product space,S =

∏

i∈I Si, is pathwise connected if and only if every factor, Si, is pathwise connected.

Proof : We are given that Si : i ∈ I is a family of topological spaces and that S =∏

i∈I Si

is the corresponding product space.

For ( ⇒ ), if S is pathwise connected, since each projection map, πi, is continuous,then each Si is pathwise connected.

For ( ⇐ ), suppose Si is pathwise connected for each i ∈ I .

Let <ai>i∈I and <bi>i∈I be distinct points in S. Since each Si is pathwise connected,

then, for each i ∈ I , there is a continuous function, fi : [0, 1] → Si, such that fi(0) = ai

and fi(1) = bi. Let f : [0, 1] → S be defined as

f(u) =<fi(u)>i∈I

Then fi(u) = (πif)(u). By lemma 7.11, the function, f : [0, 1] → S, is continuous on

[ 0, 1 ] if and only if each fi is continuous on [ 0, 1 ]. Since each fi maps 0 to ai and 1to bi then

f(0) = <fi(0)>ı∈I = <ai>i∈I

f(1) = <fi(1)>i∈I = <bi>i∈I

So S is pathwise connected, as required.


Example 15. It follows from the previous theorem that the space, S = RR, equippedwith the product topology is pathwise connected since each factor, R, is pathwise

connected.

Example 16. Let u be a point in a topological space, S. Show that S is pathwise

connected if and only if there is a path joining each point, x ∈ S, to u.

Solution : For the direction ( ⇒ ) the statement is obviously true.

For the direction ( ⇐ ), we are given that every point in S can be joined by a path

to u. Suppose a and b are distinct points in S. We are required to show that there isa path linking a to b.

By hypothesis, there are two paths, fa : I → S and fb : I → S, such that fa(0) = a

and fa(1) = u and fb(1) = b and fb(0) = u. Consider the function g(x) = fa(2x) onthe interval [0, 1/2] and g(x) = fb(2x−1) on the interval [1/2, 1]. Then g continuously

joins a to u on [0, 1/2] and joins u to b on [1/2, 1], where g(0) = a and g(1/2) = uand g(1) = b. So g : [0, 1] → S forms a path from a to b. Then S is pathwise connected.

Theorem 20.16 Let Si : i ∈ I be a family of pathwise connected topological spaces

which have the point u in common. Then space S = ∪Si : i ∈ I, is pathwise connected.

Proof : The statement follows immediately from the argument presented in the previousexample.

Definition 20.17 If S is a topological space and u ∈ S, then we define the

“pathwise component of u”

as being the union of all pathwise connected spaces that contain the point u. Hence, if Cis the pathwise component containing u, it is the largest subspace containing u which is

pathwise connected.

Example 17. Consider the space, T = ∪Ln : n ∈ N\0 ∪ (1, 0) as presented

in the example on page 376 where we showed that T is connected but not pathwiseconnected. Show that a pathwise component in T need not be closed.

Solution : Consider the subspace, S = T \(1, 0). The space, S, is an infinite union

of straight lines in R2 each of which is pathwise connected and contains the point(0, 0). From this we deduce that S is pathwise connected. It can also be seen to be a


pathwise component of T . But since S is missing the limit point (1, 0) in T , it is notclosed in T . So a pathwise components of a space need not be closed.

Under particular conditions it does occur that a pathwise component is a clopen sub-set. This condition is described in the following theorem.

Theorem 20.18 Let S be a topological space. The pathwise components of S are clopenif and only if every point in S has a pathwise connected neighbourhood.


( ⇒ ) Suppose each pathwise component is clopen. Let x ∈ S. We are required toshow that x has a pathwise connected neighbourhood. Let C be a pathwise compo-

nent containing x. By hypothesis, C is open in S, so C is a pathwise connected openneighbourhood of x. We are done with this direction.

( ⇐ ) Let x ∈ S and C be the (unique) pathwise component containing x. By hy-

pothesis, the point, x, has a pathwise connected neighbourhood, say U .

We are required to show that C is clopen.

Then since C is the largest pathwise connected neighbourhood which contains x, then

x ∈ intSU ⊆ C. So C is open. We claim it is also closed: Since the components parti-tion the space S, then C is the complement of the union of all other open components

of S and so is also closed. So all pathwise components are clopen.

The equivalent conditions, “pathwise components are clopen” and “every point in

S has a pathwise connected neighbourhood” described in the above theorem, whencombined to the connected property characterizes pathwise connectedness, as shown

in the next theorem.

Theorem 20.19 Let S be a topological space. The space, S, is pathwise connected if andonly if S is both connected and each of its points, has a pathwise connected neighbourhood.


( ⇒ ) We are given that S is a pathwise connected space. The fact that S is a con-

nected space follows from theorem 20.13. That every point has a pathwise connectedneighbourhood follows from the the fact that the pathwise connected space is a neigh-

bourhood of every point in S.


( ⇐ ) Suppose S is connected and every one of its points has a pathwise connectedneighbourhood. We are required to show that S is pathwise connected. Let x ∈ S.

By theorem 20.18, the (unique) pathwise component, C, of x is clopen. Since S isconnected, the clopen set, C, cannot be a proper subset. So the pathwise component,

C, must be S itself. So S is pathwise connected.

In Rn, any connected open subspace is a pathwise connected subspace as the followingtheorem shows.

Theorem 20.20 Let Rn be equipped with the usual topology. An open subspace, U , ofRn is connected if and only if U is a pathwise connected subspace of Rn.

Proof : The space Rn be equipped with the usual topology and U is an open subspace of Rn.

( ⇒ ) Suppose U is a connected subspace of Rn. We are required to show that U isa pathwise connected subspace. By theorem 20.19 it suffices to show that each point

in U has a pathwise connected neighbourhood

Well, let x ∈ U . Since U is open in Rn there is an ε > 0 such that x ∈ Bε(x) ⊆ U . Leta =<ai>i=1..n and b =<bi>i=1..n be distinct points inBε(x). If f(x) = (b−a)(x)+athen f [ 0, 1 ] is a line segment entirely contained in Bε(x). So Bε(x) is pathwise con-nected. From this we can conclude that every point in U has a pathwise connected

neighbourhood, so by the above theorem, U is pathwise connected, as required.

( ⇐ ) Suppose U is pathwise connected. By theorem 20.13, U is a connected subspace.Done.

20.7 Totally disconnected spaces

For some spaces, the only connected subspaces are singleton sets. We briefly discusssome basic properties of these types of spaces.

Definition 20.21 A topological space, S, is said to be totally disconnected if and only ifthe only connected components of S are its points.


Discrete spaces, the subspace Q, the Cantor set (viewed as a subspace of [0, 1]) andR\Q all previously discussed sets in this chapter are standard examples of totally

disconnected spaces.

We verify that the totally disconnected property is preserved over arbitrary products.

Theorem 20.22 The product space of totally disconnected spaces itself totally discon-

nected.

Proof : Let Si : i ∈ I be a family of totally disconnected spaces and S =∏

i∈I Si bethe corresponding Cartesian product space. We are required to show that S is totally

disconnected.

Let C be a connected component in S. It suffices to show that C is a singleton set.Suppose a =<ai>i∈I and b =<bi>i∈I are two points in C. If a 6= b then aj 6= bjfor some j. Then aj, bj ⊆ πj [C], a continuous image of the connected set, C, in S.Since the only connected sets in Sj are singleton sets, then aj = bj. So a = b in C.

So C is a singleton set. So S is totally disconnected, as required.

Example 18. Show that if S is totally disconnected and f : S → T is continuous thenf [S] need not be totally disconnected.

Solution : In theorem 7.20, we showed that there is a function ψ : C → [0, 1] whichmaps the totally disconnected Cantor set C onto the connected set, [0, 1]. So the

continuous image of a totally disconnected set need not by totally disconnected.

Total disconnectedness versus the zero-dimensional property.

Recall from definition 5.16 that zero-dimensional spaces are those spaces that have an

open base of clopen sets. Equivalently, each point has a neighbourhood base of clopensets. We know that a non-empty clopen set allows us to express a space as a union oftwo open sets. So we surmise that zero-dimensional spaces and totally disconnected

spaces are similar in many ways. For certain types of topological spaces they mayeven be equivalent, as we shall soon see.

Example 19. Show that a T1 zero-dimensional space, S, is totally disconnected.

Solution : Since S is T1, any point in S is closed. Given distinct points p and y theset S \y is an open set containing p. Since p has a neighbourhood base of clopen


sets then there exists a clopen set U such that p ∈ U ⊂ S\y. So p and y cannotbelong to the same component. So the only connected component which contains p

is p. This means that S is totally disconnected.

Example 20. Suppose S is a compact T2 space satisfying the property: For distinctpairs of points, p and x, there is a clopen set Ux such that x ∈ Ux ⊂ S\p. Show

that S must then be zero-dimensional.

Solution : Given: S is a compact T2 space. Suppose V is a proper open subset contain-ing the point p. It suffices to show that there is a clopen set W such that p ∈W ⊆ V .

By hypothesis, for each x ∈ S\V , there is a clopen set Ux such that x ∈ Ux ⊂ S\p.Now S\V is closed and so is a compact subset of S. Then

U = Ux : x ∈ S\V

covers S\V with clopen sets each of which misses p. Let

UF = Uxi : i ∈ F

be a finite subfamily of U which covers S\V . If M = ∪Uxi : i ∈ F and W = S\M ,W is a clopen subset of S such that p ∈W ⊆ V . So S is zero-dimensional, as required.

Suppose F = Ki : i ∈ I where each Ki is compact in a Hausdorff space S and

A and B are disjoint compact subsets of S such that A ∪ B = K = ∩Ki : i ∈ I.Then there exists disjoint open subsets U and V which contain A and B, respec-

tively. Then K = A ∪ B ⊆ U ∪ V . Then there exists a finite subset F ⊆ I such thatKF = ∩Ki : i ∈ F ⊆ U ∪ V . Let UF = KF ∩ U and VF = KF ∩ V . Then KF is the

disjoint union of UF and VF , where A ⊆ UF and B ⊆ VF .

Lemma 20.23 Let S be a compact Hausdorff space, x ∈ S and Cx be a connected com-ponent containing x. Then Cx = ∩D : D is clopen in S, x ∈.

Proof : Given: That S is compact Hausdorff and Cx is a connected component which

contains x. If U is a clopen neighbourhood of x then Cx ⊆ U (since Cx is connected.)Let D = D : D is clopen in S, x ∈ D. If

M = ∩D : D is clopen in S, x ∈ D

then Cx ⊆M . Then M is closed in S. We claim that M is connected in S.

To see this, suppose there exists non-empty clopen subsets A and B of M , such thatM = A ∪ B where x ∈ A. Since A is clopen in M it is compact and hence is closed


in S. Then there exists disjoint open subsets U and V of S which contain A and B,respectively. Then

M = A ∪ B ⊆ U ∪ Vwhere x ∈ A ⊆ U . Then there exists a finite subset, DF = Di : Di is clopen inS, x ∈ Di, of D such that

MF = ∩Di : Di is clopen in S, x ∈ Di ⊆ U ∪ V

otherwise M would intersect S\(U∪V ) (since S is compact). Then MF is both closed

(compact) and open in S. Then MF ∩ U is open in S.Also MF \V = MF ∩ S\V = MF ∩U is closed in S. Then MF ∩U ∈ D . This implies

that M ⊆MF ∩ U . This can only be if B = ∅. So M is connected, as claimed.

Since Cx is the largest connected set containing x then Cx = M , as required.

Theorem 20.24 A locally compact Hausdorff space is totally disconnected if and only ifit is zero-dimensional.

Proof : ( ⇐ ) We have shown in the example above that T1 zero-dimensional spaces aretotally disconnected. Since Hausdorff spaces are T1 we are done.

( ⇒ ) Suppose S is a locally compact Hausdorff totally disconnected space, p ∈ S and

V is an open open neighbourhood of p in S. We are required to show that S is zero-dimensional. To do this it suffices to show the existence of a clopen neighbourhoodH , in S such that p ∈ H ⊆ V .

Since S is locally compact there is an open subset M whose closure, clSM , is compact

such thatp ∈M ⊆ clSM ⊆ V

Since S is totally disconnected then p is a connected component in S.

Let

D = U : U is clopen in S, p ∈ UD1 = U ∩M : U is clopen in S, p ∈ UD2 = V : V is clopen in M, p ∈ V

Clearly, D1 ⊆ D2. On the other hand, if V ∈ D2 then V is compact and so is closedin S. Also V = W ∩M where W is open in S. So V is clopen in S. So V ∈ D1. So

D2 = D1.

If Q = ∩D , Q1 = ∩D1 and Q2 = ∩D2, then

Q = Q1 = Q2 ⊆M ⊆ clSM


Then Q is compact. By the previous theorem Q2 is connected in M and so is con-nected in S.

Now, the connected component of p in S is p (since S is totally disconnected). Then

Q = p. Since clSM is compact, there exists a finite subset of

DF = Di : Di is clopen in S, p ∈ Di

such that H = ∩DF ⊆ M . Note that H is the required clopen in set in S entirelycontained in M . From this we can conclude that S is zero-dimensional.

Example 21. Show that a subspace of a locally compact totally disconnected Haus-dorff space is also totally disconnected

Solution : Suppose T is a subspace of a totally disconnected locally compact Haus-

dorff space S. Then S is zero-dimensional (by the above theorem 20.24). Let C be aconnected component of T . To show that T is totally disconnected it suffices to show

that C is a singleton set.

If a and b are points in C then, since S is zero-dimensional Hausdorff, there exists a

clopen neighbourhood U (in S) of a which does not contain b. So the clopen subsetU ∩ T (of T ) separates a and b and so, if a 6= b, C cannot be connected. So a = b.

Then connected components of T are singleton sets. So T is totally disconnected.

Example 22. Let S be a compact Hausdorff topological space which is not connected.We showed in an example on page 369, that if we collapse the connected components

of S to points by the quotient map θ : S → θ[S] we obtain a decomposition space θ[S]whose points are connected components. This means that the compact space θ[S] is

totally disconnected and hence zero-dimensional.

Concepts review:

1. What does it mean to say that a topological space is connected?

2. Is the connected property invariant with respect to the continuous functions?

3. Under what conditions, if any, are unions of connected sets connected?

4. If U is a connected dense subset of V what can say about V ?

5. Under what conditions, if any, is the product space, S =∏

i∈I Si, of connected, Si’s,connected?


6. Define a connected component of a topological space.

7. Briefly argue that a connected component of a topological space is closed.

8. Provide a simple example which illustrates that a connected component need not be

open.

9. Describe the connected components of a discrete space.

10. Describe the connected components of an infinite product of discrete spaces all of

which have more than one point.

11. Is an infinite product of discrete spaces (all of which have more than one point)discrete?

12. Define a locally connected space.

13. What kind of locally connected spaces are guaranteed to have clopen connected com-ponents.

14. Describe the conditions under which the locally connected property carries over on a

product space, in both directions.

15. Suppose a locally connected space is shown to be compact. What can we say aboutits connected components?

16. If a person speaks of a path joining a point, a, to a point, b, in a space S, what is thisperson talking about?

17. If you are familiar with the notion of a “curve” in a space S, is there a subtle difference

between a curve and a path?

18. What does it mean to say that a space S is pathwise connected?

19. When is a connected space a pathwise connected space?

20. Are continuous images of pathwise connected spaces necessarily pathwise connected?

21. Under what conditions is the pathwise connected property carried over products inboth directions?

22. Under what conditions, if any, are unions of pathwise connected spaces pathwise

connected?

23. Define the pathwise component of a point in a space S.

24. Are pathwise components necessarily closed?

25. Describe a particular condition under which a pathwise component is clopen.


26. Describe a condition under which connected spaces are pathwise connected.

27. Define the property called totally disconnected.

28. Give a characterization of the totally disconnected property and describe the condi-tions under which it holds true.

29. What can we say about the product of totally disconnected spaces?

30. What can we say about the subspace of a totally disconnected space?

EXERCISES

1. Suppose U is subset of R which is not connected and V ⊆ U . Is it possible for V tobe connected?

2. Let (S, τS) and (T, τT) be two topological spaces where T is connected. If T has a

strictly stronger topology then S, determine whether S is necessarily connected.

3. Consider the subset

T = (x sin (1/x)) : x ∈ [0, 2π]∪ (0, 0)

of R2. Is T a connected set?

4. Is the set T described in the previous question locally connected?

5. Is the set T described in the previous question pathwise connected?

6. Let Si : i ∈ I be an infinite family of connected sets. Suppose that, if i 6= k, thenSi ∩ Sk 6= ∅. Determine whether the space S = ∪Si : i ∈ I is connected or not.

7. Let S be an infinite set equipped with the cofinite topology (i.e., ∅ union the familyof all sets with a finite complement). Determine whether S is a connected set or not.

8. Suppose the space (S, τ) is the topological space where

τ = ∅ ∪ U : S\U is finite

Let the function, f : [0, 1] → S, be continuous. Show that f [ [0, 1] ] in S, is a singleton

set.

9. Let S = Rn be the space equipped with the usual topology. Suppose U is a non-empty

open subspace of S. How many connected components can S have? If the possiblecardinality of all components is infinite, is it countable or uncountable?


10. Are open subspaces of locally connected spaces necessarily locally connected?

11. Suppose S is a T1-space which has, at each point, a neighbourhood base of clopensets. Show that the connected components of S are all singleton sets.

Part VII

Topics

389

Part VII: Topics 391

21 / Compactifications of completely regular spaces

Summary. In this section, we discuss those spaces, S, which can be denselyembedded in a compact Hausdorff space. Only completely regular spaces can pos-

sess this property. The process by which we determine such a compact space, αS,for S, is called compactifying S. The space, αS, is called the compactification

of S. The family of all compactifications of a completely regular space can bepartially ordered. The maximal compactification of S with respect to the chosen

partial ordering is called the Stone-Cech compactification. We discuss methodsfor its construction. We will show that only locally compact spaces have a mini-mal compactification with respect to the chosen partial ordering. It is called the

one-point compactification.

21.1 Compactifying a space

In this section we will briefly talk about methods for “compactifying a space (S, τS)”.

This essentially means adding a set, F , of points to S, to obtain a larger set, T = S∪F ,and topologizing T so that (T, τT) is a compact Hausdorff space in which a homeo-

morphic copy of S appears as a dense subspace of T .

With certain bounded subspaces of Rn, this can, sometimes, be quite easy to do. Forexample, if S = [−1, 3)∪(3, 7) is equipped with the subspace topology, then simply by

adding the points, 3, 7, to S we obtain a set T = S ∪ 3, 7 = [−1, 7] which, whenequipped with the subspace topology, is a compact Hausdorff space which densely

contains a homeomorphic copy of S. In such a case, we will say that T is a compacti-fication of S. A compactification of a bounded subset of Rn can always be obtainedby taking its closure. This does not mean, however, that there are not others. If we

are given a space such as N or Q, it is not at all obvious how one would go aboutcompactifying such spaces. We will show techniques which allow us to achieve this

objective.

For what follows, recall that the evaluation map e : S →∏

i∈J [ai, bi] induced by C∗(S)(the set of all continuous bounded real-valued functions on S) is defined as

e(x) =<fi(x)>i∈J ∈ ∏

i∈J [ai, bi]

where fi ∈ C∗(S) and fi[S] ⊆ [ai, bi].

Definition 21.1 Let (S, τS) be a topological space and (T, τT) be a compact Hausdorff

space. We will say that T is a compactification of S if S is densely embedded in T .1

1When we say “compactification of S” we always mean a Hausdorff compactification.

392 Section 21: Compactifications of completely regular spaces

If S is a compact Hausdorff space, then S can be viewed as being a compactification ofitself. Recall that a compact Hausdorff space is normal, and so is completely regular.

Since subspaces of completely regular spaces are completely regular then

. . . only a completely regular space can have a compactification.

In theorem 14.7, we showed how any completely regular space can be compactified.In the proof of that theorem, we witness how an evaluation map, e : S → T , induced

by C∗(S) embeds S into a cube,

T =∏

i∈J [ai, bi]

There may be different ways of describing this type of compactification of S. Since

each interval [ai, bi] is homeomorphic to [0, 1], then there is a homeomorphism, h :T → ∏

i∈J [0, 1], which maps T onto P =∏

i∈J [0, 1]. By Tychonoff’s theorem, P is

guaranteed to be compact. So the function, q : S → P , defined as, q = he, embeds Sinto

∏

i∈J [0, 1]. Hence clP q[S] is a compact subspace of the product space, P , which

densely contains the homeomorphic image, q[S], of S.1 So, even common topologicalspaces such at R, Q, and N have at least the compactification obtained by the method

just described.

21.2 The Stone-Cech compactification.

We have described only one of the various ways to obtain a homeomorphic copy of thecompactification, clT e[S], of S. This particular compactification has a special name.

Definition 21.2 Let (S, τS) be a completely regular topological space. Let

e : S →∏

i∈I [ai, bi]

be the evaluation map induced by C∗(S) which embeds S in the product space, T =∏

i∈I[ai, bi].

The subspace,clT e[S]

is called the Stone-Cech compactification of S. The Stone-Cech compactification of S isuniquely (and universally) denoted by, βS.

1This is just one small example which shows why Tychonoff theorem deserves to be titled and why it issuch an important theorem in topology.


So we see that a non-compact completely regular space, S, always has at least one

compactification, namely, it’s Stone-Cech compactification, βS. We will soon see thatthere can be more than one compactification, for the same space.

Equivalent compactifications

Given a completely regular space S, there is nothing stated up to now which wouldlead us to conclude that S has only one compactification. In fact most spaces we will

consider will have many compactifications. Suppose we are given two compactifica-tions for a space, S, say αS and γS. If there is a homeomorphism

h : αS → γS

mapping αS onto γS such that h(x) = x for all x ∈ S, then αS and γS will be consid-

ered to be equivalent compactifications of S. Two compactifications of the same spacewhich have been determined to be “equivalent” in this way will be assumed to be the

same compactification, topologically speaking. This equivalence is often expressed bythe symbol, αS ≡ γS. For convenience, it is sometimes expressed by, αS = γS, eventhough, strictly speaking, αS and γS may not necessarily be equal sets.

The outgrowth of a topological space.

Given a topological space, S, and a compactification, αS, the set

αS\S

is referred to as being the outgrowth of S with respect to this particular compactifi-

cation. Equivalent compactifications will be considered to have the same outgrowth.We will sometimes be interested in determining whether an outgrowth satisfies cer-

tain topological properties. If we are given a compactification αS of S, αS\S is alsoreferred to as the remainder of αS.

Example 1 : Compactifications of R. Construct two compactifications of the space, R,

one whose outgrowth contains a single point and another one whose outgrowth con-tains two points. We refer to these as a “one-point compactification” and “two-point

compactification” of R.

Solution : A two-point compactification of R. We know that arctan : R → (−π/2, π/2)maps R homeomorphically onto (−π/2, π/2). Hence R is embedded in the compact

space

γR = clRarctan[R] = clR(−π/2, π/2) = [−π/2, π/2]

with outgrowth −π/2, π/2


A one-point compactification of R. We know that there is a homeomorphism h : R →(0, 2π) which maps R onto (0, 2π). The function,

h(x) = 2[arctan(x) + π/2]

is a simple example. Define the homeomorphic function f : (0, 2π) → R2 as

f(x) = (sin(x), cos(x))

Then fh : R → R2 densely embeds R into the (closed and bounded) compact set

αR = K = (sin(x), cos(x)) : x ∈ (0, 2π) ∪ (1, 0)

with one-point outgrowth (1, 0).

Example 2. Let R+ = x ∈ R : x ≥ 0. Is there a two-point compactification of R+?

Solution : No. Suppose αR+ = R ∪ a, b. Then, since αR+ is Hausdorff, there mustbe disjoint open neighbourhoods Ba and Bb in αR+ of a and b, respectively. Then

αR+\R+ ⊆ Ba ∪ Bb. Since Ba ∪ Bb is open in αR+ containing αR+\R+ then

K = αR\(Ba ∪Bb)

is a compact subset of R+. Then there exists k ∈ R+ such that K ⊆ [0, k]. Then

[k,∞) = [(Ba ∪ Bb) ∩ R+]\ [0, k). We see that [k,∞) is a connected subspace of R+

while [(Ba ∪ Bb) ∩ R+]\[0, k) is not. So we have a contradiction. So R+ cannot havea two-point compactification.

Example 3. There is no three-point compactification of R. Prove this statement.

Solution : Suppose thatαR = R ∪ a, b, c

represents a compactification of R with outgrowth αR\R = a, b, c. Let R+ = x ∈R : x ≥ 0 and R− = x ∈ R : x ≤ 0. Then

αR = clαRR

= clαR(R+ ∪ R−)

= clαRR+ ∪ clαRR−

= R+ ∪ R− ∪ a, b, c

By the previous example neither R+ nor R− can have a two-point (nor a three-point)

outgrowth. So there is no three-point compactification of R.

We can easily generalize the statement in the previous example to “If the compact-ification αR has a finite outgrowth then αR\R must be either a singleton set or a

doubleton set.”


21.3 A partial ordering of Hausdorff compactifications of a space.

Let’s gather together all compactifications of a completely regular space, S, as follows,

C = αiS : i ∈ I

We will partially order the family, C , by defining “” in the following way:

If αiS and αjS belong to C , we will write

αiS αjS

if and only if there is a continuous function f : αjS → αiS, mapping αjS\Sonto αiS\S which fixes the points of S.

Notation. Suppose two compactifications αS and γS are such that αS γS, then bydefinition, there is a continuous function f : γS → αS, mapping γS \S onto αS \Swhich fixes the points of S. We will represent this continuous function f as,

πγ→α : γS → αS

The function πγ→α explicitly expresses which compactification is larger than (or equalto) the other. Note that a pair of compactifications need not necessarily be compa-rable in the sense that one need not necessarily by “less than” the other.

21.4 On C∗-embedded subsets of a topological space.

Given a subset T of a topological space, S, and a continuous bounded real-valued

function f : T → R, it is not guaranteed that there is a bounded continuous functiong : S → R such that g|T = f on T . If there is, then we will say that

“g is a continuous extension of f from T to S”

This motivates the following definition.

Definition 21.3 Let (S, τ) be a topological space and U be a proper non-empty subset ofS. We say that U is C∗-embedded in S if every real-valued bounded function, f ∈ C∗(U),extends to a function, g ∈ C∗(S), in the sense that g|U = f .1 In this case we will say that. . .

g : S → R is a continuous extension of f : U → R from U to S

1There is an analogous definition for “C-embedded” studied later: We say that U is C-embedded in S ifevery real-valued function, f ∈ C(U), continuously extends to a function, f∗ ∈ C(S)


The notion of “C∗-embedding” is closely related to completely regular spaces andtheir Stone-Cech compactification. For this reason, we will discuss this property in

depth now (even though C∗-embeddings may be discussed in other contexts). Thefollowing theorem shows that, for any completely regular space S, S is C∗-embedded

in its Stone-Cech compactification, βS. That is,

...every function, f ∈ C∗(S), extends continuously to a function, fβ ∈ C(βS)

Theorem 21.4 Let S be a completely regular space. Then S is C∗-embedded in βS.

Proof : If f ∈ C∗(S), let If be the range of f . Let

T =∏

f∈C∗(S)clRIf

Then the evaluation map, e : S → T embeds S in T . Recall that, by definition,

βS = clT e[S] ⊆∏f∈C∗(S)clRIf

Suppose g ∈ C∗(S).

We are required to show that g : S → R extends continuously to some function,gβ : βS → R.

If πg is the gth-projection map then

πg :∏

f∈C∗(S)clRIf → clRIg

where βS = clT e[S] and so,πg|βS : βS → clRIg

maps βS into clRIg. Let gβ = πg|βS . Then gβ[βS] = gβ[clT e[S]] = clRg[S] ⊆ clRIg.

It follows that gβ : βS → clRIg and, since g[S] ⊆ Ig, for x ∈ S, gβ|S(x) = g(x).

So gβ is a continuous extension of g from S to βS.

Note that, in the case where S is a compact space, e[S] is a compact space densely

embedded in βS and so βS\S = ∅. Then S and βS are homeomorphic.

The above theorem guarantees that every real-valued continuous bounded function,

f , on a completely regular space, S, extends to a continuous function fβ : βS → R.The function fβ is the extension of f from S to βS. Recall (from theorem 9.8) that

continuity guarantees that two continuous functions which agree on a dense subset Dof a Hausdorff space, S, must agree on all of S. So there can only be one extension,


fβ, of f from S to βS.

We will soon show (in theorem 21.9) that, if αS is a compactification of S and S isC∗-embedded in αS then αS must be the compactification, βS. That is,

...βS is the only compactification in which S is C∗-embedded

A generalization of the extension, f → fβ .

The above theorem can be generalized a step further. Suppose C(S,K) denotes all

continuous functions mapping S into a compact set K. We show that every functionf in C(S,K) extends to a function fβ(K) ∈ C(βS,K). Note that neither f or fβ(K)

need be real-valued. The space, K, represents any compact set which contains theimage of S under f .

Theorem 21.5 Let S be a completely regular (non-compact) space and g : S → K bea continuous function mapping S into a compact Hausdorff space, K. Then g extends

uniquely to a continuous function, gβ(K) : βS → K.

Proof : We are given that g : S → K continuously maps the completely regular space,S, into the compact Hausdorff space K. We are required to show that g extends to

gβ(K) : βS → K.

Since K is compact Hausdorff it is completely regular; hence there exists a function(the evaluation map) which embeds the compact set K in V =

∏

i∈J [0, 1]. Since V

contains a topological copy of K let us view K as a subset of V .

Since g : S → K then, for every x ∈ S, g(x) =<gi(x)>i∈J ∈ K ⊆ [0, 1]J.

Since S is C∗-embedded in βS then, for each i ∈ J, gi : S → [0, 1] extends to

gβi : βS → [0, 1].

We define the function gβ(K) : βS → V as

gβ(K)(x) =<gi(x)>i∈J ∈ V =∏

i∈J [0, 1]

Since gβi is continuous on βS, for each i, then gβ(K) : βS → ∏

i∈J [0, 1] is continuouson βS.

See that gβ(K)[βS] = gβ(K)[clβS(S)] = clV (g[S]) ⊆ clV (K) = K.

Since gβ(K)|S = g, then g : S → K continuously extends to gβ(K) : βS → K on βS.

As required.


Given a completely regular topological space S, and T =∏

i∈I [ai, bi] we now see thatthe evaluation function eC∗(S) : S → ∏

i∈I[ai, bi] (which homeomorphically embeds a

copy of S into T ) then continuously extends to βS as follows:

eβC∗(S)

[βS] = eβC∗(S)

[clβSS]

= clT eC∗(S)[S]

whereeβC∗(S)

(x) =<fβ(x)>f∈C∗(S)

So we can represent the Stone-Cech compactification, βS, of S as being equivalent to

eβC∗(S)[βS]

The maximal compactification, βS.

Suppose αS is any compactification of S possibly distinct from βS. We will now

show that, in the partially ordered family, C , of all compactifications of S, αS βS.Showing this requires that we produce a continuous function πβ→α : βS → αS such

that πβ→α(x) = x, for all x ∈ S. If we can prove this, then we will have shown thatβS is the unique maximal compactification of a completely regular space, S.

Theorem 21.6 If αS is a compactification of S, then αS βS.

Proof : By theorem 21.5, the identity map i : S → αS, extends to a continuous function,

i∗ : βS → αS

Then S ⊆ i∗[βS] ⊆ αS, where i∗[βS] is compact, hence closed in αS. Since S is densein αS, then the open set αS\i∗[βS] must be empty. So the continuous function,

i∗[βS] = i∗[clβSS] = clαSi∗[S] = clαSS = αS

maps βS onto αS. So αS βS, as required.

Then for any compactification αS, there is the continuous function

πβ→α : βS → αS

which maps βS onto αS where πβ→α fixes the points of S.


More on C∗-embedded subsets of a space S.

We present a miscellany of results which will help us more easily recognize a C∗-

embedded subset, T of a space S. We will return to our discussion of compactificationimmediately following this.

The simplest example of a C∗-embedded subset of R is a compact subset of R.

Theorem 21.7 If K is a compact subset of R then K is C∗-embedded in R.

Proof : Let K be a compact subset of R and f : K → R be a continuous function onK. Since every continuous real-valued function is bounded on a compact subset then

f ∈ C∗(K) = C(K). Suppose

u = sup K and v = inf K

Since K is closed and bounded in R then u and v belong to K. Suppose g : R → R is

a function such that

g = f on K

g(x) = f(u), if x ≥ u

g(x) = f(v), if x ≤ v

It is easily verified that g is a continuous extension of f from K to R. Then K isC∗-embedded in R.

Example 4. The set N is C*-embedded in R. One way of visualizing this is to plot

the points of (n, f(n)) : n ∈ N of a function f ∈ C∗(N) in the Cartesian plane R2

and join every pair of successive points (n, f(n)) and (n+ 1, f(n+ 1) ) by a straightline. This results in a continuous curve representing a continuous function g on R

which extends f .

Urysohn’s extension theorem.

The following theorem often referred to as Urysohn’s extension theorem provides animportant and useful tool for recognizing C∗-embedded sets.1

1The Urysohn’s extension theorem should not be confused with the Urysohn’s lemma. Urysohn’s lemmastates that “The topological space (S, τS) is normal if and only if given a pair of disjoint non-empty closed sets,F and W , in S there exists a continuous function f : S → [0, 1] such that, F ⊆ f←[0] and W ⊆ f←[1]”


Theorem 21.8 Urysohn’s extension theorem: Let T be a subset of the completely regularspace S. Then T is C∗-embedded in S if and only if pairs of sets which can be completely

separated by some function in C∗(T ) can also be separated by some function in C∗(S).

Proof : ( ⇒ ) Suppose T is C∗-embedded in S and U and V are completely separatedsubsets of T . Then there exists f ∈ C∗(T ) such that U ⊆ f←(0) and V ⊆ f←(1).

Then by hypothesis f extends to f∗ ∈ C∗(S). Then U ⊆ f←∗(0) and V ⊆ f←∗(1).So U and V are completely separated in S.

( ⇐ ) Suppose that pairs of sets which can be completely separated by some functionin C∗(T ) can also be separated by some function in C∗(S).

Let f1 be a function in C∗(T ). We are required to show that there exists a function

g ∈ C∗(S) such that g|T = f1.

Since f1 is bounded on T then there exists, k ∈ R, such that |f1(x)| ≤ k for all x ∈ T .

Then f1 ≤ k = 3r1 = 3 ·[

k2 ·(

23

)1]

= k (Where r1 = k

2·(

2

3

)1

)

We now inductively define a sequence of functions fn ⊆ C∗(T ). For n = 1, 2, 3, . . .,

there exists fn ∈ C∗(T ) such that −3rn ≤ fn(x) ≤ 3rn where,

3rn = 3 ·[

k

2·(

2

3

)n]

= k ·(

2

3

)n−1

(Where rn = k

2·(

2

3

)n)

For this n, let Un = f←n[ [−3rn,−rn] ] and Vn = f←n[ [rn, 3rn] ].

We see that Un and Vn are completely separated in T . 1

By hypothesis, Un and Vn are completely separated in S. This means there exists

gn ∈ C∗(S) such that, gn[S] ⊆ [−rn, rn], gn[Un] = −rn and gn[Vn] = rn(where rn =

[

k2 ·(

23

)n]). So the sequence, gn : n = 1, 2, 3, . . ., thus constructed is

well-defined in C∗(S).

We now inductively define the sequence hn ⊆ C∗(T ) initiating the process withh1 = f1 and continuing with

hn+1 = hn − gn|TThen for each n,

|hn+1| ≤ 2r1 = 2 · k2·(

2

3

)n

= 3 · k2·(

2

3

)n+1

= 3rn+1

1To see this: the function hn = (−rn ∨ fn) ∧ rn) has Un ⊆ Z(hn − (−rn) ) and Vn ⊆ Z(hn − rn).


So gn|T = hn − hn+1. Define g : S → R as the series

g(x) =∑

n∈N\0

gn(x)

See that g(x) is continuous on S: Since |gn(x)| ≤ k2

(

23

)n, and

∑

n∈N\0k2

(

23

)nis a

converging geometric series, then∑

n∈N\0 gn(x) converges uniformly to g(x). Since

each gn(x) is continuous on S then g ∈ C∗(S). So g is a continuous on S.

Also see that, g|T = f1:

g|T (x) = limm→∞

m∑

n=1

gn|T (x)

= limm→∞

(h1(x)− h2(x)) + (h2(x)− h3(x)) + · · ·+ (hm(x)− hm+1(x))

= limm→∞

h1(x)− hm+1(x)

= h1(x) = f1(x) (Since limm→∞ 3rm+1 = 0)

Then, g|T = f1 so f1 extends continuously from T to S. We are done.

Example 5. Use Urysohn’s extension lemma to show that any compact subset, K, of

a completely regular space, S, is C*-embedded in S.

Solution: Let U and V be disjoint subsets of the compact set, K, which are completely

separated in K. Then there is a function f ∈ C(K) such that U ⊆ A = f←[0] andV ⊆ B = f←[1]. Both A and B are disjoint closed subsets of compact K and so are

compact sets. Then A and B are compact in the completely regular space, S. ThenA and B are completely separated in S. So U and V are completely separated in S.

By Urysohn’s extension lemma, K is C∗-embedded in S. We conclude that . . .

. . . any compact subset, K, of a completely regular space, S, is C*-embedded

in S.

Uniqueness of βS.

We are now able to prove that, up to equivalence, the Stone-Cech compactification of

S is the only compactification in which S is C∗-embedded. By this we mean that, ifS is C∗-embedded in the compactification, γS, of S, then γS is equivalent to βS. So

the symbol, βS, is strictly reserved for the Stone-Cech compactification of S.


Theorem 21.9 The completely regular space S is C∗-embedded in the compactification,γS, if and only if γS ≡ βS.

Proof : We are given that S is completely regular.

( ⇐ ) Suppose γS ≡ βS. Then there is a homeomorphism, h : γS → βS, mappingγS onto βS such that h(x) = x, for all x ∈ S. We are required to show that S is

C∗-embedded in γS.

If f ∈ C∗(S) then f extends to fβ : βS → R. Then fβh : γS → R. Define fγ = fβ

h.We see that fγ : γS → R is the continuous extension of f from S to γS. Then S is

C∗-embedded in γS.

( ⇒ ) Suppose S is C∗-embedded in γS. We are required to show that γS and βS

are equivalent compactifications.

Let i : S → S be the identity map. Then, by theorem 21.5, i extends to i∗ : βS → γS.Also, just as shown in the proof of theorem 21.5, i← extends to i←ˆ : γS → βS. Theni←i and ii← are both identity maps on S and, since S is dense in both βS and γS,

respectively, then i∗i←ˆ and i←ˆi∗ are identity maps on γS and on βS, respectively.Then both i∗ and i←ˆ are homeomorphisms. Hence γS and βS are equivalent com-

pactifications.

Example 6. Show that if F is a closed subset of a metric space S then F is C∗-embedded in S.

Solution : Let F be a closed subset of the metric space S. We will set up the solutionso that we can invoke the Urysohn extension theorem.

Let A and B be completely separated in F . Then, by definition, there is a function f in

C∗(F ) such that A ⊆ Z(f) and B ⊆ Z(f−1). Then clFA ⊆ Z(f) and clFB ⊆ Z(f−1).Since F is closed in S then so are the disjoint sets clFA and clFB. It is shown on

page 210 that in metric spaces closed subsets are the same as zero-sets. So clFA andclFB are disjoint zero-sets in S, say clFA = Z(k) and clFB = Z(g) in S. If

h =|k|

|k|+ |g|on S, clFA = Z(h) and clFB = Z(h − 1) in S. So A and B are completely sepa-rated in S. By Urysohn’s extension lemma every closed subset of a metric space is

C∗-embedded.

Because of this, it is useful to remember that,

. . . any closed subset of R is C∗-embedded in R


21.5 Associating compactifications to subalgebras of C∗(S).

To each compactification αS we can associate a subset, Cα(S), of C∗(S) defined as

Cα(S) = f |S ∈ C∗(S) : f ∈ C(αS)

That is, f ∈ Cα(S) if and only if f extends to fα ∈ C(αS). If αS and γS are two

compactifications of S such that αS γS it is normal to wonder how Cα(S) comparesto Cγ(S) in C∗(S).

Theorem 21.10 Let αS and γS be two compactifications of S such that αS γS. LetCα(S) denote the set of all real-valued continuous bounded functions on S that extend to

αS. Let Cγ(S) denote the set of all real-valued continuous bounded functions on S thatextend to γS. Then Cα(S) ⊆ Cγ(S).

Proof : We are given that αS γS. Then there is a continuous function πγ→α : γS → αS

such that πγ→α(x) = x on S.

Suppose t ∈ Cα(S). It suffices to show that t ∈ Cγ(S). Then there a function

tα : αS → R is such that tα|S = t. Define the function g : γS → R as

g = tαπγ→α

Since πγ→α : γS → αS and tα : αS → R are both continuous then g is continuous onγS and

g|S(x) = (tπγ→α)(x) = t(x)

So t = g|S ∈ Cγ(S). Hence

αS γS ⇒ Cα(S) ⊆ Cγ(S)

as required.

Equivalent functions in C∗(S).

In what follows we will refer to pairs of functions in C∗(S) which are “equivalent”.

We define a particular subset, Cω(S) , of C∗(S) as follows:

Cω(S) = f ∈ C∗(S) : fβ is constant on βS\S


The subset Cω(S) is easily seen to be closed under addition, multiplication, scalarmultiplication and contains all constant functions. We say that it is a subalgebra of

C∗(S).

We will say that two functions f and g in C∗(S) are equivalent functions in C∗(S) if

f − g ∈ Cω(S)

Equivalent functions f and g are sometimes expressed by the notation

f ∼= g

The next theorem shows that there are as many compactifications of S as there aresubalgebras of C∗(S) of a certain type.

Suppose (S, τ) is a completely regular space and F is a subalgebra of C∗(S) which

contains Cω(S) and separates points and closed sets. We can invoke theorem 7.17 andtheorem 10.16 to obtain a compactification,

αS = eβF

[βS] = clT eF [S]

of S, generated by F . We can associate to the compactification αS, the set Cα(S) ⊆C∗(S) of real-valued continuous functions. In the following theorem statement weshow conditions that F must satisfy to guarantee that F = Cα(S).

Theorem 21.11 Let (S, τ) be a completely regular space and F be a subalgebra of C∗(S)which separates points and closed sets in S. Suppose F generates the compactification

αS = eβF

[βS] = clT eF [S]

If F satisfies the properties:

1) F contains Cω(S).

2) every f ∈ Cα(S) is equivalent to some function g ∈ F .

then F = Cα(S).

Proof : We are given that F separates points and closed sets of S and Cω(S) ⊆ F . Then

eβF

[βS] = clT eF [S] = αS

is a compactification of S (generated by F ) where Cα(S) = f |S : f ∈ C(αS).We are also given that every function in Cα(S) is equivalent to some function in F .


We are required to show that F = Cα(S).

Claim #1: F ⊆ Cα(S). Note that eβF

← : αS → βS. Let f ∈ F . We define thereal-valued function fα : αS → R as

fα = fβ eβ

F

←

We see that fα ∈ C(αS) and so is a continuous extension of f ∈ F . We then concludethat F ⊆ Cα(S), as claimed.

Claim #2: Cα(S) ⊆ F . Suppose g ∈ Cα(S). Then, by hypothesis, there is a function

f ∈ F such that g − f = h ∈ Cω(S). Then g = f + h. Since both f and h belong tothe subalgebra F then g ∈ F . We can then conclude that Cα(S) ⊆ F , as claimed.

Then Cα(S) = F , as required.

The converse is easily seen to be true. That is, if F = Cα(S), then F satisfies the

two given properties.

The reader can expect to encounter, a bit later, similar theorem statements, one ofwhich is called The Stone-Weierstrass theorem in 30.3, the other is theorem 30.4.1

Suppose γS is a compactification of S and Cγ(S) = f |S ∈ C(S) : f ∈ C(γS). That

is, Cγ(S) is the set of all function, f , in C∗(S) which extend to fγ : γS → R. Wehave shown that γS βS and Cγ(S) is a subalgebra of C∗(S). We have also seenthat there is a continuous map πβ→γ : βS → γS which fixes the points of S.

In the following lemma we show that we can express the function πβ→γ : βS → γS in

a form which better describes the mechanism behind the function itself.

Lemma 21.12 Let γS be a compactification of the space S. Let G = Cγ(S). Then

πβ→γ = eGγ←eG

β

where eG is the evaluation map generated by G .

1The Stone-Weierstrass theorem states: “Let S be a compact topological space. Let F be a completesubring of C∗(S) which contains the constant functions. If F separates the points of S then F = C∗(S)”.

A consequence of the Stone-Weierstrass statement is the theorem 30.4 which roughly states that:“If the set, C∗(S) contains a subring, F , which is complete and contains Cω(S) then F = Cα(S) for somecompactification, αS, of S.”


Proof : If f ∈ Cγ(S), for x ∈ βS, fβ(x) = (fγπβ→γ)(x). Then, for x ∈ βS,

eGβ(x) = < fβ(x) >f∈Cγ(S)

= < (fγπβ→γ)(x) >f∈Cγ(S)

= < fγ(πβ→γ(x)) >f∈Cγ(S)

= eGγ(πβ→γ(x))

= (eGγπβ→γ)(x)

Then eGβ = (eG

γπβ→γ) on βS. Then πβ→γ = eG

γ←eG

β.

21.6 Limits of z-ultrafilters in βS.

For what follows, recall the definitions of terms related to z-filters in 14.10.

Suppose Z = Z(f) : f ∈ M ⊆ C∗(S) is a free z-ultrafilter in the locally compactHausdorff space S, where M is the corresponding free maximal ideal in C∗(S). Let

Z ∗ = clβSZ(f) : f ∈M

denote a family of subsets of βS corresponding to Z . Since βS is compact Hausdorffand Z is a filter, Z ∗ satisfies the finite intersection property. Then Z ∗ must have

non-empty intersection in βS. Then it is fixed and so must have a unique limit point,

p = ∩clβSZ(f) : f ∈M

in βS\S. We clearly have clβSZ(f) ⊆ Z(fβ) for each f ∈M ⊆ C∗(S). Since fβ |Z(f)

agrees with fβ on clβSZ(f) then its extension to Z(fβ) agrees with fβ on Z(fβ). So

clβSZ(f) = Z(fβ)So,

“. . . for any free z-ultrafilter, Z = Z(f) : f ∈ M ⊆ C∗(S) in Z[S], wecan write,

p = ∩clβSZ(f) : f ∈M = ∩Z(fβ) : p ∈ Z(fβ)

where p ∈ βS\S. So the points in βS\S aree precisely the unique limits ofa unique free z-ultrafilter in Z[S].”

Of course, if Z is a fixed z-ultrafilter in Z[S] then

p = ∩Z(f) : f ∈M

for some p in S.


Theorem 21.13 Suppose T is a dense subset of the completely regular space S. Then thefollowing are equivalent:

a) The subset, T , is C∗-embedded in the space S

b) Disjoint zero-sets in T have disjoint closures in S.

c) Every point of S is the limit of a unique z-ultrafilter on T .

Proof : We are given that S is completely regular.

( a ⇒ b ) Suppose T is a C∗-embedded dense subset of S. Let h, g ∈ C∗(T ) be suchthat Z(h) and Z(g) are disjoint zero-sets of T . Then Z(h) and Z(g) are completelyseparated in T .1

By Urysohn’s extension theorem, Z(h) and Z(g) are completely separated in S. Then

there exists t ∈ C∗(S) such that Z(h) ⊆ Z(t) and Z(g) ⊆ Z(t− 1). Then clSZ(h) ⊆Z(t) and clSZ(g) ⊆ Z(t− 1). We can then conclude that clSZ(h) ∩ clSZ(g) = ∅, as

required.

( b ⇒ c ) Suppose that disjoint zero-sets in T have disjoint closures in S. Let Z1 and

Z2 be disjoint zero-sets in T .

Claim #1: We claim that clS(Z1 ∩ Z2) = clSZ1 ∩ clSZ2.

Proof of claim: Of course, LHS ⊆ RHS is always true. Suppose, on the other hand,

that x ∈ clSZ1 ∩ clSZ2.

We are required to show that x ∈ clS(Z1 ∩ Z2). For any zero-set neighbourhood, Z,of x, Z ∩ Z1 is dense in Z ∩ clSZ1 so

x ∈ clS(Z ∩ Z1) ∩ clS(Z ∩ Z2) 6= ∅ (∗)

Now both Z ∩ Z1 and Z ∩ Z2 are zero-sets so, by hypothesis, (Z ∩ Z1) ∩ (Z ∩ Z2)cannot be empty. So Z ∩ (Z1 ∩ Z2) 6= ∅. Since Z is any zero-set neighbourhood of x,this means that x ∈ clS(Z1 ∩ Z2).

We conclude clS(Z1 ∩ Z2) = clSZ1 ∩ clSZ2, as claimed.

Claim #2: We claim that each point in S\T is the limit of a unique z-ultrafilter on

T .

Proof of claim: Let y ∈ T . Then y belongs to the closure of a zero-set, Z, in T . Hencey is the limit point of a z-ultrafilter, Z , in T . Now, if y is also the limit of another

z-ultrafilter, Z1, in T then Z1, will contain a zero-set Z1 which will not intersect

1To see this, note that Z(|h| + |g|) = ∅ in T and so for the function, k(x) = |h(x)|/[|h(x)| + g(x)|],Z(h) ⊆ Z(k) and Z(g) ⊆ Z(k − 1).


some zero-set, Z2, of Z . Then y ∈ clSZ1 ∩ clSZ2 = clS(Z1 ∩ Z2) = ∅. We can onlyconclude that every point in S\T is the limit of a unique z-ultrafilter in T . As claimed.

( c ⇒ a ) We are given that S\T is a set of limits of free z-ultrafilters on T . SinceβT \T is the set of all limit points of free z-ultrafilters on T , we can say that

S\T ⊆ βT \T

Now, since S is dense in βS and T is dense in S, then T is dense in βS so we can

view βS as a compactification, say γT , of T with outgrowth

γT \T = (βS\S) ∪ (S\T )

Then for the function πβ→γ : βT → γT , we have, πβ→γ [βT \T ] = γT \T where,

πβ→γ |S(x) = x

Then, for f ∈ C∗(T ) define f∗ : S → R as

f∗(x) = fβπβ→γ |←S (x)

where f∗ is seen to be continuous on S and f∗|S = f on T . So f∗ is a continuous

extension of f from T to S. We conclude that T is C∗-embedded in S.

21.7 Pseudocompact spaces revisited.

Recall (from definition 17.10) that a topological space is said to be pseudocompact if

every continuous real-valued function on S is bounded. That is, if C(S) = C∗(S).

Although the pseudocompact property has a simple and easily understood definition,

it turns out that, when not compact, such spaces are not easily recognizable. It willbe helpful to obtain a few characterizations. In the following theorem, we show that

pseudocompact spaces are precisely those spaces, S, where βS\S does not contain azero-set.

Theorem 21.14 A locally compact Hausdorff space S is pseudocompact if and only if no

zero set Z in Z[βS] is entirely contained in βS\S.


Proof : Let S be a locally compact Hausdorff space.

( ⇒ ) Suppose S is a pseudocompact space and Z(fβ) ∈ Z[βS]. We are required to

show that Z(fβ) ∩ S 6= ∅.

Suppose not. That is, suppose, Z(h) ⊆ βS\S (where h ∈ C(βS). Then, since h|S is

not zero in S, we can define the function g = 1/h|S and so g ∈ C(S). Let z ∈ Z(h).Since z belongs to clβSS then there is a sequence, xi in S, which converges to z. By

continuity, the corresponding sequence, h(xi) in R, must converge to h(z) = 0. So gis unbounded on S, which contradicts the hypothesis which states that all real-valued

continuous functions on S are bounded. So Z(h) must intersect with S, as required.

( ⇐ ) Suppose now that, for any Z ∈ Z[βS], Z ∩ S 6= ∅. We are required to show

that S is pseudocompact.

Suppose S is not pseudocompact. Then C(S) contains an unbounded function g. Let

f = |g| ∧ k

where k > 0. Then f is continuous real-valued unbounded on S. Then for each n ∈ N,

there exists xn ∈ S such that f(xn) ∈ (n,∞). Then h = 1/f is a continuous well-defined real-valued bounded function on S. Since βS is compact xn : n ∈ N has a

converging subsequence xni : i ∈ N with limit, say x ∈ βS\S. Since h is continuousand bounded on S, h extends to hβ : βS → R with hβ(x) = 0. Then Z(hβ) ⊆ βS\S,

a contradiction of our hypothesis. Then S is pseudocompact.

It is interesting to note that, in the above theorem, a property of the outgrowth, βS\Scharacterizes a property of the space S.

We point out one easy consequence of the above theorem.

Suppose S is locally compact and completely regular. If S is pseudocompact andk = fβ(x) ∈ fβ[βS\S], then there exists y ∈ Z(fβ − k) ∩ S, so fβ(y) = k ∈ f [S]. Sofβ[βS\S] ⊆ f [S].

Conversely, suppose fβ [βS \S] ⊆ f [S]. Suppose Z(fβ − t) ∩ βS \S 6= ∅. Then

there exists y ∈ βS \S such that fβ(y) = t and x ∈ S such that f(x) = t. Thenx ∈ Z(fβ − t) ∩ S 6= ∅. So S is pseudocompact.

“The locally compact completely regular space, S, is pseudocompact if andonly if, for every f ∈ C∗(S), fβ[βS\S] ⊆ f [S].”


21.8 The one-point compactification.

It was shown in theorem 18.8, that every locally compact Hausdorff space is completely

regular. Hence a locally compact Hausdorff space, S, has at least one compactifica-tion, namely, βS. This compactification is maximal when compared to all others in

the family, C , of all compactifications. Does the family C contain a minimal element?That is, does C have a compactification, γS, such that γS αS, for all compactifi-

cations, αS in C ? The answer will depend on the space, S. In a previous chapter ofthe book, we, in fact, provided an answer to this question, as we shall soon see.

In theorem 18.7, we showed that given a locally compact Hausdorff space (S, τ) anda point ω 6∈ S, we can construct a larger set,

ωS = S ∪ ω

By first defining, Bω = U ∪ ω : U ∈ τ and S \U is compact , we then define a

topology, τω, on ωS as follows:

τω = τ ∪ Bω

We then showed that (ωS, τω) is a compact space which densely contains S. Then

(ωS, τω) satisfies the definition of a compactification of S. Furthermore, and quiteimportantly, we show in theorem 18.7 that ωS is Hausdorff if and only if S is locally

compact.

So a non-compact, locally compact Hausdorff space, S, has a compactification, ωS,

which may be different from βS. We formally define it.

Definition 21.15 Let (S, τ) be a locally compact Hausdorff topological space, ω be a pointnot in S and ωS = S ∪ ω. If

Bω = U ∪ ω : U ∈ τ and S\U is compact

and τω = τ ∪ Bω then (ωS, τω) is called...

...the one-point compactification of S 1

We will, more succinctly, denote the one-compactification of S by,

ωS = S ∪ ω

1The one-point compactification of S is also referred to as the Alexandrov compactification of S, namedafter the soviet mathematician, Pavel Alexandrov, (1896-1982).


Since we have chosen the symbol ωS as notation then, to be consistent with thenotation used up to now, we should let

Cω(S) = f |S : f ∈ C(ωS)

But the symbol, Cω(S), has already been used in another sense on page 404 where

Cω(S) is used to represent the set of all f ∈ C∗(S) such that fβ is constant on βS\S.We verify that these are the same set. For f ∈ Cω(S) if and only if,

fβ [βS\S] = fβ [ π←β→ω[ω] ] = fω(ω) ⊂ R

Then Cω(S) separates points and closed sets of S.2

It is also worth emphasizing the fact that

“...for a space S to have a one-point compactification which is Hausdorff, Smust be locally compact”.

We can even say more. Amongst all the the completely regular spaces, S, the only

ones that are open in any compactification are the ones where S is locally compact.We will prove this now.

Theorem 21.16 Let S be a completely regular topological space and αS be any compact-ification of S. Then S is open in αS if and only if S is locally compact.

Proof : We are given that S is completely regular and αS is a compactification of S.

( ⇒ ) Suppose S is open in αS. Since the space S is the intersection of the open setS and the closed set αS, by theorem 18.3, S is locally compact.

( ⇐ ) Suppose S is locally compact in αS. By theorem 18.3 part d), S is the inter-section of an open subset, U , and a closed subset, F . Since F = αS and S is dense in

αS, then S is open in αS, as required.

If S is locally compact then the map, πβ→ω : βS → ωS, collapses the set βS\S down

to the singleton set ω. More generally, if αS is any compactification of S, thenπα→ω continuously collapses the outgrowth αS\S down to ω and fixes the points

of S. So for any compactification αS,

ωS αS and Cω(S) ⊆ Cα(S)

2Had we known this fact before we would has seen that, in the statement of theorem 21.11, the condition1) is redundant.


Theorem 21.17 Uniqueness of ωS. Let S be a locally compact completely regular topo-logical space. Suppose αS and γS are both compactifications of S which contain only onepoint in their compact extension. Then they are equivalent compactifications.

Proof : We are given that S is locally compact completely regular. Suppose αS\S = ωαand γS\S = ωγ, both singleton sets.

Consider the map, h : αS → γS, where h(x) = x on S and h(ωα) = ωγ . Then h is

one-to-one and onto. It will suffice to show that h maps open neighbourhoods to openneighbourhoods.

Let Uα be any open neighbourhood of a point in αS. If Uα ⊆ S then Uα is open in

S. Then h[Uα] = Uα. Since S is locally compact it is open in γS and Uα = Uα ∩ S isopen in γS.

Suppose ωα ∈ Uα. Since αS\Uα is compact and h|S is continuous, then h[αS\Uα] is a

compact. Since h is one-to-one, h[αS\Uα] = h[αS]\h[Uα] = γS\h[Uα], a compact setwhich doesn’t contain ωγ . So h[Uα] is open. Hence h : αS → γS is a homeomorphism.

We can conclude that αS ≡ γS.

From this theorem we can conclude that the one-point compactification is unique, upto equivalence. We now consider a few examples involving compactifications of a space.

Example 7. Suppose that S is locally compact and its one-point compactification,ωS, of S is metrizable. Show that S must be second countable.

Solution : Suppose ωS is metrizable. In the proof of theorem 15.8 it is shown that

countably compact metric spaces are separable. By theorem 5.11, a separable metricspace is second countable. Since ωS is compact and so is countably compact then, by

combining these two results we obtain that ωS is second countable. By theorem, 5.13,subspaces of second countable spaces are second countable. So S is second countable.

As required.1

It may happen that βS and ωS are the same compactification. We provide an exam-ple where they are not equivalent.

Example 8. Consider the set S = (0, 1] equipped with the usual subspace topology.

Determine ωS. Show that the one-point compactification, ωS = [0, 1], of S is not

1The converse of the statement in this example is true. That is, “Locally compact second countablespaces have a metrizable one-point compactification” has been proven. But its proof is fairly involved. Sowe will not show it here.


equivalent to βS.

Solution : Since [0, 1] is a compact set which densely contains S, and the one-point

compactification is unique, then ωS = [0, 1]. Note that, since the function f(x) = sin 1x

is a bounded continuous function on S, then it extends to βS. (Plot f(x) = sin 1x .)

Verify that f does not extend continuously to [0, 1].

Then [0, 1] cannot be the Stone-Cech compactification of S.

The following theorem provides an example of a space, S, such that βS = ωS.

Theorem 21.18 If S is the ordinal space [0, ω1) (where ω1 is the first uncountable ordinal)then βS\S = ω1 and so βS = ωS = [0, ω1], the one-point compactification of S.

Proof : Given: The space, S, is the ordinal space [0, ω1). Then ωS = [0, ω1] is its one-point

compactification. So S is completely regular and locally compact.

We are required to show that βS = ωS = [0, ω1].

In the example on page 305, it is shown that S = [0, ω1) is countably compact butnon-compact. In theorem 15.9 it is shown that, if S is countably compact every func-

tion in C(S) is bounded and so has a compact image in R.

Let f ∈ C∗(S) = C(S). Then

fβ[βS] = fβ[clβSS] = clRf [S] = f [S] ⊆ R

To show that βS = ωS it suffices to show that fβ[βS \S] is a singleton set in f [S].Suppose q and q∗ are two points in fβ[βS\S] ⊆ f [S] ⊆ R.

We claim: That q = q∗.

Proof of claim: Express f [S] as a net

N = f(α) : α ∈ S = [0, ω1)

Both q and q∗ are accumulation points of the net N = f [S]. So, for each n ∈ N, bothB1/2n(q) and B1/2n+1(q

∗) each contain a cofinal subset of the tail end of the net N .

We can then choose, αn : n ∈ N strictly increasing in S such that f(α2n) ∈ B1/2n(q)and f(α2n+1) ∈ B1/2n+1(q

∗). If sup αn = κ then sup α2n = κ = sup α2n+1.Hence

limn→∞

f(α2n) = q = f(κ) = q∗ = limn→∞

f(α2n+1)

So q = q∗ as claimed.

From this we can conclude that, for all f ∈ C∗(S), fβ is constant on βS \S. Then

βS = [0, ω1] = ωS, the one-point compactification of S.


In the following statement an n-sphere , Sn, refers to a subset of Rn+1 defined as

~x ∈ Rn+1 : ‖~x‖ = 1. For example, S2 is a subset of R3 where (x, y, z) ∈ S2 if andonly if ‖(x, y, z)‖ = 1. We will show that, the 2-sphere, S2 is homeomorphic to ωR2.

Theorem 21.19 Let Sn denote the n-sphere in Rn+1. Then the “punctured n-sphere”,Sn minus a single point, (Sn\p), is homeomorphic to Rn. Then, Sn is homeomorphic to

the one-point compactification, ωRn, of Rn.

Proof : We are given that Sn is the n-sphere which is a subset of Rn+1 and that p =

(0, 0, 0, . . . , 0, 1) is a point in Rn+1 which belongs to Sn. We will first show that Sn\pis homeomorphic to Rn. We achieve this by showing that the function g : Sn\p → Rn

defined as

g(x1, x2, . . . , xn, xn+1) =1

1 − xn+1(x1, x2, . . . , xn−1, xn) 2

maps Sn\p homeomorphically onto Rn.

Since p = (0, 0, . . . , 1), then 1 − xn+1 6= 0, so g is well-defined.

Let a = (a1, a2, . . . , an+1) and b = (b1, b2, . . . , bn+1) be distinct points in Sn\p. Then

ak 6= bk for at least one k ∈ 1, 2, . . . , n+ 1. Then

1

1− an+1(a1, a2, . . . , an−1, an) 6= 1

1 − bn+1(b1, b2, . . . , bn−1, bn)

So g is one-to-one. The function g is also verified to be onto Rn.

To prove continuity of g : Sn\p → Rn it suffices to show that πig is continuous foreach i ∈ 1, . . . , n, and invoke 7.11. See that

(πig)(x1, x2, . . . , xn, xn+1) = πi

[

1

1 − xn+1(x1, x2, . . . , xn−1, xn)

]

=xi

1 − xn+1

=πi(x1, x2, . . . , xn)

1− xn+1

=

(

πi

1 − xn+1

)

(x1, x2, . . . , xn)

2The function, g, is known as a stereographic projection


so πig is continuous for each i.

Also, since π is an open map, the function g is an open map.

Then g maps Sn\p homeomorphically onto Rn, as we claimed earlier.

We now define a function g∗ : Sn → ωRn and show it is a homeomorphism.

Let g∗ : Sn → ωRn be a function mapping the one-point compactification, Sn, ofSn\p onto the one-point compactification, ωRn = R ∪ ω, of Rn which is defined

as

g∗|Sn\p(x) = g(x)

g∗(p) = ω

If U is an open subset of ωRn which doesn’t contain ω then g∗←[U ] is open in Sn\p,and so is open in Sn.

If U is an open subset of ωRn which contains ω then g∗←[U ] = p ∪ g←[U ∩ Rn].Now, ωRn\U is compact in ωRn and so is compact in Rn. Then g←[ωRn\U ] is com-pact in Sn. So, g∗←[U ] is an open neighbourhood of p in Sn. So g∗ is continuous on Sn.

Similarly, g∗ maps open subsets of Sn to open subsets of ωRn.

So g∗ is a homeomorphism between the compact sets Sn and ωRn.

We also showed along the way that g∗| \Sn\p = g maps the punctured n-sphere,Sn\p, homeomorphically onto Rn.

21.9 Topic: Cardinality of some common Stone-Cech compactifications.

We know the cardinality of the most common sets we encounter (such as N, Q andRn). We can sometimes determine the cardinality of associated sets such as their

Stone-Cech compactification. We know the cardinality, |N|, of the set N is ℵ0 while|R| = c = 2ℵ0. In the following theorem we compute the cardinalities of βN, βQ and

βR. This is good practice in working with these particular compactifications.

Theorem 21.20 The cardinality, |βN|, of the set βN is 2c.

Proof : We are given the compactification, βN, of N.

Claim#1. We first claim that |βN| ≥ 2c.


Proof of claim#1: In theorem 7.10, it is shown that, since [0, 1] is separable, then theproduct space, K =

∏

i∈R[0, 1], with |R| = c factors, is also separable. This means

that there is a countably infinite set, D, contained in K. There then exist a function

g : N → D ⊂ K

which indexes the elements of D. Note that g is continuous on N and densely embedsD in K. By Tychonoff’s theorem, K is compact. See that g extends continuouslyfrom N to

gβ(K) : βN → K

Then

gβ(K)[βN] = gβ(K)[clβNN]

= clKgβ(K)[N]

= clKD

= K

Now |K| = |∏i∈R[0, 1]| = cc = 2c. (See footnote)1. Since βN is the domain of the

function gβ(K) (which could possibly not be one-to-one), then

|βN| ≥ |K| = 2c

as claimed.

Claim#2. That |βN| ≤ |K| = 2c.

Proof of claim#2: We know that each function in C∗(N) can be seen as a sequencexi : i ∈ N in RN. Then

|C∗(N)| = |RN| = cℵ0 = c (See footnote)2

That is, if I = |C∗(N)|, then

C∗(N) = fi : N → R : i ∈ I

contains I = c distinct functions.

Let T =∏

i∈I[ai, bi] ⊆∏

i∈IR where, by Tychonoff theorem, T is compact.

Recall that eC∗(N) : N →∏

i∈I R is the evaluation map generated by C∗(N), explicitlydefined as

eC∗(N)(n) = <fi(n)>fi∈C∗(N) ∈ eC∗(N)[N] ⊆ T ⊆∏i∈IR

1The proof of |Q

i∈R[0, 1]| = cc = 2c is shown in an example of Section 24.2 of Axioms and Set theory, R.

Andre, in which we compute the cardinality of RR

2The proof of |RN| = c is shown in theorem 25.2 of Axioms and Set theory, R. Andre


Then eC∗(N)[N] ⊆ T =∏

i∈I[ai, bi], and |T | ≤ |R|I = cc = 2c. So,

eβC∗(N)[βN] = e

βC∗(N)[clβSN]

= clT eC∗(N)[N]

⊆ T

So |βN| ≤ |T | ≤ 2c. This establishes our second claim.

Since |βN| ≤ 2c and |βN| ≥ 2c then |βN| = 2c. We are done.

Now, N contains countably many points and so |βN\N| = 2c. Since we can associate

to each point in βN\N a unique free z-ultrafilter in Z[N], the above theorem confirmsthat there are 2c free z-ultrafilters in Z[N].

The cardinality of βN will help us determine the cardinalities of βR and βQ.

Theorem 21.21 The sets βN, βR and βQ each have a cardinality equal to 2c.

Proof : We have already shown that |βN| = 2c.

Claim #1 : |βQ| ≤ 2c.

Proof of claim #1. We know N and Q are countable so both have cardinality ℵ0.

Then there exists a one-to-one function, f : N → βQ, mapping N onto Q ⊆ βQ. SinceN is discrete f is continuous on N. By theorem 21.5, f : N → βQ extends to

fβ(βQ) : βN → clβQf [N] = clβQQ = βQ

Then

fβ(βQ)[βN] = fβ(βQ)[clβNN] = clβQf [N] = clβQQ = βQ

So |βQ| ≤ |βN| = 2c. This establishes claim #1.

Claim #2 : |βR| ≤ |βQ|.Proof of claim #2. Since Q is dense in R and R is dense in βR, then it is dense in βR

then clβRQ = βR. Consider the continuous inclusion function

i : Q → clβRQ = βR


By theorem 21.5, i : Q → clβRQ = βR extends to

iβ(βR) : βQ → βR

Theniβ(βR)[βQ] = iβ(βR)[clβQQ] = clβRi[Q] = clβRQ = βR

So |βR| ≤ |βQ|. This establishes claim #2.

Up to now we have shown that |βR| ≤ |βQ| ≤ |βN| = 2c.

Claim #3 : |βR| ≥ |βN|.Proof of claim #3. We know that N is C∗-embedded in R. (See example on page 399

or theorem 21.8.) Then, if i : N → βN is the continuous inclusion map, since N isC∗-embedded in R, i : N → βN extends continuously to

i∗ : R → βN

Also, i∗ : R → βN extends continuously to i∗β : βR → βN. Then

βN = clβNN

= clβNi[N]

⊆ clβNi∗[R] (Since i embeds N in i∗ [R])

= i∗β[clβRR]

= i∗β[βR]

Since i∗β[βR] contains βN, then |βN| ≤ |βR|. This establishes claim #3.

Combining the results in the three claims above we conclude that |βR| = |βQ| =

|βN| = 2c as required.

21.10 Compactifying a subset T of S ⊆ βS.

If T is a non-compact subset of S, it is interesting to reflect on how clβST compareswith βT . Does it make sense to say that βT ⊆ βS? We examine this question in the

following example.

Example 9. Let T be a non-empty subspace of a completely regular space, S. Show

thatclβST is equivalent to βT

Solution: We are given that T ⊆ S. Since subspaces of completely regular spacesare completely regular then T is completely regular. Let i : T → βS be the identity


function which embeds T into βS. By theorem 21.5, i : T → βS extends continuouslyto iβ(βS) : βT → βS. Then

βT = iβ(βS)[βT ]

= iβ(βS)[clβTT ]

= clβSi[T ]

= clβST

So clβST is equivalent to βT .

Suppose now that F is a closed non-empty subset of the completely regular space S.

The statement in the above example will allow us to say something more about F .

To see this, suppose f ∈ C∗(S). As stated in the example,

clβSF = βF

So F is C∗-embedded in clβSF . Since F is closed in S then

(clβSS)\S ⊆ βS\S

So f : F → R extends to fβ : βF → R. Since βF is compact in βS then fβ extends

to fβ∗ : βS → R. So

fβ∗|S : S → R

is a continuous extension of f on F to fβ∗|S on S (via fβ∗ on βS).

Then F is C∗-embedded in S. We conclude that,

“If F is a closed non-empty subset of a completely regular space S then Fis C∗-embedded in S.”

21.11 The zero-sets of βN are clopen.

Zero-sets in βN play a role in the solution of the following example. If Z is a zero-setin N then it is clopen in N. It is easily seen that any zero-set in N is a zero-set of a

characteristic function, g : N → 0, 1, on N. Since g extends to gβ : βN → 0, 1,gβ←(0) = Z(gβ) = clβNZ is a clopen zero-set in βN. As well, gβ←(1) = βN \ Z(gβ)

is a clopen zero-set in βN. So, for every zero-set Z(g) in N, Z(gβ) is clopen in βN.Hence zero-sets of βN are clopen in βN.

Example 10. The compactification, βN, is easily seen to be separable (N is a dense

subset of βN.). Show that βN\N is not separable.


Solution: Suppose that βN\N is separable. Then, βN\N contains a dense countablesubset

D = xi : i ∈ NSince the cardinality of βN is 2c (shown above), we can fix two points

n∗ ∈ (βN\N)\D and n ∈ N

Now, for each i ∈ N, xi and n∗, n form disjoint closed subsets of βN.

Since βN is normal, for each xi ∈ D, there is a clopen zero-set, Zi = Z(gβ) such that

xi ∈ Zi and n∗, n ⊆ βN\Zi

Then βN\Zi : i ∈ N is a family of zero-set clopen neighbourhoods of n∗, n.If W = ∩βN\Zi : i ∈ N then W is a countable intersection of zero-sets and so is,itself, a zero-set which contains n∗, n (see page 204) which does not intersect D.

Since n∗ ∈ βN\N ∩W then W ∩ βN\N is a non-empty clopen subset of βN\N whichdoes not intersect D. Since D is dense in βN\N we have a contradiction. So βN\N isnot separable.

Concepts review:

1. Suppose S is a topological space and T is a compact Hausdorff space. What does itmean to say that T is a compactification of S?

2. If S has a compactification, αS, what separation axiom is guaranteed to be satisfied

by S?

3. Given a completely regular space S let e : S → πi∈I [ai, bi] be the evaluation map onS induced by C∗(S). Give a definition of the Stone-Cech compactification of S which

involves this evaluation map.

4. What does it mean to say that the two compactifications of S, αS and γS, are

equivalent compactifications?

5. If C = αiS : i ∈ I denotes the family of all compactifications of S. Define a partialordering of C .

6. If U is a subset of the topological space S, what does it mean to say that U is C∗-embedded in S?

7. If S is C∗-embedded in the compactification, αS, of S what can we say about αS?


8. Suppose S is completely regular and g : S → K is a continuous function mappingS into a compact Hausdorff space K. For which compactifications, αS, does the

following statement hold true: “the function g extends to a continuous function g∗ :αS → K”?

9. Suppose S is locally compact and Hausdorff. Define the one-point compactification,

ωS, of S.

10. What can we says about those subspaces of a compactification, αS, which are locally

compact? What can we says about those subspaces of a compactification, αS, whichare open in αS?

11. What is the Stone-Cech compactification of the ordinal space [0, ω1)?

12. State a characterization of the pseudocompact property stated in this chapter.

422 Section 22: Singular sets and singular compactifications

22 / Singular sets and singular compactifications.

Summary. In this chapter we introduce an alternate method to construct a

compactification of a locally compact Hausdorff space. We define the notion ofthe singular set, S(f), of a function, f : S → T . We will show that we can

always use S(f) to construct a compactification, αS = S ∪ S(f), by applying asuitable topology on αS. If the singular set, S(f), contains the image, f [S], of

f , we refer to f as a singular map. When f is singular the resulting compacti-fication, S ∪f S(f), is called a singular compactification.

22.1 Singular sets and functions: definitions.

We begin by formally defining a “singular set” of a continuous function on a locallycompact non-compact Hausdorff space. We also define what is meant when a function

is referred to as being a “singular map”.

Definition 22.1 Let (S, τ) be a locally compact non-compact Hausdorff topological space

and f : S → T be a continuous function mapping S into some compact space T . We definethe singular set, S(f), of f as follows:

S(f) = x ∈ clT f [S] : clSf←[U ] is not compact ∀ clT f [S]-open nbhd, U , of x

If S(f) = clTf [S] then f : S → T is said to be a singular function or singular map.

We make the following few remarks about the two concepts we have just introduced.

1. If S is a non-compact locally compact Hausdorff, the singular set S(f) is never emptyin the compact codomain, T . This is so, whether f is a singular map or not. To see

this, recall that, by theorem 21.5, f : S → T , extends to fβ(T ) : βS → T . Supposeu ∈ βS \S and

x = fβ(T )(u) ∈ fβ(T )[βS \S]

If U is an open neighbourhood of x in T , then u ∈ fβ(T )←[U ] an open subset of βS.If clSf

←[U ] is a compact subset of S, then fβ(T )←[U ]\clSf←[U ] is an open neigh-

bourhood of u contained in βS\S, a contradiction. So clSf←[U ] is not compact. By

definition, x ∈ S(f). So S(f) is non-empty.


2. The singular set S(f) is always closed, and hence compact, in T . This is so whether fis a singular map or not. To see this, suppose x ∈ T \S(f). Then there exists an open

neighbourhood U of x in T , such that clSf←[U ] is compact in S. Then every point

p ∈ U also belongs to T \S(f). So x ∈ U ⊆ T \S(f). Hence S(f) is a closed (and so is

a compact) subset of the compact space T .

3. If f : S → T is a singular map, then f [S] is a dense subset of S(f) (since, by definitionof singular function, S(f) = clT f [S]).

22.2 Singular compactifications: definitions.

We now show how the singular set, S(f), of a function, f : S → T , can be used as theoutgrowth of a compactification of S.

Definition 22.2 Let (S, τ) be a locally compact non-compact Hausdorff topological spaceand f : S → T be a continuous function mapping S into a compact space T . Then S(f)

denotes its singular set. If f is a singular map then, by definition, S(f) = clTf [S] ⊆ T . Weconstruct a new set by adjoining S(f) to S to obtain a larger set,

γS = S ∪f S(f)

The basic open neighbourhoods, B1, of points in S will be the same as the ones in S whenviewed as a topological space on its own.1 If x ∈ S(f), F is a compact subset of S and U

an open neighbourhood of x in S(f), we define

Bx = U ∪ f←[U ]\F

as a basic open neighbourhood of x. Let B2 = Bx : x ∈ S(f). Let B = B1 ∪ B2. This

defines a base for a topology on γS which is easily seen to be a Hausdorff compactificationof S. We will refer to γS as the singular compactification induced by the singular mapf : S → T .

The definition of a singular compactification is not a simple one to grasp nor to visu-alize. It invites many questions. For example, if given a singular compactification γS,

one may want to know which function induces it. Can there be more than one suchfunction which induces it? Are there compactifications which are non-singular?

To help us answer such questions it will be helpful to find simpler characterizationsof a singular compactification.

Retractions and retracts. We remind the reader that, for a topological space S, . . .

1Remember that, if S is locally compact Hausdorff, then S is open in any compactification of S.


. . . if A ⊂ S and r : S → A is a continuous function which fixes the pointsof A, then r is referred to as a “retraction” of S onto A. In such a case, A

is called a “retract” of S.

We will see that those compactifications, αS, whose outgrowth, αS\S, is a retract of

αS characterize singular compactifications.

Theorem 22.3 Let S be locally compact and Hausdorff. Let αS be a Hausdorff com-pactification of S. Then αS is a singular compactification of S if and only if αS \S is a

retract of αS.

Proof : We are given that αS is a Hausdorff compactification of S.

( ⇒ ) Suppose αS = S ∪f S(f) is a singular compactification of S induced by thecontinuous function,

f : S → clTf [S] = S(f) = αS \SWe are required to show that αS \S is a retract of αS.

By definition, f [S] is dense in S(f). Let

fα : αS → S(f)

be a function which agrees with f on S and fixes the points of S(f).

We claim, that fα is continuous on αS: Let x ∈ S(f) and U be an open neighbour-

hood of x. Then fα←[U ] = U ∪ f←[U ], by definition of a basic open neighbourhoodin αS. So fα is continuous on αS, as claimed.

Hence αS \S is a retract of αS, as required.

( ⇐ ) Suppose αS \S is a retract of αS. We are required to show that αS is a singularcompactification.

By hypothesis, there is a continuous function r : αS → αS \S which fixes the pointsof αS \S. By definition of retract, r[S] ⊆ αS \S.

We claim that r|S is a singular function on S: Let x ∈ αS \S and U be an openneighbourhood of x in αS\S. Now, U must intersect r|S[S], for if not, r←[U ] = U ⊆αS \S which is not open in αS. So r←[U ] = U ∪ (r←[U ]∩S). Then, since clSr|S←[S]is not compact in S, then αS \S = S(r|S) is the singular set of r|S, and so

αS = S ∪r S(r|S)

a singular compactification induced by the map r|S. So

r|S : S → αS\Sis a singular, map as claimed.


We now have another way of recognizing a singular compactification:

Singular compactifications are precisely those compactifications, αS, whoseoutgrowth, αS\S, is a retract of the whole space.

In the following example we verify that, if we have one singular compactificationof S, then every compactification “below” it in the partially ordered family of all

compactifications will also be a singular compactification.

Example 1. Show that if αS is a singular compactification and γS is another com-

pactification such that γS αS then γS is also a singular compactification.

Solution : Suppose αS is a singular compactification and γS is another compactifi-cation such that γS αS. Then αS = S ∪f S(f) where f : S → T is continuous

and f [S] ⊆ S(f) = clTf [S]. Also, there exists πα→γ : αS → γS such that πα→γ iscontinuous and onto and fixes the points of S. If g = πα→γf then

clγSg[S] = clγS(πα→γf)[S]

= πα→γ [clαS(f [S])]

= πα→γ [αS\S]

= γS\S

If U is an open subset of γS\S

clSg←[U ] = clS(πα→γf)←[U ]

= clSf←(π←α→γ[U ])

= clSf←[V ] (V open in αS\S)

a non-compact set in S.

So γS\S = S(g). This means that γS\S is a retract of γS and so γS is a singularcompactification. We are done with the solution.

Does every locally compact Hausdorff space, S, have at least one singular compact-ification? Well, we know that such a space S, has a one-point compactification,

ωS = S ∪ ω. Consider the constant function r : ωS → ω. It is a retraction onωS. So r|S : S → ω is a singular map on S. So we can answer this question in the

affirmative.


22.2 Compactifications induced by non-singular functions.

Suppose (S, τ) is a topological space and f : S → T is a continuous function mapping

a non-compact locally compact Hausdorff space S into a compact space T . For whatfollows, f may or may not be a singular map.

Definition of a non-singular compactification induced by f .

Suppose f is not a singular map on S. Recall that

S(f) = x ∈ clT f [S] : clSf←[U ] is not compact for any neighbourhood U of x

Since f is not singular, then S(f) does not satisfy the condition, f [S] ⊆ S(f), requiredto construct a singular compactification. Despite this, we can still adjoin the singular

set S(f) to S to form a larger set, K = S ∪ S(f).

We define a topology on K as follows: The set, BS , of basic open neighbourhoods of

points in S will be the same as the ones in S when viewed as a topological space onits own. For x ∈ S(f), F a compact subset of S and U an open neighbourhood of xin T , we define

Bx = (U ∩ S(f)) ∪ f←[U ]\Fas a basic open neighbourhood of x. This defines a topology on K = S ∪ S(f)generated by the basic open sets,

B = BS ∪ Bx : x ∈ S(f)

The set, K, will be seen to be a compact set which contains S as a dense subset.These facts will be verified following 22.4 below. In the case where f is not singular

(that is, f [S] ∩ T \S(f) 6= ∅), the set S(f) will however not be a retract of thecompactification K = S ∪ S(f). 1 We would like to adopt notation which will help

distinguish those compactifications (induced by a function) which are singular fromthose that are not.

Notation 22.4 Suppose S(f) is a singular set of a function f : S → T where T is compact.If f is not a singular map we will represent the compactification γS induced by f as

γS = S ∪ S(f)

Notice how we distinguish it from those compactifications where f is a singular map:

γS = S ∪f S(f)

1If f is not singular, f [S] may still intersect S(f), or may not even intersect S(f) at all; it is just thatf [S] cannot be entirely contained in S(f)


We first justify a few statements made in the introductory paragraph above. All properties

are for a continuous function f : S → T mapping the locally compact Hausdorff space intothe compact space T .

1. We verify that γS = S ∪ S(f) is a well-defined topological space. To do this we willshow that the given family, B, of sets satisfies the “base property”(See page 73). If

so it generates a topology on γS. Clearly γS = ∪B ∈ B. Given

A ∩B = (Ux ∪ f←[Ux]\FU) ∩ (Vx ∪ f←[Vx]\FV )

we have,

x ∈ A ∩B = (Ux ∩ Vx) ∪ [ (f←[Ux]\FU) ∩ (f←[Vx]\FV ) ]

= (Ux ∩ Vx) ∪ [ f←[Ux] ∩ f←[Vx] ]\(FU ∪ FV )

impliesx ∈ (Ux ∩ Vx) ∪ [ f←[Ux ∩ Vx] ]\(FU ∪ FV ) = A ∩ B

So B satisfies the base property and so generates a topology on γS.

2. The set S is dense in γS = S∪S(f). See that very open neighbourhood, U∪f←[U ]\F ,of a point in S(f) intersects S.

3. Fact: Suppose f : S → T is a continuous function, where T is compact.

Case 1: If f [S] is compact, S(f) ⊆ f [S].

Case 2: If f [S] is non-compact, f also induces a compactification of f [S]:

fβ(T )[βS] = f [S] ∪ S(f) = clTf [S]

with compactification outgrowth, S(f)\f [S].

Proof : We are given that f : S → T is a continuous function mapping S into thecompact space T . The function f extends to fβ(T ) : βS → T .

By definition of S(f), f [S] ∪ S(f) ⊆ clT f [S] = fβ(T )[βS].

Case 1: Suppose f [S] is compact. Then f [S]∪S(f) ⊆ clT f [S] = f [S], so S(f) ⊆ f [S],as required.

Case 2: Suppose f [S] is non-compact. We claim that fβ(T )[βS] ⊆ f [S] ∪ S(f). Sup-

pose not. Suppose there is a point u ∈ fβ(T )[βS]\S(f)). Then, since u 6∈ S(f), thereis a clTf [S]-open neighbourhood, U , of u such that clSf

←[U ] is a compact subset of S.

Then fβ(T )←(u) ⊆ clSf←[U ] ⊆ S. So u ∈ f [S]. Then fβ(T )[βS] ⊆ f [S] ∪ S(f) as

claimed.

We conclude that fβ(T )[βS] = f [S] ∪ S(f).

So f [S]∪S(f) is a compactification of f [S] with compactification outgrowth S(f)\f [S].


4. Fact: For f : S → T , fβ(T )[βS\S] ⊆ S(f).

Proof : Let fβ(T )(u) ∈ fβ(T )[βS\S] and U be an open neighbourhood of fβ(T )(u) inclT f [S]. Then

u ∈ fβ(T )←[U ] ∩ βS\S 6= ∅

Then clSf←[U ] cannot be compact.1 So fβ(T )(u) ∈ S(f).

We conclude fβ(T )[βS\S] ⊆ S(f).

5. We verify that γS = S ∪ S(f) is indeed compact.

Proof : Given: γS = S ∪ S(f) and f : S → T a continuous function mapping com-pletely regular, S, into the compact space T . Then f extends to fβ(T ) : βS → T .

Define πβ(γ) : βS → γS = S ∪ S(f) as:

πβ(γ)|βS\S = fβ(T )|βS\S and πβ(γ)|S(x) = x

We claim that πβ(γ) : βS → S ∪ S(f) is continuous. Let U ∪ f←[U ]\F be an openneighbourhood of y ∈ S(f) ⊆ γS (where F is compact in S). Then

π←β(γ)[U ∪ f←[U ]\F ] = π←β(γ)[U ] ∪ π←β(γ)[f←[U ]\F ]

= πβ(γ)|←βS\S[U ] ∪ πβ(γ)|←S [f←[U ]\F ]

= fβ(T )|←βS\S[U ] ∪ f←[U ]\F= fβ(T )|←βS\S[U ] ∪ fβ(T )|←S [U ]\F= fβ(T )←[U ]\F

an open subset of βS. So γS = S ∪ S(f) is the continuous image of βS. Then

γS = S ∪ S(f) is compact.

Then S ∪ S(f) is a compactification of S.

6. Fact: For f : S → T , fβ(T )[βS\S] = S(f).

Proof : We have shown above that fβ(T )[βS \S] ⊆ S(f). It suffices to show that

S(f) ⊆ fβ(T )[βS\S]. Let x ∈ S(f). To show that x ∈ fβ(T )[βS\S] it suffices to showfβ(T )(y) = x for some y ∈ βS\S. Then

π←β(γ)(x) = fβ(T )←(x) ∩ βS\S ⊆ βS\S

So, for y ∈ fβ(T )←(x) ∩ βS\S,fβ(T )(y) = x

Sox ∈ fβ(T )[βS\S]

1For if clSf←[U ] is compact and V is an open neighbourhood of u which misses clSf

←[U ] then V ∩fβ(T )←[U ] is an open subset of βS contained in βS\S.


Then S(f) ⊆ fβ(T )[βS\S]. Then

S(f) = fβ(T )[βS\S]

7. Suppose f : S → T is continuous (not necessarily singular) where T is compact and

γS = S ∪ S(f) is the compactification of S induced by f .

Fact: f : S → T extends continuously to fγ : γS → clTf [S].

Proof of fact: Let πβ(γ) : βS → γS be the continuous function defined as above. That

is, πβ(γ)|βS\S = fβ(T )|βS\S and πβ(γ)|S(x) = x.

We define fγ : γS → clT f [S] = fβ(T )[βS] as

fγ(x) = fβ(T )[π←β(γ)(x)]

Then for x ∈ S(f), fγ(x) = fβ(T )[π←β(γ)(x)] = fβ(T )[fβ(T )←(x)] = x.

Then fγ : γS → clT f [S] is easily seen to be continuous on its domain.

We can then say, f extends to fγ : γS → clTf [S] where

fγ [γS] = fγ [S ∪ S(f)]

= fγ [S(f)] ∪ f [S]

= S(f) ∪ f [S]

= clTf [S]

Summary of the above results.

We are given a continuous function f : S → T mapping S into a compact space T(not necessarily a singular map). We obtain a compactification of S,

γS = S ∪ S(f)

(not necessarily a singular compactification). Then, if f [S] is non-compact,

fβ(T )[βS] = f [S] ∪ S(f) = clTf [S]

fγ [γS] = f [S] ∪ S(f) = clTf [S]

fβ(T )[βS\S] = S(f) = fγ [S(f)]

where f [S] ∪ S(f) may be disjoint, but not necessarily. If f is a singular map then

f [S] ∪ S(f) = S(f)

If f [S] is compact, fβ(T )[βS] = f [S] = f [S] ∪ S(f).


22.3 A few examples.

The above definition shows that any continuous function, f : S → T from S into a

compact space, T , can be used to construct a compactification of a locally compactHausdorff space S. We consider a few examples to better visualize how this is done.

Example 2. Consider the space R equipped with the usual topology. The space R

is known to be locally compact non-compact Hausdorff. Consider the continuousfunctions

sin : R → [−1, 1] and cos : R → [−1, 1]

both mapping R into the compact subspace [−1, 1]. Show that sin and cos are bothsingular maps on R and so each induce a singular compactification of R.

Solution : Case sine: See that, for any x ∈ [−1, 1] and open neighbourhood, U of x,in [−1, 1], sin← [U ] is unbounded and so its closure, clR sin← [U ], in R is not compact.

Then x ∈ S(sin), and so [−1, 1] ⊆ S(sin). By definition, S(sin) ⊆ cl[−1,1] sin[R], so

S(sin) = clR[sin [R]] = [−1, 1]

Then sin : R → [−1, 1] is a singular map. We can then use the sine function toconstruct the singular compactification

R ∪sin S(sin) = R ∪sin [−1, 1]

of R with outgrowth [−1, 1].

Case cosine: Proceed similarly to show that R ∪cos S(cos) = R ∪cos [−1, 1] is also asingular compactification of R.

The above example produces two compactifications of R with identical outgrowth. Itmay be tempting to conclude that they are equivalent compactifications. We will latershow (on page 431) that this is not the case.

Example 3. Consider the space R equipped with the usual topology. The space S is

known to be a locally compact non-compact Hausdorff. Let T = [−π/2, π/2]. Showthat R ∪ S(arctan) is not a singular compactification of R. Then find the compactifi-

cation induced by arctan.

Solution : We consider the function, arctan : R → T . We then verify which points

in clT [arctan [R] ] = [−π/2, π/2] belong to the singular set S(arctan). We see thatarctan, pulls back open intervals of the form (a, b) in [−π/2, π/2], to intervals whoseclosure is compact in R. Then arctan[R] = (−π/2, π/2) 6⊆ S(arctan). So arctan is not

a singular map. Thus R ∪ S(arctan) is not a singular compactification of R

We now determine S(arctan). Since the curve of y = arctan (x) is asymptotic to the

horizontal lines y = −π/2 and y = π/2, the function arctan pulls backs open intervalsof the form (a, π/2] and [−π/2, b) to unbounded sets and so the “pull backs” of these


have non-compact closures in R. So S(arctan) = −π/2, π/2. So, even though arctanis not a singular map on R it does induce a “two-point compactification”

R ∪ S(arctan) = R ∪ −π/2, π/2

of R.

Not surprisingly,

arctan[R] ∪ S(arctan) = (−π/2, π/2)∪ −π/2, π/2 = [−π/2, π/2]

is the two-point compactification of arctan[R].

Example 4. Show that βR is not a singular compactification of R.

Solution : We have shown that R ∪ S(arctan) = R ∪ −π/2, π/2 is a compactifi-

cation of R but not a singular one. Since R ∪ S(arctan) βR then, by the exampleon page 425, βR cannot be singular.

In the following example we show that two compactifications of the same set with thesame outgrowth need not be equivalent compactifications.

Example 5. Given the two singular compactifications,

αR = R ∪cos S(cos)

γR = R ∪sin S(sin)

show that, in spite of S(sin) = [−1, 1] = S(cos), αR and γR are not equivalent com-

pactifications.

Solution : Suppose αR and γR are equivalent compactifications. We will show thatthis will lead to a contradiction.

Then, by definition of “equivalent compactifications”, there exists a homeomorphism,

πγ→α : γR → αR

such that for x ∈ R, πγ→α(x) = x.

By definition, the function, πγ→α|γR\R, is a homeomorphism mapping [−1, 1] onto[−1, 1]. This means that πγ→α|γR\R is monotone and maps endpoints to endpoints.

Suppose, without loss of generality, that πγ→α is increasing and so πγ→α|γR\R(−1) =−1. Then πγ→α|γR\R(1) = 1.

If U = (a, 1] ⊆ S(sin) and V = πγ→α|γR\RU ⊆ S(cos) see that can choose a small

enough so that sin←[U ] ∩ cos←[V ]∩ [−π, π] is empty. Then sin←[U ] ∩ cos←[V ] = ∅.


Now cos : R → [−1, 1] extends to cosα : αR → [−1, 1].

Then U ∪ sin←[U ] is open in γR. Then by continuity, the two sets

πγ→α[U ∪ sin←[U ] ] = V ∪ sin←[U ]

cosα←[V ] = V ∪ cos←[V ]

are both open subsets of αR.

So (V ∪ sin←[U ]) ∩ (V ∪ cos←[V ]) = V , a non-empty open subset of αR which is

entirely contained in αR\R = [−1, 1]. Since R is dense in αR = R ∪ [−1, 1], this isimpossible. The source of our contradiction is our supposition that αR and γR areequivalent. We are done.

Example 6. Determine whether f(x) = sin (1/x) with domain R \ 0 and range

T = [−1, 1] = f [R\0] induces a singular compactification. If not determine thenon-singular compactification it induces.

Figure 3: The graph of y = sin (1/x).

Solution : We are given f(x) = sin (1/x) where f : S → T , maps S = R\0 ontothe compact set T = [−1, 1].

Suppose a ∈ [−1, 0)∪ (0, 1] and y ∈M = [−1, a)∪ (−a, 1]. Then, for an open intervalU in M , f←[U ] produces a bounded sequence of open intervals converging to zero.If we take the closure, clSf

←[U ], we obtain a bounded sequence of closed intervals

converging to zero. Since 0 6∈ S then clSf←[U ] is not compact. If y ∈ U = (−a, a)

then f←[U ] is unbounded on both ends of R so clSf←[U ] is not compact. So f : S → T

is a singular map and, for S(f) = [−1, 1],

S ∪f S(f) = R\0 ∪f [−1, 1]

is a singular compactification of R\0 induced by f(x) = sin (1/x).


22.4 More on equivalent singular compactifications.

We will now produce a characterization of pairs of singular compactifications which

are equivalent. But first we must present a few lemmas involving singular sets S(f).

Lemma 22.5 Let f : S → K be a continuous function mapping a locally compactHausdorff space into a compact Hausdorff space, K, and Y = clKf [S]. Then

S(f) = ∩clY f [S\F ] : F is compact in SProof : We are given that f : S → K is a continuous function mapping S into the compact

space K and Y = clKf [S]. In the proof, F will always represent a compact set in S.

Claim #1 : We claim that S(f) ⊆ clY f [S \F ] for all compact F . To prove this, it

suffices to show that Y \clY f [S\F ] ∩ S(f) = ∅.

Let p ∈ Y\clY f [S\F ]. Then there exists an open neighbourhood U of p in Y such that

f←[U ] ⊆ F . Then clSf←[U ] is compact so p 6∈ S(f). Then Y \clY f [S\F ] ∩ S(f) = ∅;

so S(f) ⊆ clY f [S\F ], as claimed. We can deduce that

S(f) ⊆ ∩clY f [S\F ] : F is compact in S

Claim #2: ∩clY f [S\F ] : F is compact in S ⊆ S(f).

Let p ∈ ∩clY f [S\F ] : F is compact in S. Suppose p 6∈ S(f). Then there is an open

neighbourhood U1 of p in Y = clKf [S] such that clSf←[U1] is compact. But

p ∈ ∩clY f [S\F ] : F is compact in S⊆ clY f [S\clSf

←[U1]]

⊆ clY f [S\f←[U1]]

⊆ clY ff←[Y \U1]]

= Y \U

The statement “p ∈ Y \U” contradicts the fact that U is a neighbourhood of p. Con-

sequently, ∩clY f [S\F ] : F is compact in S ⊆ S(f) as claimed.

So S(f) = ∩clY f [S\F ] : F is compact in S. This completes the proof of the lemma.

Lemma 22.6 Let S be a locally compact Hausdorff space and αS be a compactificationof S. Let K be a compact space. If f ∈ Cα(S,K)1 then fα[αS\S] = S(f).

1Where Cα(S,K) is the set of continuous functions mapping S into K which extend to fαC(αS,K).


Proof : If F is compact in S then, since αS\S ⊆ αS\K, αS\S ⊆ clαS(S\F ), and so

fα[αS\S] ⊆ fα[clαS(S\F )] ⊆ clclKf [S]f [S\F ]

Then fα[αS \S] ⊆ ∩clclKf [S]f [S \F ] : F is compact in S. By the previous lemma

S(f) = ∩clKf [S]f [S\F ] : F is compact in S, so

fα[αS\S] ⊆ S(f)

On the other hand, if p 6∈ fα[αS\S] and U is an open neighbourhood of p in K such

that clKU misses fα[αS\S] then clSf←[U ] ⊆ f←[clKU ], a compact subset of S. Hence

clSf←[U ] is compact and so p 6∈ S(f). So

S(f) ⊆ fα[αS\S]

We conclude that, if f extends to fα : αS → K, then fα[αS\S] = S(f). This provesthe lemma.

Lemma 22.7 Let f : S → K be a continuous function in C(S,K) mapping the locally

compact Hausdorff space S into a compact Hausdorff space, K. Suppose f extends tofα(K) ∈ C(αS,K) for some compactification αS such that fα(K) is one-to-one on αS \S.

Then

αS ≡ S ∪ S(f)

Proof : Let S(f) be the singular set of the continuous map f : S → K (not necessarily asingular function). We are given a compactification, αS, such that f extends contin-

uously to fα(K) : αS → K in such a way that fα is one-to-one on αS\S.

By lemma 22.6, fα(K)[αS\S] = S(f).

If γS = S ∪ S(f), we define πα→γ : αS → S ∪ S(f) as follows:

πα→γ(x) =

fα(x) if x ∈ αS\Sx if x ∈ S

Claim : That πα→γ is continuous on αS. It suffices to show that πα→γ pulls back openneighbourhoods in S ∪ S(f) to open sets in αS.

Suppose p ∈ S ∪S(f). Clearly, if p ∈ S and p ∈ U is open in S, then π←α→γ [U ] is open

in αS. If p ∈ S(f) then an open neighbourhood of p is, by definition, of the form[U ∩ S(f) ] ∪ f←[U ]\F where U is an open subset of K and F some compact set in


S. See that,

π←α→γ [ [U ∩ S(f) ] ∪ f←[U ]\F ] = π←α→γ [U ∩ S(f) ] ∪ π←α→γ [ f←[U ]\F ] ]

= π←α→γ [U ] ∩ π←α→γ [S(f)] ∪ f←[U ]\F= (fα←[U ] ∩ fα←[S(f)]) ∪ f←[U ]\F= (fα←[U ∩ S(f)] ∪ f←[U ]\F= fα←[U ]\F= fα←[U ] ∩ S\F

We see that πα→γ pulls back open neighbourhoods of points in S(f) to open sets. Soπα→γ : αS → S ∪ S(f) is continuous, as claimed.

By definition, αS and S ∪ S(f) are equivalent compactifications.

Theorem 22.8 Let S be a completely regular topological space. Suppose f : S → K and

g : S → K are two continuous singular functions mapping S into a compact space K. Sup-pose S(f) and S(g) are homeomorphic. Then the two induced singular compactifications,

αS = S ∪f S(f)

γS = S ∪g S(g)

are equivalent if and only if the singular function f : S → S(f) of αS extends continuouslyto fγ : γS → S(g) such that fγ separates the points of γS\S = S(g).

Proof : We are given that f : S → K and g : S → K are two continuous singular func-tions mapping S into a compact space K inducing the two singular compactifications,

αS = S ∪f S(f) and γS = S ∪g S(g). Also S(f) and S(g) are seen to be homeomor-phic.

( ⇒ ) Suppose αS = S ∪f S(f) and γS = S ∪g S(g) are equivalent.

We are required to show that the singular function f : S → S(f) extends continuouslyto fγ : γS → S(g) such that fγ separates the points of γS\S = S(g).

Recall that fα : αS → αS \S acts as the identity map on αS \S. Also, since αSand γS are equivalent, then there is a continuous map πγ→α : γS → αS such that

πγ→α(x) = x on S and πγ→α maps γS\S homeomorphically onto αS\S.

Let fγ : γS → S(f) be defined as follows:

fγ = fαπγ→α

Then fγ is continuous, fγ |S = f and fγ |S(g) = πγ→α|S(g). This shows that f : S →S(f) extends continuously to the function fγ : S ∪g S(g) → S(f). Since πγ→α is a


homeomorphism on S(g) and fα is the identity function on S(f) then fγ separatespoints of S(g), as required.

( ⇐ ) We are given that both f and g are singular maps on S and αS = S ∪f S(f)and γS = S ∪g S(g). We are also given that the singular function f : S → S(f) ex-

tends continuously to fγ : S ∪gS(g) → S(g) such that fγ separates the points of S(g).

Then by lemma 22.7, S ∪g S(g) is a compactification which is equivalent to S ∪ S(f).

Since f is singular, S ∪ S(f) = S ∪f S(f). So S ∪f S(f) and S ∪g S(g) are equivalent.

Theorem 22.9 Two continuous functions, f : S → K and g : S → K, will be said tobe homeomorphically related if there exists a homeomorphic function h : clKg[S] → clKh[S]

such that h(g(x)) = f(x) for all x ∈ S. Suppose that f : S → K and g : S → K aretwo singular maps such that S(f) = S(g). If f and g are homeomorphically related then

S ∪f S(f) and S ∪g S(g) are equivalent compactifications.

Proof : We are given two continuous functions, f : S → K and g : S → K, mapping S into

the compact space K.

Suppose f and g are homeomorphically related singular maps which, respectively,induce the singular compactifications,

αS = S ∪f S(f)

γS = S ∪g S(g)

where S(f) = S(g). By definition, there exists a homeomorphism h : S(g) → S(f)such that h(g(x)) = f(x).

See that g : S → S(g) extends continuously to gγ : S ∪g S(g) → S(g) where gγ is theidentity map when restricted to S(g). Then hg : S → S(f) extends to

(hg)γ : S ∪g S(g) → S(f)

where (hgγ)|S(g)(x) = h(x). Since (hg)(x) = f(x) on S, f extends to (hg)γ = fγ

where

fγ : S ∪g S(g) → S(f)

and fγ = h on S(g). So fγ separates points of S(g).

By theorem 22.8, S ∪f S(f) and S ∪g S(g) are equivalent, as required.


We can now present an example where this particular concept plays a key role.

Example 7. The two compactifications R ∪sin2 S(sin2) and R ∪cos2 S(cos2) are easilyseen to be singular compactifications with outgrowth [0, 1]. Show that they are equiv-

alent compactifications.

Solution : Consider the function h(x) = 1 − x mapping [0, 1] onto [0, 1]. It is aone-to-one continuous function. Also note that

h(sin2(x)) = 1 − sin2(x) = (1− sin2)(x) = cos2(x)

on [0, 1]. Then sin2 and cos2 are homeomorphically related. By the above theoremR ∪sin2 S(sin2) and R ∪cos2 S(cos2) are equivalent compactifications. This is what we

were required to show.

There can be various ways of showing that . . .

. . . the compactification βN is not a singular compactification.

In the next example we propose one method.

Example 8. Show that βN cannot be a singular compactification.

Solution : Suppose βN is a singular compactification, Then there is a retraction

function r : βN → βN\N which maps βN onto βN\N. We know that βN is separable,but on page 420, we showed that βN\N is not a separable space. By theorem 6.11 we

know that continuous images of separable spaces are separable. So βN\N cannot bethe continuous image of βN. So βN is not a singular compactification.

22.5 What kind of space has only singular compactifications?

We have already shown a compactification “less than” a singular compactification

must be singular. So, if βS is a singular compactification of S, then all compactifi-cations of S are singular. We wonder what class of topological spaces satisfies this

property?

Theorem 22.10 Let S be locally compact, non-compact and Hausdorff. If βS is asingular compactification then S is pseudocompact.

Proof : The proof of the statement is differed to the end of the next section in theorem

23.8 and 23.9.


22.5 Compact spaces as a β-outgrowth, βS\S, for some S.

If we are given a compact Hausdorff space, T , we may wonder whether it is theoutgrowth of some topological space. We can answer is question in the affirmative.

Moreover, we can show that all such spaces will appear as the outgrowth, βS\S, ofat least one space S. We prove this statement now. The proof involves a few notionsinvolving uncountably infinite limit ordinals, ω1, ω2, ω3, . . . , ωn, . . .

Theorem 22.11 Every compact Hausdorff topological space, T , is the β-outgrowth of

some locally compact non-compact Hausdorff topological space, S. That is, βS is equivalentto some compactification, S ∪ T , of S, with βS\S homeomorphic to T .

Proof : Let T be a compact Hausdorff space. Let ωn be a limit ordinal such that n ∈ N

and |ωn| > |T |. Let

S = [0, ωn)× T 1

See that [0, ωn] = ωn + 1 is compact and, since ωn is a limit ordinal, [0, ωn) is dense

in [0, ωn] (see example on page 305). Since both T and [0, ωn] = ωn + 1 are compactthen so is [0, ωn] × T . Also

S = ωn × T = [0, ωn) × T ⊆dense (ωn + 1) × T = [0, ωn] × T

See that ωn + 1 × T is a homeomorphic copy of T .

We claim that βS = [0, ωn] × T . If so, then T can be seen at the β-outgrowth of S.

Proof of claim: Let i : S → [0, ωn]×T denote the identity map which densely embeds

S into the compact set [0, ωn]×T . It is easily verified that ωn +1×T ⊆ S(i). Then,when equipped with the topology described earlier

αS = S ∪ S(i) = [0, ωn] × T

1The limit ordinal (equipped with the interval topology), ωn is the set of all ordinals less than ωn. It isalso expressed as ωn = [0, ωn). For example, ω0 = [0, ω0) = N, is the set of all finite ordinals and [0, ω0] isthe successor, ω0 + 1, of ω0. The first uncountable ordinal, ω1 = [0, ω1), is the set of all countable ordinalsand [0, ω1] is the successor ω1 + 1 of ω1.

If |T | ≥ |ωn| for all n ∈ N, we can always choose a non-limit ordinal, α, such that |ωα| > |T |. Thingswould then follow identically.


is the compactification whose topology is induced by i.

To establish the claim it suffices to show that αS is equivalent to βS. To do this weneed to show that every bounded function, f , in C∗(S) extends to fα on αS.

Let f ∈ C∗(S).

If p ∈ T , see that f |[0,ωn)×p must be constant on some tail end, say

[αp, ωn) × p 2

of [0, ωn) × p.Since |ωn| ≥ |T |, then

|T | = | [αp, ωn) : p ∈ T | < |ωn|

So sup αp : p ∈ T = κ < ωn.

So, for this ordinal, κ, f : S → R is constant on [ κ, ωn)× p for all p ∈ T .

For p ∈ T , say f [ [ κ, ωn)× p] = kp.

For each p ∈ T we set fα(ωn + 1 × p) = kp.

Then fα : αS → R is continuous. We can conclude that every function f in C∗(S)

extends to fα ∈ C(αS). So αS = βS as claimed.

So every compact Hausdorff space T is the β-outgrowth, βS\S, of some locally com-pact non-compact Hausdorff space, S.

Concepts review:

1. Given a continuous function f : S → T from the completely regular set S into the

compact set T , define the singular set S(f).

2. Given a continuous function f : S → T from the completely regular set S into the

compact set T , what does it mean to say that f is a singular map?

3. Given a continuous function f : S → T from the completely regular set S into thecompact set T , define a compactification induced by f .

4. Given a continuous function f : S → T from the completely regular set S into thecompact set T , define a singular compactification induced by f .

5. Show that arctan: R → [−π/2, π/2] is not a singular map. Find a compactification ofR induced by arctan.

2Since | [0, ωn) × p | = |ωn| > |ω1| = |R| ≥ | f [ [0, ωn) × p ] |. Then sup | f←[ f [ [0, ωn) × p ] ] | <|ωn|.


6. Produce an example of a pair of singular compactifications with the same singularsets but which are not equivalent.

7. Produce an example of a Stone-Cech compactification which is not singular.

8. State one way of recognizing a pair of singular compactifications which are equivalent.


23 / On C-embedded subsets of S.

Summary. In this section we define the C-embedded property. We describe a

characterization associated to the existence of a C-embedded subset. We thenpresent a few applications associated to these kind of subsets.

23.1 Definition: C-embedded subsets.

The notion of C∗-embedded subsets was introduced in the last section in the context ofan important characterization of the Stone-Cech compactification. By “C∗-embeddedsubset U of a space S” we mean that all bounded real-valued functions in C∗(U) can

be continuously extended to a function in C∗(S). This allowed us to characterize βSas:

“The compactification αS is the Stone-Cech compactification of S if and

only if S is C∗-embedded in αS”.

We presented a characterization of a C∗-embedded set T in S in the form of theUrysohn extension theorem:

“The set T is C∗-embedded in S ⇔ completely separated sets in T are com-

pletely separated in S”.

From this we deduced that “compact subsets of a completely regular set, S, are C∗-embedded in S” and “closed subsets of a metric space S are C∗-embedded in S”.

We now want to extend our study of C∗-embedded subsets to C-embedded subsets.

Definition 23.1 A subset U of topological space S is C-embedded in S if every continuousfunction f in C(U) can be continuously extended to a function f∗ in C(S).

The notion of a C-embedded subset will play an important role in the study of anotherextension of the space S denoted by, υS, (in the chapter called “realcompact spaces”).

Clearly, it always true that, for any subset U of S, C∗(U) ⊆ C(U). But one may won-

der: Is a C∗-embedded subset, U , in S necessarily C-embedded in S, and vice-versa?To answer such a question we must compare the definitions of each of these two types

of subsets carefully.

442 Section 23: On C-embedded subsets of S

We first verify the easier part of this question. We show that a C-embedded set in S isC∗-embedded in S. Suppose U is C-embedded in S. Consider a function, f ∈ C∗(U).

Then there exists a number, M , such that f [M ] ∈ [−M,M ]. Since U is C-embeddedthen f extends to f∗ ∈ C(S). Define the function g : S → R as

g = (f∗ ∨ −M) ∧M

We see that g is both a continuous and bounded function on S such that g|U = f . So

f in C∗(U) will extend to g ∈ C∗(S). So...

“...whenever U is C-embedded in S then it is also C∗-embedded in S.”

That settles this particular part of the question.

But it is not true in general that a C∗-embedded set U in S is C-embedded in S

(unless certain conditions on U and S are satisfied). That is, there are subsets U ofS, for which every function f in C∗(U) extends to a function f∗ ∈ C∗(S), but not

every function f ∈ C(U) extends to a function f∗ ∈ C(S).

The following theorem shows under what conditions “...if U is C∗-embedded in S then

U is C-embedded in S”.

Theorem 23.2 Suppose U is a non-empty subset of the topological space, S.

a) If U is C-embedded in S then U is C∗-embedded in S

b) Let U be C∗-embedded in S. The set U is also C-embedded in S if and only if U iscompletely separated from any disjoint zero-set in S.

Proof : We are given that U is a subset in S.

a) That “U is C-embedded in S” ⇒ “U is C∗-embedded in S” is shown above.

b) We are given that U is a subset of S.

( ⇐ ) Suppose that U is C∗-embedded in S and is completely separated from any

disjoint zero-set in S.

We are required to show that U is also a C-embedded subset of S.

Let f : U → R be a function in C(U). To prove this direction of the statement, itsuffices to find a function t ∈ C(S) such that t|U = f .

Step #1. We first construct a zero-set, Q, in S which is disjoint from U . Using the

given function, f ∈ C(U), we consider the continuous function, g : U → (−π/2, π/2)defined as

g(x) = (arctan f)(x)


Then |g| : U → [0, π/2) is a bounded continuous function on U . Since U is C∗-embedded in S, |g| extends to |g|∗ : S → R in C∗(S).

The set [π2 ,∞) is a closed Gδ in the normal space, R, so by 10.10, it is a zero-set in

R, say Z(h). We thus obtain the zero-set

Q = Z(h|g|∗) = (h|g|∗)←(0) = |g|∗←[h←(0)] = |g|∗←[π

2,∞)

in S. For the zero-set, Q = Z(h|g|∗), Q ∩ U = ∅, since

x ∈ U ⇒ |g|∗(x) ∈ [0, π/2) ⇒ x 6∈ |g|∗←[π

2,∞)

= Q

So Q is the desired zero-set disjoint from U .

Step #2 . Given a zero-set Q = Z(h|g|∗) in S disjoint from U , we now apply thehypothesis stating that U andQ are completely separated. That is, there is a function,

k : S → [0, 1] such that k[Q] = 0 and k[U ] = 1.Let t(x) = tan(g(x)k(x)), a continuous function in C(S). We claim that t : S → R issuch that t|U = f .

Proof of claim. See that, if x ∈ U ,

t(x) = tan[(gk)(x)]]

= tan[g(x)(1)]

= tan[arctan(f(x))]

= f(x)

So t|U = f , as claimed. Then t : S → R is a continuous extension of f : U → R.

We are done with this direction.

( ⇒ ) Suppose that U is a C∗-embedded set which is also C-embedded in S. Supposethat Z(f) is the zero-set of a function f ∈ C(S) which is disjoint from U . We are

required to show that U and Z(f) are completely separated in S.

Let

h =1

f

on U . Then h ∈ C(U). Since U is C-embedded in S, then h : U → R extends to

h∗ ∈ C(S). Then t = h∗f ∈ C(S) where U ⊆ Z(h∗f − 1) and Z(f) ⊆ Z(h∗f). So Uand Z(f) are completely separated by t, as required.

We are done with part b).


Example 1. Show that every closed subset of R is C-embedded.

Solution: Let F be a closed subset of R.

By the previous example on page 402 F is C∗-embedded in R. Let Z(f) be a zero-set

which does not intersect F . Since R is known to be normal F and Z(f) are completelyseparated (by 10.1). By the theorem above, F is C-embedded in R.

Theorem 23.3 Suppose U is a subspace of a topological space S and let f ∈ C(S). If f

maps U homeomorphically onto a closed subset, f [U ], of R then U is C-embedded in S.

Proof : We are given a function, f , in C(S) which maps U homeomorphically onto a closedsubset, A = f [U ], in R. We are required to show that U is C-embedded in S.

Let h ∈ C(U). It suffices to show that there is a function t ∈ C(S) such that t|U = h.

See that, the function hf |U← : A→ R is continuous on A and that

[hf |U←][A] = h [f |U←[A] ] = h[U ]

in R. So hf |U ∈ C(A). Since A is a closed subset of R it is C-embedded in R (asshown in the example above). This means there is a function g ∈ C(S) such that

g|A = hf |U←. Let t = gf ∈ C(S). Then for x ∈ U

t|U (x) = (gf)|U(x) (By definition of t)

= g[f(x)] ( x ∈ U)

= (hf |U←)(f(x)) ( f(x) ∈ A)

= h(x) (f is a homeomorphism on U)

So t|U = h. Then t ∈ C(S) is a continuous extensison of h ∈ C(U). We conclude Uis C-embedded in S.

23.2 C-embeddings and the pseudocompact property.

One consequence of the previous theorem is the statement proven in the followingexample.

Example 2. Suppose S is a non-pseudocompact space. Show that S contains a C-

embedded copy of N.


Solution : We are given that S is non-pseudocompact and so there is a real-valuedcontinuous function, f ∈ C(S), which is unbounded.

For each i ∈ N, choose mi ∈ [2i−1, 2i+1]∩f [S], if non-empty and not equal to mi−1.Since f is unbounded we can choose infinitely many such mi’s to construct the subset

N = mi : i ∈ N

in f [S] ⊆ R. See that . . .

– N is an unbounded subset of R,

– Each mi is a clopen subset of N

– The set N is a closed subset of R: Since, for any q ∈ R\N , there an open ball,

Bε(q), which misses N .

So N is a closed unbounded copy of N in f [S].

We claim that, N contains a C-embedded copy, A, of N.

Note that f←[N ] ⊆ S. For each i ∈ N, choose ni ∈ f←(mi) in S. If A = ni : i ∈ N,then the function, f |A : A→ N ⊆ f [S], maps A homeomorphically onto the copy, N ,of N in f [S].

Since f [A] = N is a homeomorphic copy of a closed subset of R then, by the theorem23.3, A is C-embedded copy of N in S, as required.

It follows from this statement that . . .

“If S does not contain a C-embedded copy of N then C(S) cannot contain

an unbounded function and so S must be pseudocompact.”

The converse also holds true as we shall show in the form of a theorem.

Theorem 23.4 The space S is pseudocompact if and only if S does not contain a C-embedded copy of N.

Proof : ( ⇐ ) We have shown in the example above that if S does not contain a C-embeddedcopy of N then S must be pseudocompact.

( ⇒ ) Suppose S is pseudocompact. We are required to show that S does not containa C-embedded copy of N. Suppose S contains a C-embedded copy, A, of N. Letg : A→ N ⊆ R be one-to-one and onto N. Then g ∈ C(A) and is unbounded. Since A

is C-embedded in S, there exists a function f ∈ C(S) such that f |A = g. Since suchan f is unbounded in C(S) this contradicts the fact that S is pseudocompact. So S

does not contain a C-embedded copy of N, as required.


We present a few consequences of the previous theorems.

Corollary 23.5 Let S be a locally compact Hausdorff space and N be a C-embedded copyof N in S. Let U be an open neighbourhood of N in S. Then clβS(S\U) ∩ clβSN = ∅.

Proof : We are given that N = ni : i ∈ N is a C-embedded copy of N in S and U is an

open neighbourhood of N in S.

By local compactness, S is open in βS, so we can construct the family, clSVi : i ∈ N,of pairwise disjoint compact sets such that Vi is an open neighbourhood of ni andVi ⊆ clSVi ⊆ U .

Then S\U ⊆ S\clSVi ⊆ S\Vi, for each i. So

S\U ⊆ ∩S\Vi : i ∈ N

Since S is completely regular, for each i, there a function gi ∈ C∗(S) such that

ni ∈ Z(gi − 1)

S\Vi ⊆ Z(gi)

S\U ⊆ ∩Z(gi) : i ∈ N

If Z = ∩Z(gi) : i ∈ N, it is a countable intersection of zero-sets such that, for all i,

ni 6∈ Z. So Z ∩N = ∅. Since Z is itself a zero-set1 then Z = Z(t) for some t ∈ C(S).

SoZ(t) ∩N = ∅

By theorem 23.2, since N is C-embedded in S then N is completely separated fromany zero set disjoint from N .

Then there is a continuous function h : S → [0, 1] such that

N ⊆ Z(h) and Z(t) ⊆ Z(h − 1)

It follows that,

clβSN ⊆ clβSZ(h) = Z(hβ)

clβS(S\U) ⊆ clβSZ(h − 1) = Z(hβ − 1)

Since Z(hβ − 1) ∩ Z(hβ) = ∅ then clβS(S\U) ∩ clβSN = ∅, as required.

1See justification on page 204 where it is shown that a countable intersection of zero-sets is a zero-set.


In the following statement we show that, given a C-embedded copy, N , of N in a

locally compact Hausdorff space, S, its outgrowth, clβSN \N , in βS\S is containedin the βS\S-interior of any zero-set which contains it.1

Corollary 23.6 Let S be a locally compact Hausdorff space and N be a C-embedded copyof N in S. Suppose Z(fβ) is a zero-set in βS, such that (clβSN )\N ⊆ Z(fβ)∩βS\S. Then

(clβSN )\N ⊆ intβS\S(Z(fβ) ∩ βS\S )

Proof : We are given a locally compact space, S, which contains a C-embedded copy,

N = ni : i ∈ N, of N. We are also given that (clβSN )\N ⊆ Z(fβ) ∩ βS\S. We arerequired to show that (clβSN )\N ⊆ intβS\S(Z(fβ) ∩ βS\S ).

Claim #1: There is an open U ⊆ S such that N ⊂ U and clβS(S\U)∩ clβSN = ∅.

Proof of claim: Let p ∈ (clβSN )\N . Since (clβSN )\N ⊆ Z(fβ) ∩ βS\S, fβ(p) = 0and p is a limit point of N . Then there is some subsequence nj(i) : i ∈ N ⊆ N ,

converging to p. Since fβ is continuous on βS,

fβ(nj(i)) : i ∈ N −→ fβ(p) = 0

For each i ∈ N and open ball B1/i+1(fβ(nj(i))) in R, by virtue of continuity of fβ on

βS (and that S is open in βS), there exists a βS-open neighbourhood, Vi, of nj(i),

contained in S such that

fβ(nj(i)) ∈ fβ[Vi] ⊆ fβ[clβSVi] ⊆ B1/i+1(fβ(nj(i)))

We can choose the Vi’s such that clβSVi : i ∈ N forms a family of pairwise disjoint

compact subsets of S. Then we define, U = ∪Vi : i ∈ N, a βS-open neighbourhoodof the C-embedded set N in S.

We can now invoke the corollary 23.5 above, to obtain

clβS(S\U) ∩ clβSN = ∅ (†)

This establishes claim #1. Three facts will now follow.

1This statement expresses a property related to the concept of “p-points”.

“A p-point in a space S, is a point p ∈ S such that p belongs to the interior of any Gδ whichcontains it. It is easily verified that, if S is completely regular and p is a p-point in S, thenp ∈ intSZ for any zero-set Z in Z[S] such that p ∈ Z.”


Fact # 1: Suppose xi ∈ Vi, for each i ∈ N. Then fβ(xi) : i ∈ N must converge tozero.

Proof of this fact: To see this, if xi ∈ Vi, for each i ∈ N then,

|fβ(xi)| = |fβ(xi) − fβ(nj(i)) + fβ(nj(i))|≤ | fβ(xi)− fβ(nj(i))(x) | + |fβ(nj(i))|< 1/(i+ 1) + |fβ(nj(i))|

Since fβ(nj(i)) : i ∈ N −→ 0 then fβ(xi) : i ∈ N −→ 0. This proves fact #1.

Fact #2: There exists a function a function hβ : βS → [0, 1] such that clβSN ⊆Z(hβ − 1) and clβS(S\U) ⊆ Z(hβ).

Proof of this fact : This follows immediately from (†) where it is shown that clβS(S\U)

and clβSN are disjoint compact sets in the normal space βS, and so are completely sep-arated. So a function, hβ : βS → [0, 1], exists such that such that clβS(S\U) ⊆ Z(hβ)

and clβSN ⊆ Z(hβ − 1).

We are done with Fact #2.

Then we have the following chain of containments,

clβSN ⊆ Z(hβ − 1) ⊆ coz(hβ) ⊆ βS\Z(hβ) ⊆ βS\clβS(S\U). (∗)

Fact #3 : βS\S ∩ coz(hβ) ⊆ βS\S ∩ Z(fβ).

Proof of this fact: Let q ∈ βS \S ∩ coz(hβ). To show that q ∈ βS \S ∩ Z(fβ) it

suffices to show that fβ(q) = 0.

By (∗), coz(hβ) ⊆ βS\clβS(S\U).

Then we can write,q ∈ coz(hβ)

⊆ βS\clβS(S\U) (**)

= intβS[ βS\(S\U) ]

= intβS[ βS\(S\∪Vi : i ∈ N) ]

= intβS[∪Vi : i ∈ N ]

⊆ intβS[∪clβSVi : i ∈ N ]

= intβS[∪clSVi : i ∈ N ]

Then q ∈ intβS[∪clSVi : i ∈ N ].

To complete the proof of fact #3 we need to first prove the statement: Any βS-openneighbourhood, Vq, of q must intersect infinitely many of the compact clSVi’s and so

must meet infinitely many Vi’s.


Proof: Suppose K is the union of finitely many clSVi’s. Let N1 = N\K. If Vq∩N1 = ∅

then N1 ⊆ βS\Vq. So clβSN1 ⊆ βS\Vq. Then Vq ∩ clβSN1 = ∅. But, from (**), we

have q ∈ βS\clβS(S\U) ∩ Vq. So Vq ∩ βS\clβS(S\U) is a non-empty open subset ofβS contained in βS\S. Contradiction.

So Vq must intersect infinitely many of the compact clSVi’s and so must meet infinitelymany Vi’s. This completes the proof of the above statement.

This statement shows that, for each j = 1, 2, 3, . . ., the open neighbourhood of q,

fβ←[ ( f(q)− 1/j, f(q) + 1/j ) ] intersects infinitely many Vi’s. Then, for each j, wecan choose uj such that

uj ∈ fβ←[ ( f(q)− 1/j, f(q) + 1/j ) ] ∩ Vj

By fact #1, fβ(uj) : j ∈ N −→ fβ(q) = 0. So q ∈ Z(fβ).

We conclude that, if q ∈ βS\S ∩ coz(hβ) then q ∈ βS\S ∩ Z(fβ).

Then

βS\S ∩ coz(hβ) ⊆ βS\S ∩ Z(fβ)

This completes the proof of the fact #3.

By (*), clβSN ⊆ coz(hβ) so

clβSN \N ⊆ coz(hβ) ∩ βS\S ⊆ Z(fβ) ∩ βS\S

Then clβSN \N ⊆ intβS\S(Z(fβ) ∩ βS\S), as required.

Recall that (from definition 10.17), for a completely regular space, a P -point in S is

a point which belongs to the interior of any Gδ-set that contains it. Also, every spacein which all points are P -points is a P -space. In theorem 10.18, we proved that com-

pletely regular spaces P -spaces are zero-dimensional spaces. We also stated withoutproof that pseudocompact P -spaces are finite. We prove that particular statement

now.

Theorem 23.7 A pseudocompact P -spaces is finite.

Proof : Suppose S is a pseudocompact P -space.

Then every continuous real-valued function on S is bounded.


If S is infinite then we can inductively construct a sequence A = xi : i ∈ N for whichthere is a clopen neighbourhood Bi of xi which does not intersect a clopen neighbour-

hood Bj of xj for j < i. So A is a discrete subset. Also, S\A = ∩S\xi : i ∈ N soS\A is a Gδ of S. Then both S\A and A are clopen in S. Then A is C∗-embedded in S.

Let f : A→ R be the continuous function defined as f(xi) = i.

Suppose Z is a zero-set in S which is disjoint from A. Since zero-sets in a P -space

are clopen then Z and A are completely separated. By theorem 23.2 part b) A isC-embedded in S. So f : A → R extends to f∗ : S → R. But f∗ is unbounded; thiscontradicts pseudocompactness of S. Then S cannot be infinite.

23.3 When is βS a singular compactification?

With the tools presented above we are now able to prove an interesting statementpresented in the last chapter whose proof was differed to this time. We first show that

those spaces S, for which βS is a singular compactification must be pseudocompact. 1

Theorem 23.8 Let S be locally compact, non-compact and Hausdorff. If βS is a singularcompactification then S is pseudocompact.

Proof : We are given that S is locally compact and Hausdorff and that βS is a singularcompactification.

Since βS is singular there exists a continuous function r : βS → βS \S wherer[βS] = r[βS\S] = βS\S and r(x) = x for x ∈ βS\S.

We are required to show that S is pseudocompact.

Suppose S is not pseudocompact.

By theorem 23.4, since the space S is not pseudocompact, it must contain a C-embedded copy, N , of N.

From the example found on page 418, we know that clβSN = βN ⊆ βS.

Since βN is separable, then r[βN ] is separable.1 Since βN \N was shown to be non-

separable (see page 420) r[βN ]\(βN \N ) in βS \S must contain countably infiniteelements say,

T = ti : i ∈ N 2

1Recall that βS is a singular compactification if and only if βS\S is a retract of βS.1By theorem 6.11 the continuous image of a separable set is separable2For if T is finite, then βN \N is separable.


For each i ∈ N, choose mi ∈ r←(ti), to obtain the subset, M , of N , described as,

M = mi : i ∈ N

So r[M ] = T . Note that, since T ⊆ βS \S and M ⊆ N ⊆ S, T and M are disjoint

subsets of βS. Again we have clβSM = βM .

Claim: That (clβS\ST )\T = (clβSM)\M . (This would mean that the disjoint subsets

T and M have a common outgrowth in βS\S.)

Proof of claim: First see we have two representations of r[clβSM ]:

r[clβSM ] = clβS\S(r[M ]) = clβS\ST (Since r(mi) = ti)

r[clβSM ] = r[(clβSM)\M ]∪ r[M ] = (clβSM)\M ∪ T

So clβS\ST = (clβSM)\M ∪ T . (*)

Now

T ⊆ r[βN ]\(βN \N ) ⇒ T ∩ βN \N = ∅

and

M ⊆ N ⇒ clβSM ⊆ clβSN = βN

⇒ clβSM \M ⊆ clβSN \N = βN \N⇒ clβSM \M ⊆ βN \N⇒ T ∩ clβSM \M = ∅ (Since T ∩ βN \N = ∅)

Now clβS\ST = (clβSM)\M ∪ T and T ∩ clβSM \M = ∅ implies

(clβS\ST )\T = (clβSM)\M (†)

This establishes the claim.

We now define the function g : clβS\ST → R as

g(ti) =1

i+ 1g[ (clβS\ST )\T ] = 0

So g(ti) −→ 0. The function g is continuous and bounded on the compact domain,clβS\ST , and so the function (gr) ∈ C∗(M) where (gr)(mi) = g(ti) = 1/(i+1). Then

(gr) extends to (gr)∗ ∈ C∗(S).3 So (gr) extends continuously to (gr)β ∈ C(βS).

So

(clβSM)\M = (clβS\ST )\T ⊆ Z((gr)β) ∩ βS\S (By †)

3Every function is continuous on N so M is C-embedded in N . Now N was declared to be C-embeddedin S so M is C-embedded in S. Then M is C∗-embedded in S.


By corollary 23.6,

(clβSM)\M ⊆ intβS\S[Z((gr)β) ∩ βS\S ]

Since, by (†), (clβS\ST )\T = (clβSM)\M ,

(clβS\ST )\T ⊆ intβS\S[Z((gr)β) ∩ βS\S ]

But this is impossible since it implies there exists infinitely many g(mi) = ti terms

such that ti = 1/(i+ 1) ∈ Z((gr)β).

The contradiction is the result of our assumption that S is not pseudocompact.So S is pseudocompact. as required. This completes the proof of the theorem.

Theorem 23.9 Let S be locally compact, non-compact and Hausdorff and Cs(S) be set

of all f ∈ C(S) such that f is a singular map. Then the following are equivalent:

a) The space βS is a singular compactification of S.

b) The space S is pseudocompact and every f ∈ C(S) is equivalent to a singular function

fs ∈ C(S). (That is, fβ − fβs is constant on βS\S.)

c) The set Cs(βS\S) = C(βS\S) and

eCs(S)[S] = eβCs(S)[βS\S]

Proof : We are given that S is locally compact and Hausdorff.

( a ⇒ b ) Suppose βS is a singular compactification. Let Cs(S) denote the set of all

singular functions in C(S).

We have already shown in theorem 23.8 that S must then be pseudocompact.

Claim: For every f ∈ C(S), f is equivalent to some singular function in Cs(S).

Proof of claim: Since βS is a singular compactification, then every compactification

of S is a singular compactification (see example on page 425). So, if αS is a compact-ification, there exists a retraction r : αS → αS\S.1

Let f ∈ Cα(S). Then f extends to fα : αS → R. Also, fα|αS\S : αS\S → R, extendsto the function

fα|αS\S r : αS → fα[αS\S]

1Where r[αS] = clαSr[S] = αS\S and r(x) = x on αS\S


Let fs ∈ C(S) such that fαs = fα|αS\S r on αS. Then fs[S] = fα|αS\S[ r[S] ] ⊆

fαs [αS\S] = S(fs).

So fs ∈ Cs(S). Since

fαs [αS\S] = fα[ r[αS\S] ] = fα[αS\S]

then (fα − fαs )[αS\S] = 0, so f and fs are equivalent functions.

This establishes the claim.

We are done with the proof of a ⇒ b.

( b ⇒ c ) For this direction, we are given that S is pseudocompact and every f ∈ C(S)

is equivalent to a singular function fs ∈ C(S).

Since S is pseudocompact then, for each f ∈ C(S), f : S → R is bounded and so

fβ [βS] = fβ[clβSS] = clRf [S] ⊆ R

Claim #1: For every f ∈ C(S), f [S] is closed in R.

Proof of claim. Suppose q ∈ fβ [βS\S]. Then Z(fβ − q) ⊆ βS. Since S is pseudocom-

pact, then Z(fβ − q) ∩ S 6= ∅ 1, so there is a point x ∈ S such that f(x) = q.So q ∈ f [S]. Then fβ[βS\S] ⊆ f [S]; hence

f [S] ⊆ fβ[βS]

= clRf [S]

⊆ f [S]

Then fβ [βS] = f [S]. This implies that, for any f ∈ C(S), f [S] is closed in R. We are

done with claim #1.

Note that, since fs ∈ C(S), then fs[S] is closed in R.

Furthermore, since fs[S] is dense in S(fs) (by definition of singular map) and fs[S] is

compact thenfs[S] = S(fs) = fβ

s [βS\S]

By hypothesis, for f ∈ C(S), there is a singular map, f∗s ∈ Cs(S), such that

f − f∗s = fω , for some fω in the set, Cω(S). 1 That is, for every f ∈ C(S) there issome f∗s ∈ Cs(S) and k ∈ R such that

(f − f∗s )β|βS\S[βS\S] = k

Then, if we let fs = f∗s + k, then fs ∈ Cs(S)1 and so f and fs are two functions in

C(S) such that fβ and fβs agree on βS\S.

1By theorem 21.141Where Cω(S) is the set of all functions g ∈ C(S) such that gβ is constant on βS\S.1f∗s [S] ⊆ fβ

s [βS\S] = S(f∗s ) ⇒ (f∗s + k)[S] ⊆ (fβs + k)[βS\S] = S(f∗s + k)


This means that, for every function g ∈ C(S), the function, gβ|βS\S, extends to a

function gβs ∈ Cs(βS) where gs ∈ Cs(S). So

C(βS\S) = Cs(βS\S)

Then Cs(βS\S) separates the points of βS\S. Then

eβCs(S) : βS →∏

f∈Cs(S) Rf is one-to-one on βS\S

Claim #2: We claim eCs(S) : S →∏

fs∈Cs(S) Rf is a singular map.

Proof of claim: To prove the claim it suffices to show that

eCs(S)[S] ⊆ eβCs(S)[βS\S] = S(eCs(S))

Suppose p = eCs(S)[S].p = eCs(S)(x) For some x ∈ S

= <fs(x)>f∈C(S)

∈ <fs[S]>f∈C(S)

= <fβs [βS\S]>f∈C(S)

= eβCs(S)[βS\S]

= S(eCs(S))

So eCs(S)[S] = eβCs(S)

[βS\S] = S(eCs(S)).

We conclude eCs(S) is a singular map. We are done with claim #2.

We are done with b ⇒ c.

( c ⇒ a) We are given that C(βS\S) = Cs(βS\S) and eCs(S)[S] = eβCs(S)[βS\S] =

S(eCs(S)) where eCs(S) is a homeomorphism.

We are required to show that βS is a singular compactification.

Since eβCs(S) : βS →∏

f∈Cs(S) Rf is one-to-one on βS\S we have

eβCs(S)

← eβ

Cs(S)[S] = eβ

Cs(S)

←[

eβCs(S)

[βS\S]]

= βS\S

andeβCs(S)

← eβ

Cs(S)[βS\S] = i[βS\S] = βS\S

So eβCs(S)

← eβ

Cs(S): βS → βS\S is a retraction on βS to the retract βS\S.

We conclude βS is a singular compactification as claimed.

This proves the direction c ⇒ a.


23.4 Another characterization of pseudocompactness.

A space, S, was originally said to satisfy the “pseudocompact property” if and only ifevery continuous real-valued function on S is bounded. Any compact Hausdorff space

is easily seen to be pseudocompact. But a pseudocompact space need not be com-pact: We showed, for example, that any countably compact spaces is pseudocompact

(15.9). For example, the ordinal space, S = [0, ω1), was shown to be pseudocompactbut not compact. We have also seen that every sequentially compact space satisfies

the pseudocompact property (17.4).

We saw other ways of describing those spaces which are pseudocompact.

− In theorem 21.14, we saw that pseudocompact spaces are precisely those spaces

for which zero-sets, Z, in Z[βS] cannot be entirely contained in βS \S. Thatis, for any zero-set, Z ∈ Z[βS], Z ∩ S 6= ∅. Equivalently, for any continuous,

f : S → ωR,

fβ(ω)[βS\S] ⊆ f [S]

This is interesting, since it shows that we can characterize a property of S interms of a property of its β-outgrowth. By itself, this can motivate the study of

compactifications of S.

− Also, a space S is pseudocompact if and only if S does not contain a C-embeddedcopy of N (23.4). 1

We also have the following results associated to the pseudocompact property.

− Let S be locally compact, non-compact and Hausdorff. If βS is a singular com-

pactification then S is pseudocompact (23.8).2

We add to the list of characterizations for the pseudocompactness property, the fol-lowing statement.

Theorem 23.10 Let S be a locally compact Hausdorff space. Then the following state-ments are equivalent:

a) The space S is pseudocompact.

1We will add later to this list the following characterization: Pseudocompact spaces, S, are preciselythose spaces whose realcompactification, υS, is βS (24.8).

2We will add later: A Hausdorff topological space, S, is compact if and only if S is both realcompact andpseudocompact (24.7).


b) The space S does not admit an infinite family, U , of open sets which satisfies theproperty “ every point has an open neighbourhood which will intersect at most finitely

many elements of U ”. Note.1

Proof : We are given S is a locally compact Hausdorff space. Then S is completely regular(by 18.8).

( a ⇒ b ) Suppose S is pseudocompact. Suppose S admits an infinite family, U , ofopen sets such that every point has an open neighbourhood which will intersect at

most finitely many elements of U .

− Let U1 ∈ U . Then there is a point p1 and open V1 such that p1 ∈ V1 ⊆ U1 whereV1 intersects at most finitely many elements of U .

− There exists U2 ∈ U such that U2 ∩ V1 = ∅. Then there is a point p2 and openV2 such that p2 ∈ V2 ⊆ U2 where V2 intersects at most finitely many elements ofU .

− There exists U3 ∈ U such that U3 ∩ (V1 ∪ V2) = ∅. Then there is a point p3

and open V3 such that p3 ∈ V3 ⊆ U3 where V3 intersects at most finitely many

elements of U ....

− There exists Un ∈ U such that Un ∩ (V1 ∪ · · ·∪Vn−1) = ∅. Then there is a pointpn and open Vn such that pn ∈ Vn ⊆ Un where Vn intersects at most finitely

many elements of U .

We thus obtain the sequences pi : i = 1, 2, 3, . . . and Vi : i = 1, 2, 3, . . . wherepi ∈ Vi and the Vi’s are open and pairwise disjoint.

Then, for each i ∈ N\0, there exists a continuous fi : S → [0, i] such that fi(pi) = iand fi[S\Vi] = 0. Since the Vi’s are pairwise disjoint then the function g : S → R

defined as,

g(x) =∞∑

i=1

fi(x)

is continuous and unbounded on S. This contradicts our hypothesis. So, if S is pseu-

docompact, it cannot admit an infinite family, U , of open sets such that every pointhas an open neighbourhood which will intersect at most finitely many elements of U .

This concludes a ⇒ b.

( b ⇒ a) Suppose S does not admit an infinite family, U , of open sets such that every

point has an open neighbourhood which will intersect at most finitely many elementsof U . We are required to show that S is pseudocompact.

Suppose not. Then S contains a C-embedded copy, N = ni : i ∈ N, of N. Letg : N → N be a function defined as g(ni) = i. Then g is continuous on N . Since N

1When U satisfies this property we say that U is ”locally finite”.


is C-embedded, g extends to a continuous function, f : S → R, where f |N = g. LetVi : i ∈ N be a pairwise disjoint family of open sets such that f(ni) ∈ Vi. Then

U = f←[Vi] : i ∈ N forms a pairwise disjoint family of open sets in S such thatni ∈ f←[Vi].

We claim that every point of S has an open neighbourhood which intersects at most

finitely many elements of U (thus contradicting our hypothesis).

Proof of claim: Let p ∈ S, q = f(p) and B = B1/2(q). Then f←[B] is an openneighbourhood of p in S which intersects at most two elements of U . Then U isan infinite family of open subsets of S such every point has an open neighbourhood

which intersects at most finitely many points. Since this contradicts our hypothesis,S must be pseudocompact. This concludes b ⇒ a.

23.5 The Stone-Cech compactification of a product: The Glicksberg theorem.1

Suppose we are given two completely regular spaces S and T and we consider theStone-Cech compactification of their corresponding product space, S × T . By Ty-

chonoff theorem, βS×βT is a compactification of S ×T . Under what conditions is itequivalent to β(S × T )? We investigate ths question below.

In the proof of the following theorem we invoke in various steps the statement 18.8,

23.2, 21.14 and 23.5.

Theorem 23.11 Let S and T be locally compact Hausdorff infinite spaces. If S × T is

C∗-embedded in βS × βT , then S × T must be pseudocompact.

Proof : We are given that S and T are locally compact Hausdorff infinite spaces such S×Tis C∗-embedded in βS × βT .

Notation. We start by introducing some notation. Let

X = S × T

A = S × βT \TB = βS\S × T

C = βS\S × βT \T (compact)

Let αX = X ∪ A ∪ B ∪ C where B ∪ C and A ∪ C are compact and where αX =βS × βT ≡ βX .

1Refers to Irving Glicksberg. Careful not to confuse with other theorems sometimes referred to asGlicksberg’s theorem and Glicksberg existence theorem.


For a ∈ S and b ∈ T , we set

Ta = a × T

Sb = S × b

βTa = a × βT

βSb = βS × b

Claim: We first show that β(X ∪A ∪ B) ≡ αX .

Proof of claim: SinceβX ≡ αX

f ∈ C∗(X ∪ A ∪ B) then f |X ∈ C∗(X), so f |X extends to (f |X)α ∈ C(αX) =C(X ∪ A ∪B ∪ C). So

X ∪A ∪ B is C∗-embedded in αX (∗)

Then clβXX = β(X ∪ A ∪B) ≡ αX , as claimed.

We prove the theorem statement in two steps: In step one we prove that X ∪ A ∪ Bis pseudocompact. Once the statement of step one is established we show in step twothat X is pseudocompact (which is the required property).

STEP 1. We will begin by showing that X ∪A ∪B is pseudocompact.

Suppose X ∪A∪B is not pseudocompact. Since αX is equivalent β(X ∪A∪B), then

(by theorem 21.14) C contains a zero-set Z(f), where f ∈ C∗(αX).

We know that a zero-set is a Gδ. Then

Z(f) = ∩Bi : i ∈ N

where Bi is open in αX

Since X is dense in αX , for each i ∈ N, we can choose (ni, mi) ∈ Bi ∩X .

Then the setN = (ni, mi) : i ∈ N

in X has a limit point say (u, v), in Z(f) ⊆ C. Then N has a subsequence N ∗ =

(nij , mij) : j ∈ N which converges to

(u, v) ∈ βS\S × βT \T

Then π1 : N ∗ → nij : j ∈ N converges to π1(u, v) = u ∈ βS\S and

π2 : N ∗ → mij : j ∈ N converges to π2(u, v) = v ∈ βT \T .


See that, for each j, clαXTnij= βTnij

. Since limj→∞ nij = u, for v ∈ βT , (nij , v) :

j ∈ N converges to (u, v) ∈ clαXTu = βTu ⊆ B ∪C.

So

(u, v) ∈ clαXN∗ ∩ clαXTu (†)

Let U = αX\(B∪C), an open subset of αX . Then N ∗ ⊆ U and Tu ⊆ (X ∪A∪B)\U .

By corollary 23.5,

clβX [(X ∪ A ∪B)\U ] ∩ clβXN∗ = ∅

But Tu ⊆ (X ∪A ∪ B)\U .

So

clαXN∗ ∩ clαXTu ⊆ clαXN

∗ ∩ clαX [(X ∪A ∪ B)\U ] = ∅

This contradicts (†). The source of the contradiction is the supposition thatX∪A∪Bis not pseudocompact.

So X ∪ A ∪B is pseudocompact, as claimed. This completes Step 1.

STEP 2. In this part we conclude the proof by showing that X is pseudocompact.

We are given that αX and βX are equivalent compactifications and so X is C∗-

embedded in αX . Also we know (from step 1) that X ∪A ∪ B is pseudocompact.

Suppose X is not pseudocompact. Then the space X contains a copy, N , of N whichis C-embedded in X . Then, by theorem 23.2 part a), since N is C-embedded, N is

C∗-embedded in X . Since X is C∗-embedded in αX , N is C∗-embedded in X ∪A∪B.

Claim: We claim that N is C-embedded in X∪A∪B. (If so, this contradictsX∪A∪Bis pseudocompact). To prove this claim it suffices to show that N is completely sep-arated from any zero-set in X ∪A ∪B and invoke theorem 23.2 part b).

Proof of claim: Suppose Z(f) is a zero-set in X ∪ A ∪ B which is disjoint from N .

Since f ∈ C∗(X ∪ A ∪ B), then f |X ∈ C∗(X), so Z(f |X) is a zero-set in X disjointfrom N . Since N is C-embedded in X , by theorem 23.2, N and Z(f |X) are com-

pletely separated in X . This means that there is a continuous function g : X → [0, 1]such that N ⊆ Z(g) and Z(f |X) ⊆ Z(g − 1). Since X is C∗-embedded in αX , g

extends to gα ∈ C(αX). Then N ⊆ Z(gα|X∪A∪B) and Z(f) = clX∪A∪BZ(f |X) ⊆clX∪A∪BZ(g − 1) = Z(gα|X∪A∪B − 1). So N and Z(f) are completely separated in

X ∪A ∪B. By theorem 23.2, N is C-embedded in X ∪A ∪ B, as claimed.

But, since X ∪A∪B is pseudocompact, this is impossible. So to avoid the contradic-tion we can only assume that X is pseudocompact.

We are done with the proof.


We have a simple example illustrating the above result.

Example 3. Since R × R not pseudocompact we can invoke the above theorem todeduce that β(R × R) 6≡ βR × βR.

In the following theorem we show that pseudocompactness of S × T is sufficient toguaranty that β(S × T ) ≡ βS × βT .

Theorem 23.12 If S and T are completely regular infinite spaces such that S × T ispseudocompact then S × T is C∗-embedded in βS × βT .

Proof : Given: That S and T are completely regular infinite spaces such that S × T is

pseudocompact.

Notation. We start by setting up some notation.

Let X = S × T and αX = βS × βT (viewed as a compactification of X). Let

A = S × βT \TB = βS\S × T

C = βS\S × βT \T (compact)

FACT 1. We verify that clαXA = A∪C: Note that clαXA ⊆ A ∪C = βS × βT \T .Let x ∈ C. Then x ∈ βS \S × k for some k ∈ βT \T . Then x is a limit point of

S × k ⊆ A and hence is a limit point of A. So x ∈ clαXA. Then A ∪ C ⊆ clαXA.From which we can conclude that clαXA = A ∪C. This establishes the Fact 1.

By applying similar arguments clαXB = B ∪C.

We let αX = X ∪A∪B∪C, a union of pairwise disjoint subsets of αX which densely

contains X . So clαXX = αX . Let πβ→α : βX → αX be the projection map whichfixes the points of X . Note that,

βX = π←β→α[X ∪A ∪B ∪C] = X ∪ π←β→α[A] ∪ π←β→α[B] ∪ π←β→α[C]

a union of pairwise disjoint sets.

We begin the proof with the following claim.

Claim #1. Given that X is pseudocompact, we claim that the subspace, X ∪ A ispseudocompact.

Proof of claim #1: Suppose not. Then we can find a copy, N = ni : i ∈ N, of

N where N is C-embedded in X ∪ A. Since X is pseudocompact, without loss ofgenerality, we can assume N ⊆ A.1

1For, if N ⊆ X then, the function f(ni) = i (on N) extends to f∗ : X ∪ A → R, so f∗[N ] is unboundedwhen f∗ is restricted to X . By hypothesis, this cannot be.


Let f ∈ C(N ) be defined as f(ni) = i. Since N is C-embedded in X ∪ A, then fextends to f∗ ∈ C(X ∪ A). For some k > 0, let h represent the unbounded nowhere

zero function, h = |f∗| ∨ k in C(X ∪A) and let

g = 1/h in C∗(X ∪A)

whereg(ni) = 1/h(ni) = 1/[ |f∗|(ni) ∨ k ]

and Z(g) ∩ (X ∪A) = ∅.

The function, g|X , in C∗(X) extends continuously to g|βX ∈ C(βX) mapping π←β→α[A]

into R whereg|βX[π←β→α(ni)] = g(ni) = 1/h(ni)

Since X is pseudocompact (by hypothesis), every non-empty zero-set in Z[βX ] inter-

sects X . Then, for each i, Z(g|βX − g(ni)) ∩X 6= ∅. Let

mi ∈ Z(g|βX − g(ni)) ∩X

Since g(ni) : i ∈ N −→ 0, then the sequence, g(mi) : i ∈ N converges to 0. Thereis then some y ∈ βX\X where 0 = gβ(y). So y ∈ Z(gβ), a zero-set in Z[βX ] which is

entirely contained in βX\X . This contradicts that X is pseudocompact. The sourceof our contradiction is our assumption that X ∪A is not pseudocompact.

So, given that X is pseudocompact, X ∪A must also be pseudocompact. This estab-

lishes Claim #1.

To show that X is C∗-embedded in αX , it suffices to show that C∗(X) ⊆ Cα(X).

We will show that C∗(X) ⊆ CX∪A(X)∪ CX∪B(X) ⊆ Cα(X).

The following corollary combines the two above theorems 1.

Corollary 23.13 Glicksberg theorem. The completely regular space, S × T , is C∗-embedded in βS × βT if and only if one of the spaces S or T is finite or S × T is pseudo-

compact.

1The statement is attributed to Irving Glicksberg. His proof was published in Transactions of the Amer-ican Mathematical Society Vol. 90, No. 3 (Mar., 1959), pp. 369-382 (14 pages)


23.6 Tietze’s Extension theorem.

The proof of the following theorem invokes the Urysohn lemma in an important wayto deduce that closed subspaces of a normal space are C-embedded.

Theorem 23.14 Let S be a normal topological space which contains a non-empty closedsubset, A. If f : A → [a, b] is a continuous real-valued function then f extends to acontinuous function which maps all of S into [a, b].

Proof : We are given that S is normal and contains a non-empty closed subset A.

Fact #1. If f is a continuous function which maps A into I = [−k, k] there is afunction g ∈ C∗(S) such that g[S] ⊆

[

−13k,

13k]

and

| g|A(x)− f(x) | ≤ 23k

for x ∈ A. That is, there is a function g ∈ C(S) which “approximates” f on A.

Proof: Subdivide the interval I into

I1 = [−k,−13k], I2 = [−1

3k,13k], I3 = [ 13k, k].

Then f←[I1] and f←[I3] are disjoint closed subsets of S (since f is continuous on Aand A is closed in S). We can then invoke the Urysohn lemma to justify the existence

of a function g : S →[

−13k,

13k]

such that g[ f←[I1] ] = −13k and g[ f←[I3] ] = 1

3k.Then g[S] ⊆

[

−13k,

13k]

and |g|A(x) − f(x)| ≤ 23k. This establishes Fact #1.

Without loss of generality, suppose f : A → [−1, 1] is continuous. We apply the gen-eral principle illustrated in Fact #1 replacing k with 1:

We then know of the existence of a continuous function g1 ∈ C(S):

g1[S] ⊆[

−13 ,

13

]

, where | f(x) − (g1)|A(x) | ≤ 23 on A

Applying the same principle, replacing f with f − (g1)|A, we know of the existence ofa continuous function g2 ∈ C(S):

g2[S] ⊆[

−13

(

23

)

, 13

(

23

)]

where |f(x)− (g1)|A(x)− (g2)|A(x)| ≤ [2/3]2 on A

We can inductively construct a sequence of continuous functions g1, g2, g3, . . . , gn, . . .where, if given g1, g2, . . . , gn ⊆ C(S) where,

|gn(x)| ≤ 13

(

23

)n−1and |f(x)− (g1)|A(x) − · · · − (gn)|A(x)| ≤ [2/3]n on A


there exists a continuous function gn+1 ∈ C(S) such that

|gn+1(x)| ≤ 13

(

23

)nand |f(x)− (g1)|A(x) − · · · − (gn+1)|A(x)| ≤ [2/3]n+1 on A

We can then consider the series

g(x) =

∞∑

n=1

gn(x)

on S. This series converges uniformly to g(x), 1 so g is continuous on S.

Claim. We claim that g|A = f .

Proof of claim. To see this simply witness the inequality,

| f(x)− (g1)|A(x)− · · · − (gn+1)|A(x) | ≤ [2/3]n+1

on A. Taking the limit as n goes to ∞ on each side produces 0. Then the sequenceof partial sums ∑m

n=1 gn(x) : m ∈ N converges uniformly to f(x) at each x ∈ A.

We can conclude that, if S is normal, a continuous, f ∈ C(A), mapping a closedsubset, A, into [−1, 1] extends to a continuous function, g : S → [−1, 1].

Corollary 23.15 Let S be a normal topological space which contains a non-empty closedsubset, A. If f : A→ R is a continuous real-valued function then f extends to a continuous

function h : S → R, on all of S.

Proof : We are given that S is normal and contains a non-empty closed subset A.

We know that the open interval (−1, 1) is homeomorphic to R, so we will more simplyconsider a continuous function f : A → (−1, 1) mapping A into (−1, 1). Then we

will show that it extends to a continuous function h : S → (−1, 1) mapping S inside(−1, 1).

The theorem guarantees that f extends to a function g : S → [−1, 1] mapping S into[−1, 1]. But we need a continuous function h : S → (−1, 1) mapping S into (−1, 1)

such that h|A = f . We will use the function g : S → [−1, 1] to construct h.

We define the subset, D, of S as,

D = g←[−1] ∪ g←[1]

We see that D is a closed subset of S which entirely misses the closed subset, A (since

f [A] ⊆ (−1, 1)).

1By comparison with a geometric series and the WeierstrassM -test. Note that, since 13

P∞n=1

23

nconverges

to 1 the range of g is contained in [−1, 1].


Since D ∩ A = ∅, by the Urysohn lemma there exists a continuous function k : S →[0, 1] such that

D ⊆ Z(k) and A ⊆ Z(k − 1)

We let

h(x) = k(x)g(x)

where h is seen to be continuous on S. Since

A ⊆ Z(k − 1) ⇒ h[A] = k[A]f [A] = 1 · f [A] (So h = f on A)

D ⊆ Z(k) ⇒ h[D] = k[D]g[D] = 0 · g[D] = 0a 6∈ D ⇒ |g(a)|< 1

So h[S] ⊆ (−1, 1). We are done.

Concepts review:

1. Define a C-embedded subset of a space S.

2. Describe a property which characterizes those C∗-embedded which are C-embedded.

3. Show that closed subsets of R are C-embedded.

4. State a characterization of pseudocompactness involving a copy N of N.

5. Give a characterization of those spaces S such that βS is singular.

6. State a characterization of pseudocompactness involving a locally finite family of sets.

7. State the Glicksberg theorem.

8. State Tietze’s theorem.


24 / Realcompact spaces

Summary. In this section we introduce the set of all points in βS called the

“real points” of a space S. The set of all real points of S is denoted by υS.Given a space S, the extension of S, υS, is referred to as the Hewitt-Nachbinrealcompactification of S. If υS = S then S is said to be realcompact. We pro-

vide a characterization as well as a few examples of realcompact spaces.

24.1 Realcompact space: Definitions and characterizations.

Suppose S is a locally compact Hausdorff topological space and f : S → R is a function

in the set, C(S), of all real-valued continuous functions (including unbounded ones)on S. Let i : R → ωR be the inclusion function (identity map) which embeds R into

the one-point compactification,

ωR = R ∪ ∞

defined as, i(x) = x (where ∞ represents a point not in R). We can combine the

two functions f and i to define the function f+ = if , where f+ : S → ωR mapsS into the compact set ωR. In this case, we have practically identical functions, f+and f , except that the codomain of f+ is ωR. We have shown in theorem 21.5 that

any continuous function, g : S → K, mapping a locally compact Hausdorff space, S,into a compact space K extends to a function, gβ(K) : βS → K. We will denote the

extension of f+ : S → ωR to βS as

fβ(ω) : βS → ωR

That is, if f ∈ C(S) (possibly unbounded), f extends to fβ(ω) ∈ C(βS, ωR).

In the case where f is bounded, the extension fβ(ω) : βS → ωR maps the compactspace βS onto the compact space

fβ(ω)[βS] = fβ(ω)[clβSS] = clωRf [S] = clRf [S]

a compact set in R ⊂ ωR. So we could write, C(βS) ⊆ C(βS, ωR). So in the case ofa bounded function, f , fβ(ω) : βS → ωR and the usual, fβ : βS → R, both represent

the same function.

Of course, if f : S → R is unbounded,

fβ(ω)[βS] = fβ(ω)[βS\S ∪ S]

= fβ(ω)[βS\S] ∪ f [S]

⊆ fβ(ω)[βS\S] ∪ R

466 Section 24: Realcompact spaces

where f [S] is non-compact in R (since f [S] is unbounded). So fβ(ω) must map atleast one point in βS\S to ∞. We stress at least one point, since fβ(ω)←(∞) need not

be a singleton set in βS. We then have the disjoint union,

βS = fβ(ω)←[R] ∪ fβ(ω)←(∞)

So, if C(S) has some unbounded functions,

C(βS, ωR) 6⊆ C(βS)

Real points in βS.

We will want to distinguish those points in βS which belong to fβ(ω)←(∞) from thosewhich belong to fβ(ω)←[R]. For a given f ∈ C(S) we then define υfS

1 as,

υfS = βS\fβ(ω)←(∞) = fβ(ω)←[R]

Clearly, S ⊆ υfS ⊆ βS, for all f .

For a bounded function f ∈ C∗(S),

υfS = fβ(ω)←[R]

= fβ(ω)←[clRf [S]]

= fβ←[clRf [S]]

= βS

So, in the particular case, f ∈ C∗(S)

υfS = βS

The function, f : S → R, in C(S) is one which continuously extends to a real-valued

function, fβ(ω) : υfS → R in C(υfS). The points in υfS are commonly referred to asbeing the set of all real points of f (in the sense that fβ(ω)(x) has a real number value

at each of point x in υfS). To consider the “real points” associated to all functions,f ∈ C(S), we define.

υS = ∩υfS : f ∈ C(S) = ∩fβ(ω)←[R] : f ∈ C(S)

Note that every function, f ∈ C(S), extends continuously to a real-valued function,

fβ(ω)|υS, on υS. More specifically, each function in C(S) is associated to a uniquefunction in C(υS).

With these facts in mind, we can now formally define the following related concepts.

1The Greek letter, υ, is pronounced upsilon.


Definition 24.1 Let S be locally compact and Hausdorff.

a) If f ∈ C(S), the function fβ(ω) : S → ωR represents the extension of f to βS withrange ωR. We define the subset, υfS, of βS as

υfS = fβ(ω)←[R]

If p ∈ βS\υfS, then fβ(ω)(p) = ∞. The points in υfS are referred to as real points off . We define, the subspace, υS of βS, as

υS = ∩υfS : f ∈ C(S) = ∩fβ(ω)←[R] : f ∈ C(S)

b) The points p ∈ υS ⊆ βS are referred to as real points of S. If p in βS is not “a realpoint of S” then there is at least one function f ∈ C(S) such that fβ(ω)(p) = ∞.

c) If the space S is such that S = υS, then we refer to S as being a realcompact space.

Based on the above definition, for any locally compact Hausdorff space S,

S ⊆ υS ⊆ βS

Furthermore, if f is any function in C(S), f can be viewed as belonging to C(S, ωR)

and so extends to a continuous function fβ(ω)|υS : υS → ωR. In fact,

f ∈ C(S) ⇒ fβ(ω)|υS ∈ C(υS)

We will more succinctly denote the function fβ(ω)|υS by

fυ = fβ(ω)|υS

We will require the following concept for our first characterization of a “realcompactspace”. It refers to a slightly stronger condition on z-ultrafilters.

Definition 24.2 Let S be locally compact and Hausdorff. A

real z-ultrafilter

Z , in Z[S] is a z-ultrafilter which is closed under countable intersections. That is, countablesubfamilies of Z have non-empty intersection.1

1In some books the author may denote this property by the anagram CIP for “countable intersectionproperty”.


Clearly, not all z-ultrafilters are “real”. But if a family of zero-sets satisfies the CIP

it satisfies the FIP and so all “real z-ultrafilters” are z-ultrafilters, albeit special ones.

The following characterizations of realcompact spaces will make it easier to recognize

those spaces which are realcompact.

In what follows, if hβ(ω) : βS → ωR then Z(hβ(ω)) = hβ(ω)←(0).

Theorem 24.3 Let S be a locally compact Hausdorff space. Then the following three

statements are equivalent.

a) The space S is realcompact. That is, S = υS.

b) For any p ∈ βS \ S, there is a function, h ∈ C∗(S), such that p ∈ Z(hβ) ⊆ βS\S.

c) Every real z-ultrafilter in Z[S] converges to a point p which belongs to S.

Proof : We are given that S is a locally compact Hausdorff space.

( a ⇒ b ) Suppose S is realcompact. Then βS\υS = βS\S. Let p ∈ βS\S. We arerequired to show that p ∈ Z(hβ) ⊆ βS\S for some h ∈ C∗(S).

By definition, S = υS. Since p ∈ βS\υS then p is not a real point of S, in the sense

that, for some function, say g ∈ C(S), gβ(ω)(p) = ∞. Then g is unbounded on S (for,if g was bounded on S, gβ(ω)[βS] = clRg[S] ⊆ R).

Let f = g2 ∨ 1. Then f : S → [1,∞) is continuous and unbounded on S and so

extends to fβ(ω) : βS → ωR.

Define h : S → R as

h(x) = (1/f)(x) = 1/f(x)

Then h ∈ C∗(S) such that h[S] ⊆ (0, 1].

The function h : S → (0, 1] then extends continuously to hβ : βS → R where

hβ [βS] = clRh[S] ⊆ clR(0, 1] = [0, 1]

Since p ∈ βS\S there is sequence (net) xi : i ∈ N in S converging to p. Then

hβ(xi) : i ∈ N −→ hβ

(

limxi→p

xi

)

= hβ(p) 6∈ (0, 1]

(by continuity of hβ). So hβ(p) = 0.


So p ∈ Z(hβ) ⊆ βS\S. Then, if S is realcompact, for every point p in βS\S,

p ∈ Z(hβ) ⊆ βS\S

for some hβ , as required.

( b ⇒ a ) We are given that every point in βS\S belongs to a zero-set entirely containedin βS\S. We are required to show that S is realcompact. Let p ∈ βS\S. It sufficesto show that p is not a real point. To say that p is not a real point means that there

is a function h ∈ C(S) such that hβ(ω)(p) = ∞. By hypothesis, p ∈ Z(fβ) ⊆ βS\Sfor some f ∈ C(S). Then f(x) 6= 0 on S.

We can then define h : S → ωR as a function as h = 1/f . Then h extends tohβ(ω) : βS → ωR. There exists a sequence (net), xi : i ∈ N, in S which converges to

p. Then f(xi) : i ∈ N approaches zero as xi approaches p. By continuity of hβ(ω),

hβ(ω)(p) = hβ(ω)

(

limxi→p

xi)

= limxi→p

h(xi)

= limxi→p

1/f(xi)

= ∞ ∈ ωR

So p is not a real point. So real points can only belong to S. So S = υS.

( b ⇒ c ) We are given that every point in βS\S belongs to a zero-set entirely con-

tained in βS\S. We are required to show that limits of real z-ultrafilters all belongto S. If p ∈ βS \S then, by hypothesis, p ∈ Z(hβ) ⊆ βS \S, for some h ∈ C∗(S).

Since p 6∈ S it corresponds to a free z-ultrafilter, Z , in Z[S]. It suffices to prove that,Z is not a “real” z-ultrafilter. To do this we will show that Z does not satisfy the

“countable intersection property”.

See that,

Z(hβ) = ∩hβ←[−1/n, 1/n] : n ∈ N\0where each hβ←[−1/n, 1/n] is a zero-set.1 So Z(hβ) is a countable intersection of zero

sets. Then, by the paragraph titled “On zero-sets and z-ultrafilters in βS” on page406, we can write, hβ←[−1/n, 1/n] = clβSh

←[−1/n, 1/n] so

Z(hβ) = ∩clβSh←[−1/n, 1/n] : n ∈ N\0

Since ∩h←[−1/n, 1/n] : n ∈ N\0 ⊆ βS \S, it is empty in S. By definition, Zdoes not satisfy the countable intersection property and so it is not real. So there are

1Recall that βS is a normal space. By theorem 10.10, “In a normal space, a closed set F is a Gδ if andonly if F is a zero-set”. See that hβ←[ [−a, b] ] = ∩hβ←[ (a−1/n, b+1/n) ] is a Gδ. Then hβ←[−1/n, 1/n]is a zero-set, for each n.


no free real z-ultrafilters in Z[S]. Then real z-ultrafilters in Z[S] must be fixed, asrequired.

( c ⇒ b ) We are given that every real z-ultrafilter on S is fixed. Let p ∈ βS\S. We

are required to produce a zero-set Z containing p where Z ⊂ βS\S.

Suppose Z β = Z(fβ) : f ∈ M ⊆ C(S) is the unique z-ultrafilter in Z[βS] whichconverges to p. By hypothesis, the corresponding free z-ultrafilter, Z = Z(f) : f ∈M ⊆ C(S), in Z[S] is not real. Then, by definition of “real” z-ultrafilter, Z contains

a countable subfamily Z(fn) : fn ∈ D ⊆M such that ∩Z(fn) : fn ∈ D ⊆M = ∅.

So p belongs to W = ∩Z(fβn ) : fn ∈ D ⊆M ⊆ βS\S.

Since W is a countable intersection of zero-sets, then as explained on page 204, W is

a zero-set.

So βS\S contains a zero-set W , as required.

We add a few remarks.

1) Compare statement b)

“. . .S = υS if and only if, for any p ∈ βS \S = βS\υS, there is a function,

h ∈ C∗(S), such that p ∈ Z(hβ) ⊆ βS\S.”

above to the theorem statement in 21.14 which states:

“.... . .S is pseudocompact if and only if no zero set Z in Z[βS] is entirely

contained in βS\S.”

When viewed together, see what this implies.2

2) The statement above also says that S = υS if and only if every real z-ultrafilter

in Z[S] is fixed (that is, converges in S = υS). By definition of real z-ultrafilter, thismeans that a space S is realcompact if and only if every z-ultrafilter in Z[S] which

satisfies the countable intersection property must converge must in S = υS.3

Further on in the book, we will be presenting still another characterization of real-

compact spaces (in corollary 25.6).

Example 1. Show that R equipped with the usual topology is a realcompact space.

2In the first case, the zero-set will be contained in the region bounded by S = υS and βS. It may be that,in the second case, the region bounded by υS and βS is empty, or we may be referring to the non-emptyregion bounded by S and υS = βS. Can a realcompact space be pseudocompact?

3We will confirm later that real z-ultrafilters are precisely those z-ultrafilters which converge to real pointsin υS.


Solution : Consider the function f : R → R defined as, f(x) = 1/(1 + |x|). Thefunction f is continuous, bounded and nowhere zero on R, with range, (0, 1]. Recall

from theorem 10.10 that . . .

“. . . , closed Gδ’s in a normal space are zero-sets.”

Then, since R is normal, for each n ∈ N\0, fβ←[0, 1/n] is a zero-set neighbourhood

of βR\R. ThenβR\R = ∩fβ←[0, 1/n] : n > 0

Being a countable intersection of Gδ’s in βR, βR\R is a βR zero-set disjoint from R.

By theorem 24.3, R is realcompact.1

24.2 Extending functions to υS.

Let S and T both be locally compact and Hausdorff. Then, being completely regular,S and T can be densely embedded into βS and βT , respectively.

Suppose f : S → T is a continuous function which maps S onto T . By theorem21.5, the function f : S → βT extends to fβ(βT ) : βS → βT . We know that υS is a

subspace of βS, hence the restriction, fβ(βT )|υS : υS → βT , maps υS into βT .

In the following theorem we will show a little bit more. We show that fυ[υS] ⊆ υT .

That is, if f : S → T is any continuous function mapping S to T , (both locallycompact and Hausdorff) f will extend to a continuous function fυ which maps υS

into υT .

Theorem 24.4 If S and T are locally compact Hausdorff spaces and f : S → T is a

continuous function then f extends continuously to a unique continuous function

fυ : υS → υT

Proof : We are given that S and T are locally compact Hausdorff spaces and f : S → T is

a continuous function. Since f extends to fβ : βS → βT and S ⊆ υS ⊆ βS, then fextends to fυ : υS → βT . We are required to show that fυ[υS] ⊆ υT .

Suppose h ∈ C(T ). Then, if ωR = R ∪ ∞,

h : T → R will continuously extend to hω : βT → ωR

Furthermore,

(hf) : S → R will continuously extend to (hf)ω : βS → ωR

1In theorem 25.7 of the next chapter where better tools are available, we show that any product of R’sis realcompact.


For x ∈ S, (hf)ω|S(x) = h(f(x)) = (hωfβ)|S(x). Since S is dense in βS, then

(hf)ω : βS → ωR

(hωfβ) : βS → ωR

are equal functions on βS.

Let r ∈ υS. By definition, r is a real point of the function hf : S → ωR. This means

that (hf)ω(r) = hω[fβ(r)] 6= ∞. Then fβ(r) is a real point of h. That is,

fυ(r) ∈ υhT

But h was an arbitrarily chosen function in C(T ). The point r was also arbitrarily

chosen in υS. We can then conclude that

fυ [υS] ⊆ ∩υhT : h ∈ C(T ) = υT

So fυ[υS] ⊆ υT . This is what we were required to prove.

Corollary 24.5 Suppose S and T are locally compact Hausdorff spaces. If T is real-compact and f : S → T is a continuous function then f extends continuously to a unique

continuous function fυ : υS → T . In particular, if f ∈ C(S), f extends continuously tofυ ∈ C(υS).

Proof : This is a special case of the theorem where T = υT . The second part follows from

the fact that R is realcompact as shown in the above example.

From the above corollary, we see that, if S is completely regular, every continuous

function f : S → R in C(S) extends to fυ : υS → R. We know that S is C∗-embeddedin βS, but now we can confidently say that

“S is C-embedded in υS”

The function

φ : C(S) → C(υS)

defined as φ(f) = fυ is then an isomorphism from the ring C(S) onto C(υS).

We can then generalize a statement enunciated earlier:

“Every real z-ultrafilter in Z[S] converges to a real point in υS.”

Equivalently,


“Every z-ultrafilter in Z(υS) which satisfies the countable intersection property has anon-empty intersection in υS.”

24.3 The Hewitt-Nachbin realcompactification of a space S.

We have seen that S ⊆ υS ⊆ βS. We have also seen that, for every function f ∈ C(S),the function, fβ(ω)|υS, continuously maps υS into R. That is, every function f ∈ C(S)

extends to a function, fυ , in C(υS).

We also make the following observation. Since υS ⊆ βS then clβSυS ⊆ βS. Giventhat clβSS ⊆ clβSυS, then

β(υS) = βS

Theorem 24.6 Suppose S is locally compact and Hausdorff. Then υS is a realcompact

subspace of βS.

Proof : To show that υS is realcompact, it suffices to show that υ(υS) = υS. To do this,it suffices to show that the set of all real points of υS is υS.

First note that υS ⊆ υ(υS). Then it suffices to show that υ(υS) ⊆ υS. Suppose

p ∈ υ(υS). Then p is a real point of υS. Let f ∈ C(S). Then there is g ∈ C(υS) suchthat g = fυ. Since p is a real point of υS then gβ(ω)(p) ∈ R. See that, since g and fυ

agree on υS, gβ(ω) = fβ(ω) on βS. So fβ(ω)(p) ∈ R. So p is a real point of S. That is,p ∈ υS. We have shown that υ(υS) ⊆ υS.

So

υ(υS) = υS

Then we can view υS as a realcompact space which densely contains the locally com-pact Hausdorff space S.

Given a space S, the topological space, υS, is normally referred to as the

Hewitt-Nachbin realcompactification of S.1

So, instead of referring to υS as being “the set of all real points of S”, we will now

speak of υS as being “the realcompactification of S”. If a space, S, is realcompactthen it is its own realcompactification.

1Named after the American mathematician, Edwin Hewitt (1920-1999) and the Jewish-Brazilian mathe-matician Leopoldo Nachbin (1922-1993).


If S is pseudocompact then C(S) = C∗(S) and so every point of βS is a real pointof S. Hence, if S is pseudocompact, βS is both the Stone-Cech compactification and

the Hewitt-Nachbin realcompactification of S.

24.4 Another characterization of compactness and of pseudocompactness.

The realcompact property provides us with another characterization of the compactproperty.

Theorem 24.7 A Hausdorff topological space, S, is compact if and only if S is both

realcompact and pseudocompact.

Proof : ( ⇒ ) If S is compact then S = βS; hence every real-valued function on S is

bounded. Then, by definition, S is pseudocompact. Since S ⊆ υS ⊆ βS, thenS = υS, so S is realcompact.

( ⇐ ) Suppose S is both pseudocompact and realcompact. We are required to showthat S is compact.

Since S is pseudocompact every f ∈ C(S) is bounded. Then, for such an f ,

fβ(ω)[βS] ⊆ clRf [S] ⊆ R

ThenβS ⊆ ∩fβ(ω)←[R] : f ∈ C(S) = υS

So υS = βS. Since S is realcompact then S = υS. So S = βS. So S is compact, asrequired.

Theorem 24.8 Let S be a non-compact locally compact Hausdorff space. Then the

following are equivalent.

a) The space S is pseudocompact.

b) The equality, βS = υS, holds true.

c) Every z-ultrafilter is a real z-ultrafilter.

Proof : We are given that S is a non-compact locally compact Hausdorff space.

( a ⇒ b ) We are given that, S is pseudocompact.

We are required to show that βS = υS.


Then, for any f ∈ C(S), fβ(ω) : βS → ωR maps βS onto clωRf [S] ⊆ clRf [S] ⊆ R.Then, for any f ∈ C(S), we have υfS = βS. So

βS = ∩υfS : f ∈ C(S) = υS

as required.

( b ⇒ a ) We are given that, βS = υS.

We are required to prove that S is pseudocompact.

Suppose not. Then there is an unbounded function, f , in C(S). Then f : S → R

extends to fβ(ω) : βS → ωR. If fβ(ω)[βS] ⊆ R then fβ(ω)[βS] is non-compact sinceit is unbounded. So there is an x ∈ βS\S such that fβ(ω)(x) = ∞. In such a case,

βS 6= υS. So S must be pseudocompact.

( b ⇔ c ) Given that S is a non-compact locally compact Hausdorff space.

See that βS = υS if and only if every point in βS is a real point. A z-ultrafilterconverges to a real point if and only if it is a real z-ultrafilter. So βS = υS if and

only if every z-ultrafilter is a real z-ultrafilter. 1

24.5 A few examples of realcompact spaces.

In theorem 16.3, we showed that in any Lindelof space, S, every filter of closed sets

in P(S) which satisfies the “countable intersection property” has a non-empty inter-section (in S). In fact, this property characterizes Lindelof spaces. It follows that,

in such spaces, every real z-ultrafilter in Z[S] has a non-empty intersection. That is,every real z-ultrafilter converges to a point in S. Then S = υS. Then, by theorem

24.3, Lindelof spaces are realcompact. We state this as a formal result.

Theorem 24.9 If S is a Lindelof space then S is realcompact.

Proof : Given in the paragraph above.

1We can also conclude that “...if and only if every z-ultrafilter satisfies the countable intersection prop-erty”.


Example 2. By definition 18.9, a σ-compact space is the countable union of compactsets. State conditions under which a σ-compact space is guaranteed to be realcom-

pact.

Solution : In theorem 18.10 it is shown that in a locally compact Hausdorff space the

σ-compact property is equivalent to the Lindelof property. So any locally compactHausdorff σ-compact space is realcompact.

Example 3. Show that N and Q are realcompact.

Solution : Any countable space is the union of singleton sets and hence is σ-compact.

Since both N and Q are locally compact Hausdorff countable spaces they are realcom-pact.

Example 4. Show that any separable metric space is realcompact.

Solution : By definition, a space that has a countable open base is Lindelof. By the-

orem 16.3, for metric spaces, the following are equivalent: 1) the second countableproperty, 2) the separable property and 3) the Lindelof property. By theorem 24.9, a

separable metric space is realcompact.

Example 5. Show that every subspace of Rn equipped with the usual topology is

realcompact.

Solution : The Euclidean space Rn is a metric space which is a finite product of the

separable space R and hence is separable. Then it is second countable. So every oneof its subspaces is a second countable metric space. Then every one of its subspaces

is Lindelof. By theorem 24.9, every one of its subspaces is realcompact.

24.6 Properties of realcompact spaces.

We now examine when a particular space inherits the realcompact property fromother spaces known to be realcompact. We show, in particular, that the realcompactproperty is closed under arbitrary intersections and arbitrary products. We also show

that a closed subspace inherits the realcompact property from its superset.

Theorem 24.10 Let Sα : α ∈ I be a family of non-empty completely regular realcom-pact spaces.


a) If the Sα’s are all subspaces of a realcompact space S and T = ∩Sα : α ∈ I isnon-empty, then T is a realcompact subspace of S.

b) The product space,∏

α∈I Sα, is realcompact.

Proof : We are given that Sα : α ∈ I is a family of completely regular non-empty

realcompact spaces.

a) Suppose the Sα’s are all subspaces of a realcompact space S and T = ∩Sα : α ∈I is non-empty.

Then, for each α, Sα = υSα. Since subspaces of completely regular spaces are

completely regular, T = ∩Sα : α ∈ I is a completely regular subspace. Asshown in an example on page 418, βT ⊆ βSα. Since υT ⊆ βT , υT ⊆ βSα.

We claim that υT ⊆ υSα. For each α, let iα : υT → βSα denote the identity map

embedding the subspace υT into βSα. By theorem 24.4, for each α, iα|T : T → Sextends continuously to the function

(iα|T )υ : υT → υSα = Sα

Since T is dense in υT , (iα|T )υ = iα on υT , so (iα|T )υ embeds υT into υSα,

establishing the claim.

Since (iα|T )υ[υT ] = υT ⊆ υSα = Sα for all α ∈ I , then υT ⊆ ∩Sα : α ∈ I = T .

Given that T ⊆ υT , we can only conclude that T = υT and so T is realcompact.

b) Suppose S =∏

α∈I Sα where each Sα is realcompact. For γ ∈ I , the projection

map, πγ : S → Sγ is continuous and hence extends continuously to

πγυ : υS → υSγ

where Sγ = υSγ.

Let G = πα : α ∈ I where πα : S → Sα. Then the evaluation map,eG : S → ∏

α∈I Sα extends to eGυ : υS → ∏

α∈I Sα = S a function which

embeds υS into S.

Then υS ⊆ S. We conclude that S = υS and so S is realcompact.

The subspace, T , of a realcompact space, S, need not be realcompact unless T is

closed in S.

Theorem 24.11 If F is a closed non-empty subspace of a realcompact space S then Fis realcompact.


Proof : We are given that F is closed in realcompact S. Then S = υS and clυSF = clSF =F . We are required to show that F = υF .

Let i : F → S denote the inclusion map embedding F into S. Then i : F → F ⊆ S

extends continuously to iυ : υF → υS where

iυ[υF ] = υF ⊆ υS = S

Now F is dense in υF hence

υF ⊆ clυSF = clSF = F

Then υF ⊆ F . So F = υF .

24.7 Example of a non-realcompact space.

The above few examples may lead the reader to suspect that non-realcompact spaces

are uncommon. This motivates us to construct such a space. We do so in the followingexample.

Example 6. Recall that ω1 denotes the first uncountable ordinal while ω0 denotesthe first countable infinite ordinal. Let X = [0, ω1) and Y = [0, ω0) both be equipped

with the ordinal topology, and T = X × Y be the corresponding product space.1

In the example on page 305, it is shown that X = [0, ω1) is countably compact. Bytheorem 15.9, since X is countably compact, for any f ∈ C(X), f [X ] is compact in R.

This implies that [0, ω1) is pseudocompact. But the space X is non-compact since theopen cover [0, γ) : γ ∈ S has no finite subcover. By theorem 24.7, pseudocompactspaces which are realcompact must be compact. So

[0, ω1) is not realcompact

But the space X , when viewed as a homeomorphic image of the subspace, [0, ω1)×0,of the product, T , is closed in T . By theorem 24.11, closed subspaces of realcompactspaces are realcompact. So,

...the product space, [0, ω1) × [0, ω0), cannot be realcompact

1In theorem 21.18, it is shown that βX = [0, ω1] = ωX .


Concepts review.

1. If f is a function in C(S) define set, υfS, of all the real points of f .

2. Define υS in terms of all the sets υfS.

3. What does it mean to say that the space, S, is realcompact?

4. Define a real z-ultrafilter in Z[S].

5. Give a characterization of the realcompactness property in terms of zero-sets in Z[βS].

6. Give a characterization of the realcompactness property in terms of z-ultrafilters ofzero-sets in Z[S].

7. If f : S → T is a continuous function mapping S into T what can be said about aparticular continuous extension of f?

8. What does it mean to say that f ∈ C(S) is C-embedded in υS?

9. What can we say about a space that is both realcompact and pseudocompact?

10. What can we say about Lindelof spaces in the context discussed in this chapter?

11. What can we say about arbitrary intersections of realcompact space?

12. What can we say about arbitrary products of realcompact space?

13. What kind of subsets of a realcompact space are guaranteed to be realcompact?

14. Give an example of a realcompact space.

15. Give an example of a space which is not realcompact.

480 Section 25: Perfect functions

25 / Perfect functions

Summary. In this section we introduce the notion of a “perfect function”. Afterproviding a formal definition we derive two of its most well-known characteri-

zations. Our brief discussion of perfect functions will involve notions discussedin our study of singular functions and of realcompact spaces.

25.1 Introduction.

Our study of realcompact spaces started with a brief discussion of a continuous func-

tion, f : S → R mapping a completely regular space, S, into R. So f could be seen asa map, f+ : S → ωR, continuously mapping S into the one-point compactification de-

noted as, ωR = R∪∞, where f+ = f on S. By theorem 21.5, f+ : S → ωR extendscontinuously to a function, fβ(ω) : βS → ωR. Even if it was clear that fβ(ω)[S] ⊆ R,

there was no reason to assume that, in the case where f is unbounded, fβ(ω) wouldmap all of βS \ S onto ∞. In fact, fβ(ω) could still map some points, x, in βS \ Sto fβ(ω)(x) ∈ R.

We defined υfS as

υfS = x ∈ βS : fβ(ω)(x) ∈ R = fβ(ω)←[R]

The points in υfS were referred to as the “real points of f”. When υfS was not

entirely contained in S this was viewed as a property of S rather than as a propertyof f . The set

υS = ∩υfS : f ∈ C(S)was called the set of “all real points of S”. When υS ∩ βS \S = ∅, we referred toS as satisfying the “realcompact” property and referred to S as being a realcompactspace.

In this section we will generalize the procedure used to construct υfS and study thosefunctions f : S → T such that f+ : S → αT , for any compactification αT of T (instead

of only for ωR).

25.2 Perfect function: Definition.

Recall that, for a function, f : S → T , the fibres of f refers to the elements of the set

f←(y) : y ∈ f [S] ⊆ P(S)


Definition 25.1 Let S and T be completely regular spaces and f : S → T be a contin-uous function mapping S into T . We say that f is a compact mapping if its fibres are all

compact (subsets of S). That is, f←(y) is compact for each y in the range of f .

We say that the continuous function, f : S → T , is a perfect function if f is both a closedfunction and a compact mapping.1

In our introductory paragraph, we considered a continuous function f : S → R to con-struct the function fβ(ω) : βS → ωR. Here we will start with a function, f : S → T ,

to construct, by using a similar procedure, a function fβ(α) : βS → αT .

Suppose f : S → T continuously maps S into T (both completely regular). Leti : T → αT be the identity map embedding T into αT . Let f+ : S → αT be defined

as

f+(x) = i(f(x))

Then, by theorem 21.5, f+ : S → αT extends continuously to

fβ(α) : βS → αT

With this in mind we will describe a perfect function, f : S → T by referring to a

property of, fβ(α) : βS → αT .

Theorem 25.2 Let f : S → T be a continuous function mapping a non-compact com-pletely regular space, S, into a non-compact completely regular space T . Let αT be any

Hausdorff compactification of T . Then the following are equivalent:

a) The function f : S → T is perfect.

b) If f : S → T , and fβ(α) : βS → αT is its continuous extension to βS, then

fβ(α)[βS\S] ⊆ clαTf [S]\f [S]

c) If f : S → T , and fβ(α) : βS → αT is its continuous extension to βS, then

fβ(α)[βS\S]∩ f [S] = ∅

1Some authors may not require that perfect functions be continuous.


Proof : ( a ⇒ b ) We are given that f : S → T is perfect, αT is any Hausdorff compactifi-cation of T and fβ(α) : βS → αT is the continuous extension of f to βS.

We are required to show that fβ(α)[βS\S] ⊆ clαTf [S]\f [S].

Suppose fβ(α)[βS\S] 6⊆ clαTf [S]\f [S]. Then there is a u ∈ βS\S such that fβ(α)(u) ∈f [S]. Let

M = S ∪ uso that S is a dense subspace of M . Then, given that f : S → T is perfect,

f←(fβ(α)(u)) is a compact subset of S.

Let K = f←(fβ(α)(u)). Since K is compact in S then K is a compact subset of M

which does not contain u.

Then there is an open neighbourhood, U , of M such that K ⊆ U and u 6∈ clMU . Now

u ∈ clM (S\U).1 We then have,

f(u) ∈ fβ(α)[clM [S\U ]]

⊆ clTf [S\U ] (By continuity and theorem 6.4)

= f [S\U ] (Since f is a closed function.)

Then, there is a point t in S\U such that f(u) = f(t). Then t ∈ f←(fβ(α)(u)) = K.

So t ∈ K ∩ S\U , contradicting K ⊆ U .

So fβ(α)[βS\S] ⊆ clαTf [S]\f [S], as required. We are done with ( a ⇒ b ).

( b ⇒ a ) We are given that f : S → T is a continuous function which extends tofβ(α) : βS → fβ(α)[βS] = clαTf [S] ⊆ αT such that

fβ(α)[βS\S] ⊆ clαTf [S]\f [S] (†)

We are required to show that f is both a compact and closed function.

Step #1. The function f : S → T is compact : Let y ∈ f [S] ⊆ T . Then fβ(α)←(y) isclosed in βS.

Case a) : fβ(α)←(y) ⊆ S. Then it is a closed subset of βS so it is a compact subset of S.

Case b) : fβ(α)←(y) ∩ βS\S 6∈ ∅. Then, by hypothesis,

fβ(α)[

[ fβ(α)←(y)] ∩ βS\S]

= y ⊆ clαTf [S]\f [S]

Since y ∈ f [S] (by hypothesis), this case cannot occur. So fβ(α)←(y) ⊆ S. Then

f←(y) is compact in S. So f : S → T is a compact function. This concludes step #1.

1Since u ∈ clMS = clM (S\U ∪ U) = clM (S\U) ∪ clMU ; given that u 6∈ clMU , then u ∈ clM (S\U).


Step #2. The function f : S → T is a closed function : Let F be a closed subset of S.See that clβSF is compact in βS and so fβ(α)[clβSF ] ∩ T is a closed subset of T .

We claim that fβ(α)[clβSF ] ∩ T = f [F ].

fβ(α)[clβSF ] ∩ T = fβ(α) [ (clβSF ∩ βS\S) ∪ F ] ∩ T=

[

fβ(α)[ (clβSF ∩ βS\S) ] ∪ fβ(α)[F ]]

∩ T

=[

fβ(α)[ (clβSF ∩ βS\S) ] ∩ T]

∪[

fβ(α)[F ] ∩ T]

⊆ [ [clαTf [S]\f [S] ] ∩ T ] ∪[

fβ(α)[F ] ∩ T]

(By (†)).

= ∅ ∪ [ f [F ] ∩ T ] (By our hypothesis)

= f [F ]

So fβ(α)[clβSF ] ∩ T ⊆ f [F ]. Since f [F ] ⊆ fβ(α)[clβSF ] ∩ T , then

fβ(α)[clβSF ] ∩ T = f [F ]

a closed subset of T , as claimed.

So f is a closed map.

By definition, f is a perfect function. We are done with ( b ⇒ a ).

( b ⇔ c ) Proof is left as an exercise.

Example 1. If αS γS the function πγ→αγS → αS is a perfect function.

To see this first note that πβ→α is clearly closed and compact and so is perfect. Notethat πβ→α = πγ→απβ→γ . Then πβ→απ

←β→γ = πγ→α, is easily verified to be closed

and compact and so is perfect.

25.3 Relating a perfect function, f to its singular set, S(f).

Recall (from page 429) that, if we are given a continuous function f : S → T mapping

S into a compact space T (not necessarily a singular map), we obtain a compactifica-tion of S,

γS = S ∪ S(f)

(not necessarily a singular compactification). We showed that,

fβ(T )[βS] = f [S] ∪ S(f) = clTf [S]

fγ [γS] = f [S] ∪ S(f) = clTf [S]

fβ(T )[βS\S] = S(f) = fγ [S(f)]


where f [S] and S(f) may be disjoint, but not necessarily.

In the case where f is a singular map (that is, f [S] is dense in S(f)) then

f [S] ∪ S(f) = S(f)

So should f be singular and f [S] be a compact subset of T ,

S(f) = f [S]

In the case where f is not singular f [S] will intersect T \S(f). It may even happen

that f [S] ∩ S(f) is empty. Whether f is singular or not (and f [S] is not compact),the set

S(f)\f [S]

is the outgrowth of the compactification, clTf [S], of the image, f [S], of S under f .

We now show that the function f : S → T is perfect depending, specifically, on howits range, f [S], intersects its singular set, S(f).

Theorem 25.3 A characterization of perfect functions. Let S and T be non-compact

locally compact Hausdorff spaces and f : S → T be a continuous function mapping S intoT . Then f is perfect if and only if

f [S] ∩ S(f) = ∅

Proof : We are given that f : S → T is a continuous function mapping S into T , where S

and T are both locally compact, non-compact and Hausdorff. Let αT be a compacti-fication of T so that f maps S onto the subset, f [S], of αT .

By theorem 21.5 and the fact stated on page 427, the function f : S → αT , extends

tofβ(α) : βS → clαTf [S] = f [S] ∪ S(f) ⊆ αT

where fβ(α)[βS\S] = S(f). Let

γS = S ∪ S(f)

As shown on page 429, the function f : S → αT , extends to

fγ(α) : S ∪ S(f) → f [S] ∪ S(f) = clαTf [S] = fβ(α)[βS]

where fγ(α)[S(f)] = S(f) appears in the role of the identity map on S(f).

( ⇒ ) Suppose f is perfect. Then, by theorem 25.2,

fβ(α)[βS\S] ⊆ fβ(α)[βS]\f [S] = [ f [S]∪ S(f) ]\f [S]


Since fβ(α)[βS\S] = S(f) and [ f [S]∪ S(f) ] \ f [S] = S(f)\f [S] we obtain

S(f) ⊆ S(f)\f [S] which implies f [S] ∪ S(f) = ∅

as required.

( ⇐ ) Conversely, suppose f [S] ∩ S(f) = ∅.

fβ(α)[βS\S] = S(f)

= S(f)\f [S] (Since f [S] ∩ S(f) = ∅)

⊆ clαTf [S]\f [S]

This implies fβ(α)[βS\S]∩ f [S] = ∅.

Hence f is perfect, as required.

So intersection of S(f) with f [S] helps recognize both perfect functions and singularfunctions. Since a function f : S → T is singular when f [S] ⊆ S(f) a singular func-

tion can never be perfect. A perfect function is one such that f [S] ∩ S(f) is empty.So perfect functions and singular functions are at opposite ends of a “spectrum of

functions” each characterized by the contents of the set f [S] ∩ S(f).

Example 2. Suppose S is non-compact completely regular and connected. Describethe range, f [S], if the bounded continuous function, f : S → R is to be a perfect

function.

Solution : We are given that S is connected and f : S → R belongs to C∗(S).

Since f is bounded, the function f : S → R extends to fβ : βS → clRf [S] wherefβ[βS] = clRf [S]. Since S is connected then, by theorem 20.2, f [S] is a connected

bounded interval and fβ [βS] is a closed bounded interval in R. This implies

clRf [S] = f [S] ∪ S(f) = [a, b]

for some a and b.

Case f [S] = [a, b] : Then fβ [βS] = S(f) = f [S] so f is singular and so is not perfect.

Other cases, f [S] = [a, b) or f [S] = (a, b] or f [S] = (a, b) : If f [S] = [a, b) thenb ∈ S[f ]. Then S(f) ∩ f [S] = ∅ and so f is perfect. Similarly, if f [S] = (a, b] or

f [S] = (a, b) then S(f) ∩ f [S] = ∅ and so f is perfect.

Example 3. Let f =(

2π

)

arctan. We then obtain the function f : R → [−1, 1]. Show

that f is perfect.


Solution : The function, f , maps the connected interval (−∞,∞) one-to-one ontoT = (−1, 1). Since f : R → [−1, 1] is real-valued and bounded then, by the statement

the previous example, f is perfect.

25.4 The realcompact property and perfect evaluation maps.

We will review a few facts related to the realcompact property. Suppose S is com-pletely regular and F = C(S). If f ∈ F , we have seen that f : S → R extendscontinuously to fβ(ω) : βS → ωR (where ωR = R ∪ ∞).Furthermore,

υfS = x ∈ βS : fβ(ω)(x) ∈ R = fβ(ω)←[R]

represents all the “real points in βS associated to f ”. If a point p in βS is a real

point associated to all f ∈ C(S) then we say that p is a “real point of βS”. Then theset

υS = ∩fβ(ω)←[R] : f ∈ F = ∩υfS : f ∈ Fis a way of representing the set of all real points of βS.

So, if x ∈ βS \υS there is some g ∈ C(S) such that gβ(ω)(x) = ∞. If υS = S then S is

said to be “realcompact”. That is, for every point, p, in βS\S there is some f ∈ C(S)such that fβ(ω)(p) = ∞.

In the special case where f ∈ C∗(S), then

fβ(ω)[βS] = fβ(ω)[clβSS]

= clωRf [S]

= clRR

⊆ R

So, in this case, υfS = fβ(ω)←[R] = βS. Then ∩fβ(ω)←[R] : f ∈ C∗(S) = βS.When S is not pseudocompact1 the functions, f , in C∗(S) play no role in distinguish-

ing the real points in βS\S from the “non-real points”. Only unbounded functions,f , will extend to a function, fβ(ω), so that fβ(ω)←(∞) is non-empty.

So, if S is not pseudocompact and F = C(S) we will denote the non-empty set of all

unbounded real-valued functions on S by H = F \C∗(S). Then

υS = ∩fβ(ω)←[R] : f ∈ H = ∩fβ(ω)←[R] : f ∈ F

If g ∈ H , then gβ(ω)[clβSS] = clωRg[S]. For this function g, the image, g[S], of S isunbounded in R and so is not a compact subset of R; this means that there must be

some point q in βS\S such that gβ(ω)(q) = ∞.

1A pseudocompact space is one for which C(S) = C∗(S).


We will now consider these few notions from a slightly different point of view.

First, some notation.

If F = C(S), H = F \C∗(S), and ωRf is the codomain of f : S → ωR.

H =∏

f∈FωRf

K =∏

f∈FclωRf [S]

T = eF [S]

αT = clKT

By Tychonoff’s theorem, H and K are compact with T dense in αT ⊆ K ⊆ H .

For each f : S → R, we obtain f+ : S → ωR which extends to fβ(ω) : βS → clωRf [S].

The evaluation function, eF : S → T , maps S onto T , where

eF (x) =<f(x)>f∈F ∈ T

It extends to

eβ(α)F

: βS → αT

where eβ(α)F

maps βS onto clKT = αT .

In the following theorem we establish a fundamental relationship between the real-compact property and properties of eF .

Theorem 25.4 Let S be both a completely regular and non-pseudocompact space andF = C(S). Let Rf denote the codomain of f : S → Rf . If S is realcompact then the

evaluation map,

eF : S →∏

f∈FRf

is a perfect function which maps S onto eF [S] in∏

f∈FRf .

Proof : We are given that S is both a completely regular and non-pseudocompact spaceand F = C(S).

Let H = F \C∗(S) and KH =∏

f∈HclωRf [S].

Since S is non-pseudocompact then H is non-empty.

Let function, eH : S → eH [S], be the evaluation map generated by H defined as

eH (x) =<f(x)>f∈H ∈ eH [S]


The function eH extends to eβ(α)H

: βS → eβ(α)H

[βS] = clKHeH [S] defined as

eβ(α)H

(x) =<fβ(ω)(x)>f∈H ∈ clKHeH [S]

Suppose S is realcompact.

We are required to show that eF : S → eF [S] ⊆ ∏

f∈FR is a perfect function. To

do this it will suffice to show that eβ(α)F

[βS \S] = (clKT ) \T and invoke theorem 25.2part b).

Since S is realcompact, βS\S = βS\υS, so, for every x ∈ βS\S, there is some f ∈ Hsuch that fβ(ω)(x) = ∞.

Then

∩fβ(ω)←[R] : f ∈ H = S

Then, for every x ∈ βS\S, eβ(α)H

(x) = < fβ(ω)(x) >f∈H has an entry which is ∞.

So

eβ(α)H

[βS\S] ∩ ∏

f∈HRf = ∅

hence,

eβ(α)F

[βS\S] ∩ ∏

f∈FRf = ∅ (∗)

Let T = eF [S]. Since T ⊆∏f∈FRf then (*) implies

eβ(α)F

[βS\S]∩ T = ∅ (∗∗)

For K =∏

f∈FclωRf [S].

eβ(α)F

[βS] = eβ(α)F

[βS\S] ∪ eF [S]

= eβ(α)F

[βS\S] ∪ T (By (**), disjoint union.)

= clKT

Since[

eβ(α)

F[βS\S] ∪ T

]

\T = eβ(α)F

[βS\S] then,

eβ(α)F

[βS\S] = (clKT )\T

So, if S is realcompact, then, by 25.2, eF : S →∏

f∈Ff [S] is perfect.


We now show that the converse of the above theorem statement holds true. That is,if eF : S →∏

f∈Ff [S] is perfect then S must be realcompact.

Theorem 25.5 Let S be both a completely regular non-pseudocompact space and F =

C(S). If eF : S →∏

f∈FRf is a perfect function then S is realcompact.

Proof : Let R =∏

f∈FRf , K =

∏

f∈FωRf and T = eF [S]. Then T ⊆ R ⊆ K.

Given: eF : S → T is perfect.

We are required to show that S is realcompact. Let αT = clKT . Then eF extends to

eβ(α)F : βS → αT

To prove S is realcompact it suffices to show that, if y ∈ βS \S, then

eβ(α)F (y) =<fβ(ω)(y)>f∈F

in∏

f∈FωRf has at least one entry, gβ(ω)(y) = ∞.

Since eF : S →∏

f∈FRf is perfect it is, by definition, a closed function. Then eF [S] =

T is a closed subset of R. This means that clRT = T . So clKT ∩ R = clRT = T .Then

(clKT )\T ∩ R ⊆ (clKT ∩ R)\T= T \T= ∅

implies

(clKT )\T ∩ R = ∅ (†)

Since eF is perfect, by theorem 25.2,

eβ(α)F [βS\S] ⊆ (clKT )\T

= [(clKT )\T ] \R (By (†).)

⊆ K\R

So,

eβ(α)F [βS\S] ⊆

(

∏

f∈FωR

)

\(

∏

f∈FR

)

If y ∈ βS\S then eβ(α)F (y) has at least one entry which is ∞.

So y is not a real point of S. So the only real points of βS belong to S. So S is

realcompact, as required.


Corollary 25.6 Let S be both a completely regular non-pseudocompact space and F =

C(S). The space S is realcompact if and only if the evaluation map, eF : S → ∏

f∈Ff [S],

is a perfect function which homeomorphically maps S into∏

f∈FRf .

Proof : This simply summarizes the two previous theorems into a single “if and only if”

statement.

From the previous results we restate the general statement . . .

“Realcompact spaces are precisely those spaces, S, such that

eC(S) : S →∏

f∈C(S)Rf

is a perfect function.”

We will use the techniques illustrated in the proof of the above characterization ofrealcompactness to prove the following statement about arbitrary products of R’s.

Theorem 25.7 Any closed subspace, S, of a product of R’s is realcompact.

Proof : Let R =∏

j∈J Rj, and K =∏

j∈J ωRj.

We are required to show that every closed subset of R =∏

j∈J Rj is realcompact.

Since closed subsets of realcompact spaces are realcompact (by theorem 24.11) it willsuffice to show that R is realcompact.

To do this, it suffices to show that eC(R) :∏

j∈J Rj → ∏

f∈C(R) Rf is perfect andinvoke corollary 25.6.

Consider the identity function i : R → R expressed as, i : R → K, where it is now

seen as an inclusion map which embeds R =∏

j∈J R into K =∏

j∈J ωR.

Then i extends to iβ(K) : βR → clKi[R] ⊆ K. See that

clKi[R] = iβ(K)[βR] = iβ(K)[βR\R] ∪ i[R]


Now i pulls back points to points and so is a compact function. If F is closed in Rthen i[F ] is closed in R and so it is a closed function. So i is a perfect function.

Also see that

iβ(K)[βR\R] ⊆ clKi[R]\i[R] (Since i : r → R is a perfect map)

⊆ K\R=

∏

j∈JωR\∏

j∈JR

Let πi :∏

j∈J Rj → Ri denote the function defined as πi(<xj >j∈J ) = xi ∈ Ri.

Essentially πi is the real-valued continuous ith projecting R into R. So

P = πj : j ∈ J ⊆ C(R)

The family P generates the evaluation map eP :∏

j∈J Rj →∏

j∈J Rj defined as,

eP (<xj>j∈J ) = < πj(<xi>i∈J) >j∈J

= <xj>j∈J

∈ ∏

j∈JRj

Then eP(<xj >j∈J) =<xj >j∈J . The function, eP , is simply another way of repre-

senting the identity function mappingR onto R.

Since iβ(K)[βR\R] ⊆∏j∈JωRj\∏

j∈JRj, if y ∈ βR\R, then

ePβ(K)(y) =<π

β(ω)j (y)>πj∈P ∈ ∏

πj∈PωRπj \

∏

j∈JRπj

has at least one entry which is ∞.

Since P ⊆ C(R), for y ∈ βR\R,

eC(R)β(K)(y) =<fβ(ω)(y)>f∈C(R) ∈

∏

f∈C(R)ωRf \∏

j∈JRj

has at least one entry which is ∞.

So y is not a real point of R. Then,

eC(R)β(K)[βR\R] ⊆∏f∈C(R)ωRf \

∏

f∈C(R)Rf

This means that eC(R) : R → ∏

f∈C(R)Rf is a perfect function. By corollary 25.6, Ris realcompact.

We have shown that R is realcompact. Then all closed subsets of R are realcompact,

as required.


Concepts review:

1. Define a compact function.

2. Define a perfect function.

3. Provide two characterizations of a perfect function.

4. Provide a characterization of the realcompact property involving a perfect function.

5. Provide a characterization of a perfect function f : S → K in terms of a singular set.

6. If eC(S) : S →∏

f∈C(S) f [S] is an evaluation map what can we say about S if eC(S) is

perfect?

7. If S =∏

i∈J Si what property must S satisfy if we want the projection map to be

perfect?


26 / Perfect and Freudenthal compactifications

Summary. In this chapter we define both the Freudenthal compactification andthe family of “perfect compactifications”. In spite of its name, a perfect com-

pactification has very little to do with “perfect functions”. A perfect functionis one which is closed and whose fibres are compact, while a perfect compact-ification, αS, is one for which the fibres of πβ→α are connected. We produce

an algebraic characterization of those compactifications we call “perfect”. Wealso present examples of both the Freudenthal compactification of a space and a

perfect compactification. This chapter can be studied immediately after chapterintroducing compactifications without loss of continuity.

26.1 The Freudenthal compactification.

We introduce a particular compactification for locally compact Hausdorff spaces called

the Freudenthal compactification.1 Just like the Stone-Cech compactification we willsee that a space can only have one Freudenthal compactification. We begin by re-minding ourselves that,

. . . a space is said to be zero-dimensional if every point has an open neigh-

bourhood base of clopen sets.

While,

. . . a space is totally disconnected if every connected component is a singletonset.

Both of these are related to either connectedness or disconnectedness properties. Bytheorem 20.24 “...a locally compact Hausdorff space is zero-dimensional if and only if

it is totally disconnected”.

Definition 26.1 If S is a locally compact Hausdorff compactification of S the Freudenthal

compactification, φS, is the maximal compactification whose outgrowth, φS \S, is totallydisconnected − equivalently, whose outgrowth is zero-dimensional. By “maximal” we mean

that, if Z = αiS : i ∈ I is the family of all compactifications of S with zero-dimensionalremainder, φS ∈ Z and αiS φS, for all i.

1Hans Freudenthal (1905-1990) was a Jewish-German-born Dutch mathematician at the University ofAmsterdam. He was suspended from his duties by the Nazis during the war. After the war he was reinstatedto his former position.

494 Section 26: Perfect and Freudenthal compactifications

There is another way to visualize the Freudenthal compactification, φS, of a spaceS. In the following theorem we show that φS is the unique compactification of S

obtained by collapsing the connected components of βS\S to points. This theoremguarantees that a maximal compactification with zero-dimensional outgrowth exists

in the partially ordered set of all compactifications. Furthermore, there is only onesuch compactification (up to equivalence).

Theorem 26.2 Let S be a locally compact Hausdorff space and let αS be a compacti-fication of S. Let πβ→α : βS → αS be the corresponding continuous map which fixes the

points of S. Suppose φS denotes the Freudenthal compactification of S.

a) If αS\S is zero-dimensional then, for any p ∈ αS\S, πβ→α←(p) is a union of connected

components of βS\S.

b) Suppose that, for each p ∈ αS\S, πβ→α←(p) is a connected components of βS\S. Then

αS is equivalent to the Freudenthal compactification, φS, of S.

Proof : Let S be a locally compact Hausdorff space and αS be a compactification of S.

a) We are given that the outgrowth, αS\S, is zero-dimensional. We are required to

show that, for each point p ∈ αS\S, πβ→α←(p) is a union of connected components of

βS\S.

Let p ∈ αS\S and C be a connected component in βS\S which intersects πβ→α←(p).

It suffices to show that C ⊆ πβ→α←(p).

Suppose t ∈ C \πβ→α←(p). Then there exists a point, q, in αS\S distinct from p such

that t ∈ πβ→α←(q). Since αS\S is zero-dimensional, there is some clopen neighbour-

hood U of q in αS\S which does not contain p.

Then

t ∈ πβ→α←(q) ⊆ πβ→α

←[U ]

where πβ→α←[U ] ∩ πβ→α

←(p) = ∅. Since πβ→α is continuous πβ→α←[U ] is a clopen set

in βS \S which only intersects a portion of C. This contradicts the fact that C isa connected component of βS\S. So C\πβ→α

←(p) must be empty. This means that

C ⊆ πβ→α←(p). Since every point of πβ→α

←(p) belongs to a component of βS\S whichis entirely contained in πβ→α

←(p), πβ→α←(p) must be a union of connected components.

We are done for part a).

b) We are given that, for each p ∈ αS \S, πβ→α←(p) is a connected components of

βS\S. We are required to show that αS and φS are equivalent compactifications.


A function which collapses the connected components of a compact Hausdorff set,βS \S, to points in αS \S produces a totally disconnected set (equivalently a zero-

dimensional space) (see the example on page 369). Then αS\S is zero-dimensional.

We have shown in part a) that πβ→α pulls back points to unions of components. Then,

for each p ∈ αS\S, πβ→α←(p) is the “union” of a single component. Then αS\S, cannot

be partitioned any further without slicing into a connected component. Then αS must

be the maximal compactification with zero-dimensional outgrowth. So αS ≡ φS. Asrequired.

26.2 Perfect compactification: Definition.

We have seen above that, if αS has zero-dimensional outgrowth, αS\S, then

“for all p ∈ αS\S, πβ→α←(p) is the union of connected components of βS\S”

From this fact we concluded that there is only one maximal compactification, φS, with

zero-dimensional outgrowth, namely the compactification, αS, for which the fibers ofπβ→α|βS\S is a single connected component of βS \S. It is called the Freudenthalcompactification.

Those compactifications, αS, with zero-dimensional outgrowth must then be “lessthan or equal to” the Freudenthal compactification, φS.

There may be compactifications, αS, whose outgrowth, αS \S, satisfies the property,

“for all p ∈ αS\S, πβ→α←(p) is connected”

but is not zero-dimensional. This could occur only if πβ→α←(p) is a connected subset

of at most one component.

Such compactifications are referred to as “perfect” compactifications. We formallydefine this.

Definition 26.3 Let γS be a Hausdorff compactification of S and πβ→γ : βS → γS be

the continuous map which projects βS onto γS. The compactification, γS, is said to be aperfect compactification provided πβ→γ

←(p) is connected for every p ∈ γS\S.


The Freudenthal compactification, φS, is an example of a perfect compactification. Inthe family of all perfect compactifications of S, it is in fact the smallest one. So per-

fect compactifications form a family of compactifications which are gathered “above”φS, while the ones with zero-dimensional outgrowth form a family which are gathered

“below” φS.

The main result in this chapter is an algebraic characterization of a perfect compact-ification.

26.3 Algebraic subrings of C∗(S).

Our algebraic characterization will involve properties of a particular type of subring

of C∗(S), called an “algebraic subring”. Setting up the stage for the proof of the char-acterization first requires a proper understanding the concept of an algebraic subringand some of its properties. Characterizing the topological property of “perfect com-

pactification”, αS, with an algebraic property of a subring of Cα(S), requires a bitof mathematical prowess. We begin with the following definition. Three lemmas will

follow.

Definition 26.4 Let S be a locally compact Hausdorff space and C∗(S) be the ring ofall bounded continuous real-valued functions on S.

a) We will say that A is an algebraic subring of C∗(S) if A is a subring which contains

all of the constant functions and all functions, f , such that f2 = f · f ∈ A .

b) Given a subring, A , of C∗(S), a subset T of S on which every function in A isconstant is called a stationary set of the subring, A . If for any p ∈ S \T there is a

function f ∈ A which is not constant on T ∪ p then we say that T is a maximalstationary set of A . Equivalently, if T is a stationary set of A containing y andT = x ∈ S : f(x) = f(y), for all f ∈ A then T is a maximal stationary set of A .

By definition, if T is a maximal stationary set of the subring, A , of C(S) then every

function in A is constant on T . Then, for f ∈ A , f [T ] is some singleton set, say tf,and so T ⊆ Z(f − tf ). Then T ⊆ ∩Z(f − tf ) : f ∈ A . Since T is maximal with

respect to this property, then

T = ∩Z(f − tf ) : f ∈ A ,

a closed subset of S.

The following lemma offers yet another way of describing the maximal stationary set

of subring.


Lemma 26.5 Given a space, S, let G be a subring of C(S). For any point p ∈ eG [S],eG←(p) is a maximal stationary set of G .

Proof : Let G is a subring of C(S) and

p = <pf >f∈G ∈ eG [S] ⊆∏f∈GRf

Since

eG←(p) = x ∈ S : eG (x) =<f(x)>f∈G =<pf >f∈G

Then, for g ∈ G , g[eG←(p)] = pg. Then eG

←(p) ⊆ Z(g − pg). It follows that

eG←(p) ⊆ ∩f∈GZ(f − pf)

Then every function in G is constant on eG←(p). Let T be the maximal stationary

set which contains eG←(p). Let x ∈ eG

←(p).

Suppose t ∈ S\eG←(p). Then there exists h ∈ G such that h(t) 6= h(x) = h[T ].Then t 6∈ T . So eG

←(p) = T . Then eG←(p) is a maximal stationary set of G .

Lemma 26.6 Let T be a connected subset of a topological space, S, and G be the subsetof C∗(S) which contains only those functions which are constant on T . Then G is an

algebraic subring which is closed with respect to the uniform norm topology.

Proof : We are given that G contains precisely all those functions which are constant on

the connected subspace T .

Clearly, G is closed under multiplication and addition and so G is a subring of C∗(S).

Claim: That G is algebraic. Let g be a function on S such that g2[T ] = k. It suffices

to show that g ∈ G . For every x ∈ T , g(x) is√k or −

√k. By hypothesis, T is

connected and so the range g[T ] cannot contain two elements. So g must be constant

on T and so belongs to G . We deduce that G is algebraic, as claimed.

Claim: That G is closed with respect to the uniform norm. Suppose h is a limit point

of G . Then there is a sequence ti : i ∈ N in G which converges to h with respectto the uniform norm topology. Since each ti is constant on T , h is constant on T and

so belongs to G . So G contains it’s limit points and so is closed with respect to theuniform norm.


In the next lemma we establish a relationship between a topological property of amaximal stationary set of a subring, A , and an algebraic property of this subring.

Lemma 26.7 If S is a compact Hausdorff space and A is an algebraic subring of C∗(S)

then any maximal stationary set, T , of A is connected.

Proof : We are given that S is compact and Hausdorff and T is a maximal stationary setof the algebraic subring, A , of C∗(S).

For f ∈ A , let tf = f [T ]. Then we can express the maximal stationary set, T , as,

T = ∩Z(f − tf ) : f ∈ A If t =< tf >f∈A and eA : S →∏

f∈ARf , is the evaluation map induced by A then

T = eA←(t) = ∩Z(f − tf ) : f ∈ A

We are required to show that T is connected.

Suppose T is not connected. That is, suppose T = F1 ∪ F2 where F1 and F2 are dis-joint non-empty closed subsets of T . We will show that this leads to a contradiction.

Since S is compact, they are both compact, so there exist disjoint open subsets, U1

and U2, of S, containing F1 and F2, respectively. So T ⊆ U1 ∪ U2. Now S \(U1 ∪ U2)

is closed and hence compact (since S is compact).

Consider a basic open neighbourhood of t = < tf >f∈A , U =∏

f∈AUf , where Uf is

R for all but finitely many factors Ufi : i ∈ F where Ufi is of the form

Ufi = (tfi − εi, tfi + εi)

There exists at least one function, h : S → R, in A and one ε such that

eA←(t) = F1 ∪ F2 ⊆ h←[th, th + ε] ⊆ U1 ∪ U2

1

LetW1 = h←[th, th + ε] ∩ U1 (Which contains F1.)

W2 = h←[th, th + ε] ∩ U2 (Which contains F2.)

Since U1 and U2 are disjoint then W1 and W2 are disjoint and

W1 ∪W2 = h←[th, th + ε]

So there exists k ∈W1 ∪W2 such that h(k) = th + ε.

1If f←[tf , tf + ε] ∩ S \ (U1 ∪ U2) 6= ∅ for all f and ε then, by the “FIP compactness property” ofS \(U1 ∪ U2), F1 ∪ F2 = T = eA

←[U ] ∩ S \(U1 ∪ U2) cannot be empty, a contradiction.


We define a real-valued function r : S → R on S, as follows

r(x) = (th + ε) − h(x) if x ∈W1

r(x) = h(x) − (th + ε) if x ∈ S\W1

Then r(x) ≥ 0 on W1 and r(x) ≤ 0 on S\W1.

Since h(k) = th + ε, then r(k) = 0.

See that r is continuous on S. 2

We now claim that r ∈ A : Note that r2 = ((th + ε) − h(x))2. Since A is a subring

then r2 ∈ A . Since A is algebraic and r2 ∈ A , then r ∈ A . As claimed.

Now we have a contradiction (since r ∈ A is not constant on T and every functionof A was hypothesized to be constant on T ). So T cannot be partitioned by twonon-empty closed sets F1 and F2. So T is connected. We are done.

26.4 An algebraic characterization of “perfect compactification”.

Having established that (for a compact space S),

“ . . . all maximal stationary sets of an algebraic subring are connected”

we are ready to roll out the proof of the following characterization of perfect com-pactifications.

Theorem 26.8 Let γS be a compactification of S. Then γS is a perfect compactification

if and only if Cγ(S) is an algebraic subring of C∗(S).

Proof : ( ⇐ ) We are given that G = Cγ(S) where Cγ(S) is an algebraic subring of C∗(S).

We are required to show that, for any p ∈ γS, πβ→γ←(p) is connected.

Given that Cγ(S) is algebraic, it is easily verified, that Cγ(S)β must also be an alge-braic subring of C(βS). By lemma 21.12, πβ→γ = eG

γ←eG

β. So, for p ∈ γS\S,

πβ→γ←(p) = [eG

γ←eG

β ]←(p)

= eGβ←(eG

γ(p))

= a maximal stationary set of G β in βS

2Clearly, f is continuous both on W1 and on S\W1 . But r(k) = 0 and r(x) ≥ 0 on W1 and r(x) ≤ 0 onS\W1 , and so k is a boundary point of the piecewise defined function r at which r(k) = 0. So r : S → R isa continuous function, as claimed


Since G β is an algebraic subring of C(βS), then by lemma 26.7, the maximal station-ary set, πβ→γ

←(p), of G β is connected. So, for any p ∈ γS, πβ→γ←(p) is connected.

By definition, γS is a perfect compactification. As required for the direction, ⇐.

( ⇒ ) Suppose γS is a perfect compactification. By definition of perfect compactifi-cation, πβ→γ

←(p) is connected for every p ∈ γS\S.

Let G = Cγ(S). To show that G is an algebraic subring of C∗(S) it suffices to showthat G β is an algebraic subring of C(βS).

For p ∈ γS, let Fp denote the set of all functions in C(βS) which are constant on the

connected set, πβ→γ←(p). By lemma 26.6, Fp is an algebraic subring of C(βS).

Since πβ→γ = eGγ←eG

β (by lemma 21.12), then πβ→γ←(p) = eβ

G←(eG

γ(p)). So

eβG[πβ→γ

←(p)] = eGγ(p)

It follows that G β ⊆ Fp. It is easily verified that the intersection of a family ofalgebraic subrings is an algebraic subring. So ∩Fp : p ∈ γS is an algebraic subring

of C(βS). So

G β ⊆ ∩Fp : p ∈ γS

If we can show that ∩Fp : p ∈ γS ⊆ G β, then we are done.

Let f ∈ ∩Fp : p ∈ γS. If g ∈ C(γS) is such that g(πβ→γ(x)) = f(x) for all x ∈ βS,then g|S = f |S then f |S ∈ Cγ(S) = G . Hence ∩Fp : p ∈ γS ⊆ G β.

We conclude that G β = Cγ(S)β = ∩Fp : p ∈ γS, an algebraic subring of C(βS).

So Cγ(S) is an algebraic subring of C∗(S). We are done with the direction ⇒.

26.5 Example.

Recall that the Freudenthal compactification is the maximal compactification whoseoutgrowth is zero-dimensional (equivalently, totally disconnected). The Freudenthal

compactification can also be seen as the minimal “perfect compactification” of a space.In the following proposition we provide an example of a space whose Freudenthal com-pactification is βS. In such a case S has only one perfect compactification, βS.

The proposition involves the space, N, in which any subset A is clopen and so is a

zero-set. Then there is a continuous function, f : N → 0, 1, such that A = Z(f −1).Then f extends continuously to fβ : clβNA→ 0, 1 where

fβ [clβNA] = clRf [A] = 1


ThenclβNA = Z(fβ − 1) = βN\Z(fβ − 0)

is clopen in βN. Thus ultrafilters of sets and z-ultrafilters in Z[N] represent the samefamilies of sets. Every point p in βN\N is associated to a free ultrafilter of subsets of N.

Proposition 26.9 The Stone-Cech compactification, βN, of N is both the Freudenthalcompactification and the minimal perfect compactification.

Proof : To show that βN is the Freudenthal compactification of N, it suffices to show thatβN\N is zero-dimensional, equivalently, totally disconnected. We have shown that

subspaces of totally disconnected spaces are themselves totally disconnected. So itwill suffice to show that βN is totally disconnected. Suppose p and q are points in

a connected component C of βN. If p 6= q then they are limits of two distinct freez-ultrafilters Zp and Zq in Z[N]. Then Zp contains a zero-set Z(f − 1) which does

not belong to Zq . Then p belongs to the clopen zero-set Z(fβ − 1) which does notcontain q. Since C is connected then p must equal q. We conclude that βN is totally

disconnected. It is the maximal compactification with zero-dimensional outgrowthand so is Freudenthal. It also is the smallest compactification which is perfect.

Since the Freudenthal compactification is the smallest of perfect compactifications, no

compactification αN ≺ βN is perfect. So, by the above characterization, no propersubring of C∗(N) is algebraic.

Concepts review:

1. Define the Freudenthal compactification of a space.

2. How can βS be used to construct the Freudenthal compactification φS?

3. What does it mean to say that a compact space is a perfect compactification of S?

4. State what we mean when we that A is an algebraic subring of C∗(S).

5. If A is an algebraic subring of C∗(S) what is a stationary set of A ? What does itmean to say that a subset, T , is a maximal stationary set of A ?


6. State an algebraic characterization (involving C∗(S)) of a perfect compactification.


27 / Spaces whose elements are sequences

Summary. In this chapter we investigate a family of topological spaces whoseelements are sequences. These spaces of sequences are vector spaces equippedwith a topology which is determined by norms. The norms we will study are

`p-norms and the `∞-norm. We will verify when the `p-spaces are separable,second countable, Lindelof and sequentially compact, but generally non-compact.

The `∞-spaces of sequences will be shown to be complete. Along the way we willprove two important inequalities: Holder’s inequality and Minkowski’s inequality

for the `p-spaces.

27.1 Definitions: `p-spaces.

To begin we remind ourselves that a sequence of points can be viewed as a function;one whose domain is N. If S is a normed topological space, the functions studied here

will be S-valued. We will represent the set of all S-valued sequences as

F (N, S)

These are more commonly expressed as

F (N, S) = s0, s1, s2, s3 . . . : si ∈ S = < si >i∈N: si ∈ S

the family of all sequences whose elements belong to the space S. The topology de-fined on F (N, S) depends on the topology of S.

Given a normed vector space (S,+, ·, ‖ ‖) and the set F (N, S), “`p-space” and the“`p-norm” are defined as follows.

Definition 27.1 Let S be a normed vector space (S,+, ·, ‖ ‖) and F (N, S) denote the set

of all sequences, f = s0, s1, s2, s3, . . . = < si >i∈N, whose elements, si, belong to S. Let pbe any real number greater than or equal to 1. The `p-norm of f = < si >i∈N is defined as,

‖f‖p = ‖ < si >i∈N ‖p =

[

∞∑

i=0

‖si‖p

]1/p

Let (Sb, ‖ ‖p) denote the family of those sequences, < si >i∈N, in F (N, S) such that

‖ < si >i∈N ‖p <∞

The expression, “`p-space”, refers to the (bounded) normed space, (Sb, ‖ ‖p).

504 Section 27: Spaces whose elements are sequences

The use of the lower-case cursive form, `, in the definition, is standard. When we refer

to “`p-norm” we are speaking of an infinite family of norms on F (N, S). These aredistinguished from each other by the chosen value of p in [1,∞). They are viewed as

belonging to the same family of norms because the same formula to used to computethem.

The space S may be the space, R, or the complex numbers, C, equipped with theusual norm. We should be careful not to confuse the norm, ‖ ‖, on S with the chosen

`p-norm, ‖ ‖p, on F (N, S).

The definition of the `p-norm can be modified so as to include the set of n-tuples,

F (1, 2, , 3 . . . , n, S) = Sn where

‖ < si >i=1...n ‖p =

[

n∑

i=1

‖si‖p

]1/p

The only significant difference is the symbol, n, which replaces, “∞”. Everything elseflows identically. The case where, p = 2, turns out to be the Euclidean norm on Sn.

When applied to R, the `p-norm is a generalization of the absolute value since

‖x‖2 = [x2]1/2 = |x|

On R2,

‖(a, b)‖2 =√

a2 + b2

The `p-norms on Rn flow naturally to `p-norms on R∞ (viewed as infinite sequencesof real numbers).

The `p-norm is a valid norm for all p ≥ 1.

Having defined the `p-norm on the family of S-valued sequences, F (N, S), in this

way we have yet to verify that it satisfies the three norm axioms for all values of p.Proving that any `p-norm satisfies the first two norm axioms is straightforward. Letf = < si >i∈N.

N1: ‖f‖p <∞ and ‖f‖p = [∑∞

i=0 ‖si‖p ]1/p = 0 ⇔ si = 0 for all i ∈ N.


N2:‖αf‖p =

[

∞∑

i=0

‖αsi‖p

]1/p

=

[

∞∑

i=0

|α|p‖si‖p

]1/p

=

[

limn→∞

n∑

i=0

|α|p‖si‖p

]1/p

=

[

limn→∞

|α|pn∑

i=0

‖si‖p

]1/p

= [|α|p]1/p

[

limn→∞

n∑

i=0

‖si‖p

]1/p

= |α|[

∞∑

i=0

‖si‖p

]1/p

= |α|‖f‖p

N3: The triangle inequality for `p-norms, also called the Minkowski inequality, is moredifficult to prove. It is in fact quite tricky. It’s proof is the main subject of the next

subsection on inequalities.

27.2 Two fundamental inequalities: Holder’s and Minkowski’s inequalities.

To following inequalities are important. The proof of Minkowski’s inequality (the

triangle inequality for ‖ ‖p) must be preceded by the “Holder’s inequality”.

The following lemma helps set up the proof of Holder’s inequality.

Lemma 27.2 Lemma for Holder’s Inequality For any t ∈ (0, 1) and any A > 0 , B > 0

AtB1−t ≤ tA+ (1− t)B

Proof : We are given that A > 0, B > 0. For values of t ∈ (0, 1) we define the functions

f(t) and g(t) as

f(t) = tA+ (1− t)B = (A−B)t+ B

g(t) = AtB1−t

To prove the lemma it suffices to show that g(t) ≤ f(t) on (0, 1).


If we let a = lnA and b = lnB, then

A = ea

B = eb

The functions f(t) and g(t) then take on the form,

f(t) = tea + (1 − t)eb = (ea − eb)t+ eb

g(t) = (ea)t(eb)(1−t) = eta+(1−t)b = e(a−b)t+b

where f(t) represents a straight line joining the points (0, eb) and (1, ea) and g(t) rep-

resents an exponential function joining the same points (0, eb) and (1, ea). Regardlessof whether a > b or b > a the function g(t) is concave upwards. In order for g(t) to

intersect f(t) at the points (0, eb) and (1, ea) it must be that g(t) ≤ f(t) on (0, 1).

So AtB1−t ≤ tA + (1 − t)B on (0, 1), as required.

Theorem 27.3 Holder’s Inequality Let S be a normed vector space (S,+, ·, ‖ ‖) andF (N, S) denote the set of all sequences, f = < si >i∈N, whose elements, si, belong to S.

Let p be any real number greater than or equal to 1. The `p-norm of f = < si >i∈N isdefined as,

‖f‖p = ‖ < si >i∈N ‖p =

[

∞∑

i=0

‖si‖p

]1/p

Let p be any real number > 1 and f = < si >i∈N, g = < ti >i∈N ∈ `p. Let q = pp−1 . Then,

∞∑

i=0

‖si‖‖ti‖ ≤ ‖f‖p‖g‖q

Proof : Note that q = pp−1 implies 1

p + 1q = 1. If either f or g is 0 then both sides are zero

and so the theorem holds true. Suppose then that neither is the zero sequence.

Suppose A = ‖si‖p/‖f‖pp and B = ‖ti‖q/‖g‖q

q.

Let t = 1/p. Then q = pp−1 implies 1 − t = 1/q. Then AtB1−t ≤ tA + (1 − t)B (by

lemma).


AtB1−t ≤ tA + (1 − t)B ⇒ A1/pB1/q ≤ A

p+B

q

⇒ ‖si‖‖f‖p

‖ti‖‖g‖q

≤ ‖si‖p

p‖f‖pp

+‖ti‖q

q‖g‖qq

implies

∞∑

i=0

‖si‖‖f‖p

‖ti‖‖g‖q

≤ 1

p

∞∑

i=0

‖si‖p

‖f‖pp

+1

q

∞∑

i=0

‖ti‖q

‖g‖qq

=1

p

∑∞i=0 ‖si‖p

‖f‖pp

+1

q

∑∞i=0 ‖ti‖q

‖g‖qq

=1

p

‖f‖pp

‖f‖pp

+1

q

‖g‖qq

‖g‖qq

=1

p+

1

q= 1 (By hypothesis)

From which we obtain∞∑

i=0

‖si‖‖ti‖ ≤ ‖f‖p‖g‖q

This proves Holder’s inequality .

Theorem 27.4 Minkowski Inequality Let p be any real number greater or equal to 1 and(S, ‖ ‖) be a normed vector space. Let f = < si >i∈N, g = < ti >i∈N ∈ `p where the

entries belong to S. Then,‖f + g‖p ≤ ‖f‖p + ‖g‖p

Proof : Given: That p ∈ [1,∞) and (S, ‖ ‖) be a normed vector space. and f = < si >i∈N

, g = < ti >i∈N ∈ `p.

Case p = 1: If p = 1, then, since∑n

i=0 ‖si + ti‖ ≤ ∑ni=0 ‖si‖ +

∑ni=0 ‖ti‖, for all n,

‖f + g‖1 ≤ ‖f‖1 + ‖g‖1. Then the statement holds true for this particular case.

Case p > 1: Suppose now that p > 1. Let q = pp−1 .

Claim: We claim that if f = < si >i∈N,

∥

∥ < ‖si‖p−1 >i∈N

∥

∥

q= ‖f‖p−1

p (∗)


Proof of claim. Suppose f = < si >i∈N and we are given q = p/(p− 1).

∥

∥ < ‖si‖p−1 >i∈N

∥

∥

q=

[

∞∑

i=0

∣

∣‖si‖p−1∣

∣

q

]1/q

=

[

∞∑

i=0

‖si‖q(p−1)

]1/q

=

[

∞∑

i=0

‖si‖p

]1/q

=

[

∞∑

i=0

‖si‖p

]1/p

p/q

= ‖f‖p/qp

= ‖f‖p−1p (Since q = p/(p− 1))

So∥

∥ < ‖si‖p−1 >i∈N

∥

∥

q= ‖f‖p−1

p , as claimed.

The inequality is now proven in the following chain of inequalities.

‖f + g‖pp =

[

∞∑

i=0

‖si + ti‖p

]1/p

p

=

∞∑

i=0

‖si + ti‖p−1‖si + ti‖

≤∞∑

i=0

‖si + ti‖p−1 (‖si‖ + ‖ti‖) (Triangle inequality for ‖ ‖.)

=

[

∞∑

i=0

‖si + ti‖p−1‖si‖]

+

[

∞∑

i=0

‖si + ti‖p−1‖ti‖]

≤(

∥

∥ < ‖si + ti‖p−1 >i∈N

∥

∥

q· ‖f‖p

)

+(

∥

∥ < ‖si + ti‖p−1 >i∈N

∥

∥

q· ‖g‖p

)

(By Holder’s twice)

=(

∥

∥ < ‖si + ti‖p−1 >i∈N

∥

∥

q

)

· [ ‖f‖p + ‖g‖p ]

= ‖ f + g ‖p−1p · [ ‖f‖p + ‖g‖p ] (By claim *)

Then

‖f + g‖p =

(

1

‖ f + g ‖p−1p

)

‖f + g‖pp ≤ ‖f‖p + ‖g‖p


from which we obtain the desired triangle inequality.

27.3 A few simple topological properties of `p-spaces.

We now investigate topological properties possessed by `p-spaces. If we are given an`p-space, S = F (N, S), the norm of a sequence, f = < si >i∈N, in S is defined as,

‖f‖p =

[

∞∑

i=0

‖si‖p

]1/p

where the computation of ‖f‖p involves the norm of its entries and so the topologyon S will depend on the topology of S.

Since normed spaces are metrizable spaces then (S , ‖ ‖p) can be viewed as a metricspace. Then S is normal (by theorem 9.19) and first countable (by theorem 5.9).

The following theorem illustrates why S is separable whenever S is separable. If

equipped with an `p-norm, (S , ‖ ‖p) is an `p-space whose elements are sequenceswith entries belonging to S.

Theorem 27.5 Suppose (S, ‖ ‖) is a normed vector space and S represents the `p-space,(F (N, S), ‖ ‖p ) of all infinite sequences with entries from S. If S is separable and p > 1,

then S is a separable topological space (with respect to the `p-norm).

Proof : Since S is separable, S contains a dense countable subset, D = di : i ∈ N, of S.Let

D = < xi >i∈N : xi ∈ D for finitely many i’s and xi = 0 otherwise ⊆ S

We claim that D is dense in S . Let f = < si >i∈N ∈ S . It suffices to show that, for

any ε > 0, the ε-ball centered at f ,

Bε(f) = y ∈ S : ‖y − f‖p < ε

contains a sequence in D .

Since ‖f‖p = [∑∞

i=0 ‖si‖p]1/p is a number, then

limn→∞

∞∑

i=n

‖si‖p = 0


Let ε > 0. We can then choose n such that∑∞

i=n+1 ‖si‖p < εp/2.

Since D is dense in S, every open ball, Bε(si), with center si meets D, and so, for agiven n, for each i = 0 to n, we can find di ∈ D such that ‖si − di‖ < εp/2n.

For i > n, set di = 0. Let d = di : i ∈ N = d0, d1, d2, . . .dn, 0, 0, 0, . . .. Thend ∈ D .

Then

‖f − d‖pp =

∞∑

i=0

‖si − di‖p

=

n∑

i=0

‖si − di‖p +

∞∑

i=n+1

‖si − 0‖p

<

n∑

i=0

εp

2n+εp

2

=nεp

2n+εp

2= εp

So, ‖f − d‖p < ε. Since d ∈ D , then Bε(f) ∩D 6= ∅. We conclude that D is dense in

S , so the `p-space, S , is separable. This completes the proof.

If S is separable, since `p-spaces are both metrizable and separable, then (by theorem16.3) they are both second countable and Lindelof. Also since S is separable metriz-

able it is realcompact (see the example on page 476).

Example 1. Show that an `p-space is not, generally, sequentially compact.

Solution : Let S = F (N, S) be an `p-space and choose a κ ∈ S such that ‖κ‖ = 1.Let

U = f ∈ S : all entries are 0, except one term is κWe will construct a sequence in F (N, S) with no converging subsequence.

For each n let fn = < s(n,j) >j∈N be the sequence in U whose nth term is κ. Then

U = fn : n ∈ N

If n 6= k,

‖fn − fk‖p =

∞∑

j=0

‖s(n,j) − s(k,j)‖p

1/p

= (1p + 1p)1/p

= 21/p


Let ε = (21/p)/2. If g ∈ U is the limit of a subsequence of U , then ‖fn − g‖p = 21/p

must be < ε for infinitely many terms, fn. This can obviously not be. Then there is

no converging subsequence for fn : n ∈ N. So S cannot be sequentially compact.

By theorem 17.2, in metric spaces, sequentially compact, compact and countably com-pact are equivalent topological properties. So S is not compact.

27.4 The sup-norm on F (N, S).

We know that for each real value of p ∈ [1,∞), ‖f‖p is a valid norm on F (N, S).We now wonder if there is a natural way we can extend the value of p so as to include

p = ∞, In a sense, we are wondering whether the limit, limp→∞ ‖f‖p is a number. Ifso, is this number a norm of f which we could represent as “‖f‖∞”. How would we

define it?

To investigate this, let (S, ‖ ‖) be a normed vector space, let f = < si >i∈N ∈ F (N, S)

and p ≥ 1, let

k(n) = sup‖si‖ : i = 0, 1, 2, . . . , n

Then k(n) : n ∈ N is an increasing sequence of numbers.

Let

β = sup‖si‖ : i ∈ N

To ensure that β is a number we will assume that each sequence, < si >i∈N, is boundedwith respect to ‖ ‖. Now k(n) ≤ β for all values of n. If supk(n) : n ∈ N < β thenthere exists t such that supk(n) : n ∈ N < t < β. Then k(n) < t for all n ∈ N.

Then ‖si‖ < t < β for all i ∈ N contradicting the definition of β.

So

limn→∞

k(n) = β

We claim that, if f is bounded in F (N, S), limp→∞ ‖f‖p = β.

Proof of claim: Let ε > 0. Let f =< si >i∈N be a bounded element of F (N, S). Since

limn→∞ k(n) = β, then there exists an M ∈ N such that

n > M ⇒ |k(n)− β| < ε/3

Let

g(n, p) =

[

n∑

i=0

‖si‖p

]1/p


See that, for any choice of n,

limp→∞

g(n, p) = limp→∞

[

n∑

i=0

‖si‖p

]1/p

= limp→∞

[

n−m∑

i=0

‖si‖p

k(n)p+m

]1/p

k(n) (Where m ≥ 1)

= (1)k(n)

= k(n)

Then there exists κ > 0 such that p > κ implies, for any choice of n

| g(n, p)− k(n)| < ε/3

Then, if n > M and p > κ,

|g(n, p)− β| = |g(n, p)− k(n) + k(n) − β|≤ |g(n, p)− k(n)|+ |k(n) − β|< ε/3 − ε/3

< ε

If α = maxM, κ, when n, κ > α, |g(n, p)− β| < ε.

Then

limp→∞

limn→∞

g(n, p) = limp→∞

β = β

This implies,

limp→∞

‖f‖p = limp→∞

[

∞∑

i=0

‖si‖p

]1/p

= limp→∞

[

limn→∞

n∑

i=0

‖si‖p

]1/p

= limp→∞

limn→∞

[

n∑

i=0

‖si‖p

]1/p

= limp→∞

limn→∞

g(n, p)

= β

So limp→∞ ‖f‖p = β. This establishes the claim.


If f = < si >i∈N∈ F (N, S), it will be convenient to represent β = sup‖si‖ : i ∈ Nas

limp→∞

‖f‖p = ‖f‖∞

We formally define the sup-norm (also called the ∞-norm), ‖f‖∞.

Definition 27.6 Let S be a normed vector space. We say a sequence < si >i∈N in

F (N, S) is bounded if there is a k such that ‖si‖ ≤ k for all i ∈ N. Let Sb be the set of allbounded sequences with elements from S. If f = < si >i∈N, we define the sup-norm, ‖f‖∞,

of f as‖f‖∞ = sup‖si‖ : i ∈ N

The sup-norm on Sb is also referred as the `∞-norm. If Sb is equipped with the norm ‖ ‖∞we refer to Sb as being an `∞-space.

Verifying that ‖ ‖∞ is a valid norm is straightforward. Verifying this is left as an

exercise for the reader.

27.5 Topologizing F (N, S) by defining convergence.

Once chosen, a norm on F (N, S) precisely determines how sequences of sequences in

F (N, S) will converge. The norm will topologize F (N, S). But defining norms onF (N, S) is not the only way to topologize F (N, S).

We can also view F (N, S) as an infinite product, SN (see definition 12.6 on page

247). We normally assign on infinite products either the “product topology” or the“box topology”. The box topology on SN is equivalent to the topology obtained by

assigning the `∞-norm on F (N, S) = SN. That is, if limn→∞ ‖g − fn‖∞ = 0 then thesequence, fn : n ∈ N, will ...

converge uniformly to the sequence g

On the other hand, assigning the product topology on F (N, S) = SN produces a

topological space independent of any norm defined on F (N, S). It is the topologywith which the closed subsets are defined by “pointwise convergence”. This means

that, if fn = < sni >i∈N, and for each i,

limn→∞

sni = ti

then the sequence g = < ti >i∈N represents the ...

“pointwise limit of the sequence fn : n ∈ N


Unless explicitly specified otherwise, the default topology on F (N, S) is the one de-termined by uniform convergence. Note that, if a sequence in F (N, S) converges

uniformly, then it also converges pointwise. The converse does not hold true.

27.6 Completeness of (F (N, S), ‖ ‖∞).

Recall that a normed vector space, (S, ‖ ‖), is said to be complete if every Cauchysequence in (S, ‖ ‖) converges to a point which is contained in S.

In the following example we verify that, whenever (S, ‖ ‖) is complete, the normedvector space (F (N, S)b, ‖ ‖∞) is complete.1

Example 2. Verify that, if (S, ‖ ‖) is a complete normed space, Sb = (F (N, S)b, ‖ ‖∞)

is a complete normed vector space.

Solution : Let C = fn : n ∈ N be a Cauchy sequence of sequences in the `∞-space, Sb. Since C is Cauchy, it is bounded, so there is a real number, K, such that

‖fn‖∞ < K, for all n ∈ N (as shown in the above example).

For each n, let fn = <sni >i∈N. We are required to show that fn : n ∈ N converges

to some sequence, h, in the `∞-space, Sb.

Let ε > 0. Since fn : n ∈ N is Cauchy, there exists N ∈ N such that, m, n > N ⇒‖fn − fm‖∞ < ε.

Then, when m, n > N , and each i,

‖ sni − smi‖ ≤ supi∈N

‖sni − smi‖

= ‖ <sni − smi >i∈N ‖∞= ‖ <sni >i∈N − <smi >i∈N ‖∞= ‖fn − fm‖∞< ε

Then, for each i, when m, n > N , ‖ sni − smi‖ < ε. Then for each i, sni : n ∈ Nis a Cauchy sequence in S. Since S is a complete space, for each i, sni : n ∈ Nconverges to some ti ∈ S (with respect to ‖ ‖). That is, for each i, there exists Ni,such that,

n > Ni ⇒ ‖ti − sni‖ < ε

Let h = ti : i ∈ N.We view h as a candidate for the limit of the Cauchy sequence, fn : n ∈ N. Toprove this we must first show that ‖h‖∞ ∈ R, and then show that ‖h− fn‖∞ < ε for

all n > N for some N .

1The subscript b indicates “bounded” to ensure that the sup-norm is never ∞.


For our chosen ε, for i ∈ N, and any n > Ni,

‖ti‖ = ‖ti − sni + sni‖≤ ‖ti − sni‖+ ‖sni‖< ε+ ‖fn‖∞ (Since n > Ni .)

≤ ε+K

Then ‖h‖∞ ≤ sup‖t0‖, ‖t1‖, . . .‖tNi‖, ε+K ∈ R. So ‖h‖∞ is a number.

We now claim that fn : n ∈ N converges to h.

Suppose m, n > N . See that, independently of the value of i,

‖ti − smi‖ = ‖( limn→∞

sni) − smi‖= lim

n→∞‖sni − smi‖ (Since norms are continuous.)

≤ supi∈N

‖sni − smi‖

= ‖fn − fm‖∞< ε

Then supi∈N‖ti − smi‖ ≤ ε. So ‖h− fm‖∞ ≤ ε. Therefore, fn : n ∈ N convergesto h.

27.5 Topologizing other families of functions.

For two normed topological spaces S and T , let F (S, T ) denote the family of all func-tions mapping S into T . Whenever the functions on S are continuous and real-valued

the expression, C(S,R), specifies this fact. It is standard to express this set moreconcisely as, C(S). We can topologize C(S) in a way that is analogous to the methodused to topologize the family of all sequences, F (N, S). We do this by defining a simi-

lar norm on C[a, b] called, an Lp-norm. The symbol, `p−norm, is reserved specificallyfor the set, F (N, S), or the family of all n-tuples, Sn, of elements from S.

Definition 27.7 Let C[a, b] denote the set of all real-valued continuous functions on [a, b]and p ∈ [1,∞). If f ∈ C[a, b] we define,

‖f‖p =

[∫ b

a

|f(x)|p]1/p

We refer to this norm on C[a, b] as an Lp-norm and we refer to (C[a, b], ‖ ‖p) as an “Lp-space”.


We must again verify that ‖ ‖p satisfies the three norm axioms. Verifying that it

satisfies the N1 and N2 axioms is straightforward. Just as for `p-norms the triangleinequality, ‖f + g‖p ≤ ‖f‖p + ‖g‖p, for Lp-norms is referred to as the Minkowski

inequality. It is proven by first obtaining an equivalent version of Holders inequality.

Theorem 27.8 Holder’s inequality. Let p be any real number strictly greater than 1 andf, g ∈ (C[a, b], ‖ ‖p). Let q = p

p−1 . Then,

∫ b

a|f(x)| · |g(x)| dx≤ ‖f‖p‖g‖q

Proof : The proof mimics the proof of Holder’s inequality for sequences replacing thesummation sign,

∑∞i=0 with the integration sign.

Theorem 27.9 Minkowski’s inequality Let p be any real number strictly greater than 1

and f, g ∈ (C[a, b], ‖ ‖p). Then, ‖f + g‖p ≤ ‖f‖p + ‖g‖p. That is,

[∫ b

a|f(x) + g(x)|p dx

]1/p

≤[∫ b

a|f(x)|p dx

]1/p

+

[∫ b

a|g(x)|p dx

]1/p

Proof : The proof mimics the proof of Minkowski’s inequality for sequences replacing thesummation sign,

∑∞i=0 with the integration sign.

Definition 27.10 Let C(K,R) be the set of all (bounded) continuous real-valued function

on the compact set K. If f ∈ C(K,R), we define the sup-norm, ‖f‖∞, of f as

‖f‖∞ = sup|f(x)| : x ∈ K

Example 3. Consider the expression

α(x) = xxxxx·· ·

How should we interpret this expression? Are there values of x, for which this expres-

sion has meaning?


Proposed solution : Let C = fn : n ∈ N\0 be a sequence of functions recursivelydefined as follows. f1(x) = x and for n > 1

fn+1(x) = xfn(x)

Then

f1(x) = x

f2(x) = xx

f3(x) = xxx

f4(x) = xxxx

...

forms an infinite family of continuous functions on the interval (0,∞).

We now reformulate the question: Are there values of x such that the sequence

fn(x) : n ∈ N\0 converges pointwise to α(x)?

There is at least one such value for x. Observe that fn(1) = 1 for all n; hence

limn→∞ fn(1) = 1. Are there others?

Case 1. We analyze the behaviour of the functions, fn(x) : n > 0, for values of x ≥ 1.

Fact #1: For each n, fn(x) is increasing on the part of its domain where x ≥ 1.This can be verified by seeing that f2(x) has a positive derivative whenever x ≥ 1.

Then by mathematical induction, we can conclude that, for all n, fn(x) has a positivederivative, whenever x ≥ 1.

Fact #2: For each a ≥ 1, Ma = fn(a) : n ∈ N\0 is an increasing sequence ofnumbers. This is verified by applying mathematical induction on Ma.

We first determine the interval [1, a) such that

limn→∞

fn(x)

is a number for each x in [1, a). At such points

y = limn→∞

fn+1(x)

= limn→∞

xfn(x)

= xlimn→∞ fn(x)

= xy

So the expression, y = xy, represents a curve which is the pointwise limit of the se-

quence fn(x) : n > 0 for certain values of x greater than or equal to 1.


For what values of x ≥ 1 can we accept y = xy as a representation of the expressionα(x)? We verify if the curve, y = xy, has an unbounded derivative with respect to x.

Observe that y = xy if and only if x = eln yy .

Differentiating both sides with respect to x we obtain

dy

dx=

y2

(1 − ln y)elnyy

We see that the curve of y = xy has a defined derivative whenever y belongs to [1, e)

but gets steeper as it approaches e from the left.1 Solving for x in the expression,e = xe, we obtain x = e1/e.2

So y = xy can suitably represent α(x) for values of x in [1, e1/e).

Case 2. We analyze the behaviour of the functions, fn(x) : n > 0, for values of x in(0, 1].

The functions fn(x) : n > 1 are not defined at x = 0 but converge pointwise to the

curve y = xy for x in (0, 1]. The derivative of the curve, y = xy, with respect to xshows that it is smooth and is positive with respect to x on (0, 1]. So y = xy suitablyrepresents α(x) on (0, 1].

The limit expression,

limn→∞

fn(a) : n > 0

has no (unique) numerical value for a ∈ [e1/e,∞).

Concepts review:

1. Let F (N,R) denote the set of all sequences of real numbers. Define the `p-norm and

`p-space.

2. State the Minkowski inequality for the `p-space, F (N,R).

3. State Holder’s inequality for `p-spaces.

4. Define the sup-norm, ‖f‖∞, on Sb the set of all bounded sequences in F (N,R).

1Also see that the the slope of the tangent line is negative for y > e.2Note that e does not bound all values of fn(x), just the values of fn(x) where fn(x) : n > 0 converges.


5. Describe a condition on S which guarantees that (F (N, S)b, ‖ ‖∞) is complete.

520 Section 28: Completing incomplete metric spaces

28 / Completing incomplete metric spaces

Summary. Some metric spaces are not complete. The space Q is the examplewhich immediately comes to mind. Every irrational number is the limit of a

Cauchy sequence in Q, so R is the smallest non-compact complete metric spacewhich densely contains Q. We investigate how we can complete a metric space

by isometrically and densely embedding it in a complete metric space.

28.1 Introduction.

Recall that a metric space, S, is said to be complete if every Cauchy sequence in Sconverges to a point in S. Not all metric spaces are complete. Simply witness the

subspace of rational numbers. But we can densely embed the rationals in the completemetric space R. We will show that, likewise, any incomplete metric space, S can be

densely embedded in some complete metric space.

28.2 A space whose elements are Cauchy sequences of S.

To achieve this, we will first gather together all Cauchy sequences of the metric space(S, ρ) to form a set,

S = α ∈ P(S) : α = αn : n ∈ N is a Cauchy sequence in S

An equivalence relation on S : If α = αn : n ∈ N and γ = γn : n ∈ N are bothCauchy sequences in S, we will say “α is equivalent to γ” and write α ∼ γ, if and only

if limn→∞ ρ(αn, γn) = 0. The relation ∼ is easily seen to be an equivalence relation onS where [α] = γ ∈ S : γ ∼ α forms an equivalence class and the family of subsets

[S ] = [α] : α ∈ S

partition S . For any η ∈ [α], [η] = [α].

Observations. We are given that α = αn and γ = γn are both Cauchy sequencesin (S, ρ)

1) We claim that if γ ∈ [α] and ε > 0 then there exist M such that n,m > M implies

ρ(αn, γm) < ε. Let ε > 0 and γ ∈ [α] ∈ [S ]. Since γ ∼ α, then limn→∞ ρ(γn, αn) = 0.There exists, M > 0 such that n,m > M implies, ρ(αn, γn) < ε/2 and ρ(γn, γm) < ε/2

(since γ is Cauchy). Then, if n,m > M ,

ρ(αn, γm) ≤ ρ(αn, γn) + ρ(γn, γm)

< ε/2 + ε/2

< ε


So, if γ ∈ [α], there exists M such that

ρ(αn, γm) < ε when n,m > M (∗∗)

as claimed.

2) Also, suppose limn→∞ αn = x ∈ S, and γ ∈ [α]. Then, if ε > 0, there exists, M ,such that if n,m > M ,

ρ(γn, x) ≤ ρ(γn, αm) + ρ(αm, x))

< ε/2 + ε/2 (By (∗∗))

< ε

So, “if α converges to x in S, all Cauchy sequences in [α] converge to x”.

28.3 Defining suitable metric on the set [S ].

We now define a metric δ on [S ] = [α] : α ∈ S as follows: For [α], [β] ∈ [S ],

δ([α], [β]) = inf limn→N

ρ(ηn, γn) : η ∈ [α], γ ∈ [β]

A) We claim that δ is well-defined.

Proof of claim. We prove this in two steps.

1) We begin by showing that δ([α], [β]), has a numerical value. To do this we show that,for the Cauchy sequences α, η in [α] and β, γ in [β], the sequence ρ(ηn, γn) : n ∈ N is

Cauchy in R. If so, then limn→N ρ(ηn, γn) is a positive number (since R is complete);hence inflimn→N ρ(ηn, γn) : η ∈ [α], γ ∈ [β] is a number and so δ([α], [β]) ∈ R.Let ε > 0. Then there exists N such that k, j > N implies ρ(ηj, ηk) < ε/2 and

ρ(γk, γj) < ε/2.

ρ(ηj, γj) ≤ ρ(ηj, ηk) + ρ(ηk, γj)

≤ ρ(ηj, ηk) + [ ρ(ηk, γk) + ρ(γk, γj) ]

ρ(ηj, γj) − ρ(ηk, γk) ≤ ρ(ηj, ηk) + ρ(γk, γj) (Subtract ρ(ηk, γk) on both sides)

< ε/2 + ε/2 = ε (By (∗∗))

ρ(ηk, γk) ≤ ρ(ηk, ηj) + ρ(ηj, γk)

≤ ρ(ηk, ηj) + [ρ(ηj, γj) + ρ(γj, γk)]

ρ(ηk, γk) − ρ(ηj, γj) ≤ ρ(ηk, ηj) + ρ(γj, γk) (Subtract ρ(ηj , γj) on both sides)

< ε/2 + ε/2 = ε (By (∗∗))

Then k, j > N implies,|ρ(ηj, γj) − ρ(ηk, γk)| < ε


So ρ(ηj, γj) : j ∈ N is Cauchy and so is bounded in R, say byK. So limj→∞ ρ(ηj, γj) ≤K and since R is complete limj→∞ ρ(ηj, γj) is a real number greater than zero. Then

δ([α], [β]) ∈ R.

2) Next, we verify that for the Cauchy sequences η, γ in [α] and [β], respectively,δ([η], [γ]) = δ([α], [β]). So the value of δ([α], [β]) does not depend on the choice of

representatives η and γ.

For all j, n,

ρ(η(j)n, γ(j)n

) ≤ ρ(η(j)n, α(j)n

) + ρ(α(j)n, β(j)n

) + ρ(β(j)n, γ(j)n

)

⇒lim

n→∞ρ(η(j)n

, γ(j)n) ≤ lim

n→∞ρ(η(j)n

, α(j)n) + lim

n→∞ρ(α(j)n

, β(j)n) + lim

n→∞ρ(β(j)n

, γ(j)n)

implies

limn→∞

ρ(η(j)n, γ(j)n

) ≤ 0 + limn→∞

ρ(α(j)n, β(j)n

) + 0

implies

δ([η], [γ]) ≤ δ([α], [β])

Similarly

δ([α], [β]) ≤ δ([η], [γ])

So δ([α], [β]) = δ([η], [γ]).

So δ([α], [β]) is well-defined, as claimed.

B) Next we verify that δ : [S ] × [S ] → R+, is a valid metric.

Indiscernibility of identicals. Clearly, if [α] = [β] then δ([α], [β]) = 0.

Identity of indiscernibles. Suppose δ([α], [β]) = 0. Then limn→∞ ρ(ηn, γn) = 0 for

some η ∈ [α], γ ∈ [β]. This implies η ∈ [γ], so [α] = [β].

Symmetry of δ follows immediately from the symmetry of ρ.

Triangle inequality. Let [α], [β] and [ϕ] be three elements of [S ].

Let µ = µn : n ∈ N ∈ [ϕ],

δ(µ, [α]) = inf limn→∞

ρ(µn, ηn) : η ∈ α, µ ∈ µ δ(µ, [β]) = inf lim

n→∞ρ(µn, γn) : γ ∈ β, µ ∈ µ


For i ∈ N\0 there exist η(i) ∈ [α] and γ(i) ∈ [β] such that

limn→∞

ρ(µn, η(i)n) < δ(µ, [α]) + 1/i

limn→∞

ρ(µn, γ(i)n) < δ(µ, [β]) + 1/i

Then, for all i ∈ N, and the chosen µ ∈ [ϕ],


ρ(ηn, γn) : η ∈ [α], γ ∈ [β]

≤ limn→∞

ρ(η(i)n, γ(i)n

)

≤ limn→∞

ρ(η(i)n, µn) + lim

n→∞ρ(µn, γ(i)n

) (Triangle inequality)

< δ([α], µ) + 1/i + δ(µ, [β]) + 1/i

This impliesδ([α], [β]) ≤ δ([α], µ) + δ(µ, [β])

This inequality holds true independently of the choice of µ ∈ [ϕ]. So

δ([α], [β]) ≤ δ([α], [ϕ])+ δ([ϕ], [β])

This establishes the triangle inequality property.

Then δ : [S ]× [S ] → R+, is a valid metric, as claimed. So ([S ], δ) is a metric space.

28.4 The space ([S ], δ) contains an isometric copy of (S, ρ).

If α is a sequence which converges to x ∈ S then α is Cauchy and so [α] ∈ [S ]. Moresignificantly, the constant sequence x = x, x, x, . . . , belongs to [α] and so [α] = [x].

If y = y, y, y, . . . , and x = x, x, x, . . . , are distinct sequences, then [y] 6= [x]. Itfollows that the function f : (S, ρ) → ([S ], δ) defined as

f(x) = [x]

maps S one-to-one into [S ].

We claim the function f is also isometric. It suffices to show that δ( f(x), f(y) ) =ρ(x, y). Suppose x, y ∈ S and η ∈ [x], γ ∈ [y]. Then

ρ(ηn, γn) ≤ ρ(ηn, x) + ρ(x, y) + ρ(y, γn)

implies

limn∈∞

ρ(ηn, γn) ≤ limn∈∞

ρ(ηn, x) + limn∈∞

ρ(x, y) + limn∈∞

ρ(y, γn)

= 0 + ρ(x, y) + 0

= ρ(x, y)


Then

δ([x], [y]) = inf ρ(x, y) : η ∈ [x], γ ∈ [y] = ρ(x, y)

Soδ( f(x), f(y) ) = ρ(x, y)

So the function f is isometric, as claimed.

28.5 The space ([S ], δ), is complete.

If [S ] is complete and contains a dense isometric copy of S (as we will soon show)

then it can be viewed as a completion of S.

Proposition 28.1 The metric space, ([S ], δ), is complete.

Proof : We are given a metric space (S, ρ). We defined S as being the set whose elements,α = αn : n ∈ N, are the Cauchy sequences in S. We are given the set [S ] = [α] :

α ∈ S where β ∈ [α] if and only if β = βn : n ∈ N ∈ S and limn→∞ ρ(αn, βn) = 0.If [α], [β] ∈ [S ],


ρ(ηn, γn) : η ∈ [α], γ ∈ [β]

is a metric.

Let

[U ] = [α(j)] : j ∈ N]be a Cauchy sequence in [S ] with α(j) = α(j)n

: n ∈ N as representative of theclass, [α(j)]. We are required to show that [U ] converges to some [ζ] in [S ].

Construction of a Cauchy sequence ζ. Let ε > 0.

Since [U ] is Cauchy in [S ], there exists M such that j, k > M ⇒ δ([α(j)], [α(k)]) <ε/3. That is, for any j, k > M ,

inf limn→∞

ρ(η(j)n, η(k)n

) : η(j) ∈ [α(j)], η(k) ∈ [α(k)] < ε/3

Then there exists at least one pair j, k > M such that,

limn→∞

ρ(α(j)n, α(k)n

) < ε/3. 1

1For, if limn→∞ ρ(α(j)n, α(k)n

) ≥ ε/3, for all j, k > M δ([α(j)], [α(k)]) = inflimn→∞ ρ(α(j)n, α(k)n

) :η(j) ∈ [α(j)], η(k) ∈ [α(k)] ≥ ε/3


So, for this j, k, there exists N1 such that n > N1 implies,

ρ(α(j)n, α(k)n

) < ε/3 (∗)

Consider the triangle inequality expression for ρ,

ρ(α(k)k, α(j)j

) ≤ ρ(α(k)k, α(k)n

) + ρ(α(k)n, α(j)n

) + ρ(α(j)n, α(j)j

)

By (∗) there is j, k > M and N1, such that

n > N1 ⇒ ρ(α(j)n, α(k)n

) < ε/3

Since a(k) = α(k)n: n ∈ N is Cauchy, there exists N2 such that,

n, k > N2 ⇒ ρ(α(k)k, α(k)n

) < ε/3

Since a(j) = α(j)n: n ∈ N is Cauchy in S, there exists N3 such that,

j, n > N3 ⇒ ρ(α(j)j, α(j)n

) < ε/3

Then for, k, j, n > N = max M,N1, N2, N3

ρ(α(k)k, α(j)j

) ≤ ρ(α(k)k, α(k)n

) + ρ(α(k)n, α(j)n

) + ρ(α(j)n, α(j)j

)

< ε/3 + ε/3 + ε/3

= ε

Then ∪a(j) : j ∈ N contains a sequence ζ = a(j)j: j ∈ N in S which is a Cauchy

sequence and so belongs to S . Then [ζ] ∈ [S ].

Claim. We claim that the Cauchy sequence, [U ] = [α(j)] : j ∈ N], in [S ] converges

to [ζ].

Proof of claim. It suffices to show that limj→∞ δ([α(j)], [ζ]) = 0.

Let ε > 0.

Since ζ = α(0)0, α(1)1

, α(2)2, . . . , is a Cauchy sequence in S, there exists M1 such

that j, k > M1 implies ρ(α(j)j, α(k)k

) < ε/2.

Since a(j) = α(j)n: n ∈ N is Cauchy in S, there exists M2 such that j, k > M2

implies ρ(α(j)k, α(j)j

) < ε/2.

By triangle inequality,

ρ(α(j)k, α(k)k

) ≤ ρ(α(j)k, α(j)j

) + ρ(α(j)j, α(k)k

)


If j, k > M = maxM1,M2 then

ρ(α(j)k, α(k)k

) ≤ ρ(α(j)k, α(j)j

) + ρ(α(j)j, α(k)k

)

< ε/2 + ε/2

= ε

So

j, k > M ⇒ ρ(α(j)k, α(k)k

) < ε

Since inflimk→∞ ρ(η(j)k, η(k)k

) : η(j) ∈ [α(j)], η(k) ∈ [ζ] < ε, then

limj→∞ δ([α(j)], [ζ]) = 0.

We conclude that [U ] converges to [ζ], as claimed.

With the next result we complete the proof of the statement

“Every metric space has a completion.”

Proposition 28.2 The metric space (S, ρ) is dense in ([S ], δ).

Proof : We are given a metric space (S, ρ) and S , the set of all Cauchy sequences in S. Ifα ∈ S , [α] = η ∈ S : limn→∞ ρ(ηn, αn) = 0. Let [S ] = [α] : α is Cauchy in S.For x = x, x, x, . . . , in S let

[S] = [x] : x ∈ S

in [S ]. The function f(x) = [x], embeds an isometric copy, f [S], of S into [S ].

For α = αn : n ∈ N, suppose [α] ∈ [S ] \ [S]. To show that [S] is dense in [S ], itwill suffice to show that there is a sequence, [a(n)] : n ∈ N, in [S] which converges

to [α] with respect to the metric δ.

We define [a(n)] : n ∈ N in terms of α = αn : n ∈ N as,

a(0) = α0, α0, α0, . . . .a(1) = α1, α1, α1, . . . .a(2) = α2, α2, α2, . . . .

...

a(k) = αk, αk, αk, . . . ....


We claim that [a(n)] : n ∈ N converges to [α].

Proof of claim. Let ε > 0. It suffices to show there is a N0 such that k > N0 impliesδ([α], [ak]) < ε.

Since α, is Cauchy in S, there exists N0, such that

m, n > N0 ⇒ ρ(αn, αm) < ε

Suppose k > N0 and n > N0. Then

δ([α], [ak]) = inf limn→N

ρ(ηn, γn) : η ∈ [α], γ ∈ [ak]

= inf limn→N

ρ(αn, αk) : η ∈ [α], γ ∈ [ak] = ρ(αn, αk)

< ε

Then [a(n)] : n ∈ N converges to [α].

So (S, ρ) is dense in ([S ], δ).

528 Section 29: The uniform space and the uniform topology

29 / The uniform space and the uniform topology

Summary. This section serves as a brief introduction to uniform space. Givena non-empty set S we define a family of subsets, U , of P(S×S) which satisfies

a set of axioms. The family, U , is called a uniformity on S and the pair (S,U ),is called a uniform space. We then show that any metric space, (S, ρ), generates

a uniform space, (S,Uρ), which inherits some of its metric space properties.Even though (S,U ) is not a topological space, the uniformity U , generates a

topology τU , on S. A topological space, (S, τ), which can be obtained from auniformity, U , is called a uniformizable topological space.

29.1 Introduction.

A topological space generalizes many concepts and properties associated to metric

spaces. But, some properties closely associated to the structure of a metric spacedo not transfer easily to a less structured topological space. For example, notions

associated to “Cauchy sequences” or “uniform continuity” are not meaningful in thecontext of a topological space even though such tools can be useful in describing some

topological properties. To see this, witness how the function g(x) = 1/x on R\0maps R\0 homeomorphically onto itself. The sequence T = 1, 1/2, 1/3, . . . , is

a Cauchy sequence in the domain of g. But its image, g[T ] = 1, 2, 3, . . . , , is notCauchy in R\0. So being “Cauchy” is not a topological property and so cannot be

adequately expressed in this mathematical context.

In this section we discuss a space which, like topological space, generalizes metricspaces, but in slightly different way.

We begin by reviewing a few basic notions associated with relations.

Basic definitions and notation. Recall that for a given non-empty set S, a subset Uof S × S is referred to as a (binary) relation on S. If U is a relation on S,

dom U = x ∈ S : there is a y ∈ S such that (x, y) ∈ Uim U = y ∈ S : (x, y) ∈ U for some x ∈ S

If U is a relation on S, then its inverse, U−1, is defined as,

U−1 = (x, y) : (y, x) ∈ U

Composition of relations. If U and V are relations on S

U V = (a, b) : (a, c) ∈ V and (c, b) ∈ U for some c ∈ im V


The relation∆(S) = (x, x) : x ∈ S

is referred to as the diagonal of S × S.

Symmetric relations. The relation, U , is said to be symmetric if and only if U = U−1.Alternatively, U is symmetric whenever. . .

“ (a, b) ∈ U if and only if (b, a) ∈ U ”

Definition 29.1 Let S be a non-empty set. A uniform space is a pair (S,U ) where Uis a non-empty subset of P(S × S) whose elements satisfy the following five axioms:

U1. ∆(S) ⊆ U for all U ∈ U

U2. If U, V ∈ U , then U ∩ V ∈ U . That is, U is closed under finite intersections.

U3. If U ∈ U then V V ⊆ U for some V ∈ U .

U4. If U ∈ U then E−1 ⊆ U for some E ∈ U .

U5. If U ∈ U and U ⊆ V ∈ P(S×S), then V ∈ U . That is, U is closed under supersets.

A subset, U , of P(S × S) which satisfies these five axioms is called a uniformity for S.

It may be helpful to view each element of a uniformity, U , of S, as a region in S × Ssurrounding the diagonal, ∆(S). A pair of points a, b in S may be considered as be-

ing “near” each other if and only if the pair (a, b) in S×S is “near” the diagonal, ∆(S).

In spite of its nomenclature, a uniform space, (S,U ), is not a topological space. Wewill however soon define a topology, τU , on S which is derived from the structure,

U . Remember that, even though a topology and a uniformity are both families ofsubsets, a topology is a subset of P(S) while a uniformity is a subset of P(S × S).

Like a topology on S, a uniformity, U , for S is not unique.

Also note that, since U is non-empty and is closed both under finite intersections andsupersets (U2 and U5), it is a filter of sets on S × S.

Example 1. The largest uniformity on S, is the family

Ud = U ∈ P(S × S) : ∆(S) ⊆ U

It is called the discrete uniformity.


Example 2. The smallest uniformity on S is the singleton set,

Ui = S × S

It is called the trivial uniformity or the indiscrete uniformity.

Proposition 29.2 Let S be a non-empty set and U be a uniformity on S. If U ∈ U , then

U−1 also belongs to U on S.

Proof : We are given a set S and an element, U , of a uniformity, U . By property U4, Ucontains an element V such that V −1 ⊆ U . Then V = (V −1)−1 ⊆ U−1. Since V ∈ Uand V ⊆ U−1 then, by U5, U−1 ∈ U , as required.

Note that,

U ∈ U ⇒ [U ∩ U−1 ∈ U and is symmetric] (†)To see this, note that if (a, b) ∈ U ∩U−1 then (a, b) ∈ U , so (b, a) ∈ U−1; if (b, a) 6∈ U ,

then (a, b) 6∈ U−1, contradiction. So (b, a) ∈ U . Then U ∩ U−1 is symmetric.

29.2 A base for a uniformity.

To facilitate our investigation of uniform spaces, (S,U ), we first briefly review some ofthe mechanisms used to generate various topologies on a set S. There are surprisingly

few axioms that govern the family, τ , which defines a topological space. They simplystate that τ contains both ∅ and S, is closed under finite intersections and closed

under arbitrary unions. We also know that any given topology, τ , on S, contains adistinguished subfamily, B, referred to as a base for τ , where,

“. . . for every x ∈ S and x ∈ U ∈ τ , there is a B ∈ B such that x ∈ B ⊆ U”.

It allows us to recover all open sets simply by taking arbitrary unions of elements of B.

We also saw that, given any (untopologized) set S and B ⊆ P(S), the property,

“. . . if B covers S and for A,B ∈ B, if x ∈ A∩B, there is C ∈ B such that

x ∈ C ⊆ A ∩B.

characterizes a base for a topology on S. By this we mean that, if B satisfies this

property, it generates a topology, τ , on S by collecting the union of all subsets of B.More specifically,

τ = U ∈ P(S) : U = ∪C : C ∈ C for some subset C of B


The definition we provide for the base of a uniformity will, in some ways, be similarto the one given for a base for a topology.

Definition 29.3 Let S be a non-empty set and U be a uniformity on S.

A subfamily B of U is a base for U , if,

. . . for every U ∈ U , there is a B ∈ B such that B ⊆ U

A subfamily, S , of U is a subbase for U , if the family of finite intersections of the membersof S forms a base for U .

The characterization we will obtain for the uniformity base, B, is a slightly weakened

version of axioms U1 to U4. The uniformity, U , is then generated from B by applyingthe axiom U5 to B.

Theorem 29.4 Let S be a non-empty set and B be a non-empty subset of P(S × S).

The family, B, is a base for some uniformity on S if and only if, B satisfies U1, U2∗, U3

and U4 where U2∗ is a slightly weaker version of U2:

U2∗ : “ If U, V ∈ B, then W ⊆ U ∩ V ∈ U for some W ∈ B ”

Proof : We are given a set S.

( ⇐ ) Suppose B is a subset of P(S×S) which satisfies properties, U1, U2∗, U3 andU4. We are required to show that B is a base for some uniformity on S.

LetUB = V ∈ P(S × S) : V contains some B ∈ B

- By definition of UB every V in UB contains an element of B.

- Furthermore, if B ∈ B, B ⊆ B so B is a subfamily of UB.

It now suffices to show that the family, UB generated by B, is a uniformity on S. We

do this by verifying that it satisfies the properties U1 to U5.

U1 : Since ∆(S) ⊆ B for all B ∈ B and B ⊆ V for some V in UB then every element

of UB contains ∆(S). So UB satisfies U1.

U2 : Let U, V ∈ UB. Then B ⊆ U , W ⊆ V for some B,W ∈ B. There exists Z ∈ Bsuch that Z ⊆ B ∩W (by U2∗). Then Z ⊆ U ∩ V . By definition of UB, U ∩ V ∈ UB.


So UB satisfies U2.

U3 : Let U ∈ UB. Then B ⊆ U for some B ∈ B. Since B satisfies U3, W W ⊆ Bfor some W ∈ B where W ∈ UB. So W W ⊆ U . We conclude that UB satisfies U3.

U4 : Let U ∈ UB. Then B ⊆ U for some B ∈ B. Since B satisfies U4, W−1 ⊆ B for

some W ∈ B where W ∈ UB. Then W−1 ⊆ U . So UB satisfies U4.

U5 : Let U ∈ UB and V ∈ P(S × S) such that U ⊆ V . Then B ⊆ U ⊆ V for someB ∈ B. Then B ⊆ V . By definition of UB, V ∈ UB. So UB satisfies U5.

We conclude that if B is a subset of P(S × S) which satisfies properties, U1, U2∗,

U3 and U4, then it is a base for the uniformity, UB.

( ⇒ ) Suppose now that B is a base for the uniformity U on S. Then B is a sub-

family of U where for every V ∈ U there is a B ∈ B such that B ⊆ V . It suffices toshow that B satisfies U1, U2∗, U3, U4.

That B satisfies U1 and U2∗ is immediate.

Suppose B ∈ B. Since B ∈ U there is a V ∈ U such that V V ⊆ B for some V ∈ U .

Since B is a base of U there is a W ∈ B such that W ⊆ V . Then W W ⊆ V V ⊆ B.So B satisfies U3.

Suppose B ∈ B. Since B ∈ U there is a V ∈ U such that V −1 ⊆ B for some V ∈ U .

Since B is a base of U there is a W ∈ B such that W ⊆ V . Then W−1 ⊆ V −1 ⊆ B.So B satisfies U4.

We are done.

The above result guarantees that . . .

. . . if B ⊆ P(S× S) satisfies U1, U2∗, U3 and U4 then the collection, UB,

of all supersets of elements of B is a uniformity on S.

The following result shows that, for any uniformity U on S, the family of all sym-

metric relations in U forms a base for U .

Theorem 29.5 Let S be a non-empty set and U be a uniformity on S with base B.

Suppose

Bsym = V ∈ U : V is symmetric

Then Bsym forms a base for the uniformity U .


Proof : Let U be a uniformity on S with base B and U ∈ B ⊆ U . By proposition 29.2,U−1 ∈ U . Let B = U ∩ U−1. By U2, B ∈ U . If (a, b) ∈ U ∩ U−1 then (a, b) ∈ U

and so (b, a) ∈ U−1. Since (a, b) ∈ U−1 then (b, a) ∈ U . Then (b, a) ∈ U ∩ U−1. SoB = U ∩ U−1 is symmetric. Since B ⊆ U , Bsym forms a base for the uniformity U ,

as required.

Example 3. Consider the set of real numbers R. For κ > 0, let

Bκ = (a, b) ∈ R × R : |a− b| < κ

Verify that the collection,

B = Bκ : κ > 0forms a base for some uniformity on R.

Solution : We verify that B satisfies the four base properties for a uniformity.

U1. Since (x, x) ∈ Bκ for all κ, then B satisfies U1.

U2∗. For κ1 and κ2 larger than 0, let κ3 = minκ1, κ2. If (x, y) ∈ Bκ3 then (x, y)

are strictly within a distance of κ3 from each other. So (x, y) ∈ Bκ1 ∩ Bκ2 . ThenBκ3 ⊆ Bκ1 ∩Bκ2 . It follows that B satisfies U2∗.

U3. Let Bκ ∈ B. We claim there exists λ such that BλBλ ⊆ Bκ.

Let λ = κ/4. Recall that

U V = (u, v) : (u, y) ∈ V and (y, v) ∈ U for some y ∈ im V .

Let (u, v) ∈ Bκ/4Bκ/4. We claim that |u− v| < κ.

(u, v) ∈ Bκ/4Bκ/4 ⇒ ∃ z ∈ im Bκ/4 such that (u, z) ∈ Bκ/4 and (z, v) ∈ Bκ/4

⇒ |u− z| < κ/4 and |z − v| ∈ κ/4

|u− v| ≤ |u− z| + |z − v|< κ/4 + κ/4

= κ/2 < κ

Then (u, v) ∈ Bκ. So Bκ/4Bκ/4 ⊆ Bκ. We conclude that B satisfies U3.

U4. Let Bκ ∈ B. Since Bκ = (x, y) : |x− y| < κ = (x, y) : |y − x| < κ = B−1κ ,

then B satisfies U4.

ThenB = Bκ : κ > 0

is a base which generates a uniformity on R. Then the uniformity on R is

U = V ∈ P(R × R) : Bκ ⊆ V for some κ > 0


Definition : The uniformity on R generated by the base, B = Bκ : κ > 0, illus-trated in the above example is called the usual or the standard uniformity on R.

29.3 A uniform space, (S, U ), generated by a metric, ρ, on S.

We see that a uniform space, (S,U ), has slightly more structure than a topologicalspace (S, τ). In the following example we verify that a metric, ρ, on a set S will always

generate a uniformity on S.

Example 4. Let (S, ρ) be a metric space. For each κ > 0, let

Bκ = (x, y) ∈ S × S : ρ(x, y) < κ

be a subfamily of P(S × S) generated by the metric ρ. Verify that the family,

Bρ = Bκ : κ > 0, is a base which generates a uniformity, Uρ, on S.

Solution : It will suffice to show that Bρ satisfies properties U1, U2∗, U3 and U4. Ifso, then, by theorem 29.4, Bρ generates the uniformity,

Uρ = V ∈ P(S × S) : V contains some Bκ ∈ Bρ

Since ρ(x, x) = 0 < κ for all x ∈ S, then ∆(S) ⊆ Bκ for each κ. This is property U1.

For Bκ1 , Bκ2 ∈ Bρ, let κ = minκ1, κ2/2. Then (x, y) ∈ Bκ implies ρ(x, y) < κ.Since κ < κ1 and κ < κ2, then (x, y) ∈ Bκ1 ∩ Bκ2 . So Bκ ⊆ Bκ1 ∩ Bκ2 . Then there

exists κ such that Bκ ⊆ Bκ1 ∩ Bκ2 . This is property U2∗.

Let Bκ ∈ Bρ. We claim there exists λ such that BλBλ ⊆ Bκ. Let λ = κ/4. Recallthat

U V = (u, v) : (u, y) ∈ V, (y, v) ∈ U, for some y ∈ im V .

Let (u, v) ∈ Bκ/4Bκ/4. We claim that ρ(u, v)< κ.

(u, v) ∈ Bκ/4Bκ/4 ⇒ there is z ∈ im Bκ/4 such that (u, z) ∈ Bκ/4, (z, v) ∈ Bκ/4

⇒ ρ(u, z) < κ/4 and ρ(z, v) ∈ κ/4

ρ(u, v) ≤ ρ(u, z) + ρ(z, v)

< κ/4 + κ/4

= κ/2 < κ

Then (u, v) ∈ Bκ. So Bκ/4Bκ/4 ⊆ Bκ. We conclude that B satisfies U3.

Let Bκ ∈ Bρ. Recall Bκ = (x, y) : ρ(x, y) < κ = (x, y) : ρ(y, x) < κ = B−1κ .

Then Bρ satisfies U4.

Then Bρ is a base which generates a uniformity on (S, ρ).


Definition : Given a metric space, (S, ρ), the uniformity, Uρ, generated by ρ is calledthe metric uniformity.

This example shows that every metric, ρ, on a space S, will generate a uniformity,

Uρ. In example three, the uniformity described on R is simply the metric uniformitygenerated by the metric ρ(a, b) = |a− b| on R.

Example 5. Show that the metric 2ρ generates the same uniformity on S as the metric

space ρ.

Solution : It will suffice to show that Uρ = U2ρ.Suppose V ∈ Uρ. Then V ∈ P(S×S) contains some Bλ ∈ Bρ. There exist Bµ ∈ B2ρ

such that Bµ ⊆ Bλ ⊆ V . So V ∈ U2ρ. Then Uρ ⊆ U2ρ.

Suppose V ∈ U2ρ. Then V ∈ P(S×S) contains some Bλ ∈ B2ρ. There existBµ ∈ Bρ

such that Bµ ⊆ Bλ ⊆ V . So V ∈ Uρ. Then U2ρ ⊆ Uρ.

Then ρ and 2ρ generate the same uniformity on S.

Since a single uniformity does not always distinguish distinct metric spaces uniformspaces have less structure than metric spaces.

29.4 The uniform topology, τU , generated by a uniformity, U , on a set.

It is important to keep in mind that a uniformity, U , on a set is not, by itself, atopology on S. It is not even a subset of P(S). We will show, however, that we can

use the elements of a uniformity, U , on S to define, in a natural way, a correspondingtopology, τU on S.

Recall that, if B is an element of a uniformity U on S, then B : S → S is a subset ofS×S which contains ∆(S). So, for each B ∈ U , the image of x under B is a subset

of S which contains the point x. Let B(x) denote the image of x. The collection

B(x) : B ∈ U is a family of subsets of S each containing x. Since, for each B ∈ U , the image of B

at x is B(x) = y ∈ S : (x, y) ∈ B, B(x) can also be viewed as

B(x) = π2 [ ( x × S ) ∩ B ] 1

Theorem 29.6 Let S be a non-empty set, U be a uniformity on S and B be a base forU . For x ∈ S, and B in U , let B(x) denote the image of x (in S) under the relation, B.

Let

τU = U ∈ P(S) : for each x ∈ U there exists B ∈ U such that B(x) ⊆ UThen τU is a topology on S with base Bτ = B(x) : x ∈ S and B ∈ B.

1Where π2(a, b) = b.


Proof : We are given a set S and a uniformity U on S and the family of subsets,τU = U ∈ P(S) : for every x ∈ U there exists B ∈ U such that B(x) ⊆ UO1. Clearly ∅ ∈ τU . Let x ∈ S. Since S × S ∈ U , x ∈ (S × S)(x) = S ⊆ S. So

S ∈ τU .

O2. Let U, V ∈ τU . We claim U ∩ V ∈ τU . Suppose x ∈ U ∩ V . It suffices to show

there is a W ∈ U such that W (x) ⊆ U ∩ V .

By definition of τU there exists B1, B2 in U such that x ∈ B1(x) ⊆ U and x ∈B2(x) ⊆ V . By U2, B1 ∩B2 ∈ U .

z ∈ (B1 ∩ B2)(x) ⇒ (x, z) ∈ B1 ∩B2

⇒ z ∈ B1(x) and z ∈ B2(x)

⇒ z ∈ B1(x) ∩B2(x)

⇒ z ∈ U ∩ V

Then (B1 ∩ B2)(x) ⊆ U ∩ V . So U ∩ V ∈ τU .

O3. Let Ui : i ∈ A be a family of elements of τU . If x ∈ ∪Ui : i ∈ A then x ∈ Uj

for some j ∈ A. Then there exists B ∈ U such that x ∈ B(x) ⊆ Uj ⊆ ∪Ui : i ∈ A.So τU is closed under arbitrary unions.

So τU is a topology. By definition of τU , for each U ∈ τU and x ∈ U there is aV ∈ U such that x ∈ V (x) ⊆ U . Then there is a B ∈ B such that B ⊆ V and sox ∈ B(x) ⊆ V (x) ⊆ U . We conclude that

Bτ = B(x) : x ∈ S and B ∈ B

forms a base for τU . (Note that, in the proof, we can choose as base B for the uni-formity, U , the family, B = Bsym, if necessary.)

Definition 29.7 Let S be a non-empty set and U be a uniformity on S. The topologyon S, τU , generated by the uniformity, U , is called a uniform topology.

Let (S, τ) be a topological space. If

τ = τU

for some uniformity, U , on S, then (S, τ) is said to be uniformizable topological space.


In the literature, the words “uniform space” is often used to refer to both (S,U ) and(S, τU ). It is usually clear from the context to which space the author is referring to.

Theorem 29.6 guarantees that given a uniformity on a set S we can generate a corre-

sponding topology on S:

S −→ (S,U ) −→ (S, τU )

Finally, note that, for a given set S, we can speak of a base B for the uniformity Uand a base B for the topology τU . Even though we tend to use the same symbol, B,

for both bases, the reader should determine, by the context, which one it refers to.

For convenience we juxtapose both concepts.

B is a base for U : For every U ∈ U , there is a B ∈ B such that B ⊆ U .

Bτ is a base for τU : Bτ = B(x) : x ∈ S and B ∈ B, a base for U .

By definition 29.7, a topological space, (S, τ), derived from some uniformity, U , on

S (that is, where τ = τU ) is said to be “uniformizable”. Are there any topologicalspaces that are not uniformizable? That is, is there a topology, τ , on S for which

there is no uniformity, U , such that τ = τU ? The answer is yes. We will show intheorem 29.15 that a topological space derived from a uniformity is a regular space

(i.e., T1 + T3).1 So a non-regular space is not uniformizable.

Example 6. Consider the usual uniformity, U , on R which is generated by the base,B = Bκ : κ > 0, given in example three (where Bκ = (a, b) ∈ R×R : |a−b| < κ).By theorem, 29.6, the uniformity, U , generates a topology τU on R. Verify that τUis the usual topology, τ , on R. That is,

R −→ (R,U ) −→ (R, τU ) = (R, τ)

Solution : Let U be the usual uniformity on R and τ denote the usual topology on R.

Then U is generated by the base, B = Bκ : κ > 0, where Bκ represents all pairs(a, b) such that |a− b| < κ .

Claim #1: We claim that τU ⊆ τ .

Proof of claim. Let U be an open neighbourhood of x ∈ R with respect to τU . Bydefinition of τU , there exists Bκ ∈ B such that x ∈ (Bκ)(x) ⊆ U .

See that (Bκ)(x) = (x − κ/2, x+ κ/2). So U is a union of basic open intervals. So

U ∈ τ . We conclude that τU ⊆ τ , as claimed.

Claim #2: We claim that τ ⊆ τU .

1In fact, a uniformizable space is always completely regular. We will not prove this in this book. For aproof the reader is referred to theorem 38.2 of Willard where it is shown that, for (S, τ), “uniformizable ⇔completely regular”.


Proof of claim. Let A be a non-empty open subset in R and x ∈ A. Then thereexists ε such that (x − ε, x + ε) ⊆ A. Consider Bε/2, a base element of U . Then

(Bε/2)(x) ∈ τU . By claim #1, τU ⊆ τ ; then (Bε/2)(x) ∈ τ .

See that,

x ∈ (x− ε/4, x+ ε/4)

= π2

[

(x × S ) ∩ Bε/2

]

= (Bε/2)(x)

⊆ (x− ε, x+ ε)

⊆ A

Then the family, B = B(x) : x ∈ R, B ∈ U forms a base for τ . By theorem 29.6,B is also a base for τU . So A ∈ τU . We conclude that τ ⊆ τU , as claimed.

So τU = τ .

We now examine, in a more general context, how the open subsets of the topology,

τU , relate to the uniformity, U , which generates it.

Theorem 29.8 Let S be a non-empty set and U be a uniformity on S. Let τU be the

topology on S generated by U . If T ⊆ S, then

intτUT = x ∈ T : x ∈ B(x) ⊆ T for some B ∈ U

Proof : We are given a set S and a uniformity U on S.

For each B ∈ U , B(x) = π2[ (x× S)∩B ] and τU is the topology generated by thebase B = B(x) : x ∈ S and B ∈ U .Let K = x ∈ T : x ∈ B(x) ⊆ T for some B ∈ U . By definition K ⊆ T . We arerequired to show that K = intτU

T .

We first show that K is open in S (hence K ⊆ intτUT ).

Let x ∈ K. Then there exists U ∈ U such that x ∈ U(x) ⊆ T . There exists V ∈ Usuch that V V ⊆ U . We claim V (x) ⊆ T ; if so x ∈ V (x) ⊂ K and so K is open.


Let u ∈ V (x). Then we have V (u) ⊆ (V V )(x). We then obtain the chain of inclusions,

u ∈ V (u)

⊆ (V V )(x)

⊆ U(x)

⊆ T

So u ∈ T implies V (x) ⊆ T , as claimed.

Then x ∈ V (x) ⊂ K, which implies K is open.

Then K ⊆ intτUT .

We now show intτUT ⊆ K. Let x ∈ intτU

T . Since B is a base for τU and intτUT is

open there exists B ∈ U such that x ∈ B(x) ⊆ intτUT ⊆ T . So x ∈ K.

Then intτUT = K.

The following results exhibits another way of expressing the closure of a set in (S, τU ).

In the statement, we use the following notation.

Notation. Let U be a uniformity on S. If T ⊆ S then

U [T ] = ∪x∈T U(x)


topology on S generated by U . Let T ⊆ S. Then the closure, clτUT , of T with respect to

the uniform topology τU can be expressed as, then

clτUT =

⋂

U∈U

U [T ]

Proof : We are given a set S and a uniformity U on S from which we obtain the space(S, τU ) equipped with the uniform topology. Let T ⊆ S.

We know that B = B(x) : x ∈ S and B ∈ Bsym forms a base of τU .

We are required to show that x ∈ clτUT if and only if x ∈ ⋂U∈U

U [T ].To say that “x ∈ clτU

T ” is equivalent to saying that “B(x) intersects T for allB ∈ B ”.

To say that “B(x) intersects T for all B ∈ B ” is equivalent to saying that “x belongsto B−1[T ] for all B ∈ B ”. But since B is symmetric, we can replace “B−1[T ]” with


“B[T ] ”.

So we can rewrite the second statement as:

“The set, B(x) intersects T for all B ∈ B if and only if, x ∈ ⋂U∈UU [T ]”.

The desired conclusion follows.

Example 7. Suppose (S,U ) is the uniform space equipped with the discrete unifor-mity. Show that the corresponding topological space is equipped with the discrete

topology, τU = P(S).

Solution : By definition, U = U ∈ P(S×S) : ∆(S) ⊆ U, is the discrete uniformity.

By definition, a base for U is a subfamily, B, such that for every U ∈ U , there is aB ∈ B such that B ⊆ U . We choose,

B = ∆(S)

By definition, for the given B,

Bτ = B(x) : x ∈ S and B ∈ B a base for U

forms a base for (S, τU ).

For each x ∈ S,

B(x) = π2[ (x × S) ∩ B ]

= π2[ (x × S) ∩ ∆(S) ]

= x

Then x : x ∈ S forms a base for the uniform topology. This base generates

τU = P(S).

Example 8. Suppose (S,U ) is the uniform space equipped with the single elementuniformity, U = S×S. Show that the corresponding topological space is equippedwith the trivial topology, τU = ∅, S.

Solution : We are given that U = S × S, so B = S × S is a base for U .Now, Bτ = B(x) : x ∈ S and B ∈ B a base for U . For each x ∈ S, B(x) =

π2[ (x × S) ∩ (S × S) ] = S. So τU = ∅, S.

Example 9. For each κ ∈ R, let

Bκ = ∆(R) ∪ [κ,∞)× [κ,∞)


Suppose (R,U ) is the uniform space where U has as base,

B = Bκ : κ ∈ R

Show that the corresponding topological space, (R, τU ), is the discrete space.

Solution : We are given the family of subsets B = Bκ : κ ∈ R is a base for U . It

is easily verified that B satisfies the properties required for it to serve as a base forsome uniformity, U , on S.

Now τU = U ∈ P(S) : for each x ∈ U there exists B ∈ U such that B(x) ⊆ U.If x < κ then B(x) = x; so x is an open base element of R. So (R, τU ) is a

discrete topological space.

The following result will be a useful tool in proofs to come.

Proposition 29.10 For any M ∈ U , there exists W ∈ B such that W W W ⊆M .

Proof : We are given a set S, a uniformity U on S. For any M ∈ U , by U3, there is aB ∈ B such that BB ⊆M . There is a D ∈ B such that DD ⊆ B. Since

DD ⊆ B

(D ∩B)(D ∩ B) ⊆ DD ⊆ B

(D ∩ B)(D ∩B)(D ∩ B) ⊆ (D ∩B)B ⊆ BB ⊆M

there exists W = D ∩ B ∈ B such that W W W ⊆M .

This proves the proposition.

29.5 On the product topology derived from (S, τU ) × (S, τU ).

Since a uniformity, U , is subfamily of P(S×S), every element, M , in U can be viewedas a subset of the Cartesian product, (S, τU ) × (S, τU ), equipped with the producttopology. We will see that every element, M , in U on S is a neighbourhood of the

diagonal, ∆(S) (with respect to this product topology). That is, ∆(S) ⊆ intS×SM .To confirm this fact, it will suffice to show that, for all M ∈ U , intS×SM ∈ U . For,

if this is so, it will follow from U1 that ∆(S) ⊆ intS×SM , for all M ∈ U .


Theorem 29.11 Let S be a non-empty set and U be a uniformity on S. If M ∈ U thenintS×SM ∈ U .

Proof : We are given a set S, a uniformity U on S and the uniform topology,

τU = U ∈ P(S) : for each x ∈ U there exists B ∈ U such that B(x) ⊆ U

on S. Recall (from theorem 29.6) that B = B(x) : x ∈ S and B ∈ Bsym forms abase for the topology, τU , on S (where Bsym denotes all symmetric relations in U ).

An open base element of S × S (equipped with the product topology) is of the formU(x) × V (y) for some U, V ∈ B ⊆ U . We will begin by establishing the following

three facts.

Fact #1. For any M ∈ U ,

intS×SM = (x, y) : (x, y) ∈W (x) ×W (y) ⊆M for some W ∈ B

Proof: By U2, U ∩ V ∈ U Then (U ∩ V )(x) ⊆ U(x) and (U ∩ V )(y) ⊆ V (y).

Then (U ∩ V )(x)× (U ∩ V )(y) ⊆ U(x)× V (y).

Let M ∈ U . We are required to show that intS×SM ∈ U . The set, intS×SM , can be

expressed as,

intS×SM = (x, y) : (x, y) ∈ U(x)× V (y) ⊆M for U, V ∈ B

Since (U ∩ V )(x) ⊆ U(x) and (U ∩ V )(y) ⊆ V (y), then (U ∩ V )(x) × (U ∩ V )(y) ⊆U(x)× V (y).

By U2, U ∩ V ∈ U . Also, if U, V ∈ B then U ∩ V ∈ B. If we let W = U ∩ V , the

set, intS×SM , can then be expressed as,


This proves Fact #1.

Fact #2. For any M ∈ U , there exists W ∈ B such that W W W ⊆M .

This fact is proven in proposition 29.10.

Fact #3. For any U,W ∈ U where W is known to be symmetric,

W U W =⋃

(x,y)∈U

W (x)×W (y)

Proof: Consider the composition W U W where W is symmetric in U . Recall that

W U W = (u, v) : (u, x) ∈W and (y, v) ∈W for some (x, y) ∈ U (∗)


See that, if (u, v) ∈ W , then u ∈ W (x) and v ∈ W (y) for some (x, y) ∈ U . So(u, v) ∈W (x)×W (y). Then, for some (x, y) ∈ U , W ⊆W (x) ×W (y).

Conversely, if, for some (x, y) ∈ U , (a, b) ∈ W (x) × W (y), then a ∈ W (x) and

b ∈ W (y). So (a, x) ∈ W and (b, y) ∈ W . Since W is symmetric, (y, b) ∈ W . Since(x, y) ∈ U , then (a, b) ∈W , so W (x)×W (y) ⊆W .

We conclude that, whenever W is symmetric, there is (x, y) ∈ U such that

W = W (x)×W (y)

Restating (*) we get,

W U W = (u, v) ∈ S × S : (u, v) ∈W (x) ×W (y) for some (x, y) ∈ U

Equivalently,

W U W =⋃

(x,y)∈U

W (x)×W (y)

This proves Fact #3.

We now combine these few facts, to proceed as follows. Let M ∈ U .

By Fact #1,


If we prove that W ⊆ intS×SM , then by applying U5, we will be done.

By Fact #2, there exists W ∈ B such that W W W ⊆M .

By Fact #3,

W W W =⋃

(x,y)∈W

W (x)×W (y) ⊆M

Then

. . . (x, y) ∈W ⇒ (x, y) ∈W (x) ×W (y) ⊆ intS×SM

Then every point of W belongs to intS×SM . So W ⊆ intS×SM .

Since W ∈ B, by U5, intS×SM ∈ U . We are done.

We know that B = B ∈ U : B is symmetric forms a base for U . This means that,for every U ∈ U , there is a B ∈ B such that B ⊆ U . By the above theorem, not only

is intS×SB ⊆ B, but intS×SB ∈ U . This proves the following corollary.


Corollary 29.12 Let S be a non-empty set and U be a uniformity on S. The the familyof all (S × S)-open symmetric elements of U forms a base for U .

We now describe an alternate way for representing the closure (with respect to the

product topology on S × S) of the elements of U .

Theorem 29.13 Let S be a non-empty set and U be a uniformity on S. Let Bsym denotethe base of all symmetric elements of U . If M ∈ U then

clS×SM =⋂

B∈Bsym

BM B

Proof : We are given a set S, a uniformity U on S and the base, Bsym, of all symmetricelements of U .

Let M ∈ U . Then by U5, clS×SM ∈ U . We keep in mind that M is a subset of

(S, τU )× (S, τU ), equipped with the product topology. We are required to prove thatclS×SM =

⋂

B∈BsymBM B. It suffices to show that,

B(x) × B(y) intersects M for all B ∈ Bsym if and only if (x, y) ∈⋂

B∈Bsym

BM B

Let (x, y) ∈ S × S. Let B(x) × B(y) be a basic open neighbourhood of (x, y) (whereB ∈ Bsym).

To say that “B(x) × B(y) intersects M” is equivalent to saying that “there is some(u, v) ∈M such that (x, y) ∈ B(u) ×B(v)”.

To say that “there is some (u, v) ∈ M such that (x, y) ∈ B(u) × B(v)” is equivalent

to saying that “(x, y) ∈ ⋃(u,v)∈M B(u) × B(v)”.

By Fact #3 proven in the previous theorem, we can express the right-hand side,“(x, y) ∈ ⋃(u,v)∈M B(u) × B(v)” as,

(x, y) ∈ BM B

Combining the two statements we obtain,

B(x) × B(y) intersects M for all B ∈ Bsym ⇔ (x, y) ∈⋂

B∈Bsym

BM B

We conclude that, if M ∈ U ,

clS×SM =⋂

B∈Bsym

BM B


29.6 Separation properties inherent to (S, τU ).

The uniform space, (S, τU ), inherits from the uniformity U on S some separationproperties. We can show immediately that the uniformity U on S guarantees that

(S, τU ) is T3.


topology on S generated by U . The space, (S, τU ), is T31.

Proof : We are given a set S, a uniformity U on S and the derived topological space,(S, τU ), equipped with the uniform topology.

Fact: The family of all (S × S)-closed symmetric sets in U forms a base for U .

Proof of fact: Suppose U ∈ U . By proposition 29.10, there exists V ∈ Bsym such thatV V V ⊆ U .

By theorem 29.13,

clS×SV =⋂

B∈Bsym

BV B

which is, by U5, an element of U .

Then

clS×SV ⊆ V V V ⊆ U

So we can declare that there exists a (S × S)-closed W = clS×SV ∈ U such thatW ⊆ V . Then W ∩W−1 is a closed and symmetric element of U contained in U .

We conclude that the family of all (S × S)-closed symmetric sets in U forms a basefor U . This establishes the stated fact. In what follows we will represent this base

by, Bcl-sym

Let F be closed in S (with respect to τU ) and x ∈ S\F . Then S\F is open in S.

To show that (S, τU ) is T3 it suffices to show that x has a closed neighbourhood whichmisses F . Since S\F is open, there exists an open neighbourhood V of x such that

x ∈ V ⊆ S\F .

A base Bτ for τU is of the form

Bτ = B(x) : x ∈ S and B ∈ Bcl-sym, a base for U 1Recall that a space is T3 if, given a closed set F and a point x 6∈ F , there exists disjoint open neigh-

bourhoods U, V such that F ⊆ U and x ∈ V .


We have seen that there is closed symmetric W ∈ Bcl-sym such that x ∈W (x) ⊆ S\F .Since W (x) = π2[(x× S)∩W ], and π2 is open and one-to-one on the closed subset

x × S) ∩W of x × S), then W (x) is a closed neighbourhood of x. Since W (x)misses F , we conclude that (S, τU ) is T3.

Corollary 29.15 Let S be a non-empty set and U be a uniformity on S. Let τU be thetopology on S generated by U . The space, (S, τU ), is regular (T1 + T3) if and only if

⋂

U∈U

U = ∆(S)

Proof : We are given a set S, a uniformity U on S and the derived topological space,(S, τU ), equipped with the uniform topology.

( ⇒ ) Suppose S is regular. Then S is T3 and T1. This implies that the points in

(S, τU ) are closed. By theorem 29.9,

x = clτUx =

⋂

U∈U

U(x)

Suppose (a, b) ∈ ⋂U∈UU. Then b ∈ ⋂U∈U

U(a). So b must equal a. We conclude

that⋂

U∈UU = ∆(S).

( ⇐ ) Suppose that⋂

U∈UU = ∆(S). Then for each x ∈ S, x =

⋂

U∈UU(x) =

clτUx. So x is closed and hence (S, τU ) is T1. We have shown that (S, τU ) is T3,

so, if⋂

U∈UU = ∆(S), the space is regular (T3 + T1).

29.7 Uniform continuity on a uniform space (S, U ).

The uniformity U defined on a set S will allow us to introduce the notion of “uniform

continuity” on the resulting uniform space, (S,U ).

Definition 29.16 Let (S,U ) and (T,V ) be two sets equipped with the uniformities, Uand V , respectively. Let f : S → T be a function mapping S into T . Let gf : S×S → T×Tbe the function defined as

gf(x, y) = ( f(x), f(y) )


We will say that a function f : (S,U ) → (T,V ) is uniformly continuous relative to U andV if and only if for any V ∈ V the set

g←f [V ] = (x, y) ∈ S × S : (f(x), f(y)) ∈ V

is an element of the uniformity, U .

Equivalently, f : S → T is uniformly continuous if and only if for any V ∈ V there is anelement U ∈ U such that gf [U ] ⊆ V .

Example 10. Recall that the usual uniformity on R is the uniformity with base

B = Bε : ε > 0, where Bε = (a, b) ∈ R × R : |a − b| < ε. Let (S,U ) bea uniform space and f : S → R be a function mapping S into R, where (R,V ) is

equipped with the usual uniformity, V . By definition, f : S → R is uniformly con-tinuous on S if and only if for any ε > 0, there exists U ∈ U such that gf [U ] ⊆ Bκ.That is, for any pair, (x, y) ∈ U , |f(x) − f(y)| < ε. Note that the choice of U is

independent of the location of (x, y) in S × S.

In the following example, a little bit closer to home, we examine what uniform conti-

nuity means for a function f : R → R.

Example 11. Let (R,U ) be the uniform space for R equipped with the usual uniformitywith the base B = Bε : ε > 0. Let f : R → R be a function mapping R into R. By

definition, f is uniformly continuous on R if and only if for any Bε ∈ U , there is aBδ ∈ U such that gf [Bδ] ⊆ Bε. Equivalently, for any ε > 0 there exists a δ > 0 such

that |x− y| < δ implies |f(x)− f(y)| < ε.

Note that the composition of two uniformly continuous functions is easily verified tobe uniformly continuous.

We will immediately formally define the following three concepts.

Definition 29.17 If, for a given one-to-one function, f : (S,U ) → (T,V ), both f andf← are known to be uniformly continuous then we say that the function f : S → T is a

uniform isomorphism from between S and T .

If f : S → T is a uniform isomorphism then we will say that (S,U ) and (T,V ) are uni-formly isomorphic.

A property on (S,U ) which is shared by all uniform spaces which are uniformly isomorphicto (S,U ) is said to be a uniform invariant.


29.8 Uniform continuity implies continuity.

Continuity is always defined with respect to some topology. If we are given a functionf : S → T which is known to be uniformly continuous with respect to the uniformities

U and V , respectively, what can say about the same function f : S → T mapping Sto T with respect to the corresponding topologies τU and τV . We answer this question

with the following theorem.

Theorem 29.18 Let S and T be sets with corresponding uniformities, U and V , respec-tively. If f : S → T is a uniformly continuous function with respect to the uniformities

U and V then this same function is a continuous function with respect to the uniformtopologies, τU and τV .

Proof : We are given sets S and T with respective uniformities U and V . Suppose

f : S → T is a uniformly continuous function.

To prove continuity of f it suffices to show f is continuous at each point x ∈ S.

Let x ∈ S and let V be a base element of V such that f(x) ∈ V (f(x)), a neigh-bourhood of f(x) in T . It suffices to produce a neighbourhood W , of x such that

f [W ] ⊆ V (f(x)). (See 6.2 and 6.3.) By the definition of “f : S → T is uniformlycontinuous”, g←f [V ] is an element of U (with ∆(S) in its (S × S)-interior).

See that,

f←[V (f(x))] = y ∈ S : f(y) ∈ V (f(x)) = y ∈ S : (f(x), f(y)) ∈ V = y ∈ S : gf(x, y) ∈ V =

[

g←f [V ]]

(x) (Where g←f [V ] ∈ U )

Since[

g←f [V ]]

(x) is a neighbourhood of x, then f←[V (f(x))] is a neighbourhood of x.

Since f [f←[V (f(x))]] = V (f(x)), f is continuous at x. Since x was arbitrarily chosen

in S, f is continuous on all of S.

It follows from the theorem that a function f : S → T which is a uniform isomorphismcan also be seen as a homeomorphism mapping S onto T .

Uniform continuity on a uniform space generalizes the notion of uniform continuity

from one metric space to another. In the case of a metric space, it is more oftendiscussed without explicitly introducing the concept of uniformities on a set. To see


this we present the following example.

Example 12. Suppose (M1, ρ1) and (M2, ρ2) are two metric spaces. Suppose f : M1 →M2 is a function satisfying the condition,

. . . for every ε > 0, there is a δ > 0 such that f [Bδ(x)] ⊆ Bε(f(x)), for all x ∈ S.

In this case, the condition imposed on the existence of δ is stronger than the onerequired in the definition of continuity at x (where the value of δ depends on both the

value of ε and the point, x). It easily seen here that if f satisfies the given conditionthen f is continuous at each point.

We compare this to the same function f : M1 → M2 where the sets M1 and M2 areequipped with the metric uniformities, U and V , respectively. A metric uniformity

has as base, B = Bκ : κ > 0 where Bκ = (x, y) : ρ(x, y) < κ. By using thelanguage and notation for uniformities, we can say that f is uniformly continuous if,

. . . for any ε > 0, there is δ > 0 such that, if (x, y) ∈ Bδ ∈ U , then

(f(x), f(y)) ∈ Bε ∈ V . Equivalently, f [Bδ(x)] ⊆ Bε(f(x)), for all x ∈ S.

550 Section 30: The Stone-Weierstrass theorem

30 / The Stone-Weierstrass theorem

Summary. In this section, we revisit the relationship between algebraic struc-

tures of C∗(S) and the family of Hausdorff compactifications of S. We providea detailed proof of the Stone-Weierstrass Theorem. One consequence of this im-

portant theorem is allowing us to view the family of all compactifications as alattice of topological spaces.

30.1 Introduction.

In this section, for a topological space S, the symbol Cb(S) represents all real-valuedbounded functions on S. We will view (Cb(S),+, ·) as a ring of bounded functions.

We equip Cb(S) with the sup-norm,

‖f‖∞ = sup|f(x)| : x ∈ S

It is an easy exercise to show that ‖ ‖∞ is a valid norm on Cb(S). So Cb(S) can be

viewed as a metric space, (Cb(S), ρ), where ρ(f, g) = sup|f(x) − g(x)| : x ∈ S. Intheorem 1.11, we proved that (Cb(S), ‖ ‖∞) is a complete space. That is, all Cauchy

sequences in Cb(S) converge to an element of Cb(S).

Suppose that (A,≤) is a partially ordered set. Recall that A is referred to as being alattice if, for every a, b ∈ A, a ∨ b = supa, b ∈ A and a ∧ b = infa, b ∈ A. The

symbol, ∨, is read as “join” while the symbol, ∧, is read as “meet”. The set A is saidto be a complete lattice if, for any non-empty B ⊆ A, ∨B = supB exists in A and

∧B = inf B exists in A. The lattice derived from (A,≤) is denoted as

(A,∨,∧)

The set (B,∨∗,∧∗) is said to be a sublattice of (A,∨,∧) if (B,≤∗) ⊆ (A,≤) and

(B,∨∗,∧∗) is itself a lattice with respect to the binary relations, ∨∗ and ∧∗, it inheritsfrom (A,∨,∧). That is, if a, b ∈ B, a ≤∗ b if and only if a ≤ b and a ∨∗ b = a ∨ b and

a ∧∗ b = a ∧ b.

Example 1. For a topological space S, the set C(S) denotes the set of all continuousreal-valued functions on S. For f, g ∈ C(S), we will say f ≤ g if f(x) ≤ g(x) for all

x ∈ S. Then (C(S),≤) is a partial ordering. If

h(x) = max f(x), g(x) : x ∈ S = (f ∨ g)(x)k(x) = min f(x), g(x) : x ∈ S = (f ∧ g)(x)


(C(S),∨,∧) can easily be verified to be a lattice. To show this use the identities,

f ∨ g =f + g + |f − g|

2(∗)

f ∧ g =f + g − |f − g|

2(∗∗)

For further reference, we include the following identity which is deduced by these.

f ∧ g = f + g − (f ∨ g) (∗∗∗)

Analogous definitions apply to C∗(S) (the set of all continuous bounded functions) to

obtain the lattice (C∗(S),∨,∧). 1

The subset F of C∗(S) is said to be a subring of the ring, (C∗(S),+, ·), if it closedunder + and ·. When we say that the subring, F , of the ring, C∗(S), is complete, we

mean that it is complete with respect to the sup-norm, ‖ ‖∞. We would like to provethe Stone-Weierstrass theorem. But first we will need in the following two lemmas,in which we associate the metric space notion of “completeness of F with respect to

‖ ‖∞”, to the algebraic notion of “F , a sublattice of C∗(S)”.

Lemma 30.1 The Stone-Weierstrass lemma one. Let S be a topological space. Let F be

a complete subring of (C∗(S),+, ·) which contains all constant functions. Then, if f ∈ F ,|f | ∈ F .

Proof : We are given that F is a complete subring of (C∗(S),+, ·) which contains all con-

stant functions. Let f ∈ F . We are required to show that |f | belongs F . Since f isbounded there exists k ∈ R such that |f(x)|∞ < k on S. Then ‖f/k‖∞ < 1. If we

show that |f/k| ∈ F then so does |f |. So let us suppose, without loss of generality,that ‖f‖∞ < 1. Then |f(x)| < 1 on S

Consider the binomial expansion of the expression h(x) = (1 − x)1/2,

(1 − x)1/2 =∞∑

n=0

C(1/2, n)xn

Convergence of this series to (1−x)1/2 on [0, 1] is known to be uniform. (See page 247.)

The power series∑∞

n=0 C(1/2, n)xn can be seen as a sequence of polynomials whichconverges uniformly to (1 − x)1/2 on [0, 1]. This means that there is a polynomial,

p(x), such that∣

∣

∣ (1 − x)1/2 − p(x)∣

∣

∣ < ε on [0, 1]

1Note that neither (C(S),∨,∧) nor (C∗(S),∨,∧) need be a complete lattice unless extra conditions areattached to S.


Let x = 1 − u2. Substituting this term in the above expression we obtain,

∣

∣

∣

√u2 − p(1− u2)

∣

∣

∣< ε, for u in [−1, 1]

from which we obtain∣

∣ |u| − p(1− u2)∣

∣ < ε, for u in [−1, 1]. We substitute f(x) intou, to obtain

∣

∣ |f(x)| − p(1 − f(x)2)∣

∣ < ε, for f(x) in [−1, 1]

and so,‖ |f | − p(1 − f2)‖∞ ≤ ε

Let h = p(1 − f2) ∈ F . Now since p is a polynomial and F is a subring whichcontains all constant functions, then h ∈ F . So every ε-ball, Bε(|f |) (with respectto ‖ ‖∞), contains an element, h, of F . So |f | is a limit point of F . Since F is

complete, the function, |f |, must belong to F . As required.

Lemma 30.2 The Stone-Weierstrass lemma two. Let S be a topological space. Let F bea complete subring of (C∗(S),+, ·) which contains all constant functions. Then (F ,∨,∧)

is a sublattice of (C∗(S),∨,∧).

Proof : We are given that F is a complete subring of (C∗(S),+, ·) which contains all con-

stant functions. Let f, g ∈ F . To show that F is a sublattice we are required to showthat f ∨ g and f ∧ g both belong F

Consider the two identities (*) and (***) (given above)

f ∨ g =f + g + |f − g|

2f ∧ g = f + g − (f ∨ g)

In the first lemma, we showed that if f ∈ F then |f | ∈ F . Since F is a subring then

both f∨g and f∧g both belong F . So F is a sublattice of (C∗(S),∨,∧). As required.

In the following theorem, we refer to a subset F of C∗(S) which separates points of S.This means that, “. . . for every pair of distinct points p and q in S, there is a function

h ∈ F such that h(p) 6= h(q)”.

While studying the Stone-Weierstrass theorem below, the reader may have a vague

impression of having previously seen a similar statement (presented in 21.11). Com-pare the statement 21.11 to 30.3, the conditions and the conclusions that follow from


these conditions. See the slight differences in how we approach the statements. Mostof the technical work for the proof of the following theorem is done in the preceding

two lemmas. 1

Theorem 30.3 The Stone-Weierstrass theorem. Let S be a compact topological space. Let

F be a complete subring of C∗(S) which contains the constant functions. If F separatesthe points of S then F = C∗(S).

Proof : Suppose S is a compact space and f ∈ C∗(S). Also suppose that F is a completesubring of C∗(S) which separates points of S and which contains all constant func-

tions, r ∈ C∗(S) : r(x) = r ∀x, r ∈ R. To show that F = C∗(S) it will suffice toshow that, for any ε > 0, Bε(f)∩F 6= ∅ with respect to ‖ ‖∞. (For this would imply

f is a limit point of F and since F is complete, f ∈ F .)

If f is constant then, since F contains all constant functions, then f ∈ F . So we willassume, without loss of generality, that f is not constant; then inf f(x) : x ∈ S <sup f(x) : x ∈ S.Step 1. Choose p, q ∈ S. By hypothesis, there exists hpq ∈ F such that hpq(p) 6=hpq(q). We define the function gpq as,

gpq(x) =

(

hpq(x)− hpq(q)

hpq(p) − hpq(q)

)

f(p)−(

hpq(x) − hpq(p)

hpq(p) − hpq(q)

)

f(q)

Verify that, for such choices of p and q, gpq(p) = f(p) and gpq(q) = f(q). Since F isa subring which contains the constant functions and hpq ∈ F , then gpq ∈ F .

Step 2. Let ε > 0. Let

Upq = x ∈ S : (gpq − f)(x) < εVpq = x ∈ S : (gpq − f)(x) > −ε

For what follows we will fix q ∈ S. Note that, since gpq(q) = f(q),

q ∈ Vpq (independently of the value of p)

Similarly, see that gpq(p) = f(p), so p ∈ Upq. Since Upq = (gpq − f)←[(−∞, ε)] is openfor any p, then, Upq : p ∈ S is an open cover of S. Since S is compact, there is a

subset, p1, p2, p3, . . . , pn, of S such that Up1q, Up2q, . . . , Upnq, covers all of S. Let

1The theorem 21.11 states: Suppose (S, τ) is a completely regular space and F is a subalgebra of C∗(S)which separates points and closed sets in S. Suppose F generates the compactification αS = eβ

F[βS] =

clT eF [S]. If F satisfies the properties, 1) F contains Cω(S) and 2) every f ∈ Cα(S) is equivalent to somefunction g ∈ F , then F = Cα(S).


Vq = ∩Vp1q, Vp2q, . . . , Vpnq. Then q ∈ Vq.

We define, for our fixed q, the function gq : S → R as,

gq(x) = ∧gp1q(x), gp2q(x), . . . , gpnq(x)

By the lemma 30.2, since F is a complete subring which contains the constant func-tions, F is a sublattice. So gq ∈ F .

Then for any x ∈ S, x ∈ Upkq for some k ∈ 1, 2, 3, . . . , n. For this k and x we have,

gq(x) ≤ gpkq(x) < f(x) + ε (1)

This expression is independent of the value of x in S, so we can write more generallythat, independent of the chosen value q ∈ S,

gq < f + ε, for all q ∈ S. (2)

Let y ∈ Vq where Vq = ∩Vp1q, Vp2q, . . . , Vpnq is a non-empty open subset of S. Then

there is some ptq ∈ p1q, p2q, . . . , pnq such that gq(y) = gptq(y) and so,

gq(y) > f(y)− ε (3)

Step 3. Since q ∈ Vq for each q ∈ S, then Vq : q ∈ S forms an open cover of S, sothere exist q1, q1, . . .qm such that Vq1, Vq2, . . .Vqm covers S.

Let x ∈ S. Then x ∈ Vqbfor some qb ∈ q1, q2, . . . , qm.

Let

g = ∨gq1, gq2 . . . gqmThen g ∈ F (by the lemma) and there is some qa ∈ gq1, gq2 . . . gqm such that

g(x) = gqa(x) > f(x) − ε (4)

Since this expression is independent of the value of q and x, we can conclude that

f − ε < g (5)

Step 4. From inequality (2) we have gq < f + ε (on S) for all values of q. Then

g = ∨gq1, gq2 . . . gqm < f + ε (6)

Then, combining inequalities (5) and (6),

f − ε < g < f + ε

This is equivalent to stating “for arbitrary ε > 0, g ∈ Bε(f) with respect to ‖ ‖∞”.

Since, by hypothesis, F is complete, it is closed; then f ∈ F . So C∗(S) ⊆ F ; itfollows that F = C∗(S), as required.


30.4 A consequence of the Stone-Weierstrass theorem.

Let C = αiS : i ∈ I represent the family of all compactifications of a locally finitecompletely regular space. We will partially order the family, C , by defining “” in

the following way:

If αiS and αjS belong to C , we will write

αiS αjS

if and only if there is a continuous function f : αjS → αiS, mapping αjS\Sonto αiS\S which fixes the points of S.

See that with thus defined, (C ,) forms a partially ordered set of compact topo-logical spaces.

Notation. Suppose two compactifications αS and γS are such that αS γS, then bydefinition, there is a continuous function f : γS → αS, mapping γS \S onto αS \Swhich fixes the points of S. We will represent this continuous function f as,

πγ→α : γS → αS

The subscript of the function πγ→α explicitly expresses which compactification islarger than (or equal to) the other. Note that a pair of compactifications need not

necessarily be comparable in the sense that one need not necessarily by “less than”the other.

Notation : For any compactification αS let Cα(S) = f |S : f ∈ C(αS). Clearly,Cα(S) ⊆ C∗(S). If S is locally compact and completely regular then it has a one-pointcompactification, ωS = S ∪ ω (see 21.15). We will adopt the following notation,

Cω(S) = f |S : f ∈ C(ωS)

Note that, since ωS is compact and Hausdorff then it is normal (see 14.3). Then it iscompletely regular and so C(ωS) separates points and closed sets of ωS. Clearly, italso contains all constant functions. It is also a complete ring, with respect to ‖ ‖∞(see 1.11). Then it is easily verified that Cω(S) also separates points and closed setsof S (remember that since S is locally compact it is open in ωS).

Also note that f ∈ Cω(S) ⊆ C∗(S) if and only if f is constant on the complimentof some compact subset of S (since, if it extends to ωS, it must be constant on the


outgrowth, ω).Compare the following statement to the one in theorem 21.11.1

Theorem 30.4 Let S be locally compact and completely regular. If C∗(S) contains a sub-

ring, F , which is complete and contains Cω(S) then F = Cα(S) for some compactification,αS, of S.

Proof : We are given that S is locally compact completely regular, that F is complete sub-

ring of C∗(S) and contains Cω(S). Since ωS is compact and C(ωS) separates pointsand closed subsets of ωS then Cω(S) separates points and closed sets of S. Then Fseparates points and closed sets in S. Every function f in F extends to a functionfβ ∈ C(βS). Then F β = fβ : f ∈ F ⊆ C(βS).

By the embedding theorem I (theorem 7.17) the evaluation map

eF : S →∏

f∈F

f [S]

embeds S into T =∏

f∈FR. Then eF β [βS] = clT eF [S] can be viewed as a com-

pact space which densely contains a homeomorphic copy of S. We can then represent

clT eF [S] as a compactification, αS = clT eF [S]. By the Stone-Weierstrass theorem,Fα = C∗(αS). Then F = Cα(S)

By the theorem above we can associate to each complete subring, F , of C∗(S), whichcontains Cω(S), a compactification αS of S such that F = Cα(S).

Conversely, for each compactification γS we can associate a unique complete subring,

Cα(S), of C∗(S), which contains Cω(S). (If S is locally compact, the ring Cα(S)contains Cω(S) so every function, f in ωS extends to all compactifications of S).

For compactifications αS and γS, αS γS if and only if Cα(S) ⊆ Cγ(S).

We define Cα(S) ∨ Cγ(S) = the smallest complete subring, Cδ(S), of C∗(S) which

contains both Cα(S) and Cγ(S). We can equivalently say,

Cα(S) ∨Cγ(S) =< Cα(S), Cγ(S) >

1The theorem 21.11 states: Suppose (S, τ) is a completely regular space and F is a subalgebra of C∗(S)which separates points and closed sets in S. Suppose F generates the compactification αS = eβ

F[βS] =

clT eF [S]. If F satisfies the properties, 1) F contains Cω(S) and 2) every f ∈ Cα(S) is equivalent to somefunction g ∈ F , then F = Cα(S).


to represent the subring generated by Cα(S) and Cγ(S).1

We define Cα(S)∧Cγ(S) = Cα(S)∩Cγ(S), the largest subring, Cδ(S), of C∗(S) whichis contained in both Cα(S) and Cγ(S).

We can then view C as a lattice, (C ,∨,∧) where

αS ∨ γS = δS if and only if Cα(S) ∨Cγ(S) = Cδ(S)

αS ∧ γS = δS if and only if Cα(S) ∩Cγ(S) = Cδ(S)

1It is the intersection of all subrings of C∗(S) which contains both Cα(S) and Cγ (S).

558 Section 31: Metrizability

31 / Metrizability

Summary. In this section, we prove two important metrizability theorems. TheUrysohn metrization theorem is proved for the special case where the space is

known to be separable. This is of particular interest since it exposes the strategyused to prove the more general Nagata-Smirnov metrization theorem.

31.1 Introduction.

We discussed topological spaces, (S, τ), on which we can define a metric which pro-duces a metric topology which is equivalent to τ . Such spaces are said to be metrizable.

We know that metric spaces are first countable (5.9), normal (9.19) and paracompact(19.11). But determining those specific properties on a topological space which char-

acterize its metrizability is the question we will investigate here.

31.2 The Urysohn metrization theorem.

We begin with a lemma which states that a countably infinite product space of metricsspaces is sure to be metrizable.

Lemma 31.1 Let (Sn, ρn) : n ∈ N be a countable family of metric spaces. Suppose that,for each n ∈ N, supρn(xn, yn) : xn, yn ∈ Sn ≤ 1. Let S =

∏

n∈N Sn and ρ : S × S → R be

defined as

ρ(x, y) =∑

n∈N

ρn(xn, yn)

2n

Then ρ : S × S → R is a metric on S. Furthermore, the topology on S =∏

n∈N Sn derivedby the metric ρ is the product topology.

Proof : This statement has been proven in an example which appears on page 131.

Lemma 31.2 Let S be a T1 topological space. Suppose we are given the countably infinite

indexed family of non-empty topological spaces, Sn : i ∈ N where, for all n, Sn = [0, 1]equipped with the usual topology. Let F = fn : n ∈ N be a set of continuous functions

where each fn maps S onto its range, fn[S], inside Sn = [0, 1].


a) If F separates points and closed sets of its domain, S, then the evaluation map,

e(x) =< fn(x)>n∈N ∈∏

n∈Nfn[S] ⊆∏

n∈N[0, 1]

with respect to F , is both continuous and one-to-one on S.

b) Furthermore, the function, e : S →∏

n∈N Sn, maps S homeomorphically onto

e[S] ⊆∏n∈Nfn[S] ⊆∏n∈NSn

Hence this evaluation map embeds a homeomorphic copy of S into the (metrizable1)

product space,∏

n∈N[0, 1].

Proof : This lemma is simply restating the Embedding Theorem one, 7.17, for the case

where the codomains are all [0, 1].

The two above lemmas allow to streamline the proof of Urysohn’s metrization theorem.

Theorem 31.3 The Urysohn metrization theorem. Let S be a regular (T1 + T3) secondcountable topological space. Then S is embedded in

∏

n∈N[0, 1] = [0, 1]N (with the producttopology) and so S is a metrizable space.

Proof : We are given that S is a regular and second countable topological space. Then, bytheorem 16.2, S is Lindelof and, by theorem 16.6, S is normal.

If we can show that there exists a function which embeds S into the metrizable prod-uct space,

∏

n∈N[0, 1], then S can be viewed as a subspace of a metric space and somust be metrizable.

Let B = Bi : i ∈ N be a countable base of open sets in S. Let

A = (U, V ) : U, V ∈ B, clSU ⊆ V

Now, since B is countable, A is a countably infinite family of pairs of disjointopen sets. Since S is normal, for each pair (U, V ) there is a continuous function

f(U,V ) : S → [0, 1] such that clSU ⊆ Z(f(U,V )) and S \V ⊆ Z(f(U,V ) − 1). LetF = f(U,V ) : (U, V ) ∈ A .We claim that F separates points and closed sets. Let F be a closed subset of S and

x ∈ S\F . We can choose V ∈ B such that x ∈ V ⊆ S\F and choose U ∈ B such thatx ∈ clSU ⊂ V . Then (U, V ) ∈ A . Let f(U,V ) be the member of F which corresponds

1By lemma 31.1


to this choice of (U, V ) where x ∈ Z(f) and F ⊆ S\V ⊆ Z(f − 1). So F separatesthe point x and the closed subset F of S, as claimed.

Then, by the Embedding Theorem one, 7.17, the evaluation map, eF , embeds a home-omorphic copy of S in the metrizable space,

∏

n∈N[0, 1].

It follows that S is metrizable.

Corollary 31.4 Let S be a T1-space. The following statements are equivalent.

1) The space S is regular and second countable.

2) The space S is homeomorphically embedded in [0, 1]N.

3) The space S is both metrizable and separable.

Proof : We are given that S be a T1-space.

1) ⇒ 2) That a regular and second countable space is embedded in [0, 1]N is proven inthe theorem immediately above.

2) ⇒ 3) We are given that S is embedded in [0, 1]N. In the lemma 31.1 we proved that

[0, 1]N is metrizable. Since [0, 1] is second countable (example on page 83), so is[0, 1]N (7.9). Since S is embedded in [0, 1]N, S is second countable. By theorem

5.10, S is separable. So S is both metrizable and separable.

3) ⇒ 1) Suppose S is metrizable and separable. Metrizable spaces are normal (9.19) andso are regular. A metric space which is separable is second countable (5.11). So

S is regular and second countable.

So all second countable regular spaces are separable metrizable spaces.

31.3 The Nagata-Smirnov metrization theorem.

The statement in the following theorem involves a collection of subsets referred to

as being a “locally finite collection”. In lemma 6.16 we show that it satisfies theinteresting property:

“If C is locally finite, clS[∪C ] = ∪clSC : C ∈ C ” (*)

That is, if p is an accumulation point of ∪C it is an accumulation point of some setin C . The statement, (*), is invoked in the proof of the lemma immediately below.

Although we have briefly discussed the property of locally finite families of sets before

(on page 110 and in 19.1), for convenience, we formally restate the definition of theconcept along with associated terms here.


Definition 31.5 A locally finite collection, C , of subsets of a topological space, S, is onefor which every point in S has at least one neighbourhood which meets at most finitely

many elements of C .

The collection of sets, D is said to be σ-locally finite if and only if D is the countable unionof locally finite families of locally finite sets.

The collection of sets, A is said to be a discrete family of sets is one for which every pointin S has at least one neighbourhood which meets at most one elements of A .

Lemma 31.6 Suppose W is an open subset of a regular space, S, which has a σ-locally

finite base. That is, S has a base, B = ∪n∈NBn, where each Bn is locally finite. Then

W = ∪UnW: n ∈ N = ∪clSUnW

: n ∈ N

where UnW: n ∈ N is a countable collection of open subsets of W (in S).

Proof : Let B = ∪n∈NBn be a σ-locally finite base of S. Let W be open in S. Regularity

of S guarantees that, for x ∈W , there is a Bx ∈ B such that x ∈ Bx and clSBx ⊆W .Let

UW = Bx ∈ B : x ∈W, clSBx ⊆WUnW

= UW ∩ Bn (For each n.)

UnW= ∪B : B ∈ UnW

(For each n.)

Since each Bx ∈ B, then each Bx ∈ Bn, for some n. Then, for each x ∈W ,

x ∈ UnW⊆ clSUnW

⊆ ∪clSUnW: n ∈ N

where UnWis an open subset of S. So W ⊆ ∪clSUnW

: n ∈ N. It now suffices toshow ∪clSUnW

: n ∈ N ⊆W .

For each n ∈ N

clSUnW= clS [∪B : B ∈ UnW

]= ∪clSB : B ∈ UnW

(By lemma 6.16 or (*) above.)

⊆ W (Since clSBx ⊆W .)

So ∪clSUnW: n ∈ N ⊆W .


We conclude that if S is regular, has a σ-locally finite base, B = ∪Bn : n ∈ N, andW is open in S then

W = ∪UnW: n ∈ N = ∪clSUnW

: n ∈ N (∗∗)

where each UnWis the countable union of open sets in S. This completes the proof

of the lemma.

Note . An easy (and interesting) consequence of the result (∗∗) is the following fact:

“If S is regular with a σ-locally finite base then every closed subset of S is a Gδ.”

To see this, consider a closed subset, F , of S. Then, by (∗∗), S\F = ∪clSUn : n ∈ Na countable union of closed sets. Then F = ∩S\clSUn : n ∈ N, a countable inter-

section of open sets, a Gδ.

Lemma 31.7 A regular space S with a σ-locally finite base is a normal space.

Proof : We are given that S is regular and has a σ-locally finite base, B. That is,

B = ∪n∈NBn

where each Bn is locally finite.

Suppose A and B are disjoint non-empty closed subsets. To show that S is normalit suffices to show that A and B are respectively contained in disjoint open subsets of S.

Since S is regular and has a σ-locally finite base the open subset S\B can de expressed

as

S\B = ∪Un : n ∈ N = ∪clSUn : n ∈ N

where each Un is the countable union of elements of Bn (see lemma 31.6 above).

Then A ⊆ ∪Un : n ∈ N and clSUn ∩ B = ∅.

Similarly, there is a family Vn : n ∈ N of open subsets of S such that

S\A = ∪Vn : n ∈ N = ∪clSVn : n ∈ N

where each Vn is the countable union of elements of Bn.

Then B ⊆ ∪Vn : n ∈ N and clSVn ∩A = ∅.


Let U = ∪Un : n ∈ N and V = ∪Vn : n ∈ N. Now U and V are open neighbour-hoods of A and B, . . . but they may not be disjoint.

To resolve this problem, we try to reduce the sizes of the Un’s and Vn’s while keepingthem open in S. For each n, we replace Un and Vn with U∗n and V ∗n defined as follows:

U∗n = Un\∪clSVi : i ≤ nV ∗n = Vn\∪clSUi : i ≤ n

Then, for each n,

A ⊆ U∗n

B ⊆ V ∗n

Let U∗ = ∪U∗n : n ∈ N and V ∗ = ∪V ∗n : n ∈ N. Clearly, they are both open andneighbourhoods of A and B, respectively.

We claim that U∗ ∩ V ∗ = ∅.

Proof of claim: Suppose x ∈ U∗ ∩ V ∗. Then x ∈ U∗j ∩ V ∗k for some j and k.

Then

x ∈ Uj\∪clSVi : i ≤ j ∩ Vk\∪clSUi : i ≤ kSo x ∈ Uj and x ∈ Vk. Suppose without loss of generality that j ≤ k.

But since j ≤ k,

x ∈ ∪clSUi : i ≤ kSo, by definition of V ∗k , x cannot belong to Vk, a contradiction. So U∗ ∩ V ∗ = ∅.

Then A and B are contained in the disjoint open neighbourhoods U∗ and V ∗. So Sis normal.

Lemma 31.8 A regular space S with a σ-locally finite topology is metrizable.

Proof : We are given that S is regular and has a σ-locally finite base, B. That is,

B = ∪n∈NBn

where each Bn is locally finite. Then, by the previous lemma, S is normal. To provemetrizability of S it will suffice to produce a continuous function which embeds S into

a metric space.


If B ∈ Bn, B will be assigned the unique label, (n, B), which specifies to whichsubfamily, Bn, it belongs to. So (n, B) ∈ B means that B is some element in Bn. So

we can writeB = (n, B) : n ∈ N, B ∈ Bn

Suppose (n, B) ∈ B and u ∈ (n, B). Normality of S guarantees that there is acontinuous function

f(n,B) : S → [0, 1/n) ⊆ [0, 1]

such that

f(n,B)[S\B] = 0 and f(n,B)(u) ∈ (0, 1/n) (†)With this we can construct a countable family of continuous functions

F = f(n,B) : n ∈ N, B ∈ Bn

in which every function, f(n,B), vanishes on the complement of B and is bounded by

0 and 1/n on B ∈ Bn.

It is easily verified that F separates points and closed sets in S.

Let J = (n, B) : n ∈ N, B ∈ Bn.Let

T =∏

j∈J [0, 1]j = [0, 1]J

Since F separates points and closed sets, then the evaluation map, eF : S → T ,defined as

eF (x) =<fj(x)>j∈J

is a one-to-one function mapping S into T =∏

j∈J [0, 1]j.

We could equip T with the product topology. But in this case, we will do otherwise

(since we need T to be a metric space). Rather, we will choose the uniform topologyinduced by the uniform metric1

ρ(<xj>j∈J , <aj>j∈J) = supj∈J

|xj − aj|

where, in this particular case,

ρ(eF (x), eF(a)) = ρ(<fj(x)>j∈J , <fj(a)>j∈J) = supj∈J

|fj(x) − fj(a)|

Then T can be viewed as a metric space with metric, ρ.

Since we are not using the product topology on T , the statement, “eF is a homeo-

morphism on S”, is new, and so requires proof. If this statement holds true, then themetric space, T , contains a homeomorphic copy of S and so S is metrizable.

1See the example found on page 536.


The proof that eF : S → T is one-to-one and open is precisely as shown in the proofof the Embedding theorem I, (7.17). To prove it is a homeomorphism it now suffices

to show that it is a continuous function.

Claim : We claim that the function eF : S → ∏

j∈J [0, 1]j, where∏

j∈J [0, 1]j isequipped with the uniform metric, is a continuous function.

Proof of claim. Let ε > 0 and a ∈ S. It suffices to find an open neighbourhood V (a)

in S such that x ∈ V (a) implies

ρ(eF (x), eF (a)) = sup(n,B)∈J

|f(n,B)(x) − f(n,B)(a)| < ε

We will fix n ∈ N for now. Given that Bn is a locally finite family on S, we can findan S-open neighbourhood, Un(a), of a such that B ∩ Un(a) 6= ∅ for at most m baseelements, B1, B2, . . . , Bm, in Bn.

Then, for all B 6∈ B1, B2, . . . , Bm, B ∩ Un(a) = ∅; so for B 6∈ B1, B2, . . . , Bm,f(n,B)[Un(a)] = 0.On the other hand, for x ∈ Un(a)∩B where B ∈ B1, B2, . . . , Bm, f(n,B)(x) may belarger than 0; but it is always be less than 1/n, (by †).Recall that each function in F is continuous at the given point a ∈ Un. So for thegiven ε > 0, and choice of n, we can find an open neighbourhood, Vn(a) ⊆ Un(a),

such that,

x ∈ Vn(a) implies∣

∣f(n,B)(x) − f(n,B)(a)∣

∣ ≤ min1/n, ε/2

is satisfied.

We can now choose N ∈ N, such that 1/N ≤ ε/2.

We consider two cases:

Case 1: i ≤ N .

We repeat the procedure for a ∈ S and i = 1, 2, . . . , N to prove the existence ofV1(a), V2(a), . . . , VN(a). For Vi(a) ⊆ Ui(a),

x ∈ Vi(a) implies∣

∣f(i,B)(x)− f(i,B)(a)∣

∣ ≤ min1/i, ε/2 ≤ ε/2 < ε

If V (a) = V1(a) ∩ V2(a) ∩ · · · ∩ VN(a).

x ∈ V (a) implies∣

∣f(i,B)(x) − f(i,B)(a)∣

∣ ≤ min 1/i, ε/2< ε (††)

Case 2: i > N .

For Vi(a) = V (a)

x ∈ V (a) implies∣

∣f(i,B)(x)− f(i,B)(a)∣

∣ ≤ min1/i, ε/2< 1/N ≤ ε/2 < ε († † †)


So for all values of n ∈ N,

x ∈ V (a) ⇒ ρ(eF (x), eF(a)) = sup(n,B)∈J

∣

∣f(n,B)(x) − f(n,B)(a)∣

∣ < ε

So eF is continuous at a ∈ S. Since a is arbitrary, eF : S → T is continuous on S, as

claimed.

So S is metrizable.

In the following theorem we invoke a result (19.8) previously proven in the chapteron paracompactness. It states:

If S is metrizable space and U is an open cover of S then there is an opencover, E , which both refines U and is σ-locally finite.

Recall that a collection of sets, E , refines a collection of sets, U , if every element of

E is contained in some element of U .

Theorem 31.9 The Nagata-Smirnov metrization theorem.1 The space S is metrizable if

and only if S is regular and has a σ-locally compact base of open sets.

Proof : Let S be a topological space.( ⇐ ) The proof that regular spaces which have a σ-locally compact base of open sets

are metrizable is presented in the lemma 31.8 above.

( ⇒ ) Suppose S is metrizable. Then there exists a metric, ρ, such that (S, ρ) isequivalent to S. Since metric spaces are regular, so is S. We are simply required to

find a σ-locally finite basis for S.

To do this, we will construct an open cover of S which will allow us to invokethe lemma 19.8 (cited above) to arrive at the required result. For n ∈ N, let

Un = B1/n(x) : x ∈ S; it is, clearly, an open cover of S. Then, by lemma 19.8, thereis a σ-locally finite cover, En, of S which refines Un.

Let E = ∪En : n = 1, 2, 3, . . . , , a σ-locally finite cover of S which refines U =

∪Un : n = 1, 2, 3, . . . , . We claim that E is a base for S. Suppose a ∈ S and Bε(a)is an open ε-ball centered at a with respect to ρ. Then there exists k ∈ N, such that

1/k < ε/2. Now Bk covers S so there is B ∈ Bk such that a ∈ B. The diameter of B

1Jun-iti Nagata (1925, 6 November 2007) was a Japanese mathematician specializing in topology. YuriMikhailovich Smirnov (September 19, 1921, Kaluga, September 3, 2007, Moscow) was a Soviet and Russianmathematician, specializing in topology. This theorem was proved independently by Nagata in 1950 andSmirnov in 1951.


is less than (1/k)/2 < ε/2. So B ⊆ Bε(a). Then E is a base for S as claimed. ThenE is a σ-locally finite base for S, as required.

568 Section 32: The Stone space

32 / The Stone space

Summary. In this section we discuss those partially ordered sets called latticesand Boolean algebras along with some of their properties. Given a Boolean alge-bra, B, we construct the set, S (B), of all ultrafilters on B and topologize it. We

call the resulting topological space, (S (B), τ), the Stone space. We show thatthe Stone space, (S (B), τ), is a compact, zero-dimensional Hausdorff space.

We then show that any Boolean algebra, B, is isomorphic to B(S (B)), the setof all clopen subsets of (S (B), τ).

32.1 Lattices.

Given a partially ordered set (P,≤) there may be pairs of elements a, b in P such

that a and b do not have a common upper bound or a common lower bound in P .Those partially ordered sets in which every pair of elements in P have lower and upper

bounds play an important role in mathematics. We refer to these sets as lattices. Wediscussed the notion of a lattice before on page 550. For convenience, and, hopefully, torender this section a bit more “self-contained” we review the basics of this notion here.

Definition 32.1 A partially ordered set (P,≤) is called a lattice if a ∨ b = luba, b and

a ∧ b = glba, b both exist in P for all pairs a, b in P . The lattice derived from (P,≤) isdenoted as

(P,≤,∧,∨)

Definition 32.2 If B ⊆ (P,≤,∧,∨), then ∨B denotes the least upper bound of B and ∧Bdenotes the greatest lower bound of B (both with respect to ≤). Note that ∨B and ∧Bmay or may not be an element of B. A lattice (P,≤,∧,∨) is said to be a

complete lattice

if for any non-empty subset B of P , both ∨B and ∧B exist and belong to P .

Example 1. Let S be a set and (P(S),⊆) be a partially ordered set ordered by

inclusion. Let A and B be any pair of subsets of S (read, A,B ∈ P(S)). We defineA ∨ B = A ∪ B and A ∧ B = A ∩B. For B ⊆ P(S), ∨B = ∪A ∈ P(S) : A ∈ B


(read, “∨B is the union of all subsets, A, of S such that A ∈ B ”). Also define∧B = ∩A ∈ P(S) : A ∈ B (read, “∧B is the intersection of all subsets, A, of S

such that A ∈ B ”). In this case, with ∨ and ∧ defined as ∪ and ∩, respectively,

(P(S),⊆,∪,∩)

is a complete lattice.

Example 2. Let (S, τ) be a topological space. We partially order the elements of τwith “⊆”. If A and B are both members of τ and B ⊆ τ , we define

A ∨ B = A ∪BA ∧ B = intS(A ∩ B) = A ∩B

∨B = ∪A ∈ τ : A ∈ B∧B = intS (∩A ∈ P(S) : A ∈ B)

In this case, (τ,⊆,∨,∧) is a complete lattice in (P(S),⊆).

Example 3. Let S be topological space. Let

B(S) = A ⊆ S : A is clopen in S2

partially ordered by inclusion, ⊆. If A and B are clopen subsets of S we defineA ∨ B = A ∪ B and A ∧ B = A ∩ B. Then (B(S),⊆,∪,∩) is a lattice in (P(S),⊆).

But it need not necessarily be complete.

Example 4. Let (B,≤) be a partially ordered set and (B,≤,∨,∧) be an associatedlattice. Suppose that B has a least element, say L. That is, L ≤ B for all B ∈ B.In what follows the M is a non-empty subset of B. Show that, if ∨M ∈ B for all

M ⊆ B then ∧M ∈ B for allM ⊆ B.

Solution : We are given that (B,≤) is a partially ordered set which has a least elementL. We are also given that ∨M ∈ B for all M ⊆ B.

Let M ⊆ B where M 6= ∅. We are required to show that ∧M ∈ B.

Let

N = N ∈ B : N ≤M for all M ∈ M

Since L ∈ B and L ≤ M for all M ∈ M , then L ∈ N . So N 6= ∅. Since N 6= ∅

then, by hypothesis, ∨N exists in B.

For N ∈ N , N ≤ M for all M ∈ M ; so M ≥ N for all M ∈ M . This means N is

a lower bound of M . This holds true for each N ∈ N . Then the lubN = ∨N is alower bound of M . Then

∨N ≤ glbM = ∧M

2A set is clopen if it is simultaneously open and closed in S


We claim that ∨N = glbM . For suppose A is a lower bound of M . That is, supposeA ≤ M for all M ∈ M . Then A ∈ N . This implies A ≤ ∨N . So ∨N is a lower

bound of M which greater than or equal to the lower bound A. Then ∨N is thegreatest lower bound of M . So ∨N = ∧M , as claimed.

By hypothesis, ∨N ∈ B so ∧M ∈ B, as required.

For convenience, we restate the definition of “regular open set” previously discussedon page 64.

Definition 32.3 Let S be a topological space. A subset, B, of S is said to be regular openin S if

B = intS(clS(B))

The set of all regular open subsets of S will be denoted as Ro(S).

For example, (−5, 0)∪ (0, 5) is open in R but not regular open in R.

When the elements of a set are subsets of a certain type, partially ordered by “⊆”,it does not automatically follow that ∨ and ∧ are ∪ and ∩, respectively. Considerthe following statement which verifies that (Ro(S),⊆,∨,∧) forms a complete lattice

in the partially ordered set (τ,⊆), as an example.

Theorem 32.4 Let (S, τ) be a topological space. Suppose that, for A,B ∈ Ro(S) and

M ⊆ Ro(S),

A ∧ B = A ∩ B∨M = intSclS(∪M )

Then (Ro(S),⊆,∨,∧) is a complete lattice in (τ,⊆).

Proof : Given S be a topological space.

Suppose A and B belong to Ro(S).

Step 1. We claim that A ∧ B = A ∩B = glbA,B ∈ Ro(S). By definition of Ro(S),A = intSclSA and B = intSclSB. Then

intSclS(A ∩B) ⊆ intS(clSA ∩ clSB) By theorem 4.6

= intSclSA ∩ intSclSB

= A ∩ B


Since A∩B ⊆ intSclS(A∩B) and intSclS(A∩B) ⊆ A∩B then A∩B = intSclS(A∩B).Then A ∩B is regular open. So A ∧B = glbA,B = A ∩ B ∈ Ro(S). As claimed.

Step 2. Let M ⊆ Ro(S) where M is non-empty. Then ∨M = lubM : M ∈ M . Weclaim that ∨M = intSclS(∪M ).

Proof of claim. If M ∈ M then M ⊆ ∪M , hence M ⊆ intSclS(∪M ) for all M ∈ M .Then intSclS(∪M ) is an upper bound of M . Hence ∪M ⊆ intSclS(∪M ). On the

other hand, suppose V ∈ Ro(S) is an upper bound of M . Then M ⊆ V for allM ∈ M , so ∪M ⊆ V . This implies intSclS(∪M ) ⊆ intSclS(V ) = V , for any upper

bound V of M .

Then ∨M = lubM : M ∈ M = intSclS(∪M ), as claimed.

Since intSclS(∪M ) ∈ Ro(S), then ∨M ∈ Ro(S).

So for any M ⊆ Ro(S), ∨M exists in Ro(S) and equals, intSclS(∪M ).

Step 3. We now show that, (Ro(S),⊆,∨,∧) is complete. That is, for M ⊆ Ro(S)(where M is non-empty), ∧M exists in Ro(S).

First note that Ro(S) has a smallest element: Since intSclS∅ = intS∅ = ∅ and ∅ ⊆ Ufor all U ∈ Ro(S) then ∅ is the smallest regular open set with respect to ⊆.

We know that ∨M = intSclS(∪M ) for all M ⊆ Ro(S).

We have shown in example 4 above that for any lattice, (B,≤,∨,∧), which has a least

element and ∨M ∈ B (for all non-empty M ⊆ B) then ∧M ∈ B for all non-emptyM ⊆ B. The lattice (Ro(S),⊆,∨,∧) satisfies precisely these conditions.

So ∧M ∈ Ro(S) for all non-empty M ⊆ Ro(S).

We conclude that (Ro(S),⊆,∨,∧) is a complete lattice.

Suppose (S, τ) be a given topological space. Clearly, ∅ and S belong to Ro(S). Onpage 65, we showed that Ro(S) is closed under finite intersections. Then, if x ∈ S

and x ∈ A ∩B where A,B ⊆ Ro(S), since A∩B ∈ Ro(S), then Ro(S) satisfies the“base property”. Then, by theorem 5.4,

Ro(S) is an open base for some topology on S

Then Ro(S) generates a topology, τs , on S. Note that τs is strictly weaker than τ(since there are open sets which are not regular open). So (S, τ) may differ from (S, τs).

If (S, τ) is assumed to be Hausdorff, τ separates points of S. Then, for x, y ∈ S,there exists disjoint U, V ∈ τ containing x and y, respectively. Then x, y be-

long to intSclS(U) and intSclS(V ), respectively. Since intSclS(U) ∩ intSclS(V ) =intSclS(U ∩ V ) = ∅, then τs also separates points of S. We have shown that (S, τs) a


Hausdorff on S.

Notice that both (B(S),⊆,∩,∪) and Ro(S),⊆,∨,∧) all subsets of (P(S),⊆,∩,∪),are themselves lattices. Since ∨ and ∧ in (B(S),⊆,∩,∪) agree with ∨ and ∧ in

(P(S),⊆,∩,∪) then we say that it is a

sublattice

of P(S). The lattice (Ro(S),⊆,∨,∧) cannot be viewed as a “sublattice” of P(S)since A ∨ B = intSclS(A ∪B) 6= A ∪ B.

32.2 Filters and ultrafilters in a lattice L ⊆ P(S).

Suppose S is a set and the family, P(S), of all subsets of S is partially ordered by

inclusion “⊆”. In what follows the set, L, denotes a particular subset of P(S) whichcontains ∅ and S and is partially ordered by “⊆”.

The definition of “lattice filter” is analogous to the notion of “filter of sets”.

Let (L ⊆,∨,∧) be a lattice of sets and F ⊆ L. The subfamily F of L is called an

L-filter base if it satisfies the two conditions,

1) F is non-empty,

2) F is such that, for non-empty A,B ∈ F there exists D 6= ∅ in F such that

D ⊆ A ∧ B ∈ F .

If an L-filter base F also satisfies the condition,

“ If A ∈ F and A ⊆ C for some C ∈ L then C ∈ F”

then we say that F is an L-filter. Given an L-filter base, F , it can always generate

an L-filter which we will denote as, F ∗. A filter, F , is said to be a proper L-filter ifF 6= L. Note that an L-filter, F is proper if and only if ∅ 6∈ F .1

Definition 32.5 Let (L,⊆,∨,∧) be a lattice of sets in P(S). An L-ultrafilter is a properfilter F ⊆ L which is not properly contained in any other proper filter in L. If the filter

F is such that ∧F : F ∈ F 6= ∅ then we say that the filter F is a fixed L-ultrafilter.L-ultrafilters which are not fixed are said to be free L-ultrafilters.

1Since if ∅ ∈ F , ∅ ⊆ F for all F ∈ L, so F = L.


Theorem 32.6 Let S be a topological space and L = P(S). Suppose F is a proper L-filter where

(L,⊆,∨,∧) is a lattice in (P(S),⊆).

a) The L-filter, F , can be extended to an L-ultrafilter.

b) The L-filter, F , is an L-ultrafilter if and only if, for every A ∈ P(S), either A ∈ For S\A ∈ F .

Proof : a) We are given that L = P(S) and F is a proper L-filter. Let H = M :M is a proper L-filter such that F ⊆ M . We partially order H with ⊆. Let C be

a chain in (H ,⊆). Then ∪C : C ∈ C is an upper bound of C with respect to ⊆.So every chain in H has an upper bound. By Zorn’s lemma, (H ,⊆) has a maximal

element. That is, H contains a filter, F ∗, which is not properly contained in anyother filter. Since F ∗ ∈ H , F ⊆ F ∗. Then F can be extended to an L-ultrafilter,

as required.

b) ( ⇒ ) We are given that F is an L-ultrafilter in L = P(S) and that A ⊆ S. Weare required to show that either A ∈ F or S\A ∈ F .

Suppose neither A nor S \A belongs to F . Let

HA = B ∈ L : A ∩ F ⊆ B for some F ∈ F

We claim that HA is a proper L-filter base which contains F as a proper subset.

Proof of claim. See that, if U ∈ F , A ∩ U ⊆ U , an element of F . Then U ∈ HA, soF ⊆ HA.

Also, since A ∩ F ⊆ A for all F ∈ F , A ∈ HA. Then, A ∈ HA\F .Then F is a proper subset of HA.

Suppose B,D ∈ HA. We now show that B ∧D ∈ HA.

Then B,D ∈ L, A ∩ F1 ⊆ B and A ∩ F2 ⊆ D for some F1, F2 ∈ F .

Then A ∩ (F1 ∧ F2) ⊆ A ∩ F1 ⊆ B and A ∩ (F1 ∧ F2) ⊆ A ∩ F2 ⊆ D. ThenA ∩ (F1 ∧ F2) ⊆ B ∩D ⊆ glbB,D = B ∧D. Since B ∧D ∈ L, then B ∧D ∈ HA.

Also see that ∅ 6∈ HA. For, if it was, A ∩ F = ∅ for some F and so F ⊆ S\A which

would imply S\A ∈ F , a contradiction. So HA is proper.

Hence, since HA is proper, is non-empty and is closed under finite applications of ∧,

it is an L-filter base, as claimed.

Then HA will generate an L-filter H ∗A . Then F ⊂ H ∗

A which contradicts the factthat F is an L-ultrafilter.

Then the L-ultrafilter F must intersect either A or S\A.


( ⇐ ) Conversely, for L = P(S), suppose that, for every A ⊆ S, either A ∈ F orS\A ∈ F . We are required to show that F is an ultrafilter. Suppose not. That is,

suppose that there exists a proper filter, H , such that F ⊂ H . Then there existssome non-empty A such A ∈ H \F . Then S\A cannot belong to F , for, if it did,

then A ∩ (S\A) = ∅ must belong to H , contradicting the fact that H is a properfilter. Hence F must be an L-ultrafilter.

A word of caution: For the characterization of an L-ultrafilter above, things will work

out a bit differently in the case L = Ro(S), as the following theorem shows.

Theorem 32.7 Let S be a topological space. We have shown that (Ro(S),⊆,∨,∩) is a

complete lattice in (τ,⊆). An Ro(S)-filter, F , is an Ro(S)-ultrafilter if and only if, for anyA ∈ Ro(S), either A or S\clS(A) belongs to F .

Proof : ( ⇒ ) Suppose F is an Ro(S)-ultrafilter and A ∈ Ro(S). Let F ∈ F .

Case 1: If F ∩ A = ∅ then (since F is open) F ⊆ S\clSA. Since

S\clSA = intS(S\A)

= intS(S\intSclSA)

= intSclS(S\clSA)

F ⊆ S\clSA ∈ Ro(S).

Case 2: Suppose F ∩A 6= ∅ for all F ∈ F . Suppose A 6∈ F . Let

H = B ∈ Ro(S) : A ∩ F ⊆ B

Then F ⊆ H and A ∈ H \F . As shown in the theorem above, H is a filter base

which generates a filter H ∗ in (Ro(S),⊆). Since F ⊂ H ∗ this contradicts the factthat F is an Ro(S)-ultrafilter. So A must belong to F .

( ⇐ ) Suppose that for any A ∈ Ro(S), either A or S \clS(A) belongs to F . Seethat the filter F extends to an Ro(S)-ultrafilter F ∗. Suppose A ∈ F ∗. If A 6∈ Fthen S\clSA ∈ F ⊆ F ∗ implying that A ∩ (S\clSA) = ∅ ∈ F ∗, a contradiction. SoA ∈ F . Then F ∗ ⊆ F which implies that F is an Ro(S)-ultrafilter.

It can similarly be shown that a τ -filter, F , is a τ -ultrafilter if and only if, for any

A ∈ τ , either A or S\clS(A) belongs to F .


32.3 Boolean algebras.

We now define certain special types of lattices and discuss some of their special prop-erties. In what follows, the set L need not be a set of subsets of some topological

space (unless specifically stated otherwise).

Definition 32.8 Let (L,≤) be any partially ordered set. A lattice (L,≤,∨,∧) is said to

be a distributive lattice if, for any x, y, and z in L, x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) andx ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

1) The lattice (L,∨,∧) is said to be a complemented lattice if it has a maximal element,

denoted by 1, and a minimal element, denoted by 0, and for every x ∈ L there exists aunique x′ such that

x ∨ x′ = 1 and x ∧ x′ = 01

2) A complemented distributive lattice is referred to as being a Boolean algebra. A Booleanalgebra is denoted as

(B,≤,∨,∧, 0, 1, ′)

when we explicitly want to express what the maximum and minimum elements are.

Note that, since x′ ∧ x = 0 and x′ ∨ x = 1 then

x′′ = x

Also, 0 ∧ 1 = 0 and 0 ∨ 1 = 1 implies

0′ = 1 and 1′ = 0

Example 5. The lattices (P(S),⊆,∪,∩, S,∅, ′) and (B(S),⊆,∪,∩, S,∅, ′) where

B(S) = A ⊆ S : A is clopen in S

are the simplest examples of Boolean algebras. In both cases A′ = S\A.

Recall that, for a given topological space, S, (Ro(S),⊆,∨,∩) is a complete lattice. Itis also a Boolean algebra, although this is not at all obvious. We verify this.

, See that (Ro(S),⊆,∨,∩) contains 0 = ∅ and 1 = S.

1Note that in this context it is not meaningful to say “x′ = S\x”


We claim that the set Ro(S) is a complemented lattice: Let A ∈ Ro(S). ThenA = intSclSA. See that

A ∧ (S \clSA) = A ∩ (S \clSA)

= ∅

A ∨ (S \clSA) = intSclS(A ∪ (S\clSA))

= intSS

= S

So A′ = S\clSA. (It is erroneous to interpret A′ as meaning S\A.)

The element A′ belongs to Ro(S) since

S\clSA = S\clSintSclSA

= intSclS(S\clSA)

So Ro(S) is closed under complements, as claimed.

Finally, we show that (Ro(S),⊆,∨,∩) is a distributive lattice.

Let A,B, C ∈ Ro(S).

A ∩ (B ∨C) = intSclSA ∩ intSclS(B ∪C)

= intSclS(A ∩ (B ∪C))

= intSclS( (A∩B) ∪ (A ∩ C) )

= (A ∩B) ∨ (A ∩C)

Proceed similarly to show that A ∨ (B ∩ C) = (A ∨B) ∩ (A ∨C).

Then (Ro(S),⊆,∨,∩) is a Boolean algebra.

32.4 Boolean filters and ultrafilters..

Like lattices (L,⊆,∨,∧) in P(S) a Boolean algebra (B,≤,∨,∧, 0, 1, ′) contains B-filters and B-ultrafilters. The following concepts and properties are slight generaliza-

tions of ones we have already seen, so they will seem familiar to readers.


Definition 32.9 Suppose that, for a given set, B, (B,≤,∨,∧, 0, 1, ′) is a Boolean algebra.Then F , a

B-filter base

if F is a non-empty subset of B such that, for any x, y ∈ F , there is a z ∈ F where z 6= 0and z ≤ x ∧ y.

If F is a B-filter base which also satisfies the property,

[ x ∈ F and x ≤ y ] ⇒ y ∈ F

F is called a B-filter.

We say that F is a B-ultrafilter if F is a proper filter and is not properly contained in anyother proper B-ultrafilter.

Note that a B-ultrafilter is a subset of the Boolean algebra, B, and so can be seen as a

particular element of P(B).

We will emphasize three important B-ultrafilter characterizations.

Theorem 32.10 Let F be a proper B-filter in the Boolean algebra,(B,≤,∨,∧, 0, ′). Then

the following statements are equivalent.

1) The set F is a (proper) B-ultrafilter.

2) If x ∈ B and x ∧ y 6= 0 for all y ∈ F then y ∈ F .

3) Whenever x ∨ y ∈ F , either x ∈ F or y ∈ F .

4) For any x ∈ B, either x ∈ F or x′ ∈ F .

Proof : We are given that F is a B-filter in the Boolean algebra,(B,≤,∨,∧, 0, ′).( 1 ⇒ 2 ) Suppose the set F is a B-ultrafilter and that a ∈ B is such that a ∧ y 6= 0for each y ∈ F . We are required to show that a ∈ F .

Let

Ua = z ∈ B : z ≥ a ∧ y 6= ∅ for some y ∈ FSee that, since a ≥ a ∧ y for all y ∈ F (by hypothesis), then a ∈ Ua. So Ua 6= ∅ .

In fact, since z ∈ F ⇒ z ≥ a ∧ z ⇒ z ∈ Ua.

F ⊆ Ua


We claim that Ua is a proper B-filter.

Proof of claim. See that 0 6∈ Ua (since 0 ∈ Ua ⇒ 0 ≥ a ∧ y 6= 0, for y ∈ F ).

Suppose u, v ∈ Ua. There must exist b, c ∈ F such that u ≥ a∧ b and b ≥ a∧ c. Then

u ∧ v ≥ (a ∧ b) ∧ (a ∧ c)= a ∧ (b ∧ c)

implies u ∧ v ∈ Ua (since b ∧ c ∈ F ).

For u ∈ Ua and v ≥ u, v ≥ u ≥ a ∧ b for some b ∈ F . So v ∈ Ua.

We conclude that Ua is a proper B-filter, as claimed.

Since F is an ultrafilter which was shown to be contained in the filter Ua, it must be

the case thatF = Ua

Since a ≥ a ∧ y 6= 0 for all y ∈ F , then a ∈ Ua ⊆ F , as required.

( 2 ⇒ 3 ) Suppose that, if a ∈ B is such that a ∧ y 6= 0 for each y ∈ F , then a ∈ F .

We are required to show that, whenever x ∨ y ∈ F , either x ∈ F or y ∈ F .

Suppose x ∨ y ∈ F . Suppose that neither x nor y belongs to F . Then there exists

a, b ∈ F such that a ∧ x = 0 and b ∧ y = 0. See that (x ∨ y) ∧ (a ∨ b) ∈ F .

Since

(x ∨ y) ∧ (a ∨ b) = (x ∧ a ∧ b) ∨ (y ∧ a ∧ b)≤ (x ∧ a) ∨ (y ∧ b)= 0 ∨ 0

= 0

So 0 ∈ F , contradicting that F is proper. So either x or y ∈∈ F .

( 3 ⇒ 4 ) Suppose that whenever x ∨ y ∈ F , either x ∈ F or y ∈ F .

Since x ∨ y′ = 1 ∈ F , then either x or x′ belongs to F .

( 4 ⇒ 1 ) Suppose F is a B-filter such that for x ∈ B, either x or x′ belongs toF . Suppose U is a B-ultrafilter which contains F . If x ∈ U then, by hypothesis,

either x or x′ belongs to F . If x′ ∈ F then x′ ∈ U . So x ∧ x′ = 0 ∈ U . This is acontradiction. Then x ∈ F . So F = U .

Definition 32.11 Suppose we are given two Boolean algebras, (B1, ≤1, ∨1, ∧1, 0, 1,′ ) and(B2, ≤2, ∨2, ∧2, 0, 1,′ ) and a function f which maps elements of B1 to elements of B2. We

say that f : B1 → B2 is a Boolean algebra!homomorphism if, for any x, y ∈ B,

1) f(x ∨1 y) = f(x) ∨2 f(y),


2) f(x ∧1 y) = f(x) ∧2 f(y)

3) f(x′) = f(x)′.

The function f : B1 → B2 is a Boolean isomorphism if f is a bijection and both f and f←

are Boolean homomorphisms.

Note. See that if f : (B1, ≤1, ∨1, ∧1, 0, 1,′ ) → (B2, ≤2, ∨2, ∧2, 0, 1,′ ) is a Boolean

homomorphism then

f(1) = f(1 ∨ 1′)

= f(1) ∨ f(1′)

= f(1) ∨ f(1)′

= 1

Similarly f(0) = f(0 ∧ 0′) = f(0) ∧ f(0)′ = 0.

32.5 The Stone space.

Given a Boolean algebra, (B, ≤, ∨, ∧, 0, 1), a B-ultrafilter, U , is an element of P(B)

which both satisfies the filter properties and is “maximal” with respect to other B-filters in P(B).

If B is a Boolean algebra, we define the set, S (B), corresponding to B as,

S (B) = U : U is a B-ultrafilter on B

Since any B-ultrafilter is an element of P(B), then S (B) ⊆ P(B).

For each element, x, of a Boolean algebra (B,≤,∨,∧, 0, 1,′ ), we define

ϕx = U ∈ S (B) : x ∈ U ⊆ S (B)

the set of all B-ultrafilters on B which contain the element x. Then for each x ∈ B,ϕx ∈ P(S (B)). We define the function, ϕ : B → P(S (B)), as follows:

ϕ(x) = ϕx

See that ϕ : B → P(S (B)) is a well-defined function. For suppose ϕx 6= ϕy. Without

loss of generality, suppose U ∈ ϕx and U 6∈ ϕy. Then x ∈ U and y 6∈ U . So x 6= y.

Properties of the function ϕ : B → P(S (B)). The function ϕ : B → P(S (B)) iseasily seen to satisfy the following four properties. In what follows U ∈ S (B).


a) ϕx∧y = ϕx ∩ ϕy.

b) ϕx∨y = ϕx ∪ ϕy.

c) ϕ0 = ∅ and ϕ1 = S (B).

d) ϕx′ = S (B)\ϕx.

Proof of property a): This follows from the fact x ∧ y ∈ U ⇔ x ∈ U and y ∈ U .

Proof of property b): This follows from theorem 32.10, which states that x∨y ∈ U ⇔x ∈ U or y ∈ U .

Proof of property c): Since B-ultrafilters are proper filters no ultrafilter can contain

the 0-element of B; then ϕ(0) = ϕ0 = ∅. Also, 1 belongs to all B-ultrafilters henceϕ(1) = ϕ1 = S (B).

Proof of property d): See that, ∅ = ϕ0 = ϕx∧x′ = ϕx ∩ ϕx′.

Also, S (B) = ϕ1 = ϕx∨x′ = ϕx ∪ ϕx′ . It follows that ϕx′ = S (B)\ϕx.

We will now use the set ϕx : x ∈ B to topologize the set, S (B), of all B-ultrafilters.

Theorem 32.12 Let (B,≤,∨,∧, 0, 1,′ ) be a Boolean algebra. Then the set

M = ϕx : x ∈ B

is an open base for some topology, τ , on the set S (B).

Proof : Let M = ϕx : x ∈ B ⊆ P(S (B)) where ϕx = U ∈ S (B) : x ∈ U . To showthat M is an open base topology on S (B) it suffices to show that M satisfies the

“base property”. (See theorem 5.4).

We claim that ∪M = S (B). The set ∪M is a union of sets whose elements areB-ultrafilters. So ∪M ⊆ S (B). If U ∈ S (B) and x ∈ U then U ∈ ϕx ∈ ∪M . So

S (B) ⊆ ∪M . Then ∪M = S (B), as claimed.

So the elements of M cover S (B).

Suppose ϕx and ϕy belong to M . Since B is a Boolean algebra then x ∧ y ∈ B. Soϕx∧y ∈ M . Also see that for any B-ultrafilter, U ,

x ∧ y ∈ U ⇔ x, y ∈ U

Then U ∈ ϕ(x ∧ y) if and only if U belongs to both ϕx and ϕy. Then

ϕ(x∧ y) = ϕ(x) ∩ ϕ(y)

This implies that, for any x, y ∈ B other than 0 and 1, there exists a B-ultrafilter, U ,

such thatU ∈ ϕx∧y ⊆ ϕx ∩ ϕy


We conclude that M satisfies the “base property” with respect to the set S (B).Then, by 5.4, M is an open base for a topology, say τ , on S (B), as required.

The previous theorem states that, given a Boolean algebra (B,≤,∨,∧, 0, 1,′ ), we cantopologize, S (B), the set of all B-ultrafilters, by using M as an open base. That is,

for each x ∈ B, the set, ϕx, is an open base element for a topology on S (B). Oneshould also note that both ϕx and ϕx′ = ϕ′x = S (B)\ϕx are open in S (B). Then

M = ϕx : x ∈ B forms a base of clopen sets for S (B)

Then M generates a topology, τM , on S (B).

Definition 32.13 Let (B,≤,∨,∧, 0, 1,′ ) be a Boolean algebra and let S (B) = U :

U is a B-ultrafilter . When equipped with the topology, τM , the set (S (B), τM ) is re-ferred to as the Stone space1.

For what follows, recall that, if (S, τ) is some topological space then

(B(S), ⊆, ∪, ∩, ∅, S)

represents the Boolean algebra of all clopen sets in S. Then M ⊆ B(S (B)).

Definition 32.14 A Boolean algebra (B,≤,∨,∧, 0, 1,′ ) is said to have a

topological representation

if there exists a a topological space, S, and a Boolean isomorphism f : B → B(S) whichmaps B onto the Boolean algebra, (B(S),⊆,∪,∩,∅, S), of all clopen subsets of some topo-

logical space, S.

1Named after the American mathematician, Marshall Harvey Stone (April 8, 1903 − January 9, 1989).He significantly contributed to real analysis, functional analysis, topology and the study of Boolean algebras.In 1936, he published a long paper that included Stone’s representation theorem for Boolean algebras,


The following theorem relates a Boolean algebra,B, to the topological space, (S (B), τM),which is generated from B. We know that the space, S (B), has a base of clopen sets

(with respect to the topology, τM ). From S (B) we gather together all clopen subsetsof S (B) to form the Boolean algebra,

(B(S (B)), ⊆, ∪, ∩, 0, 1)

of all clopen subsets of S (B). We show that B can be mapped isomorphically ontoB(S (B)). By definition, B(S (B)) will then serve as a “topological representation”of B.

Theorem 32.15 The Stone representation theorem. Let (B,≤,∨,∧, 0, 1,′ ) be a Booleanalgebra and S (B) = U : U is a B-ultrafilter on B. For each x ∈ B, let ϕx = U ∈S (B) : x ∈ U . The function, ϕ : B → P(S (B)), is defined as ϕ(x) = ϕx. Let τM denotethe topology on S (B) which is generated by M = ϕx : x ∈ B, where each ϕx has been

shown to be clopen in S (B).

1) The topological space (S (B), τM ) equipped with the topology, τM , generated by the

open base, M = ϕx : x ∈ B, is a compact zero-dimensional Hausdorff topologicalspace.

2) The set M = ϕx : x ∈ B = B(S (B)), where B(S (B)) is the set of all clopensubsets of S (B). So ϕ maps B onto B(S (B)).

3) The function ϕ : B → B(S (B)) is a Boolean isomorphism mapping mapping B onto

B(S (B)).

Proof : We are given that (B,≤,∨,∧, 0, 1,′ ) is a Boolean algebra.

1) We are given that (S (B), τM ) is the topological space generated by the open baseM = ϕx : x ∈ B. A topological space is zero-dimensional if it has an open baseof clopen sets. We have shown above that ϕx : x ∈ B is a set of clopen subsets of

S (B). So, by definition, S (B) is a zero-dimensional space.

We claim that (S (B), τM ) is Hausdorff. Let U1 and U2 be distinct B-ultrafilters

in S (B). Then there exists some x ∈ U2 which does not belong to U1. Then x′

must belong to U1. It follows that U2 ∈ ϕx and U1 ∈ ϕx′ . Since x ∧ x′ = 0 implies

ϕx ∩ ϕx′ = ∅ it follows that S (B) is a Hausdorff topological space, as claimed.

We claim that (S (B), τ) is a compact space. To show that S (B) is compact it suf-fices to show that a set Fii∈I of closed subsets which satisfies the finite intersection

property has non-empty intersection.


Let H be a collection of closed subsets of S (B) which satisfies the finite intersec-tion property. Then H is a filter base of closed sets. It suffices to show that ∩H 6= ∅.

Let H ∗ be the filter of closed subsets which is generated by the filter base H . It now

suffices to show that ∩H ∗ 6= ∅. Let

T = x ∈ B : F ⊆ ϕx for some F ∈ H ∗

Since M = ϕx : x ∈ B is a base of clopen sets in S (B), M is a base for closed setsin S (B) and so every closed set in S (B) is the intersection of elements in M . Then

∩F : F ∈ H ∗ = ∩ϕx : x ∈ T

We now show that T is a B-filter. To see this, suppose x, y ∈ T . There exists F1 andF2 in H ∗ such that F1 ⊆ ϕx and F2 ⊆ ϕy. Then

F1 ∩ F2 ⊆ ϕx ∩ ϕy = ϕx∧y

Since H ∗ is a B-filter, F1 ∩ F2 must be non-empty and belong to H ∗; so x ∧ y ∈ T .The T is a B-filter base. From this we deduce that T is a filter in B.

Now T extends to a B-ultrafilter, say T . Since x ∈ T for all x ∈ T , then

T ∈ ∩ϕx : x ∈ T = ∩F : F ∈ H ∗ ⊆ ∩F : F ∈ H

We conclude that ∩F : F ∈ H 6= ∅, so S (B) is compact, as required.

2) We are required to show that M = B(S (B)). We already know that M ⊆B(S (B)). So it will suffice to show that B(S (B)) ⊆ M . If U is any clopensubset of S (B) then U = ∪ϕy : y ∈ V for some V ⊆ B. We have shown that

S (B) is a compact Hausdorff topological space, hence U must be compact. Thenϕy : y ∈ V is an open cover the compact set U . So there exists a finite subcover

ϕy : y = y1, y2, . . . , yk such that U = ∪ϕy : y = y1, y2, . . . , yk. Since

U = ∪ϕy : y = y1, y2, . . . , yk = ϕy1∨y2∨···∨yk∈ M

(see properties of ϕ above). Then B(S (B)) ⊆ M . So B(S (B)) = M , as required.

3) We are required to show that the function ϕ : B → B(S (B)) is a Boolean isomor-phism onto B(S (B)). We have shown in the list of properties of ϕ, that

ϕx∧y = ϕx ∩ ϕy

ϕx∨y = ϕx ∪ ϕy

ϕx′ = S (B)\ϕx

Since we have shown that M = B(S (B)), ϕ : B → B(S (B)) is a homomorphism

mapping B onto B(S (B)).


The function ϕ : B → B(S (B)) is one-to-one : Suppose x and y are distinct pointsin B. It suffices to show that ϕx 6= ϕy.

Since x 6= y then one of the two statements x ≤ y, y ≤ x must be false. We willassume, without loss of generality, that x 6≤ y.

Claim: We claim that, since x 6≤ y, then x ∧ y′ 6= 0.

Proof of claim. Suppose that x 6≤ y and x ∧ y′ = 0. See that

x′ = x′ ∨ 0 ⇒ x′ ∨ (x∧ y′)⇒ x′ ∨ (x∧ y′)⇒ (x′ ∨ x) ∧ (x′ ∨ y′) (B is distributive)

⇒ 1 ∧ (x′ ∨ y′)⇒ (x′ ∨ y′)⇒ y′ ≤ x′

⇒ x ≤ y

We have a contradiction. The source of the contradiction is our supposition that

x ∧ y′ = 0. So x ∧ y′ 6= 0, as claimed.

Consider the collection F = u ∈ B : x ∧ y′ ≤ u. Clearly, x ∧ y′, x, y′ ∈ F .

Now, F is easily seen to be a B-filter. So there exists U ∈ S (B) such that F ⊆ U .Since x, y′ ∈ U , U ∈ ϕx ∩ ϕy′ = ϕx∧y′ . Then U belongs to ϕx but cannot belong to

ϕy.

So ϕx 6= ϕy. So ϕ : B → B(S (B)) is one-to-one.

So ϕ : B → B(S (B)) is an isomorphism onto B(S (B)), as required.

We have just shown that

. . . every Boolean algebra, (B,≤,∨,∧, 0, 1,′ ), has a topological representa-tion as a Boolean algebra, (B(S (B)),⊆,∪,∩, 0, 1,′ ), of a compact, zero-

dimensional Hausdorff space, S (B).


33 / Baire spaces

Summary. In this section we present a brief introduction to a particularly

important class of topological spaces called “Baire spaces”. The properties pos-sessed by members of this class turned out to be relevant in many applied fields,

in particular in analysis. We will define such spaces and describe a few charac-terizations and properties. This is followed by a few examples.

33.1 Definitions

On a first reading, the definition of the class of Baire spaces may appear odd or oflittle interest and may lead the reader to wonder whether it is worth investing mental

energy in studying such properties. This first impression may change once we startinvestigating characterizations and seeking examples of Baire spaces and non-Baire

spaces.

Definition 33.1 A topological space, S, is a Baire space if and only if, the union of anycountable family, Fi : i ∈ N, of closed subsets (each) with empty interior, also has empty

interior.

A subset U of S is said to be first category in S if,

“U is the union, ∪Fi : i ∈ N of a countable family of closed subsets, Fi (each)

with empty interior. ”1

A subset U of S which is not of the first category in S is said to be of the second categoryin S. 2

Note that it is quite common to describe a set, A, whose closure has empty interiorby saying that “A is nowhere dense”.

Suppose S is a Baire space. See that, if U is any non-empty open subset of the space,S, then U cannot be of the first category in S (for, if it was, U would be an open

1The expression A is first category also appears in texts in the form of the following equivalent definition:“The set, A, is first category in S if and only if S\clSA is dense in S”. A set A is said to be meager in S, ifA is the countable union of closed subsets with empty interior. The set A if first category in S if and onlyif it is meager in S.

2A set A is non-meager in S if and only if it is a set of second category. A set A is said to be residual ifit is the complement of a meager set.

586 Section 33: Baire spaces

subset of the union, ∪Fi : i ∈ N, of closed subsets (each) with empty interior, where,S being a Baire space, ∪Fi : i ∈ N must have empty interior). So U is of the second

category in S.

Conversely, suppose that for a given topological space, S, no open subset U is a

set of the first category of S. Then, for any countable collection of closed subsets,F = Fi : i ∈ N, each of which has empty interior, cannot contain a non-empty

interior (the interior being an open subset of S). We conclude that, if every open Uin S is second category in S, then S is a Baire space.

33.2 Characterizations.

The definition of a Baire space often appears in texts in various ways. The followingproposition shows that the alternate definitions are equivalent.

Proposition 33.2 Let S be a topological space. The following statements are equivalent.

1) The space S is a Baire space.

2) The intersection of all members of a countable family, Ui : i ∈ N, of dense and opensubsets of S, is a dense subset of S.

3) Every non-empty open subset of S is second category


( 1 ⇒ 2 ) Suppose S is a Baire space as defined above. Let Ui : i ∈ N be a count-able family of open dense subsets of S . We claim ∩Ui : i ∈ N is dense in S. For

each i, S \Ui is a closed subset of S with empty interior. Since S is a Baire space,E = ∪S\Ui : i ∈ N also has empty interior. Then, S\E = S\∪S\Ui : i ∈ N =

∩Ui : i ∈ N is dense in S.

( 2 ⇐ 1 ) Proceed similarly, for the converse.

( 1 ⇔ 3) The proof is presented following the definition, above.

33.3 Baire category theorems.

At the first reading, a few questions which may come to mind are “How common are

Baire spaces?”, “Are they hard to recognize?” Baire spaces are, in fact, quite commonas we shall see. The following results show how common they are.


Lemma 33.3 Suppose S is a regular space. Let F = Fi : i ∈ N be a countable collectionof closed subsets, each with empty interior. Let U be any open subset of S. Then S contains

a countably infinite nested collection, U = Ui : i ∈ N, of non-empty open subsets and aninfinite sequence, ui : i ∈ N such that, for each i ∈ N,

ui ∈ Ui ⊆ clSUi ⊆ S\Fi and clSUi ⊆ Ui−1\Fi ⊆ U \Fi

Proof : We are given that S is a regular topological space and F = Fi : i ∈ N, is a

countably infinite collection of closed subsets each with empty interior. Let U be anynon-empty open subset of S.

Since F0 has empty interior, U 6⊆ F0, so there exists u0 ∈ U\F0. Since u0 ∩F0 = ∅

and S is regular, there exists some open set U0, such that

u0 ∈ U0 ⊆ clSU0 ⊆ S\F0 and clSU0 ⊆ U \F0

Since F1 has empty interior, U0 6⊆ F1, so there exists u1 ∈ U0\F1

Since u1 ∩ F1 = ∅ and S is regular, there exists some open set, U1 such that

u1 ∈ U1 ⊆ clSU1 ⊆ S\F1 and clSU1 ⊆ U0\F1 ⊆ U \F1

We inductively construct a nested set, U = Ui : i ∈ N, of open subsets, by usingthe recipe just described. Suppose Ui−1 is an open subset such that

ui−1 ∈ Ui−1 ⊆ clSUi−1 ⊆ S\Fi−1 and clSUi−1 ⊆ Ui−2\Fi−1 ⊆ U \Fi−1

Since Fi has empty interior, Ui−1 6⊆ Fi, so there exists ui ∈ Ui\Fi

Since ui ∩ Fi = ∅ and S is regular, there exists some open set, Ui such that

ui ∈ Ui ⊆ clSUi ⊆ S\Fi and clSUi ⊆ Ui−1\Fi ⊆ U \Fi (∗)

Then U is a collection of open subsets of S which satisfies, for all n ∈ N, the property

described in (*). This is the property required for the lemma.

Theorem 33.4 Suppose S is a compact Hausdorff space. Let F = Fi : i ∈ N be

a countable collection of closed subsets, each with empty interior. Then ∪F has emptyinterior. So S is a Baire space.

Proof : We are given that S is a compact Hausdorff topological space and F = Fi : i ∈ N,is a countably infinite collection of closed subsets, each with empty interior. Since Sis compact Hausdorff, S is a normal space. Let U be any non-empty open subset of S.

To show that ∪F = ∪Fi : i ∈ N has empty interior it suffices to show that U 6⊆ ∪F .


By the lemma, the space S contains a countably infinite collection, U = Ui : i ∈ N,of non-empty open subsets and an infinite sequence, ui : i ∈ N such that, for each

i ∈ N,

ui ∈ Ui ⊆ clSUi ⊆ S\Fi and clSUi ⊆ Ui−1\Fi ⊆ U \Fi

Then the collection U ∗ = clSUi : i ∈ N, is a nested collection of closed subsets suchthat clSUi ⊆ Ui−1 ⊆ clSUi−1 for all i. Since U ∗ satisfies the FIP and S is compact

then ∩clSUi : i ∈ N 6= ∅. So there exists z ∈ ∩clSUi : i ∈ N 6= ∅.

Since z ∈ clSUi ⊆ U \Fi for all i ∈ N, then

z ∈ ∩U \Fi : i ∈ N = U \∪Fi : i ∈ N

Then U 6⊆ ∪F . Since U is arbitrarily chosen open subset of S then we conclude that

∪F has empty interior. So S is a Baire space.

Theorem 33.5 The Baire category theorem I. Complete metric spaces are Baire spaces.

Proof : We are given that S is a compact Hausdorff topological space and F = Fi : i ∈ N,is a countably infinite collection of closed subsets, each with empty interior. Since S

is a metric space, S is a normal space. Let U be any non-empty open subset of S.To show that ∪F = ∪Fi : i ∈ N has empty interior it suffices to show that U 6⊆ ∪F .

By the lemma, the space S contains a countably infinite collection, U = Ui : i ∈ N,of non-empty open subsets and an infinite sequence, ui : i ∈ N such that, for eachi ∈ N,

ui ∈ Ui ⊆ clSUi ⊆ S\Fi and clSUi ⊆ Ui−1\Fi ⊆ U \Fi (∗)

For the case where S is a metric space equipped with the metric ρ, the set U canbe constructed so that diam(U0) = supρ(x, y) : x, y ∈ U1 ≤ 1 and diam(Ui) =

supρ(x, y) : x, y ∈ Ui < 1/i, while still maintaining the property described in (*).

Then the collection U ∗ = clSUi : i ∈ N, is a nested collection of closed subsets suchthat ui ∈ clSUi ⊂ Ui−1 ⊆ clSUi−1, for all i.

Let ε > 0. We can choose N such that i > N implies diam(clSUi) ≤ 1/i < ε. Since

the collection U ∗ is nested, for i, j > N , ρ(ui, uj) < ε. Then the sequence ui : i ∈ Nis Cauchy in S. Since S is complete, the sequence ui : i ∈ N converges to some point

z in ∩U ∗. So ∩U ∗ is non-empty. For the reasons stated at the end of the previoustheorem, z ∈ U \∪F . So U 6⊆ ∪F . So ∪F has empty interior. We conclude that S

is a Baire space.


Theorem 33.6 The Baire Category theorem II. Locally compact Hausdorff spaces areBaire spaces.

Proof : We are given that S is a locally compact Hausdorff space.

Then S is completely regular (by 18.8) and S is an open subset of βS (by 21.16).Since βS is compact and Hausdorff then it is a Baire space (by 33.5).

Let F = Fi : i ∈ N, be a countably infinite collection of closed subsets, each withempty interior. We consider the collection F ∗ = clβSFi : i ∈ N. If U is an open

subset of βS such that U ⊆ clβSFi then U ∩ S is an open subset of S contained in Fi

contradicting the hypothesis that Fi has empty interior in S. So clβSFi : i ∈ N is a

collection of closed subsets of βS each of which has empty interior.

Since βS is compact Hausdorff and so is a Baire space, ∪F ∗ has empty interior. Sup-pose V is an open subset (in S) of ∪F . Since S is open in βS, V is open (in βS)

contrary to the fact that ∪F ∗ has empty interior. So ∪F has empty interior. Weconclude that S is a Baire space.

As an example of an application of the Baire Category Theorem we consider the fol-

lowing proposition.

Recall that an point, p, in a space S is called an isolated point if p is open in S.

Proposition 33.7 Suppose S is a complete metric space which contains no isolated points.

Then S is an uncountably large set.

Proof : Suppose the non-empty complete metric space, S, is countable. Then S = ∪pi :i ∈ N where each pi is a point in S. Since intSpi = ∅, for each i, and S has

been shown to be a Baire space, then S has empty interior contradicting that S isnon-empty. So S is an uncountable set.

In an example presented on page 41, we showed that the set, Q, of all rationals is anFσ (the countable union of closed sets). We verify, that Q is not a Gδ (the countable

intersection of open sets).

Proposition 33.8 The set of all rationals cannot be a Gδ.


Proof : We know that Q is dense in R. Let J denote the set of all irrationals. Suppose Q canbe expressed as Q = ∩Ui : i ∈ N where each Ui is an open neighbourhood of Q in

R. Since every open neighbourhood of a point not in Ui is an open neighbourhood ofsome irrational number, then this neighbourhood must meet Q and so must intersect

Ui. That is, each Ui is dense in R.

Since Q is countable we can also express it as Q = xi : i ∈ N and let Vi = R\xi,a dense open subset of R, for each i. Then

(∩Ui : i ∈ N) ∩ (∩Vi : i ∈ N)

is a countable intersection of dense open subsets of R. But this set turns out to beequal to Q∩J and so is empty. Since we know that R is a Baire space this intersection

of sets should be dense in R. This points to a contradiction. Then Q cannot be a Gδ.

Proposition 33.9 The open subset of a Baire space is a Baire space.

Proof : We are given that S is a Baire space and U is open subspace of S.

Let F = Fi : i ∈ N, be a countably infinite collection of closed subsets of U , eachwith empty interior (in U). We consider the collection F ∗ = clSFi : i ∈ N. If V

is an open subset of U such that V ⊆ clSFi then V ∩ U is a non-empty open subsetof U contained in Fi contradicting the hypothesis that Fi has empty interior in U .

So clSFi : i ∈ N is a collection of closed subsets each of which has empty interior in S.

Since S is a Baire space, ∪F ∗ has empty interior. Suppose V is an open subset (inU) of ∪F . Since U is open in S, V ∩ U is open (in S) contrary to the fact that ∪F ∗

has empty interior. So ∪F has empty interior. We conclude that U is a Baire space.

Example 1. Show that if T is a Baire space and T is dense in S then S is a Baire space.

Solution : Suppose that a Baire space, T , is a dense subset of a space S. SupposeF = Fi : i ∈ N, is a countably infinite collection of closed subsets of S, each withempty interior (in S). We consider the collection, F ∗ = Fi ∩ T : i ∈ N, of closed

subsets of T .

Suppose that V ∗ is an open subset of T such that V ∗ ⊆ Fi ∩ T . Then there exists anopen subset V in S such that V ∗ = V ∩ T . Then V ∗ ⊆ intSclSV

∗ ⊆ Fi.

Since V is non-empty and T is dense in S, intSclSV∗ ⊆ intSFi contradicting our hy-

pothesis. Then each Fi ∩ T has empty interior in T .

Since T is a Baire space then ∪F ∗ has empty interior.


If W is an open subset in S such that W ⊆ ∪F then W ∩ T ⊆ ∪F ∗. Since T is aBaire space W ∩ T must be empty. Since T is dense in S, intS ∪ F ∗ is empty. So S

is a Baire space.

Example 2. The space, Q, is not a Baire space.

Solution : Since Q is a countable set it can be enumerated Q = xi : i ∈ N SoQ = ∪xi : i ∈ N, the countable union of sets with empty interior in Q (with

respect to the subspace topology). Since intQ ∪ xi : i ∈ N = Q 6= ∅, Q is not aBaire space.

Example 3. Let f ∈ F (R,R). Let Qf = x ∈ R : f is continuous at x. Show that Qf

is a Gδ.

Solution : For each i ∈ N\0, let

Ui = ∪f←[B1/i(z)] : f←[B1/i(z)] is open, for all z ∈ R

Then, for each i ∈ N\0, Ui is an open subset of R.

Let U = ∩Ui : i ∈ N\0. Suppose U 6= ∅. Then f←[B1/i(z)] is open in U forall z ∈ R and each i. Then f pulls back open base elements to open sets in U . So

f is continuous on U . Conversely, f is continuous on U implies f pulls back openbase elements to open sets in U . So f is continuous at u if and only if u ∈ U . Then

Qf = ∩Ui : i ∈ N\0. So Qf is a Gδ.

Example 4. Show that no function in F (R,R) is continuous precisely on Q (and dis-

continuous on R\Q).

Solution : By the statement in Example 3, if f is continuous on Q (and discontinuous

elsewhere). then Q is a Gδ. We showed in proposition 33.8 that Q is not a Gδ. So weare done.

592

Appendix

Appendix 593

Appendix : Axioms and set theory statements.

I Axioms and classes : 1 / Classes, sets and axioms

Axiom A1 (Axiom of extent) : For the classes x, A and B, [A = B] ⇔ [x ∈ A⇔ x ∈ B]

Axiom A2 (Axiom of class construction): Let P (x) designate a statement about x which

can be expressed entirely in terms of the symbols ∈,∨,∧,¬,⇒, ∀, brackets and vari-ables x, y, z, . . . , A, B, . . . Then there exists a class C which consists of all theelements x which satisfy P (x).

Axiom A3 (Axiom of pair) : If A and B are sets , then the doubleton A, B is a set.

Axiom A4 (Axiom of subsets) : If S is a set and φ is a formula describing a particu-lar property, then the class of all sets in S which satisfy this property φ is a set.

More succinctly, every subclass of a set of sets is a set. (Also called the Axiom ofcomprehension, Axiom of separation or Axiom of specification).

Axiom A5 (Axiom of power set): If A is a set then the power set P (A) is a set.

Axiom A6 (Axiom of union): If A is a set of sets then⋃

C∈AC is a set.

Axiom A7 (Axiom of replacement): Let A be a set. Let φ(x, y) be a formula which

associates to each element x of A a set y in such a way that, whenever both φ(x, y)and φ(x, z) hold true, y = z. Then there exists a set B which contains all sets y such

that φ(x, y) holds true for some x ∈ A.

Axiom A8 (Axiom of infinity): There exists a non-empty set A that satisfies the condition:

“X ∈ A” ⇒ “X ∪ X ∈ A”. (A set satisfying this condition is called a successor setor an inductive set.)

Axiom A9 (Axiom of regularity) Every non-empty set A contains an element x whoseintersection with A is empty.

Axiom of choice : For every set A of non-empty sets there is a rule f which associatesto every set A in A an element a ∈ A.

I Axioms and classes : 2 / Constructing classes and sets

Theorem 2.1 For any class C, C = C. If it is not true that A = B we will write A 6= B.

Definition 2.2 If A and B are classes (sets) we define A ⊆ B to mean that every elementof A is an element of B. That is, A ⊆ B iff x ∈ A ⇒ x ∈ B If A ⊆ B we will say

that A is a subclass (subset) of B. If A ⊆ B and A 6= B we will say that A is a propersubclass (proper subset) of B and write A ⊂ B when we explicitly want to say A 6= B.

594 Appendix

Theorem 2.3 If C, D, and E are classes (sets) then:

a) C = C.

b) C = D ⇒ D = C.

c) C = D and D = E ⇒ C = D.

d) C ⊆ D and D ⊆ C ⇒ C = D.

e) C ⊆ D and D ⊆ E ⇒ C ⊆ E.

Theorem 2.4 There exists a class which is not an element.

Definition 2.5 The Axiom 2 states that C = x : x 6= x is a class. It contains no

elements. We will call the class with no elements the empty class and denote it by ∅.

Theorem 2.6 For any class C, ∅ ⊆ C.

Theorem 2.7 Let S be a set. Then:

a) ∅ ⊆ S and so ∅ is a set.

b) The set S is an element. Hence all sets are elements.

Definition 2.8 If A is a set then we define the power set of A as being the class P(A) ofall subsets of A. It can be described as follows: P(A) = X : X ⊆ A.

II Class operations 3 / Operations on classes and sets

Definition 3.1 Let A and B be classes (sets). We define the union A ∪ B of the class A

and the class B asA ∪B = x : (x ∈ A) ∨ (x ∈ B)

That is, the element x ∈ A ∪ B iff x ∈ A or x ∈ B. If A is a non-empty class ofclasses then we define the union of all classes in A as

⋃

C∈A

C = x : x ∈ C for some C ∈ A

That is, the element x ∈ ⋃C∈AC iff there exists C ∈ A such that x ∈ C.

Definition 3.2 Let A and B be classes (sets). We define the intersection A ∩ B of theclass A and the class B as

A ∩B = x : (x ∈ A) ∧ (x ∈ B)

That is, the element x ∈ A ∩ B iff x ∈ A and x ∈ B. If A is a non-empty class of

classes then we define the intersection of all classes in A as⋂

C∈A

C = x : x ∈ C for all C ∈ A

That is, the element x ∈ ⋂C∈AC iff x ∈ C for every class C in A .

Appendix 595

Definition 3.3 We will say that two classes (sets) C and D are disjoint if the two classeshave no elements in common. That is, the classes C and D are disjoint if and only if

C ∩D = ∅.

Definition 3.4 The complement, C′, of a class (set) C is the class of all elements which

are not in C. That is, if C is a class, then

C′ = x : x 6∈ C

Hence x ∈ C′ iff x 6∈ C. Given two classes (sets) C and D, the difference C −D, of C

and D, is the classC −D = C ∩D′

The symmetric difference, C4D, is the class

C4D = (C −D) ∪ (D −C)

Theorem 3.5 Let C and D be classes (sets). Then,

a) C ⊆ C ∪Db) C ∩D ⊆ C

Theorem 3.6 Let C and D be classes (sets). Then,

a) C ∪ (C ∩D) = C

b) C ∩ (C ∪D) = C

Theorem 3.7 Let C be a class (a set). Then (C′)′ = C.

Theorem 3.8 DeMorgan’s laws. Let C and D be classes (sets). Then,

a) (C ∪D)′ = C′ ∩D′b) (C ∩D)′ = C′ ∪D′

Theorem 3.9 Let C, D and E be classes (sets). Then,

a) C ∪D = D ∪ C and C ∩D = D ∩C (Commutative laws)

b) C ∪C = C and C ∩ C = C (Idempotent laws)

c) C ∪ (D ∪E) = (C ∪D) ∪E and C ∩ (D ∩E) = (C ∩D) ∩E (Associative laws)

d) C ∪ (D ∩E) = (C ∪D)∩ (C ∪E) and C ∩ (D ∪E) = (C ∩D)∪ (C ∩E) (Distribution)

Theorem 3.10 Let A be a class (a set) and U denote the class of all elements.

a) U ∪A = U

b) A ∩ U = A

c) U ′ = ∅

d) ∅′ = U

e) A ∪A′ = U

596 Appendix

Theorem 3.11 Let A be a non-empty class (set).

a)(⋃

C∈AC)′

=⋂

C∈AC′

b)(⋂

C∈AC)′

=⋃

C∈AC′

Theorem 3.12 Let D be a class and A be a non-empty class (set) of classes.

a) D ∩(⋃

C∈AC)

=⋃

C∈A(D ∩C)

b) D ∪(⋂

C∈AC)

=⋂

C∈A(D ∪C)

Theorem 3.13 Let B(i,j) : i = 1, 2, 3, . . . , j = 1, 2, 3, . . . be a set of sets Then ∪∞i=1(∩∞j=1B(i,j)) =∩∞j=1(∪∞i=1B(i,j)).

II Class operations 4 / Cartesian products

Definition 4.1 Let c and d be elements. We define the ordered pair (c, d) as (c, d) =c, c, d.

Theorem 4.2 Let a, b, c and d be classes (which are elements). Then (a, b) = (c, d) iffa = c and b = d.

Alternate definition 4.3 If c and d are classes define (c, d) as follows: (c, d) = c,∅, d, ∅ .

Definition 4.4 Let C and D be two classes (sets). We define the Cartesian product, C×D,

as follows: C ×D = (c, d) : c ∈ C and d ∈ D.

Lemma 4.5 Let C and D be two classes (sets). Then the Cartesian product, C ×D, of Cand D satisfies the property: C ×D ⊆ P(P(C ∪D)).

Corollary 4.6 If C and D are classes, then the Cartesian product, C ×D, is a class. If C

and D are sets, then C ×D is a set.

Theorem 4.7 Let C, D, E and F be a classes. Then

a) C × (D ∩E) = (C ×D) ∩ (C ×E)

b) C × (D ∪E) = (C ×D) ∪ (C ×E)

c) (C ∩ E)×D = (C ×D) ∩ (E ×D)

d) (C ∪ E)×D = (C ×D) ∪ (E ×D)

e) (C ∪D)× (E ∪ F ) = (C × E) ∪ (D ×E) ∪ (C × F ) ∪ (D × F )

f ) (C ∩D)× (E ∩ F ) = (C × E) ∩ (D ×E) ∩ (C × F ) ∩ (D × F )

Theorem 4.8 If C ⊆ D and E ⊆ F , then C × E ⊆ D × F .

Theorem 4.9 Given three classes (sets) S, U and V there is a one-to-one correspondencebetween the two classes (sets) S × (U × V ) and (S × U) × V .

Appendix 597

Theorem 4.10 For classes c, d, e and f , if (c, d) = c,∅, d, ∅ and (e, f) = e,∅,f, ∅ then (c, d) = (e, f) iff c = e and d = f .

III Relations 5 / Relations on a class or set

Definition 5.1 a) We will call any subset R of ordered pairs in U ×U a binary relation.

b) We will say that R is binary relation on a class C if R is a subclass (subset) of

C ×C. In such cases we will simply say that R is a relation in C or on C.

c) If A and B are classes (sets) and R is a subclass (subset) of A × B then R canbe viewed as a relation on A ∪ B.

Definitions 5.2 Let C be a class (a set). a) The relation∈C= (x, y) : x ∈ C, y ∈ C, x ∈ yis called the membership relation on C. b) The relation

IdC = (x, y) : x ∈ C, y ∈ C, x = y

is called the identity relation on C.

Definitions 5.3 Let R be a relation on a class (set) C. The domain of R is the class,domR = x : x ∈ C and (x, y) ∈ R for some y ∈ C. The image of R is the class,imR = y : y ∈ C and (x, y) ∈ R for some x ∈ C. The word range of R is often

used instead of “the image of R”. If R ⊆ A×B is viewed as a relation on A∪B, thendomR ⊆ A and imR ⊆ B.

Definition 5.4 Let C be a class (a set) and let R be a relation defined in C. The inverse,R−1, of the relation R is defined as follows:

R−1 = (x, y) : (y, x) ∈ R

Definition 5.5 Let C be a class (a set) and let R and T be two relations inC. We define therelationTR as follows : TR = (x, y) : there exists some z ∈ imR such that (x, z) ∈R and (z, y) ∈ T

III Relations 6 / Equivalence relations and order relations.

Definition 6.1 Let S be a class and R be a relation on S.

a) We say that R is a reflexive relation on S if, for every x ∈ S, (x, x) ∈ R.

b) We say that R is a symmetric relation on S if, whenever (x, y) ∈ R then (y, x) ∈ R.

c) We say that R is an anti-symmetric relation on S if, whenever (x, y) ∈ R and (y, x) ∈ R

then x = y.

d) We say that R is an asymmetric relation on S if, whenever (x, y) ∈ R then (y, x) 6∈ R.

e) We say that R is a transitive relation on S if, whenever (x, y) ∈ R and (y, z) ∈ R then(x, z) ∈ R.

598 Appendix

f) We say that R is an irreflexive relation on S if, for every x ∈ S, (x, x) 6∈ R.

g) We say that R satisfies the property of comparability on S if, for every x, y ∈ S wherex 6= y, either (x, y) ∈ R or (y, x) ∈ R.

Definition 6.2 Let S be a class and R be a relation on S. We say that R is an equivalence

relation on S if R is simultaneously 1) reflexive, 2) symmetric and 3) transitive on S.

Definition 6.3 Let S be a class.

a) Non-strict order relation. The relation R is a non-strict order relation on S ifit is simultaneously reflexive (aRa holds true for any a in S), antisymmetric (if

aRb and bRa then a = b) and transitive (aRb and bRc implies aRc) on S. Anon-strict order relation, R, on S is said to be a non-strict linear order relation

if, for every pair of elements a and b in S, either (a, b) ∈ R, (b, a) ∈ R or a = b.That is, every pair of elements are comparable under R.3 A non-strict ordering,R, on S which is not linear is said to be a non-strict partial ordering relation on

S.4

b) Strict order relation. The relation R is a strict order relation if it is simultane-ously irreflexive ((a, a) 6∈ R), asymmetric ((a, b) ∈ R ⇒ (b, a) 6∈ R) and transitive

on S. If every pair of distinct elements, a and b, in S are comparable under astrict order relation, R, then R is a strict linear ordering on S. Those strictorderings which are not linear are called strict non-linear orderings or, more

commonly, strict partial ordering relation.A non-strict partial order R on S always induces a strict partial order R∗ by

defining aR∗b⇒ [aRb and a 6= b]. Similarly, a strict partial order R on S alwaysinduces a non-strict partial order R† by defining aR†b⇒ [aRb or a = b].

Definition 6.4 Let S be a class and R be is a partial ordering or a strict ordering relation

on S. If R is a partial ordering relation (a, b) ∈ R is represented as a ≤ b and if R isa strict ordering relation (a, b) ∈ R is represented as a < b. If a ≤ b and a 6= b, wewill simply write a < b.

a) A subset of S which is linearly ordered by R is called a chain in S. If R linearlyorders S then S is a linearly ordered subset of itself and so is a chain.

b) An element a of S is called a maximal element of S if there does not exist anelement b in S such that a < b. An element a of S is called a minimal element

of S if there does not exist an element b in S such that b < a.

c) An element m in S is called the minimum element of S if m ≤ a (m < a) for

all a ∈ S. An element M in S is called the maximum element of S if a ≤ M

3A class on which is defined a linear ordering R is also said to be fully ordered or totally ordered by R.In certain branches of mathematics “linearly ordered set” is abbreviated as l.o.set or simply called loset.

4In certain branches of mathematics “partially ordered set” is abbreviated as p.o.set or simply called aposet

Appendix 599

(a < M) for all a ∈ S.

III Relations 7 / The partition of a set induced by an equivalence relation.

Notation 7.1 Let R be an equivalence relation on a set S and let x ∈ S. Then the set Sx

is defined as follows: Sx = y : (x, y) ∈ R. That is, Sx is the set of all elements y inS such that y is related to x under R.

Theorem 7.2 Let R be an equivalence relation on a set S. Let x and y be two elementsin S which are not related under R. Then any element z in S which is related to x

cannot be related to y.

Theorem 7.3 Let R be an equivalence relation on a set S. Let x and y be two elements

in S which are not related under R. Then Sx ∩ Sy = ∅.

Theorem 7.4 Let R be an equivalence relation on a set S. Let x and y be two elementsin S which are related under R. Then Sx = Sy.

Theorem 7.5 Let R be an equivalence relation on a set S. For every x ∈ S there existssome y ∈ S such that x ∈ Sy.

Theorem 7.6 Let R be an equivalence relation on a set S. Then⋃

x∈S Sx = S.

III Relations 8 / On partitions and quotient sets of a set.

Definition 8.1 Let S be a set. We say that a set C ⊆ P(S) forms a partition of S if Csatisfies the 3 properties:

1)⋃

A∈CA = S

2) If A and B ∈ C and A 6= B then A ∩B = ∅.

3) A 6= ∅ for all A ∈ C .

Definition 8.2 Let S be a set on which an equivalence relation R is defined.

a) Each element Sx of SR = Sx : x ∈ S is called an equivalence class of x under R or

an equivalence class induced by the relation R.

b) The set SR = Sx : x ∈ S of all equivalence classes induced by the relation R is calledthe quotient set of S induced by R. The set SR is more commonly represented by the

symbol S/R. So S/R = Sx : x ∈ S. From here on we will use the more commonnotation, S/R.

Theorem 8.3 Let S be a set and C be a partition of S. Let RC be the relation such that(x, y) ∈ RC iff x, y ⊆ S for some S in C . Then RC is an equivalence relation on S.

IV Functions 9 / Functions: A set-theoretic definition.

600 Appendix

Definition 9.1 A function from A to B is a triple 〈f, A, B〉 satisfying the following prop-erties:

1) A and B are classes and f ⊆ A×B

2) For every a ∈ A there exists b ∈ B such that (a, b) ∈ f .

3) If (a, b) ∈ f and (a, c) ∈ f then a = c.

Definition 9.2 If f : A → B is a function and D ⊆ A then we say that the functionf : D → C is a restriction of f to D. In this case we will use the symbol f |D to

represent the restriction of f to D. Note that, if D ⊆ A, then we can write f |D ⊆ fsince f |D = (x, y) : x ∈ D and (x, y) ∈ f ⊆ f .

Theorem 9.3 Let f : A→ B be a function and suppose A = C ∪D. Then f = f |c ∪ f |D.

Theorem 9.4 Two functions f : A→ B and g : A→ B are equal if and only if f(x) = g(x)for all x ∈ A.

Definitions 9.5 Let f : A→ B be a function.

a) We say that “f maps A onto B” if im f = B. We often use the expression “f : A→ B

is surjective” instead of the words onto B.

b) We say that “f maps A one-to-one into B” if whenever f(x) = f(y) then x = y. Weoften use the expression “f : A→ B is injective” instead of the words one-to-one into

B.

c) If the function f : A→ B is both one-to-one and onto B then we can simply say that

f is “one-to-one and onto”. Another way of conveying this is to say that f is bijective,or f is a bijection. So “injective + surjective ⇒ bijective”.

d) Two classes (or sets) A and B for which there exists some bijective function f : A→ B

are said to be in one-to-one correspondence.

IV Functions 10 / Compositions of function.

Definition 10.1 Suppose f : A→ B and g : B → C are two functions such that the image

of the function f is contained in the domain of the function g. Let h = (x, z) : y =f(x) and z = g(y) = g(f(x)) . Thus (x, z) ∈ h if and only if (x, z) = (x, g(f(x)).We will call h the composition of g and f, and denote it as g f .

Theorem 10.2 Let f : A → B and g : B → C be two functions such that the image ofthe function f is contained in the domain of the function g. Then the composition of

g and f , (g f) : A→ C, is a function.

Theorem 10.3 Let f : A → B, g : B → C and h : C → D be three functions. Then

h (g f) = (h g) f .

Theorem 10.4 Let f : A→ B. Then IB f = f and f IA = f .

Appendix 601

Definition 10.5 Let f : A → B. If g : B → A is a function satisfying g f = IA then wewill call g an “inverse of f ” and denote it as f−1.

Theorem 10.6 Let f : A→ B be a one-to-one onto function.

a) An inverse function f−1 : B → A of f exists.

b) The function f−1 is one-to-one and onto.

c) The function f−1 : B → A satisfies the property f f−1 = IB .

d) The inverse function, f−1, of f is unique.

Definition 10.7 A function f : A→ B which is one-to-one and onto is called an invertiblefunction.

Theorem 10.8 Let f : A→ B and g : B → C be two onto-to-one and onto functions.

a) The function g f is also one-to-one and onto.

b) The inverse, g f , is (g f)−1 = f−1 g−1

IV Functions 11 / Images and inverse images of sets.

Definition 11.1 Let A and B be sets and suppose f : A→ B is a function acting on A. IfS is a subset of A = dom f we define the expression f [S] as follows: f [S] = y : y =f(x) for some x ∈ S. We will say that f [S] is the image of the set S under f .

Definitions 11.2 Let f : A → B be a function where A and B are sets. We definef← : P(B) → P(A) as f←(X) = Y iff Y = y : y ∈ A, f(y) ∈ X. In particular,

f←(x) = Y iff Y = y : y ∈ A, f(y) = x. We will refer to it as the set-valuedinverse function f←.

Theorem 11.3 Let f : A→ B be a function. Then f← : P(imf) → P(A) is a one-to-one

function on its domain P(imf).

Theorem 11.4 Let f : A→ B be a function mapping the set A to the set B. Let A be a

set of subsets of A and B be a set of subsets of B. Let D ⊆ A and E ⊆ B. Then:

a) f[⋃

S∈AS]

=⋃

S∈Af [S]

b) f[⋂

S∈AS]

⊆ ⋂S∈Af [S] with equality only if f is one-to-one.

c) f [A−D] ⊆ B − f [D] with equality only if f is one-to-one and onto B.

d) f←(⋃

S∈BS)

=⋃

S∈Bf← (S)

e) f←(⋂

S∈BS)

=⋂

S∈Bf← (S)

f) f← (B − E) = A− f← (E)

IV Functions 12 / Equivalence relations defined by functions.

602 Appendix

Definition 12.1 Let f : A→ B be a function which maps a set A into a set B. We definean equivalence relationRf on A as follows: Two elements a and b are related under Rf

if and only if a, b ⊆ f←(x) for some x in im f . The quotient set of A induced by Rf

is then A/Rf = ARf= f←(x) : x ∈ f [A] We will refer to Rf as the equivalence

relation determined by f and A/Rf (or ARf) as the quotient set of A determined by f.

Theorem 12.2 Let f : S → T be a function where S and T are sets. There exists anonto function g

f: S → S/Rf and a one-to-one function h

f: S/Rf → T such that

hf gf = f . The function, hf gf = f , is called the canonical decomposition of f.

V From sets to numbers 13 / The natural numbers.

Definition 13.1 For any set x, we define the successor x+, of x as x+ = x ∪ x.

Definition 13.2 If x is a set then x+ = x ∪ x. A set A is called an inductive set if itsatisfies the following two properties:

a) ∅ ∈ A.

b) x ∈ A⇒ x+ ∈ A.

Definition 13.3 We define the set of all natural numbers, N, as the intersection of all

inductive sets. That is N = x : x ∈ I for any inductive set I.

Theorem 13.4 Let A be a subset of N. If A satisfies the two properties:

a) 0 ∈ A

b) m ∈ A⇒ m+ ∈ A

then A = N.

Corollary 13.5 (The Principle of mathematical induction.) Let P denote a particular setproperty. Suppose P (n) means “the property P is satisfied depending on the value of

the natural number n”. Let

A = n ∈ N : P (n) holds true

If A satisfies the two properties:

a) 0 ∈ A. That is P (0) holds true,

b) (n ∈ A) ⇒ (n+ ∈ A). That is, P (n) holds true ⇒ P (n+) holds true.

then A = N. That is, P (n) holds true for all natural numbers n.

Definition 13.6 A set S which satisfies the property “x ∈ S ⇒ x ⊂ S” is called a transitiveset.

Appendix 603

Theorem 13.7 The non-empty set S is a transitive set if and only if the property “x ∈ yand y ∈ S” ⇒ “x ∈ S”.

Theorem 13.8 The set N of natural numbers is a transitive set.

Theorem 13.9 a) For natural numbers n,m, m ∈ n ⇒ m ⊆ n. Hence every naturalnumber is a transitive set.

b) For any natural number n, n 6= n+.

c) For any natural number n, n 6∈ n.

d) For any distinct natural numbers n,m, m ⊂ n⇒ m ∈ n.

Theorem 13.10 Let m and n be distinct natural numbers.

a) If m ⊂ n then m+ ⊆ n.

b) All natural numbers are comparable. Either m ⊂ n or n ⊂ m. Equivalently,

m ∈ n or n ∈ m. Hence both “⊂” and “∈” linearly order N.

c) There is no natural number m such that n ⊂ m ⊂ n+.

Theorem 13.11 Every natural number has an immediate predecessor. If k and n are

natural numbers such that k+ = n then k is called an immediate predecessor of n. Forany non-zero natural number n, k =

⋃

m⊂n m is an immediate predecessor of n.

Theorem 3.12 Unique immediate predecessors. Any non-zero natural number has a unique

immediate predecessor.

Theorem 3.13 (The Principle of mathematical induction: second version.) Let P denote

a particular property. Suppose P (n) means “the property P is satisfied depending onthe value of the natural number n”. Let

A = n ∈ N : P (n) holds true

Suppose that, for any natural number n,

P (k) is true for all k < n⇒ P (n) is true

Then P (n) holds true for all natural numbers n.

V From sets to numbers 14 / The natural numbers as a well-ordered set.

Notation 14.1 We define the relation “∈=” on N as follows:

m ∈= n if and only if m = n or m ∈ n

If m ∈= n and we want to state explicitly that m 6= n we write m ∈ n.

604 Appendix

Theorem 14.2 Let (S,≤) be a linearly ordered set. Suppose T ⊆ S. The element q isa least element of T with respect to “≤” if q ∈ T and q ≤ m for all m ∈ T . If S is

equipped with a strict linear ordering “<” a least element of T with respect to < isan element q ∈ T such that q < m for all m ∈ T where m 6= q. The set (S,≤) is said

to be well-ordered with respect to “≤” if every non-empty subset T of S contains itsleast element with respect to ≤. Similarly, the set (S, <) is said to be well-ordered

with respect to “<” if every non-empty subset T of S contains its least element withrespect to <.

Theorem 14.3 The natural numbers N is a strict ∈-well-ordered set.

Corollary 14.4 Every natural numbers n is a ∈-well-ordered set.

Theorem 14.5 Any bounded non-empty subset of (N,∈) has a maximal element.

Definition 14.6 Consider the set 1, 2N of all functions mapping natural numbers to 1 or

2. We define the lexicographic order “<” on 1, 2N as follows: For any two elementsf = a0, a1, a2, a3, . . . , and g = b0, b1, b2, b3, . . . , in 1, 2N, f = g if and only if

ai = bi for all i ∈ N and f < g if and only if for the first two unequal correspondingterms ai and bi, ai ∈ bi. A lexicographic ordering can similarly be defined on SN

where S is any subset of N.

V From sets to numbers 15 / Arithmetic of the natural numbers.

Definition 15.1 Let m be a fixed natural number. Addition of a natural number n withm is defined as the function rm : N → N satisfying the two conditions

rm(0) = m

rm(n+) = [rm(n)]+

The expression m+ n as simply another way of writing rm(n). Thus

rm(0) = m ⇔ m+ 0 = m (1)

rm(n+) = [rm(n)]+ ⇔ m+ n+ = (m+ n)+ (2)

Theorem 15.2 Let m be a fixed natural number and let rm : N → N be a functionsatisfying the two properties

rm(0) = mrm(n+) = [rm(n)]+

Then rm is a well-defined function on N.

Appendix 605

Definition 15.3 For any natural number m, multiplication with the natural number m isdefined as the function sm : N → N satisfying the two conditions

sm(0) = 0

sm(n+) = sm(n) +m

We define the expression mn and m× n as alternate ways of writing sm(n). Thus

sm(0) = 0 ⇔ m0 = m× 0 = 0 (3)

sm(n+) = sm(n) +m ⇔ mn+ = mn+m = m× n+m (4)

Theorem 15.4 Let m be a fixed natural number and let sm : N → N be a function

satisfying the two properties

sm(0) = 0sm(n+) = sm(n) +m

Then sm is a well-defined function on N.

Theorem 15.5 For any two natural numbers m and n, m ∈= n if and only if there existsa unique natural number k such that n = m+ k.

Definition 15.6 For any two natural numbers m and n such that m ≤ n, the unique nat-

ural number k satisfying n = m + k is called the difference between n and m and isdenoted by n−m. The operation “−” is called subtraction.

V From sets to numbers 16 / The integers Z and the rationals Q.

Theorem 16.1 Let Z = N × N. Let Rz be a relation on Z that is defined as follows:

(a, b)Rz(c, d) if and only if a+ d = b+ c. Then Rz is an equivalence relation on Z.

Corollary 16.2 Let Z = N × N be equipped with the equivalence relation Rz defined as:

(a, b)Rz(c, d) ⇔ a+ d = b+ c

For each n ∈ N let [(0, n)] and [(n, 0)] denote the Rz-equivalence classes containingthe elements (0, n) and (n, 0) respectively. Then the quotient set induced by Rz canbe expressed as Z/Rz = [(0, n)] : n ∈ N ∪ [(n, 0)] : n ∈ N

Definitions 16.3 The set of integers, Z, is defined as:

Z = Z/Rz = [(a, b)] : a, b ∈ N = [(0, n)] : n ∈ N ∪ [(n, 0)] : n ∈ N

606 Appendix

a) Negative integers: The set of negative integers is defined as being the set

Z− = [(0, n)] : n ∈ N

Positive integers: The set of positive integers is defined as being the set

Z+ = [(n, 0)] : n ∈ N

The elements of the form [(0, n)] can be represented by −n = [(0, n)] while theelements of the form [(n, 0)] can be represented as n = [(n, 0)].

b) Order relation on Z: We define a relation ≤z on Z as follows: [(a, b)] ≤z [(c, d)]

if and only if a + d ≤ b + c. It is a routine exercise to show that ≤z is a linearordering of Z.

c) Addition on Z: We must sometimes distinguish between addition of natural

numbers and addition of integers. Where there is a risk of confusion we willuse the following notation: “+n” means addition of natural numbers while “+z”means addition of integers.

Addition +z on Z is defined as:

[(a, b)] +z [(c, d)] = [(a+n c, b+n d)]

d) Opposites of integers: The opposite −[(a, b)] of [(a, b)] is defined as

−[(a, b)] = [(b, a)]1

e) Subtraction on integers: Subtraction “−z” on Z is defined as:

[(a, b)]−z [(c, d)] = [(a, b)]+ (−[(c, d)])2

f) Multiplication of integers: Multiplication ×z on Z is defined as

[(a, b)]×z [(c, d)] = [(ac+ bd, ad+ bc)].3

In particular, [(0, n)]×z [(m, 0)] = [(0+ 0, 0+nm)] = [(0, nm)] = −[(nm, 0)] and

[(n, 0)]×z [(m, 0)] = [(nm, 0)].

g) Absolute value of an integer: The absolute value, |n|, of an integer n is definedas

|n| =

n if 0 ≤z n−n if n <z 0

1Note that −[(n, 0)] = [(0, n)] = −n.2When there is no risk of confusion with subtraction of other types of numbers we will simply use “−”.3Note that the “center dot” can be used instead of the “×′′z symbol. When there is no risk of confusion

with multiplication of other types of numbers we will simply use “×”.

Appendix 607

h) Equality of two integers: If (a, b) and (c, d) are ordered pairs which are equivalentunder the relationRz, then theRz-equivalence classes [(a, b)] and [(c, d)] are equal

sets. To emphasize that they are equal sets under the relation Rz we can write

[(a, b)] =z [(c, d)]

i) Distribution properties: If [(a, b)], [(c, d)] and [(e, f)] are integers then

[(a, b)]×z ([(c, d)]+z [(e, f)]) =z [(a, b)]×z [(c, d)]+z [(a, b)]×z [(e, f)]

and

([(c, d)]+z [(e, f)])×z [(a, b)] =z [(c, d)]×z [(a, b)]+z [(e, f)]×z [(a, b)]

Theorem 16.4 Let Q = Z×Z∗ where Z∗ = Z−0. Let Rq be a relation on Q defined asfollows: (a, b)Rq(c, d) if and only if a×z d = b×z c. Then Rq is an equivalence relationon Q.

Definitions 16.5 The set of rational numbers, Q, is defined as:

Q = Q/Rq = [(a, b)] : a ∈ Z, b ∈ Z∗1

The expression [(a, b)] is normally written in the form ab .

a) We define a relation ≤q on Q as follows: If b and d are both positive, [(a, b)] ≤q

[(c, d)] if and only if a×z d ≤z b×z c.

b) Addition +q on Q is defined as:

[(a, b)]+q [(c, d)] = [(ad+z bc, b×z d)]

c) Subtraction −q on Q is defined as:

[(a, b)]−q [(c, d)] = [(a, b)]+q [(−c, d)]

d) Multiplication ×q on Q is defined as

[(a, b)]×q [(c, d)] = [(a×z c, b×z d)]

e) Equality of two rational numbers: If (a, b) and (c, d) are ordered pairs of integers

(b, d 6= 0) which are equivalent under the relation Rq, then the Rq-equivalenceclasses [(a, b)] and [(c, d)] are equal sets. To emphasize that they are equal sets

under the relation Rq we can write

[(a, b)] =q [(c, d)]

1Recall that a and b is shorthand for expressions of the form [(0, a)] or −[(0, b)]

608 Appendix

f) Opposites of rational numbers. If (a, b) is an ordered pair of integers (b 6= 0) and[(a, b)] is its Rq-equivalence class then the opposite of the rational number [(a, b)]

is defined as [(−a, b)] and is denoted as −[(a, b)] =q [(−a, b)].

Theorem 16.6 Suppose a and b are positive integers where b 6= 0 and [(a, b)] is an Rq

equivalence class. Then

a) [(−a,−b)] = −a−b = a

b = [(a, b)].

b) −[(a, b)] =q [(−a, b)] = −ab = a

−b = [(a,−b)]

V From sets to numbers 17 / Dedekind cuts: “Real numbers are us!”

Definition 17.1 For any real number r, let RSr denote the interval (−∞, r) in R. It is

the subset of all real numbers strictly smaller than r. The subset RSr is called aninitial segment in R. For any real number r, the subset QSr = (−∞, r)∩Q = RSr∩Q

is called an initial segment in Q. For each of RSr and QSr the real number r is theleast upper bound of (−∞, r) and (−∞, r) ∩ Q respectively. Note that r may be a

non-rational number even for QSr a proper subset of Q.

Definition 17.2 The elements QSr = (−∞, r)∩Q of the set D = QSr : r ∈ R are called

Dedekind cuts

Definition 17.3 The set of all Dedekind cuts D , linearly ordered by inclusion with addition

+ and multiplication × as described above is called the real numbers. Those Dedekindcuts which do not have a least upper bound in Q are called irrational numbers.

Lemma 17.4 The union of a set of Dedekind cuts is either Q or a Dedekind cut.

Theorem 17.5 Every non-empty subset of R which has an upper bound has a least upper

bound.

VI Infinite sets 18 / Infinite sets versus finite sets.

Definition 18.1 A set S is said to be an infinite set if S has a proper subset X such that

a function f : S → X maps S one-to-one onto X . If a set S is not infinite then wewill say that it is a finite set.

Theorem 18.2 Basic properties of infinite and finite sets.

a) The empty set is a finite set.

b) Any singleton set is a finite set.

c) Any set which has a subset which is infinite must itself be infinite.

d) Any subset of a finite set must be finite.

Appendix 609

Theorem 18.3 Let f : X → Y be a one-to-one function mapping X onto Y . The set Y isinfinite if and only if the set X is infinite.

Corollary 18.4 The one-to-one image of a finite set is finite.

Lemma 18.5 Let S be an infinite set and x ∈ S. Then S − x is an infinite set.

Theorem 18.6 Every natural number n is a finite set.

Corollary 18.7 [AC] A set S is finite if and only if S is empty or in one-to-one correspon-

dence with some natural number n.

Theorem 18.8 The recursively constructed function theorem. Let S be a set, k : P(S) →S be a well-defined function on P(S) and f ⊆ N×S be a relation. We write f(n) = aif and only if (n, a) ∈ f . Let m ∈ S. Suppose the relation f satisfies the two properties

f(0) = m ⇒ (0, m) = (0, f(0)) ∈ f(n, f(n)) ∈ f ⇒ (n+ 1, k(S − f(0), f(1), . . . , f(n)) = (n+ 1, f(n+ 1)) ∈ f

Then f is a well-defined function on N.

Theorem 18.9 [AC] A set S is an infinite set if and only if it contains a one-to-one imageof the N.

Theorem 18.10 If the set S is a finite set and f : S → X is a function, then f [S] is finite.

Theorem 18.11 If a set S contains n elements then P(S) contains 2n elements. Hence,

if a set S is a finite set, then the set P(S) is finite.

VI Infinite sets 19 / Countable and uncountable sets.

Definition 19.1 Two sets A and B are said to be equipotent sets if there exists a one-to-one function f : A → B mapping one onto the other. If A and B are equipotent we

will say that “A is equipotent to B” or “A is equipotent with B”.

Definition 19.2 Countable sets are those sets that are either finite or equipotent to N. All

infinite sets which are not countable are called uncountable sets.

Theorem 19.3 A subset of a countable set is countable.

Theorem 19.4 Any finite product, N × N × · · · × N, of N is a countable set.

Lemma 19.5 Suppose f maps an infinite countable set A onto a set B = f [A]. Then B iscountable.

Theorem 19.6 Let Ai : i ∈ S ⊆ N be a countable set of non-empty countable sets Ai.Then

⋃

i∈S Ai is countable.

610 Appendix

Theorem 19.7 The set of all real numbers R is uncountable.

VI Infinite sets 20 / Properties of the equipotence relation.

Theorem 20.1 Let S be a class of sets. The equipotence relation Re on S is an equiva-

lence relation on S .

Theorem 20.2 Suppose A, B, C and D are sets such that A ∼e B and C ∼e D where

A ∩C = ∅ = B ∩D. Then (A ∪ C) ∼e (B ∪D).

Theorem 20.3 Suppose A, B, C and D are sets such that A ∼e B and C ∼e D. ThenA×C ∼e B ×D.

Corollary 20.4 Suppose A and B are infinite sets.

1) If A,B ⊂ [N]e then A ×B ∈ [N]e.

2) If A,B ⊂ [R]e then A ×B ∈ [R]e.

Theorem 20.5 If Ai : i = 0, 1, 2, ..., n is a set of n non-empty countable sets then∏n

i=0 Ai

is countable for all n.

Theorem 20.6 Suppose S is and infinite set and T is a countable set such that S∩T = ∅.Then S ∼e S ∪ T .

Theorem 20.7 If the sets A and B are equipotent, then so are their associated power sets

P(A) and P(B).

Theorem 20.8 Any non-empty set S is embedded in its power set P(S). But no subsetof S is equipotent with P(S).

Definition 20.9 We will say that the non-empty set A is properly embedded in the set Bif A is equipotent to a proper subset of B but B is not equipotent to A or any of its

subsets. To represent the relationship “A is properly embedded in B” we will write

A →e B

If A and B are sets such that A is equipotent to a subset of B (where B may, ormay not, be embedded in A), we will say that A is embedded in B. To represent the

relationship “A is embedded in B” we will write

A →e∼ B

Definition 20.10 Let S = S : S is a set and E = [S]e : S ∈ S . Let [A]e and [B]ebe elements of E . We write

[A]e <e [B]e

Appendix 611

if and only if A →e B. We write

[A]e ≤e [B]e

if and only if A →e∼ B.

Proposition 20.11 Let S be any set. Suppose P0(S) = S,P1(S) = P(S) and Pn(S) =P(Pn−1(S)) for all n ≥ 1. The set

[S]e, [P(S)]e, [P2(S)]e, [P

3(S)]e, . . . , [Pn(S)]e . . . ,

forms an infinite <e-ordered chain of distinct classes in E .

Theorem 20.12 For every any non-empty set S, 2S = χT : T ∈ P(S) ∼e P(S)

Theorem 20.13 The set R is embedded in P(N).

Theorem 20.14 The set P(N) is embedded in R.

VI Infinite sets 21 / The Shroder-Bernstein theorem.

Theorem 21.1 (The Schroder-Bernstein theorem) If S and T are infinite subsets where Sis embedded in T and T is embedded in S then S and T are equipotent.

Lemma 21.2 Let T be a proper subset of the set S and f : S → T be a one-to-one functionmapping S into T . Then there exists a one-to-one function h : S → T mapping S

onto T .

Theorem 21.3 The set R of all real numbers is equipotent to P(N).

Theorem 21.4 The sets NN and R are equipotent.

Definition 21.5 If A and B are two sets, then the symbol BA refers to the set of all func-tions mapping A into B.

VII Cardinal numbers 22 / An introduction to cardinal numbers.

Postulate 22.1 There exists a class of sets C which satisfies the following properties:

1. Every natural number n is an element of C .

2. Any set S ∈ S is equipotent with precisely one element in C

The sets in C are called cardinal numbers. When we say that a set S has cardinality

κ we mean that κ ∈ C and that S ∼e κ. If the set S has cardinality κ, we will write|S| = κ.

612 Appendix

Definition 23.2 If S and T are sets and κ = |S| and λ = |T | then we define addition“+”,multiplication “×” and exponentiation of two cardinal numbers as follows:

a) If S ∩ T = ∅,

κ+ λ = |S ∪ T |

b)

κ × λ = |S × T |

c)

κλ = |ST |where ST represents the set of all functions mapping T into S (as previouslydefined). That is, |S||T | = |ST |. For convenience we define 0λ = 0 and κ0 = 1.

Theorem 22.3 The class C of all cardinal numbers is a proper class.

VII Cardinal numbers 23 / Arithmetic of cardinal numbers.

Theorem 23.1 Addition on C is well-defined. That is, if S1, S2, T1 and T2 are sets suchthat κ = |S1| = |S2| and λ = |T1| = |T2|, then |S1 ∪ T1| = κ+ λ = |S2 ∪ T2|.

Theorem 23.2 Let κ, λ, φ and ψ be any four cardinal numbers. Then

a) κ+ λ = λ+ κ (Commutativity of addition)

b) (κ+ λ) + φ = κ+ (λ+ φ) (Associativity of addition)

c) κ ≤ κ + λ

d) κ ≤ λ and φ ≤ ψ ⇒ κ+ φ ≤ λ+ ψ.

Theorem 23.3 Multiplication on C is well-defined. That is, if S1, S2, T1 and T2 are sets

such that κ = |S1| = |S2| and λ = |T1| = |T2|, then |S1 × T1| = κ× λ = |S2 × T2|.

Theorem 23.4 Let κ, λ, φ and ψ be any three cardinal numbers. Then

a) κ× λ = λ× κ (Commutativity of multiplication)

b) (κ× λ)× φ = κ× (λ× φ) (Associativity of multiplication)

c) κ× (λ+ φ) = (κ× λ) + (κ× φ) (Left-hand distributivity)

d) λ > 0 ⇒ κ ≤ (κ× λ)

e) κ ≤ λ and φ ≤ ψ ⇒ κ× φ ≤ λ× ψ.

f) κ+ κ = 2 × κ.

g) κ+ κ ≤ κ × κ when κ ≥ 2.

Appendix 613

VII Cardinal numbers 24 / Exponentiation of cardinal numbers.

Theorem 24.1 Exponentiation on C is well-defined. That is, if S, S∗, T and T ∗ are setssuch that |S| = |S∗| and |T | = |T ∗|, then |ST | = |S∗T ∗|.

Theorem 24.2 Let κ, λ and φ be any three cardinal numbers. Then

a) κλ+φ = κλ × κφ

b) (κλ)φ = κλ×φ

c) (κ× λ)φ = κφ × λφ.

Theorem 24.3 Let κ, λ, and α be infinite cardinal numbers. Then

a) κ ≤ κλ

b) α ≤ κ⇒ αλ ≤ κλ

c) α ≤ λ⇒ κα ≤ κλ

VII Cardinal numbers 25 / Sets of cardinality c

Theorem 25.1 Let C denote the set of all complex numbers and J denote the set of all

irrational numbers. Let n denote the cardinality of a non-empty finite set.

a) The cardinality of Rn is c.

b) The cardinality of C is c.

c) The cardinality of J is c.

Theorem 25.2

a) Let SR denote the set of all countably infinite sequences of real numbers. The

cardinality of SR is c.

b) Let SN denote the set of all countably infinite sequences of natural numbers. Thecardinality of SN is c.

c) Let NN(1−1) denote the set of all one-to-one functions mapping N to N. The cardi-

nality of NN(1−1) is c.

d) Let RN(1−1) denote the set of all one-to-one functions mapping N to R. Then the

cardinality of RN(1−1) is c.

Proposition 25.3 The Cantor set has cardinality c.

VIII Ordinal numbers 26 / Well-ordered sets.

614 Appendix

Theorem 26.1 Let f : T → S be a one-to-one function mapping T onto S. If T is awell-ordered then T induces a well-ordering on S. In particular, every countable set

can be well-ordered.

Definition 26.2 Given a well-ordered set (S,≤), a proper subset U satisfying the property

[u ∈ U and x ≤ u] ⇒ [x ∈ U ]

is called an initial segment of S. In this definition the strict order relation < can beused instead of ≤ without altering the meaning of “initial segment”.

Theorem 26.3 If (S,≤) is a well-ordered set then every initial segment in S is of the form

Sa = x ∈ S : x < a for some a ∈ S.

Definition 26.4 Let f : (S,≤S) → (T,≤T ) be a function mapping a well-ordered class,(S,≤S), onto a well-ordered class, (T,≤T ). Note that the symbols ≤S and ≤T willallow us to distinguish between the order relations applied to each set S and T .

a) We will say that the function f is increasing on (S,≤S) if

(x ≤S y) ⇒ (f(x) ≤T f(y))

b) We will say that the function f is strictly increasing on (S,≤S) if

(x <Sy) ⇒ (f(x) <

Tf(y))

A strictly increasing function must be one-to-one.

c) If f : (S,≤S) → (T,≤

T) is strictly increasing then f is said to be an order

isomorphism mapping S into T .

If there exists an onto order isomorphism between the two well-ordered classes, (S,≤S)

and (T,≤T ), we will say that the classes are order isomorphic, or that a function mapsS order isomorphically onto T .

If there exists an onto order isomorphism between the two well-ordered classes (S,≤S)

and (T,≤T ) we will say that the classes are order isomorphic or that a function mapsS order isomorphically onto T .

Theorem 26.5 Let (S,≤S) and (T,≤T ) be a well-ordered sets.

a) The inverse of an order isomorphism is an order isomorphism.

b) If f : (S,≤S) → (S,≤S) is a strictly increasing function mapping S into itself

then f(x) ≥ x for all x ∈ S.

c) The set S cannot be order isomorphic to an initial segment of itself.

d) If f : (S,≤S) → (S,≤S) is an order isomorphism then f is the identity function.

Appendix 615

e) If f : (S,≤S) → (T,≤T ) and g : (S,≤S) → (T,≤T ) are order isomorphismsmapping S onto T then f = g.

f) Suppose f : (S,≤S) → (T,≤T ) is an order isomorphism mapping S onto an

initial segment of T . Then S and T cannot be order isomorphic.

Notation 26.6 Let S and T be two well-ordered sets. Then the expression

S ∼WO T

means “S and T are order isomorphic”. The expression

S <WO T

means “S ∼WO Ta” where Ta is some initial segment of T . The expression

S ≤WO T

means “S ∼WO T or S <WO T”.

Theorem 26.7 Let (S,≤S) and (T,≤

T) be two well-ordered sets. Then either S ≤WO T

or T ≤WO S.

Proposition 26.8 For every natural number n, the lexicographically ordered set S =

1, 2, . . . , n × N is well-ordered.

VIII Ordinal numbers 27 / Ordinal numbers: Definition and properties.

Definition 27.1 Let S be a set. If S satisfies the two properties,

1) S a transitive set,

2) S is strictly ∈-well-ordered

then S is called an ordinal number.

Notation 27.2 When viewed as an ordinal number, N will be represented by the lower-case

Greek letter ω.

Theorem 27.3 If α is an ordinal number then so is its successor α+ = α ∪ α.

Definition 27.4 Suppose the set S is <-ordered. We say that an element y in S is animmediate successor of the element x if x < y and there does not exist any element z

in S such that x < z < y. We say that x is an immediate predecessor of y if y is animmediate successor of x.

Theorem 27.5 Let α be an ordinal number greater than zero.

a) Every element x of the ordinal α is an initial segment of α.

616 Appendix

b) The ordinal α is an initial segment of some ordinal.

c) Every initial segment x in α is an ordinal number.

d) Every element of the ordinal α is an ordinal number.

Proposition 27.6 Any infinite ordinal not equal to ω contains ω.

Proposition 27.7 Let α and β be distinct ordinal numbers. If α ⊂ β, then α ∈ β.

Lemma 27.8 If the ordinals α and β are order isomorphic, then α = β.

Theorem 27.9 The relation “∈” linearly orders the class of all ordinals.

Definition 27.10 An ordinal α which does not contain a maximal element is called a limitordinal.

Proposition 27.11 If U is a non-empty set of ordinals which contains a maximal element

β with respect to ∈, then the union, ∪α : α ∈ U, is equal to the maximal ordinal,β, of U .

Theorem 27.12 If U is a set of ordinals which does not contain a maximal element with

respect to “∈”, then γ = ∪α : α ∈ U is a limit ordinal which is not contained in U .Furthermore, γ is the ∈-least ordinal which contains all elements of U .

Corollary 27.13 Let U be a non-empty set of ordinals which contains no maximal element.

If U satisfies the “initial segment property”, then U is the limit ordinal ∪α : α ∈ U.

Definition 27.14 Let T be a non-empty subset of an ordered set (S, <). If u is an upperbound of the set T and, for any other upper bound v of T , u ≤ v, then we say that u

is the least upper bound of T . We also abbreviate the expression by writing u = lubTor u = lub(T ).

Theorem 27.15 Let γ be a non-zero ordinal number. The following are equivalent:

1) The ordinal γ is a limit ordinal.

2) The ordinal γ is such that γ = ∪α : α ∈ γ.3) The ordinal γ is such that lub(γ) = γ.

VIII Ordinal numbers 28 / Properties of the class of ordinal numbers.

Theorem 28.1 The class, O , of ordinal numbers is a strict ∈-linearly ordered class.

Theorem 28.2 The class O of all ordinal numbers is ∈-well-ordered.

Theorem 28.3 A set S is an initial segment of O if and only if S is an ordinal number.

Theorem 28.4 The class O of all ordinal numbers is not a set.

Appendix 617

Theorem 28.5 Principle of transfinite induction. Let xα : α ∈ O be a class whoseelements are indexed by the ordinals. Let P denote a particular element property.

Suppose P (α) means “the element xα satisfies the property P”. Suppose that, forany β ∈ O ,

“P (α) is true ∀ α ∈ β” implies “P (β) is true”

Then P (α) holds true for all ordinals α ∈ O .

Corollary 28.6 Transfinite induction. Version 2. Let xα : α ∈ O be a class whose

elements are indexed by the ordinals. Let P denote a particular element property.Suppose P (α) means “the element xα satisfies the property P”. Suppose that:

1) P (0) holds true,

2) P (α) holds true implies P (α + 1) holds true,

3) If β is a limit ordinal, “P (α) is true for all α ∈ β implies P (β) is true”.

Then P (α) holds true for all ordinals α.

Theorem 28.7 Let S be a <-well-ordered set. Then S is order isomorphic to some ordinal

number α ∈ O . Furthermore this order isomorphism is unique.

Definition 28.8 Let S be a <-well-ordered set. If α is the unique ordinal which is order

isomorphic to S then we will say that S is of order type α, or of ordinality α. of S isα. If S is of ordinality α, we will write ordS = α.

Lemma 28.9 Hartogs’ lemma. Let S be any set. Then there exists an ordinal α which isnot equipotent with S or any of its subsets.

Theorem 28.10 There exists an uncountable ordinal.

Corollary 28.11 The class ω1 = α ∈ O : α is a countable ordinal is the least uncount-

able ordinal.

Definition 28.12 Let S be any set. Let

U = α ∈ O : α not equipotent to any subset of S

By Hartogs’ lemma the class U is non-empty. Since O is ∈-well-ordered, U containsa unique least ordinal h(S). We will call the ordinal h(S) the Hartogs number of S.

Then h can be viewed as a class function which associates each set S in the class ofall sets to a unique ordinal number α in the class of all ordinals O .

Theorem 28.13 There exists a strictly increasing class ωα : α ∈ O of pairwise non-equipotent infinite ordinals all of which are uncountable except for ω0 = ω.

Theorem 28.14 Let ωα : α ∈ O be the class of ordinals as defined in the previoustheorem.

618 Appendix

a) Every element of ωα : α ∈ O is a limit ordinal.

b) For every ordinal α, either α ∈ ωα or α = ωα.

Theorem 28.15 The Transfinite recursion theorem. Let W be a well-ordered class and

f : W →W be a class function mapping W into W . Let u ∈W . Then there exists aunique class function g : O →W which satisfies the following properties:

a) g(0) = u

b) g(α+) = f(g(α)), ∀α ∈ O

c) g(β) = lubg(α) : α ∈ β, ∀ limit ordinals β

VIII Ordinal numbers 29 / Initial ordinals: “Cardinal numbers are us!”

Definition 29.1 We say that β is an initial ordinal if it is the least of all ordinals equipotentwith itself. That is, β is an initial ordinal if α ∈ β ⇒ α 6∼e β.

Lemma 29.2 The class of all initial ordinals is a subclass of I .

Theorem 29.3 The class I is precisely the class of all initial ordinals.

Theorem 29.4 [AC] The Well-ordering theorem. Every set can be well-ordered.

Theorem 29.5 The class of all initial ordinals I = ω0 ∪ ωα : α ∈ O satisfies the follow-

ing properties :

1. Every set S is equipotent to exactly one element in I .

2. Two sets S and T are equipotent if and only if they are equipotent to the sameelement of I .

3. The class I is ∈-linearly ordered.

Definition 29.6 Cardinal numbers. An ordinal is called a cardinal number if and only if

this ordinal is an initial ordinal. The class, I , is also referred to as the class, C , ofall cardinal numbers.

Lemma 29.7 For any infinite cardinal number κ, define a relation <∗ on κ× κ as follows.For pairs (α, β) and (γ, ψ) of ordinal pairs in κ × κ,

(α, β) <∗ (γ, ψ)

α ∪ β ∈ γ ∪ ψor

β ∈ ψ when α ∪ β = γ ∪ ψor

α ∈ γ when α ∪ β = γ ∪ ψ and β = ψ

Then <∗ well-orders κ× κ.

Appendix 619

Theorem 29.8 [AC] For any ordinal α, ℵα × ℵα = ℵα.

Corollary 29.9 Let κ be an infinite cardinal and Aα : α ∈ β be a set of non-empty sets

indexed by the elements of the ordinal β ∈= κ where |Aα| ∈= κ for all α ∈ β. Then| ∪ Aα : α ∈ β| ∈= κ.

Corollary 29.10 For any infinite cardinal number κ, κκ = 2κ.

Definition 29.11

a) We say that an infinite cardinal number, ℵγ , is a successor cardinal if the index,γ, has an immediate predecessor (i.e., γ = β + 1, for some β). The expression,

ℵα+ = ℵα+1, denotes a successor cardinal. We say that an infinite cardinal num-ber, ℵγ , is a limit cardinal if γ is a limit ordinal (i.e., γ = lubα : α ∈ γ).

b) We say that a limit cardinal ℵγ is a strong limit cardinal if ℵγ is uncountable and

2ℵα : α ∈ γ ⊆ ℵγ .

Theorem 29.12 [GCH] Every uncountable limit cardinal is a strong limit cardinal.

Theorem 29.13 [AC] There exists a strong limit cardinal number.

Definition 29.14

a) We say that an infinite cardinal ℵγ is a singular cardinal number if ℵγ is the least

upper bound of a strictly increasing sequence of ordinals, ακ : κ ∈ β, indexedby the elements of some ordinal β in ℵγ .

b) An infinite cardinal ℵγ is said to be a regular cardinal number if it is not a singular

cardinal number. That is, there does not exist an ordinal, β, in ℵγ such thatℵγ = lubακ : κ ∈ β.

Theorem 29.15 Every infinite successor cardinal, ℵα+ = ℵα+1, is a regular cardinal.

Theorem 29.16 Let γ be an infinite cardinal number. If γ is a singular cardinal then thecardinal number ℵγ is a singular cardinal.

Definition 29.17 A regular cardinal number which a limit cardinal is called an inaccessiblecardinal. A regular cardinal number which is a strong limit cardinal is called a strongly

inaccessible cardinal.

Definition 29.18 Let ℵγ be an infinite cardinal. We say the cofinality of ℵγ is β and writecf(ℵγ) = β if β is the least ordinal in ℵγ which indexes an increasing set of ordinals

θα : α < β such that ℵγ = lubθα : α < β. If no such β exists in ℵγ then we saythe cofinality cf(ℵγ) of is ℵγ and write cf(ℵγ) = ℵγ .

620 Appendix

Theorem 29.19 The cofinality cf(ℵγ) of an infinite cardinal number ℵγ is a cardinal num-ber. Hence cf(ℵγ) is the smallest cardinality of all sets which are cofinal in ℵγ .

Theorem 29.20 The cofinality cf(ϕ) of an infinite cardinal number ϕ is a regular cardinal.

Theorem 29.21 If κ is an infinite cardinal and cf(κ) ≤ λ, then κ < κλ.

IX More on axioms: Choice, regularity and Martin’s axiom 30 / Axiom of choice

Theorem 30.1 Suppose S is a finite set of non-empty sets whose union is the set M .

Then there exists a function f : S →M which maps each set to one of its elements.

Theorem 30.2 Let AC∗ denote the statement:

“For any set S = Sα : α ∈ γ of non-empty sets, Πα∈γSα is non-empty.”

The Axiom of choice holds true if and only if AC∗ holds true. The Axiom of choice

holds true if and only if AC∗ holds true.

Theorem 30.3 The statement “Every set is well-orderable” holds true if and only if the

Axiom of choice holds true.

Theorem 30.4 [AC] Any infinite set can be expressed as the union of a pairwise disjoint

set of infinite countable sets.

Theorem 30.5 [AC] Let (X,<) be a partially ordered set. If every chain of X has an

upper bound then X has a maximal element.

Theorem 30.6 Suppose that those partially ordered sets (X,<) in which every chain hasan upper bound must have a maximal element. Then given any subset S ⊆ P(S)−∅

there exists a choice function f : S → S which maps each set in S to one of itselements.

Theorem 30.7 [ZL] Every vector space has a basis.

IX More on axioms: Choice, regularity and Martin’s axiom 31 / Axiom of regu-larity and cumulative hierarchy.

Theorem 31.1 The Axiom of regularity holds true if and only if every non-empty set Scontains a minimal element with respect to the membership relation “∈”.

Theorem 31.2 [Axiom of regularity] No set is an element of itself.

Definition 31.3 We say that a class is well-founded if it does not contain an infinite de-

scending chain of sets. That is, there does not exist an infinite sequence xn : n ∈ ωsuch that · · · ∈ x4 ∈ x3 ∈ x2 ∈ x1 ∈ x0.

Appendix 621

Theorem 31.4 [AC] The Axiom of regularity and the statement “Every set is well-founded ”are equivalent statements.

Definition 31.5 Let x be a set. The transitive closure of x is a set tx satisfying the fol-lowing three properties:

1) The set tx is a transitive set.

2) x ⊆ tx

3) tx is the ⊆-least transitive set satisfying properties 1 and 2.

Theorem 31.6 Let x be a set. Then there exists a smallest transitive set tx which containsall elements of x. That is, if s is a transitive set such that x ⊆ s, then x ⊆ tx ⊆ s.

Definition 31.7 Define the class function f : S → S as f(S) = P(S). The elements ofthe class Vα : α ∈ O belong to the image of the class function g(α) = Vα recursivelydefined as follows:

g(0) = V0 = ∅ = 0

g(1) = V1 = f(V0) = P(0) = ∅ = 1g(2) = V2 = f(V1) = P(1) = ∅,∅ = 21 = 2g(3) = V3 = f(V2) = P(2) = ∅,∅, ∅, ∅,∅g(4) = V4 = f(V3) = P(4) (16 elements)

......

g(α+) = Vα+ = f(Vα) = P(Vα)...

...If λ = limit ordinal, g(λ) = Vλ = ∪α∈λVα

......

The class Vα : α ∈ O is called the Cumulative hierarchy of sets. We define V =

∪α∈OVα.

Lemma 31.8 a) Each set Vα is a transitive set.

b) If α ∈ β then Vα ∈ Vβ. Hence Vα ⊂ Vβ .

Lemma 31.9 For any non-empty set B, B ⊂ V implies B ∈ V .

Theorem 31.10 For every set x, x ∈ V = ∪α∈OVα.

Theorem 31.11 Let V = ∪α∈OVα be the class of sets constructed as described above. IfV contains all sets then every set has a ∈-minimal element.

Theorem 31.12 Let V = ∪α∈OVα be the class of sets constructed as described above.

a) The rank of the empty set ∅ is zero.

622 Appendix

b) If U ∈ Vβ then rank(U) < β; hence U 6∈ Vrank(U ) for all sets U . Conversely,rank(U) < β ⇒ U ∈ Vβ.

c) If U and V are sets such that U ∈ V then rank(U) < rank(V ).

d) If γ is an ordinal then rank(γ) = γ.

Proposition 31.13 Let β be a limit ordinal. If a and b are two elements of Vβ. Then

a, b is an element of Vβ. That is, Vβ satisfies the property described by the Axiomof pair.

Proposition 31.14 Let β be any ordinal. If U is an element of Vβ then ∪x : x ∈ U ∈ Vβ.

That is, Vβ satisfies the property described by the Axiom of union.

Proposition 31.15 Let β be a limit ordinal. If U is an element of Vβ then Vβ also contains

a set Y = P(U) such that S ⊆ U implies S ∈ Y . That is, Vβ satisfies the propertydescribed by the Axiom of power set.

Proposition 31.16 Let α be any ordinal. For any two elements x and y of Vα, if for all

z ∈ Vα(z ∈ x⇔ z ∈ y), then x and y are the same set.

Proposition 31.17 Let α be an ordinal number such that α > ω0. Then ω0 ∈ Vα.

Proposition 31.18 Let α be an ordinal number. Then the Axiom of subsets holds true in

Vα.

Proposition 31.19 The set Vω0 satisfies the property described by the Axiom of replace-ment.

Proposition 31.20 Let γ be a limit ordinal. Then the set Vγ satisfies the property de-scribed by the Axiom of choice.

Proposition 31.21 Let α be an ordinal. Then the set Vα satisfies the property described

by the Axiom of construction.

IX More on axioms: Choice, regularity and Martin’s axiom 32 / Martin’s axiom.

Definition 32.1 Let (P,≤) be a partially ordered set. If P contains no uncountable strong

antichain then (P,≤) is said to satisfy the countable chain condition. In this case, wesay that (P,≤) satisfies the ccc or that (P,≤) is a ccc partial order.

Definition 32.2 Let (P,≤) be a partially ordered set. Let D be a subset of P such that

for every element p in P there exists an element d in D such that d ≤ p. A subset Dsatisfying this property is said to be dense in the partial ordering (P,≤).

Appendix 623

Definition 32.3 Let F be a subset of a partially ordered set (P,≤). If F is non-emptyand satisfies the two properties, 1) If x and y belong to F there exists z in F which

is less than or equal to both x and y (i.e., F is a filter base or downward directed), 2)if x belongs to F and x is less than or equal to an element y of P , then y belongs to

F (i.e., F is upward closed). A filter in (P,≤) is a proper filter if it is not all of P .If x ∈ P then the set of all elements above x is called a principal filter with principal

element x. Such a filter is the smallest filter which contains x.

Theorem 32.4 MA(κ): Let κ be an infinite cardinal and (P,≤) be a non-empty par-tially ordered set satisfying the countable chain condition. Let D = D ∈ P(P ) :

D is dense in P such that |D | ≤ κ. Then there is a proper filter F ⊆ P such that,F ∩D 6= ∅ for every set D ∈ D . The statement MA(ℵ0) holds true in ZFC.

Theorem 32.5 The statement MA(2ℵ0) fails in ZFC.

Definition 32.6 Martin’s axiom, MA, is defined as being MA(κ) where κ satisfies ℵ0 ≤κ < 2ℵ0

Theorem 32.7 [MA] Suppose κ is an infinite cardinal such that κ < 2ℵ0. If X is a Haus-dorff compact space with ccc and Uα : α ≤ κ is a family of open dense subsets of

X then ∩Uα : α 6= ∅.

X Ordinal numbers arithmetic 33 / Addition.

Definition 33.1 Let (S,≤S) and (T,≤T ) be two disjoint well-ordered sets. We define therelation “≤

S∪T” on S ∪ T as follows:

a) u ≤S∪T

v if u, v ⊆ S and u ≤Sv.

b) u ≤S∪T v if u, v ⊆ T and u ≤T v.

c) u ≤S∪T v if u ∈ S, v ∈ T .

Theorem 33.2 Let (S,≤S) and (T,≤T ) be two disjoint well-ordered sets. Then the relation≤S∪T well-orders the set S ∪ T .

Definition 33.3 Let α and β be two ordinal numbers. Let (S,≤S) and (T,≤

T) be two

disjoint well-ordered sets of order type α and β respectively.1 We define α + β asfollows:

α+ β = ord(S ∪ T, ≤S∪T

)

Theorem 33.4 Let (S,≤S), (T,≤

T) and (U,≤

U), (V,≤

V) be two pairs of disjoint well-

ordered sets such that

1Addition can also be defined inductively as follows: For all α and β, a) β + 0 = β, b) β + (α + 1) =(β + α) + 1, c) β + α =lubβ + γ : γ < α whenever α is a limit ordinal.

624 Appendix

ordS = α = ordUordT = β = ordV

Then ord(S∪T,≤S∪T ) = α+β = ord(U∪V,≤U∪V ). Hence addition of ordinal numbersis well-defined.

Theorem 33.5 Let α, β and γ be three ordinal numbers. Then:

a) (α+ β) + γ = α + (β + γ) (Addition is associative.)

b) For any ordinal γ > 0, α < α + γ

c) For any ordinal γ, γ ≤ α+ γ

d) α < β ⇒ α + γ ≤ β + γ

e) α < β ⇒ γ + α < γ + β

f) α+ β = α + γ ⇒ β = γ (Left term cancellation is acceptable.)

g) α+ 0 = α

Theorem 33.6 Let β be a limit ordinal. Then, for any ordinal, α,

α+ β = sup α+ γ : γ < β

X Ordinal numbers arithmetic 34 / Multiplication

Definition 34.1 Let (S,≤S) and (T,≤

T) be two well-ordered sets. We define the lexico-

graphic ordering on the Cartesian product S × T as follows:

(s1, t1) ≤S×T (s2, t2) provided

s1 <S s2or

s1 = s2 and t1 ≤Tt2

Theorem 34.2 Let (S,≤S) and (T,≤

T) be two well-ordered sets. The lexicographic or-

dering of the Cartesian product S × T is a well-ordering.

Theorem 34.3 If the well-ordered sets S1 and S2 are order isomorphic and the well-ordered

sets T1 and T2 are order isomorphic then the lexicographically ordered Cartesianproducts S1 × T1 and S2 × T2 are order isomorphic.

Definition 34.4 Let α and β be two ordinals with set representativesA and B respectively.We define the multiplication α × β as:

α× β = ord(B ×A)

The product α × β is equivalently written as αβ, (respecting the order). Note the

order of the terms in the Cartesian product B × A is different from the order α × βof their respective ordinalities.

Appendix 625

Theorem 34.5 Let α, β and γ be three ordinal numbers. Then:

a) (γβ)α = γ(βα) (Multiplication is associative.)

b) For any γ > 0, α < β ⇒ γα < γβ

c) γ(α+ β) = γα+ γβ (Left-hand distribution is acceptable.)

d) For any γ > 0, γα = γβ ⇒ α = β (Left-hand cancellation is acceptable.)

e) γ0 = 0

f) For any limit ordinal β 6= 0, αβ = sup αγ : γ < β

Definition 34.6 Let γ be any non-zero ordinal. We define the γ-based exponentiationfunction gγ : O → O as follows:

1) gγ(0) = 1

2) gγ(α+) = gγ(α)γ

3) gγ(α) = lubgγ(β) : β < α whenever α is a limit ordinal.

Whenever γ 6= 0 we represent gγ(α) as γα. Then γα+1 = γαγ. If γ = 0 we define

γα = 0α = 0.

Theorem 34.7 Let α, β and γ be three ordinal numbers. Then, assuming γ > 1,

α < β ⇔ γα < γβ

Theorem 34.8 Let α, β and γ be three ordinal numbers where α 6= 0.

a) γβγα = γβ+α

b) (γβ)α = γβα

Appendix A / Boolean algebras and Martin’s axiom.

Definition 0.1 A partially ordered set (P,≤) is called a lattice if a ∨ b = maxa, b and

a ∧ b = mina, b both exist in P for all pairs a, b in P .

Definition 0.2 If B is a subset of a partially ordered set, (P,≤), ∨B denotes the least

upper bound of B and ∧B denotes the greatest lower bound of B (both with respectto ≤). Note that ∨B and ∧B may or may not be an element of B. A lattice (P,≤)

is said to be a complete lattice if for any non-empty subset B of P , both ∨B and ∧Bexist and belong to P .

Definition 0.3 Let X be a topological space. A subset B is said to be regular open in

X if B = intX(clX(B)). The set of all regular open subsets of X will be denoted asRo(X).

626 Appendix

Theorem 0.4 Let X be a topological space. Then (Ro(X),⊆,∨,∩) is a complete latticein (τ(X),⊆).

Definition 0.5 Let (L,≤,∨,∧) be a lattice. An L-ultrafilter is a proper filter F in L which

is not properly contained in any other proper filter in L. If the filter F is such that∩F : F ∈ F 6= ∅ then we say that the filter F is a fixed ultrafilter. Ultrafilters

which are not fixed are said to be free ultrafilters.

Theorem 0.6 Let X be a topological space.

a) Suppose F is a proper L-filter where (L,⊆,∨,∧) is a lattice in (P(X),⊆). Then

F can be extended to an L-ultrafilter.

b) Suppose F is an L-filter in (L,⊆,∨,∧) a lattice in (P(X),⊆). Then F is an

L-ultrafilter if and only if for every A ⊆ X , either A ∈ F or X −A ∈ F .

Theorem 0.7 Let X be a topological space. Then (Ro(X),⊆,∨,∩) is a complete lattice

in (τ(X),⊆). An Ro(X)-filter, F , is an Ro(X)-ultrafilter if and only if, for anyA ∈ Ro(X), either A or X − clX(A) belongs to F .

Definition 0.8 A lattice (L,∨,∧) is said to be a distributive lattice if, for any x, y, and z

in L, x ∨ (y ∧ z) = (x ∨ y) ∧ (x∨ z) and x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

· The lattice (L,∨,∧) is said to be a complemented lattice it has a maximum

element, denoted by 1, and a minimum element, denoted by 0, and for everyx ∈ L there exists a unique x′ such that x ∨ x′ = 1 and x ∧ x′ = 0.

· A complemented distributive lattice is referred to as being a Boolean algebra.A Boolean algebra is denoted as (B,≤,∨,∧, 0, 1,′ ) when we explicitly want to

express what the maximum and minimum elements are.

Definition 0.9 Suppose we are given two lattices (B1,≤1,∨1,∧1, 0, 1,′ ) and (B2,≤2,∨2,∧2, 0, 1,

′ )and a function f which maps elements of B1 to elements of B2. We say that f : B1 →B2 is a Boolean homomorphism if, for any x, y ∈ B,

1) f(x ∨1 y) = f(x) ∨2 f(y),

2) f(x ∧1 y) = f(x) ∧2 f(y)

3) f(x′) = f(x)′.

The function f : B1 → B2 is a Boolean isomorphism if f is a bijection and both f

and f← are Boolean homomorphisms.

Definition 0.10 Let (B,≤,∨,∧, 0, 1,′ ) be a Boolean algebra. Let S (B) = U : U is a

B-ultrafilter. We define the function fB : B → P(S (B)) as follows: fB(x) = F ∈S (B) : x ∈ U .

Appendix 627

Theorem 0.11 Let (B,≤,∨,∧, 0, 1,′ ) be a Boolean algebra. Then the set fB(x) : x ∈ Bis a base for the open sets of some topology, τ(S (B)), on the set S (B) of all B-

ultrafilters.

Theorem 0.12 Let (B,≤,∨,∧, 0, 1,′ ) be a Boolean algebra.

1) The function fB : B → P(S (B)) is a Boolean homomorphism mapping B into

P(S (B)).

2) For every x ∈ B, fB(x) is clopen in S (B). Hence fB[B] ⊆ B(S (B)) (the set ofall clopen sets in S (B)).

3) The function fB : B → S (B) is a Boolean isomorphism mapping B into

B(S (B)) (the set of all clopen sets in S (B)).

4) The topological space (S (B), τ(S (B))) where τ(S (B)) is the topology gener-ated by the open base fB(x) : x ∈ B is a compact zero-dimensional Hausdorff

topological space.

5) The Boolean isomorphism fB : B → S (B) maps B onto B(S (B)) (the set ofall clopen sets in S (B)).

Theorem 0.13 Let κ be an infinite cardinal number such that κ < 2ℵ0. Then the following

are equivalent:

1) (Martin’s axiom MA) If (P,≤) is a partially ordered set satisfying ccc and D =Dα : α ≤ κ is a family of dense subsets of P , then there exists a filter F on P

such that F ∩Dα 6= ∅ for each α ≤ κ.

2) If X is a compact Hausdorff topological space satisfying ccc and D = Dα : α ≤κ is a family of dense open subsets of X , then ∩Dα : α ≤ κ 6= ∅.

3) If (B,≤,∨,∧, ′) is a Boolean algebra with the ccc property and D = Dα : α ≤ κis a family of dense subsets of B, then there exists a filter F on B such thatF ∩Dα 6= ∅ for each α ≤ κ.

4) If (P,≤) is a partially ordered set satisfying ccc and |P | ≤ κ and D = Dα : α ≤κ is a family of dense subsets of P , then there exist a filter F on P such thatF ∩Dα 6= ∅ for each α ≤ κ.

Theorem 0.14 Let κ be a cardinal such that ℵ0 ≤ κ < 2ℵ0. Let X be a Hausdorfftopological space satisfying ccc such that x ∈ X : x has a compact neighbourhoodis dense in X . Suppose that D = Dα : α ≤ κ is a family of dense open subsets ofX . Then ∩Dα : α ≤ κ is dense in X if and only if Martin’s axiom holds true.

628 Bibliography.

Bibliography(In alphabetical order based on the first letter of the family name.)

1. Andre, R., Axioms and Set Theory, Online,

www.math.uwaterloo.ca/~randre/1aaset_theory_140613.pdf, (2015).

2. Andre, R., When does the family of singular compactifications form a complete lattice?,

Rocky Mountain Journal of Mathematics, Volume 27, Number 4, Fall 1997.

3. Andre, R., Compactifying a convergence space with functions, International J. Math& Math. Sci., Volume 19, Number 4, (1996) 657-666

(https://core.ac.uk/download/pdf/208409171.pdf)

4. Bell, M.G. and Ginsburg, J., First countable Lindelof extensions of uncountable dis-

crete spaces, Canad. Math. Bull. 23(1980), 397-399.

5. Caterino, A., Faulkner, G.D., Vipera, M.C., Two applications of singular sets to the

theory of compactifications. Rendiconti Dell’Institue di Mat. dell’Univ. di Thieste.

6. Chandler, R.E., Hausdorff compactifications, Marcel Dekker, New York, 1976.

7. Chandler, R.E, Cain G.L., Faulkner, G.D. Singular sets and remainders, Trans. Amer.Math Soc. 268(1981) 161-171.

8. Chandler, R.E., Faulkner, G.D., Singular compactifications: The order structure, Pro-ceedings of the American Mathematical Society, 100, number 2 (1987) 377-382

9. Comfort, W.W. and Negropontis, S., The Theory of Ultrafilters, Springer-Verlag, NewYork, 1974.

10. Comfort, W.W., Retractions and other continuous maps, from βX onto βX\X , Trans.Amer. Math. Soc. 114 (1965)843-847

11. Devlin, K.J., Fundamentals of Contemporary Set Theory, Springer-Verlag, New York,1974.

12. Diamond, B., Algebraic subrings and perfect compactifications., Topology and its Ap-

plications, 39, (1991), 217-228.

13. Dugundji, J., Topology, Allyn and Bacon, Boston, Mass., 1967.

14. Enderton, H.B. Elements of set theory, Academic Press, Inc., 1977.

15. Engelking, R., General Topology, Polish Scientific Publishers, Warsaw, 1977.

16. Gillman, L., and Jerison, M., Rings of continuous functions, Princeton, Van Nostrand,1960.

Bibliography 629

17. Ginsburg, J and Saks, V., Some applications of ultrafilters in topology, Pac. J. Math57(1975), 403-417.

18. Glicksberg, I., Stone-Cech compactifications of products, Trans. Amer. Math. Soc.90(1959), 369-382.

19. Hall, D.W. and Spencer, G.L., Elementary topology, John Wiley and Sons, Inc., New

York, 1955.

20. Henrickson, M. and Isbell, J.r., Some properties of compactifications, Duke Math. J.25(1958), 83-106.

21. Hrbacek, K., Jech, T. Introduction to set theory, Marcel Dekker, Inc., New York and

Basel, 1984.

22. Isbell, J.R., Zero-dimensional spaces, Tohoku Math. J., 7(1955), 1-8.

23. Kelley, J.L., General topology, Springer-Verlag, New York, 1991.

24. Monk, J.D., Introduction to set theory, McGraw-Hill, New York, 1969.

25. Monkres, J.R., Topology, Prentice Hall Inc., 2000.

26. Ponomarev, V.I., On paracompact spaces and their continuous mappings, Dokl. Aked.Nauk SSSR, 143(1962), 46-49 - Soviet Math. Dokl, 3(1962), 347-350.

27. Porter, J. R., Woods, R. G. Extensions and absolutes of Hausdorff spaces, Springer-

Verlag, 1988.

28. Rayburn, M.C., On Hausdorff Compactifications, Pac. J. Math., 44(1973), 707-714.

29. Roitman, J., Introduction to modern set theory, John Wiley & sons, New York, 1990.

30. Rudin, M.E., The box product of countably many compact metric spaces, General

Topology and its Applications, 2(1972), 293-298.

31. Rudin, M. E., A new proof that metric spaces are paracompact, Proc. Amer. Math.Soc. vol.20 (1969), p.603.

32. Simmons, G.F. Introduction to Topology and Modern Analysis, Robert E. Krieger

Publishing Company, Florida, 1983.

33. Steen, L.A., and Seebach, J.A. Jr, Counterexamples in topology, Springer-Verlag, 1986.

34. Stephenson, R. M., Jr., Pseudocompact spaces, Trans. Amer. Math. Soc., 134(1968),

437-448.

35. Thomas, J., A regular space not completely regular, Amer. Math. monthly, 76(1969),181-182.

630 Bibliography.

36. Thrivikraman, T., On compactifications of Tychonoff spaces, Yokohama Math. J.,20(1972), 99-106.

37. Walker, R.C., The Stone-Cech Compactification, Springer-Verlag, New York, 1974.

38. Willard, S. General Topology, Addison-Wesley Publishing Company, 1968.

39. Weese, M., Winfried, J. Discovering Modern set theory, American mathematical soci-ety, 1998.


ISBN 978-0-9938485-1-3

Index

(u, v) ∈ Rf , 157

C-embedded, 472

a sufficient condition, 444

C(S), 109

C[a, b], 515

Cα(S), 403, 404

C∗-embedded

closed subset in compl. reg., 419

closed subset of R, 402

closed subset of a metric, 402

implies C-embedded, 442

of a dense subset, 407

C∗-embedded subset, 395

N in R, 399

compact subset, 401

compact subsets in R, 399

Urysohn’s extension theorem, 400

C∗-embedded subsets, 206

C∗(S), 218

Cα(S), 555

Cγ(S), 403

Cω(S), 403

F -space

C∗-embedded, 592

βS F -space, 594, 597

characterization, 593, 594

F (N, S) is separable, 509

Fσ

Q is, 41

Fσ , 40

examples, 41

properties, 41

Gδ-topology, 74

Gδ, 40

examples, 41

properties, 41

Lp-norm, 7, 515

P -points, 221

P -space, 221

pseudocompact, 222, 449

zero-dimensional , 221

S ∪ S(f), 426

S ∪f S(f), 426

S-open, 37

S\F , 33

S\A, 31

Sn, 165

homeomorphic to Rn, 414

SI, 117

Sx, 157

T0-space, 171

T1-space, 172

a T0-space not T1, 173

carries over products, 174

decomposition space, 174

hereditary, 174

singleton sets are closed, 173

transmitted by closed functions, 174

T2-space, 175

T3-space, 180

T4, not T3, 186

T4-space, 186

T2 12, 179

T3 12-space, 200, 201

U [T ], 539

Z(f), 204

Z[S], 209

βN is Freudenthal, 501

631

632 INDEX

β(S × T ), 457, 460, 461

β(S × T ), 457

βN cardinality, 415

`p, 503

`p-space, 503

is Lindelof, 510

is normal, 510

is sec. ctble, 510

is separable, 510

not compact, 510

not seq. compact, 510

`∞-norm, 513

≡, 393

N

number of finite subsets, 326

N, # of finite subsets of, 125, 326

F , 264

F → x, 255

F ∗, 254, 291

H = C(S)\C∗(S), 486

P(S), 32

Ro(S), 91

Ro(S) , 570

ωS, 410, 555

ωR, 465

φS, 493

πβ→α, 398

πγ→α, 395, 555

, 395, 555

RN, 117

RR, 117

σ-compact, 340

σ-locally finite, 350, 561

Gδ, 562

σ-locally finite subcover, 351

σ-ring, 43

τB extends τ over B, 39

τI , 58

τd, 32

υS, 467, 486

υfS, 467

∨, ∧, 109

Z+, 138

a(F ), 256f ∨ g, 109

f ∧ g, 109

f pulls back open sets to open sets, 98f [A], 95

f−1, 95f←, 95

f←[B], 95

fβ(ω), 465f←, 95

f←[U ], 24

fυ, 467f+, 465

g-saturated, 158n-sphere, 165

homeomorphic to, 165

homeomorphic to Rn, 414p-norm, 6

uRfv, 157

z-filter, 291base, 291

fixed, 295

free, 295properties, 292

z-ultrafilterevery z-filter is contained in, 294

real, 467

z-ultrafiltersin βS, 406

1-norm, 6

2-piadic, 18

accumulation point, 228agree, two functions, 108

algebra C(S), 109

algebraic subring, 496axiom of choice, 118

Baire category, 587

application, 589

application two, 589open-hereditary, 590

INDEX 633

theorem one, 588theorem two, 589

Baire spacecharacterization, 586

lemma, 587Baire space , 585

Banach space, 10base

zero-sets as base, 210base for a topology, 71

base for closed sets, 71base property, 73basis, 71

bdSA, 61Bolzano-Weierstrass property, 302

Bolzano-Weierstrass theorem, 302Boolean algebra, 575

filter, 576isomorphism, 579

topological representation, 581ultrafilter, 577

ultrafilter characterization, 577Boolean homomorphism, 578Borel sets, 43

properties, 44boundary of a set, 61

bounded in Rn, 285box topology, 122

Brouwer’s fixed point theorem, 365BWP, 302

BWT, 302

C-embeddedclosed subsets of R, 444

property, 446, 447C-embedded subsets, 441

implies C∗-embedded, 442Cantor set, 140

graphic construction, 141

maps continuously onto [0, 1]3, 147Cartesian product, 116

factor, 117projection map, 118

set theoretic properties, 118

Cauchy sequence, 10, 20

is not a topological property, 103

Cauchy-Schwarz inequality, 3

clopen, 37

closed function, 100

closed set axioms, 35

closed subset, 34

basic properties, 34

closed subsets, 21

closure, 48

properties, 48, 50

closure viewed as an operator, 51

cluster point, 49

coarser topology, 32

cocountable topology on R2, 54

cofinality of a cardinal number, 634

cofinite topology, 35

not Hausdorff, 175

compact, 278

σ, 340

in metric space, 320

maximum value theorem, 285

characterization, 295

closed and bounded subsets of Rn, 285

continuous image, 281

countably, 299, 302

in normed vector space, 12

non-Hausdorff, 283

one-to-one continuous image of, 283

products of compact, 283

realcompact and pseudocompact, 474

subspaces, 281

with T 2 implies normal, 281

compact function, 481

compactification, 391

cardinality, 417

equivalent, 393

induced by a non-singular function, 426

induced by a singular function, 423

one-point, 410

partial ordering, 403

634 INDEX

partial ordering of family of, 395, 555

singular, 423

Stone-Cech, 392

extends, 396

maximal, 398

uniqueness, 402

subalgebra of C∗(S), 404

two-point compactification, 431

compactness characterizations, 279

complemented lattice, 575

complete, 10

C(S), 11

R, 11

complete lattice, 568

complete metric space, 20

completely Hausdorff, 179

completely normal, 288

completely regular, 200

Z[S] as base ⇒,, 210

carry over on products, 203

characterization , 218

metrizable spaces, 201

subspace is completely regular, 202


closed base of zero-sets implies, 210

closed subset of, 419

if loc. cpct + T2, 340

zero sets are base for closed sets, 209

completely separated

if and only if, 199

completely separated in S, 199

completely separated sets (redefined), 205,207

completion of a metric space, 526

Component

in a compact set, 382

component, 366

pathwise, 378

prod. of components is a, 368

components

are closed, 367

connected

not locally connected, 371connected components, 366

prod. of components is a, 368connected space, 361

closure of, 364continuous image, 362

products, 364unions, 363

continuity, sequential, 25continuity, topological, 25

continuous function, 24, 97characterizations, 99composition, 99

iff f pulls back closed sets, 98on a set iff at every point, 98

restriction to a subset, 99countability properties, 82

countable intersection property, 311countably compact, 299, 302

[0, ω1) is, 305closed subsets of, 302

implies pseudocompact, 307, 308in second countable spaces., 306products, 303

characterizations, 299continuous image of, 303

coz, 204coz[S], 209

cozero-set, 204forms a base for open sets, 209

cube, 149, 220curve, 149

decomposition map, 161

decomposition of S, 161decomposition space, 161

by components, 369zero-dimensional, 384

deleted Tychonoff plank, 478

dense subset, 62cardinality, 85

derived set, 49directed set, 233

INDEX 635

discrete metric, 18

discrete topology, 32discrete uniformity, 529

distance

a pt to closed set, 210distributive lattice, 575

embedding, 103

embedding theorem I, 136

embedding theorem II , 220Embedding theorem III, 287

equivalent bases, 80equivalent functions, 403

equivalent topological sets, 32

Euclidean topology, 31evaluation map, 136

extending a topology over a set, 39

extension of a function, 396

feebly compact, 324characterization, 325

pseudocompact, 327

fibre, 480of a continuous function is a zero-set, 204

fibre of an element under a map, 156

filter, 253, 572closed subsets, 262

continuous functions, 263derived from a net, 257

fixed, 254

free, 254filter base, 253

accumulation point, 255

adherence, 255convergence, 255

finer, coarser, 264

generates, 254, 292limit point, 255

finer topology, 32finite intersection property, 279

first countable, 82

metrizable spaces, 83radial plane not, 84

under open fcn, 105

fixed point, 365

Brouwer’s fixed point, 365Frechet space, 172

free union, 39

disjoint union topology, 39

example, 39Freudenthal compactification, 493

βN is , 501

Glicksberg theorem, 457, 461

Hausdorff

carried over by 1-1 closed function, 177

carries over to products, 177

characterizations, 176cofinite topology is T2, 175

completely, 179

hereditary, 177

not completely Hausdorff, 180Hausdorff space, 175

Heine-Borel theorem (generalized), 12

hereditary topological property, 87

Hewitt-Nachbin realcompactification, 473Hilbert space, 10

Holder’s inequality, 506

Holder’s inequality, C[a, b], 516

homeomorphic index sets, 134

homeomorphic topological spaces, 102homeomorphism, 102

characterizations, 102

ideal, 292maximal, 292

properties, 292

identification map, 161

identity map not cts, 101identity of indiscernibles, 16

indiscernibility of identicals, 16

indiscrete topology, 33

infinity norm, 8

inner product, 1interior of a set, 54

636 INDEX

interior operator axioms, 57

interior point, 54interior properties, 55

interior viewed as an operator, 57

Kolmogorov space, 171

Kuratowski closure operator, 51, 52

lattice, 550, 568complete, 568

filter, 572

ultrafilter, 573limit point, 19

limit point of a sequence, 227

limit point of a set, 227limits, 9

limits in metric space, 19

Lindelof, 310characterizations, 312

Lindelofcarried over by continuous functions, 315

hereditarily closed, 315

regular implies normal, 316second countable implies Lindelof, 311

are realcompact, 475

is realcompact, 475local connectedness, 370

locally compact, 332

completely regular, 340dense subsets, 335

finite products are, 338hereditary properties, 334

image under open continuous function,

336neighbourhood base, 333

one point compactification, 339arbitrary products, 338

open in all compactifications, 411

Locally connectedcomponents that are open, 372

locally connected

characterization, 371products, 373

locally finite, 110, 324, 561σ-locally finite, 350

locally finite family of sets, 345lower limit topology, 78

maximal filter, 265

meager, 585metric

ρ(x, F ), 49compact is complete, 323

separable is realcompact, 476uniform on prod, 122

metric axioms, 16metric space, 16

closed sets are zero-sets, 210closed subsets of,, 210compact ⇒ totally bounded, 323

countable product of, 558is perfectly normal, 212

metric subspace, 18metric uniformity, 535

metrizabilitycharacterization, 566

hereditary, 87lemma, 561

lemma one, 562lemma two, 563with separable, 560

metrizable, 30[0, 1]3 is, 122

R3 is, 122σ-locally finite subcover, 351

and paracompact, 354closed subsets = zero-sets, 210

implies completely regular, 201implies normal, 188

is regular, 182metrizable space

is Hausdorff, 175

metrizable spacesare first countable, 83

Metrization theorem, 559Minkowski inequality, 8

INDEX 637

Minkowski’s inequality, C[a, b], 516Mobius strip, 163

Moore plane, 75completely regular, 214

is regular, 185not normal, 216

not paracompact, 350topology is stronger than usual, 75

Nagata and Smirnov, 566

neighbourhood base, 69neighbourhood base at x, 69

neighbourhood of x, 32neighbourhood of x with respect to τ , 69

neighbourhood system of x with respect toτ , 69

net generated by a filter, 258

nets, 234accumulation point, 234

convergence, 234limit point, 234

subnets, 238ultranet, 269

Niemytzki’s tangent disc topology, 75non-Borel set, 61

norm, 2`p, 503`∞, 513

norm axioms, 5normal

F closed ⇒ F C-emb., 205, 207F closed ⇒ F C∗-emb., 205, 207

and second ctble ⇒ perfectly, 211characterizations, 205, 207

completely, 288perfectly, 211

subspace of normal spaces, 191normal spaces, 186

characterizations, 187

closed G-delta’s, 208metrizable implies normal, 188

not perfectly normal, 212, 213, 289properties, 188

norms, 5

nowhere dense, 62, 585number of finite subsets of N, 125, 326

one point compactification, 412

unique, 412

one-to-one fcn, 100f← cts iff f open, 101

open iff closed, 100

open ball, 21open base for a topology, 71

open cover, 277open function, 100

iff closed if 1-1, 100

open neighbourhood, 32open set axioms, 30

open subsets, 21

ordinal space, 89, 305interval topology, 89

is normal, 189

ordinal space compactification, 413outgrowth, 393, 438

p-point, 447

paracompact

and metrizable space, 354–356Tychonoff plank (deleted) not, 350

paracompact space, 346compact spaces are paracompact, 346

continuous image of, 349

hereditarily closed, 349if Hausdorff, are normal, 348

partial order of C(S), 109

Pasting lemma, 110path, 374

direction, 374

initial point, 374terminal point, 374

pathwise connected, 374components, 378

components may be clopen, 379

equivalent to connected, 379implies connected, 375

638 INDEX

invariant over products, 377

invariant with continuity, 376

open subsets of Rn, 380

unions, 378

Peano’s curve, 147

perfect compactification, 495

βN is , 501


perfect function, 481

and realcompactness, 487, 489, 490

and singular map, 484


perfectly normal, 211

characterization, 212, 214

metric spaces are, 212

second ctble normal space is, 211

point-wise convergence, 247

power set, 32

prod. of discrete spaces

need not be discrete, 369

product topology, 80, 120

base for product topology, 121

closure, 123

convergence of functions, 246

embedding of each factor, 130

homeo. factors form homeo. prod., 130

interior, 123

product space, 120

second countable, 124

separable, 126

projection map, 118

case where closed, 286

is open, 121

proper filter, 572

pseudocompact, 328

Z(fβ) 6⊆ βS\S, 408

f [S] closed, 453

fβ [βS\S] ⊆ f [S], 409

but not compact, 328

C-embedded copy of N, 445

characterization, 328, 408, 445, 455, 456,474

locally finite, 456

not ⇒ βS\S contains zero-set, 409pulls back, 95

quotient set, 157

quotient topology

quotient map, 154quotient space, 154

recognizing a, 155

radial plane, 75

not first countable, 84not first ctble, 84

real z-ultrafilter, 467

real points of f , 466realcompact, 467

N and Q are, 476R is, 470

υS, 467

υS is, 473characterization, 468, 487, 489, 490

closed subspace, 477

closed under intersections, 476closed under intersections of, , 476

closed under products, 476

deleted Tychonoff plank is not, 478every subspace of Rn is,, 476

Lindelof spaces are, 475loc. comp. Hausd. σ-compact are, 476

property, 490

separable metric space is, 476realcompactification, 473

regular, 180

regular open, 183, 570properties, 65

regular open set

complete lattice, 570regular open subsets, 64

regular space, 180are hereditary, 182

is completely Hausdorff, 183

is semiregular, 183carries over products, 184

INDEX 639

Hausdorff and not regular, 181metrizable implies regular, 182

regular spacescharacterizations, 181

regular is Hausdorff, 181relation, 528

symmetric, 529relative topology, 36

remainder of αS, 393residual, 585

retraction, 424retract, 424

ring

subring, 551ring C(S), 109

ring Cb(S), 550

scattered line, 81second countable, 82

R is, 83R with usual topology, 84

compact iff c.c. iff seq. c, 318compact iff countbly compact, 306

hereditary, 87implies first countable, 82

implies separable, 85Lindelof, 311products, 124

under open fcn, 105second countable normal

implies perfectly, 211semiregular

assumed to be Hausdorff, 183does not imply completely Hausdorff, 184

implication chart, 184regular implies, 183

semiregular space, 91semiregularization of S, 91separable

F (N, S), 509continuous image of, 105

metric is realcompact, 476outgrowth of N is not, 420

second countable implies, 85

separable space, 63

separated

completely, 199

if and only if, 199

separates A and B, 194

separates points and closed sets, 136

also see Completely sep. pts + cl.sets,

136

sequence, 227

sequence p-norm, 503

sequentially compact, 318

continuous image of, 321

implies countably compact, 320

in second countable spaces, 318

sequentially continuous, 228

sigma compact, 340

sigma-compact


singular compactification, 423

βS, 437, 450, 452

equivalent, 435

property, 425

retract, 424, 437, 450, 452

singular function, 422

singular set, 423


is closed, 423

lemma, 433

Sorgenfrey

1st countable, 83

not 2nd countable, 83

Sorgenfrey line, 78

space , 592

space filling curve, 147

standard uniformity on R, 534

stationary set, 496

maximal, 496

connected, 498

Stone representation theorem, 582

Stone space, 581

Stone-Cech

640 INDEX

T subset of S, 418Stone-Weierstrass theorem, 553

stronger topology, 32subalgebra, 404

subalgebra of C∗(S), 404subbase for the topology τ , 77

subbase theorem, 78subcover, finite , 277

subnet, 238subring, 292

complete, 556subsequence, 230subsets of N, 326

subspace metric, 19subspace topology, 36

sup norm, 8sup norm on C∗(S), 8

sup-norm, 513sup-norm, C[a, b], 516

supremum, 8symmetric relation, 529

Tietze’s theorem, 462

topological invariant, 103topological property, 103

Cauchy sequence is not, 103topological space, 29

induced by a metric, 30

Topologizing a set S by using a closure op-erator, 52

Topologizing as set S by using an interior op-erator, 57

topology induced by a metric, 30totally bounded, 323

totally disconnected, 380product, 381

subspace, 384zero-dimensional, 383

triangle inequality, 4

Tychonoff corksrew, 200Tychonoff plank, 135, 478

deleted, 135deleted is not normal, 190

Tychonoff plank (deleted)not paracompact, 350

Tychonoff theorem, 283

ultrafilter, 265, 572every filter is contained in, 266

characterizations, 265uniform

metric, 122uniform continuity, 546

implies continuity, 548isomorphic, 547

uniform convergence, 247uniform limit, 8

uniform metricon a product, 122

uniform norm, 8

uniform norm topology, 247uniform space

T3, 545, 546uniform topology, 536

closure, 539T3, 545, 546

uniform topology on a produce, 122uniformity, 529

and products, 542base, 531, 544base characterization, 531

base of symmetric sets, 532closure, 539, 544

diagonal, 529discrete, 529

interior, 538metric, 535

property, 530topology generated by,, 535

usual on R, 534upper limit topology

first ctble not second ctble, 82

Urysohn separating functionseparates closed sets, 198

Urysohn separating function for two closedsets, 198

INDEX 641

Urysohn space, 179Urysohn’s lemma, 195

usual topology on R, 31usual uniformity on R, 534

uuu, 558

weak topology induced by functions, 107weaker topology, 32

well-ordered set, 87

xxx, 598, 600, 603

Zariski topology, 35zero dimensional, 92

if T1 is regular, 185zero-dimensional

plus T1 is tot. disconn., 381

zero-set, 204closed G-delta in normal sp., 208

countable intersection, 205G-delta need not be, 208

in metric closed sets are, 210is a Gδ, 204

642 INDEX

ISBN978-0-9938485-1-3

INDEX 643

basic general topology for graduate studies

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