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TRANSCRIPT
INVERSE LIMIT SPACES
THESIS
Presented to the Graduate Council of the
North Texas State University in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF SCIENCE
By
Stephen Boyd Williams, B. A.
Denton, Texas
December, 1974
N,,
Williams, Stephen B., Inverse Limit Spaces. Master
of Science (Mathematics), December, 1974, 107 pp., biblio-
graphy, 6 titles.
Inverse systems, inverse limit spaces, and bonding
maps are defined. An investigation of the properties that
an inverse limit space inherits, depending on the condi-
tions placed on the factor spaces and bonding maps is made.
Conditions necessary to ensure that the inverse limit space
is compact, connected, locally connected, and semi-locally
connected are examined.
A mapping from one inverse system to another is defined
and the nature of the function between the respective inverse
limits, induced by this mapping, is investigated. Certain
restrictions guarantee that the induced function is contin-
uous, onto, monotone, periodic, or open. It is also shown
that any compact metric space is the continuous image of
the cantor set.
Finally, any compact Hausdorff space is characterized
as the inverse limit of an inverse system of polyhedra.
PREFACE
The purpose of this thesis is to examine the inverse
limit spaces of inverse systems and to use inverse systems
to characterize some well known topological spaces.
In Chapter I, inverse systems, inverse limit spaces
and bonding maps are defined. An investigation of some
of the properties that the inverse limit of an inverse
system can be forced to inherit from the factor spaces
of the system is made.
A mapping from one inverse system to another is de-
fined in Chapter II and the relationship between the
inverse limits of these systems is examined. The second
chapter will culminate with the use of inverse limit
systems to prove that any compact metric space is the
continuous image of the cantor set.
In Chapter III, simplices, complexes, and polyhedra
are defined. Some preliminary theorems are proven, then
it is shown that any compact Hausdorff space is homeo-
morphic to the inverse limit of an inverse system of
polyhedra.
Many definitions and theorems of general topology
such as those found in Willard's General Topology
are used without reference. Most of the theorems in
this paper can be found in Capel's "Inverse Limit Spaces"
iii
Gentry's "Some Properties of the Induced Map" and Nagata's
Modern General Topology. However, all proofs are original.
iv
TABLE OF CONTENTS
PagePREFACE . ............. ....11l
Chapter
I. INTRODUCTION............ . 1
II. MAPPINGS BETWEEN INVERSE LIMITS,...... 41
III. AN INVERSE LIMIT CHARACTERIZATIONOF COMPACT HAUSDORFF SPACES*. ...... 79
BIBLIOGRAPHY.. . .. . . . ....... 107
V
CHAPTER I
INTRODUCTION
The intention of this chapter is to examine the proper-
ties inherited by 1lm(X ,f ,D) from the factor spaces of
HX . To accomplish this, some preliminary definitions and
theorems are needed.
Definition 1.1; Let {X }aD be a collection of topo-
logical spaces, indexed by the directed set D. For a : Sin
D, let f :X +a X be a continuous function such that
f :X -* Xa is the identity function, and for ac5y 5f in D,
f = f a0fy. Then (Xa3 f ,D) is an inverse system and
f is called a bonding map.
Let H Xa be the topological product of the X .s, here-acD
after denoted by fXca when no confusion may result. Let
P a:HX + X be the a-th projection function. For x llXa'
let P (x) =x
Definition 1.2: Let X = {x eX a|for ca:5 in D,
f (x ) = x a } and give X, the relative product topology.
Then X., is called the inverse limit space of the inverse
system (Xaf ,D). For a c D, let 7r = POaIXXOO + X .
Then Tr is continuous. For f 6 D, let
S = {x 6Xalif as 3 in D, then f (x X l) = .
Then X = sFS.D
1
2
For notational purposes, let X = lim(Xa , f D).
Whenever D is the positive integers, hereafter denoted by
Z+, then (Xt If5z +) will be called an inverse sequence.
Unless otherwise specified, the hypotheses of the theorems
in Chapter I will assume that (X3f ,D) is an inverse
system and X= lim(X ifD).
Theorem 1.1: If ao5 in D, then f o r = 7.
Proof: Let x e X . Then f 0 o TF(jx) = f ( 7(X)
Since (x) = x ,fa TrX) f 6).x But x c X*. There-
fore, f (x ) = x . Since T[O (x) = x , f o7T(x) x .
Hence, f,,07 = 7Ta
Theorem 1.2: A base for X, is
{7ra) 1 (U)a D and U is open in Xa
Proof: Let x c X. Then, for a E D, xU . X . There
exists an open set Ua of X such that xa 6 U. Therefore,
x 6 Tj(U). Hence, X = U 7j~1(U). LetacD
x 6 7T~(UT) Y 1f ~(Us)
where U a and U are open in X and XS respectively. Since
a and 6 are in D, there is 6 e D such that a:5X and 5X.
Let U = f ~(Ua) fx ~1(Us). Then U is open in X and
x U . Hence, x 6 rx1 (Ux). If y e A- 1(U), yX E Ux.
But fx(yx) = ya. Therefore, Y. 6 U. Hence, y c '7a 1(U ).
Similarly, y c Nr~1(U ). Consequently,
7TX_1(U X)C:7T ~1(U 06 n7T (U ).
3
Therefore, {T'(UQla)ea c D and U is open in Xa} is a base
for some topology on X. Let V be a basic open set of HX
Then, V = n P "(UcU) where F is a finite subset of D andaeF O
U is open in X. Let H be a basic open set of X,. Then,
H = X OrW where W is a basic open set of HX,. Then
w = nP a 1(U ). There exists f3 E D such that a:5 for allaF-E
* e E. Let U = fn f t(Ua). Then U is open in X.Let
x c H. Then x e X. W. Since x e W, xa 6 U for all a E.
Since x e X, x f ~'(Ua) for all a E. Therefore,
x U . Hence, x c TrK-(U ). Therefore, HSr Erf(U3
Let z 71(U ). Then z e XO and z6 U. Hence,
z a flfnf (Ua). Therefore, f (zn) e U for all a E.
Since z e XC,, f (z ) = z . Therefore, za C U for all
a 6 E. Hence, z s enP . (U) . Consequently, z F W nX,.
Therefore, T 1 (U ) = H. Now,
{TrO 1 (U)1ao c D and U is open in X }
is a base for X
Theorem 1.3: The collection {S }aFD is descending in
the sense that if a and are in D, then there exists \ in D
such that ScS qS,.
Proof: If a and 0 are in D, there exists A D such
that as A and 5X, Let x c S . If a F D and a A, then
f CxA) = x . Let 6 !3 in D. Then 6 A. Therefore,
14
= f X6(x )=f 0o f X (x )=f (X~)
Hence, x e S . imilarly, x 6 S . Therefore, S
Theorem 1.4: If each X is Hausdorff, then each S
is closed in Xa. Hence, X, is closed in HX.
Proof: Let x E IHX \S Then there exists "A e D such
that A <a and fa(x ) x . Since X is Hausdorff, there
exists open disjoint sets U and V in X, such that x X U
and fX (xa) E V. Now, f ~j'(u) nf ~'(V) = 4 and each is
open in X , Since f (x ) e V, xaE : fx 1 (V). Hence,
X E Pj'( faj'(V)). But xs6 U. Therefore, x e P X(U).
Let H = Px_(U)Q)P'1(f fl(V)). Then x c H and H is open
in HX . If y c S. H, then P (y) c f ~1(V). Hence,
Y = f (ya ) e V. But y e PX~ 1 (U). Therefore, y c U.
Consequently, Y c UqnV. This is a contradiction to
UqV = $. Hence, S a H = p. Therefore, IX \ S is open.
Thus, S is closed. Since X = q ns, X, is closed in RX
aF-D
Lemma 1.1: If 4 is a collection of compact, nonempty,
connected sets in a Hausdorff space, X, such that for each
A and B in 14 there is s ome C in 1 s uch that CcAqB, then
rni is compact, nonempty, and connected.
Proof: Since each C c ?( is compact and X is Hausdorff,
each C c 2 is closed. Hence nu is closed. Let C 6 2.
Then nu CC. Hence, n2 is a closed subset of a compact
set. Therefore, rn u is compact.
Suppose that n is empty. Let V 6 2. Then, for
x 6 V, there exists U x 6 such that x U . Since {x} and
5
U are compact and disjoint, there exists open disjoint sets
H and K in X such that x 6 H and U K . Letx x x x~ xH = {H |x F-V}.
Then H covers V. Since V is compact, there exists M, a
finite subset of H, covering V. Let M ='{H ,H ,...,H }1 2For each H. s M, consider the corresponding Ki and the
m m m mU cK . Then qU q K. But ( nK.)((UHi) =
= i=l i= i=l
m mHence, since Vo .UHi, V n( nK.) = . Therefore,
i=l i=l
mv ( q U) =
i=l
This violates the hypothesis that the finite intersection
of elements of 14 contains an element of -. Hence, ni4 7 .Suppose that Q2 is not connected. Then there exists
A and B, each closed in rw, such that A mB = p, A / $ / B,
and Q2= A(UB. Let C c . Then f .sC. Therefore, AE C
and B cC. Since A and B are closed in fl4 and n( is
compact, A and B are compact. Now AnB = 4 and X is Haus-
dorff. Therefore, there exist open disjoint sets W and V,
of C, such that A oW and B,=V. If x c C and x [ , then
there exists C V such that Cx _cC and x ( Cx. Hence,
x E C\C X which is open in C. Therefore,
M =*{C\Cx such that x c C\qn{}uf{(Wv)} covers C.
Since C is compact, there exists a finite subcollection N
of M covering C. Now, (W UV) must be an element of N.
6
Let the other elements of N be denoted by C\C ,C\C ,...,C\C1 2 m
mThen there exists C t c? such that C'c qn0c. If x C C',
i=1
x C\Ci for any i. Hence, x 6 W UV. Therefore, C'-.;W UV.
If C' cW, then n ?..( SW. But n.W = A UB. Hence, A UB.L=W.
However, B CV and wIv = $. Therefore, B = $. This is a
contradiction to B # (. Hence, C' .W. Similarly, C'4 V.
Now C, W (p C' rV. Therefore, since (C W) U(ca v)= C',
(C ,qw) n((Cqnv) = (p and each of C' TW and C'QV are open in
C', C' is not connected. This contradicts the hypothesis
that C1 is connected. Therefore, fv is connected.
The following example illustrates the necessity of,,
compactness.
Example 1.1: For each n c Z+, let Xn be the box in R2
with corners at (0,0), (1,0),(0,1) and (1,) such thatn n{(x,0)|0<x<l} and {(x,1 )|0<x<l} have been deleted. Direct
{X }n= by set inclusion. Then {X }= = satisfies then n=l n nlhypothesis of the lemma, except for compactness. However,
ne = {(0,0),(1,0)}. Hence, )C is not connected.
The preceding theorems now bear fruit by assisting in
the proof of Theorem 1.5.
Theorem 1.5: If each X is compact and Hausdorff, then
X is compact. If, in addition, each X (p, then X (C .
In addition, if each Xa is connected, then X is connected.
Proof: If XU is compact and Hausdorff for each ca c D,then HiXa is compact and Hausdorff. Hence, by Theorem 1.4,
7
X0 is closed in HXa. Consequently, since a closed subset
of a compact set is compact, X, is compact.
Suppose that for each a e D, X a p. Let a c D. Then
there is xa 6 Xa, For each c D such that f a,
f (xa) 6 X Let x HX such that for each f :a in D,
X f (x ) and, for y e D such that y$a, x is some ele-
ment of X . Then for 6 <a in D, f (x ) = x . Therefore
x c S . Hence, S O $ for each a. Thus, by Lemma 1.1,
XO / $.
Suppose that X is connected for each a E D. Let
a 6 D, Suppose that S is not connected. Then Sa = A UB
where A and B are closed in S . Further, A 1B = $ and
A ; $ 7 B. Since A and B are closed in Sa, they are closed
in HX . If x e A and y 6 B then, x F HX \ B, which is open.
There exists a basic neighborhood V, of x, such that V is a
subset of HX \B. Let
V = E P (V ),3sF
where V is open in X and F is a finite subset of D. If
P 6 (Y) Cv6 for all 6 e F, then y c P6 1 (V6) for all 6 EF.
Hence, y 6 n PTV ). This contradicts the suppositionSEF
that y c B. Therefore, there is some 6 c F such that
PY(x) 1 PY(y). Since X is Hausdorff, there exist disjoint
open sets H and K of X, such that P (x) E H andY Y Y Y Y
P (y) F KY.Y Y
8
Let ,Xca for all ( e F. Then fx j1 (HY) qf ~' (K ) =
and P Cx) P (71.,H ence P CA)nP (B) = $. Let s c X .
Let t be an element in IIX such that P (t) = f (s) for each
<a in D. Then t c S . Hence, P (S ) is onto X . If
h ca in D, then S _oS. Since A>a, SX.S for each a c F.
Therefore, S_ A UB. Hence,
Xx = P(S x)cPx(A UB) cPx(A) UPx(B).
Since P is continuous and A and B are compact, Px(A) and
P (B) are compact. Hence, Px(A) and Px(B) are closed in
Xx. Therefore, X is not connected. This is a contradiction
to the supposition. Hence, S is connected. Therefore,
by Lemma 1.1, X., is connected.
It is noteworthy to point out that as a result of
Theorem 1.5, if each factor space is a continuum, then X0
is a continuum.
Theorems 1.6 through 1.11 will be necessary to estab-
lish the background for proving Theorems 1.12 through 1.16.
Theorem 1.6: If X0 is compact and Hausdorff for each
a in D, then
7Ta(X) = q{f (XS)ja:5 in D}.
Hence, if each f is onto, then 7 is onto.
Proof: Let z c r (X). Then, there is z F X such
that Tr(z) = z . Since z E X , if a in D, f ~z )._x
Hence, z f(X ) . Therefore, z a E: {f (X )jac5 3 in D}.
Let x e {ff (X )|a 53 in D}. Suppose that x a y (X ),
Then P'-(xa)qnx.=0p. Therefore, if x C Pa~1(x), thereaa
9
exists 2. s D such that x { Sx. Hence, x 6 HX\S2 . Letx a x
K ='{A|x P 1(x )}.
Then,
M O= {IX \ A }x x
covers P-(x ). Since PJ1(x ) is closed in HX, it is
compact. Therefore, there exists a finite subcollection
7/ of?7 covering P a'(xa). Let
7/ =OIXS\ S such that 2 X6M},
where M is a finite subset of D. There exists a E D such
that A a for each A e and a5a. Since a , a,
xa Cfa(X ). There exists xa c X such that x = f (x ).Let y c IX such that y is f (x ) if A!5a and some point
in X if Xt;a. Then y F S and y 6 P ~1(x ). Since
y C PaI(xa), y HX1\S c ?/. But Aa. Hence, S _CS .Therefore, since y 6 S ay 6 SX. This is a contradiction
to y EHX \S . Therefore, PJ~1(x )q X 0 0 $. Thus,
Sa :a ( -Hence, q{f, (Xe)|a J z in D}cira(Xc,) .There-
fore,
Ta(X )= O {fo (X )|a 'f3 in D}.
Since each f is onto, f (X ) = X for each ca5 in
D. Therefore, 7%a (X ,) = Xa.
To settle an obvious question, it will now be shown
that if each f is onto, then f (X ) must be onto X for
each a j5. If not, there exists f 6 D such that f (X ) is
10
not onto X . Hence, there exists x c X such that
f (X ) rx} =
But each 7r is onto. Therefore, there exists y 6 X00 such
that 7T (y). =x, However, by Theorem 1.1,
foaa C7 (Y) = 70cjY-Hence,
f ( OT(y)) =X.
This contradicts the supposition that
f (X) Qn{x} =
Therefore, f (X ) is onto Xa.
Theorem 1.6 is important because, as the following
example shows, there are nonempty inverse limit spaces such
that the 7 function does not map onto each factor space.
Example 1.2: For each n E Z+, let Xn be the closed
unit interval. Define f X X by f (x) = andn+ln: n+l n n+l n 2f nn :Xn Xn to be the identity map. For m>n, let
mn Xm+Xn be fn+l n0f n+2 n+l 0' 0' f m m-1. Since each
mn is continuous, (XnfmnZ+) is an inverse system. For
each n, fn+l n(0) = 0. Hence, v = (0,,0,...) 6 X. There-
fore, XC O p. If n' l and x 6 [0,1], fn1 (x) = 2n-lx.
Let y = (p,2p, ...,2 n-1 p),...) be an element of X, such
that y # v. Then p # 0. Hence p>0. There exists me Z
such that 2ml(P)>1 . Therefore,
,(y 1= (x)>l.
This is impossible since 7Tm(Y)cE[,]. Therefore, y = v.
11
Hence, X, has only one element, namely v., Now, since
V CQ,0,.,.1, T.nCX1 ={O} for each n, Hence, TJn(XO) is
not onto for any n.
Lemma 1.2 Let 0 be a collection of compact subsets of
X, a Hausdorff space, indexed by the directed set D, such
that for a and f in D, there exists A D such that
X S (),q . Then, if U is open in X and o2cU, there is
C E a such that CSU.
Proof: Suppose that for each C 6 3, C U. Then for
each a 6 D, there exists xa ECa such that x U. Hence,
for each a, X\U nC, 4 . Since U is open, X\U is closed.
Therefore(X\U)qca is closed. But since C. is compact
(X\U)OCa is compact. For a c D, let H =(x\U)nac. Let
H = {Haja F D}.
If H. and H are elements of H, there exists 6 c D such
that C&. Ca oC ,"Let x 6 H6. Then, x s(X\U)(OC 6 . There-
fore, x c C6 and x 4 U. But since x c C6, x c C. 'Since x i U, X 6 (X\U)OC and x c (x\u)qc Therefore,
x e H .nH3 6-Hence, -c;Ha H .V By Lemma 1.1, H $.
Let y c H. Then y U. But if y c H, y e Ca for each
a. Therefore, y c n . Now qacU. Hence y s U. This
contradicts y { U. Therefore, there is C c C such that
C CU:
The following theorem is now made easy by Lemma 1.2
Theorem 1,7: If X. is compact and Hausdorff for all
a c D and K is open in i X such that X OcK, then there
12
exists a ED such that
XO cS c K.
Proof:, Since each X is compact and Hausdorff, llX
is compact and Hausdorff. Since each S. is closed in IX
each S is compact. By Lemma 1.2, the theorem is proven.
For the remaining theorems of Chapter 1 it will be
assumed that if (X 3f ,D) is an inverse system, then each
X is Hausdorff.
Theorem 1.8; Let A be a compact subset of X,
Au = 7a (A), B = HA, and g = f |A .Then the following
are true:
(a) (A , g ,D) is an inverse system and eact gq is
onto.
(b) If A = lim(A ,g D), then A = Bf'X=A.+o fca O 0
(c) X\A = U{ r 1(X \A )|ac D}.c a a
Proof:
(a) Since f is continuous, |A is continuous.
Therefore, each g is continuous. Let x c A . Then there
exists y e A such that y = x. Therefore, if Sa in D,
then
g (x) = f (y ) = y
Since y e A, y F A Hence g A + A If x c A ,
gc(x) = f (x) = x.
Therefore g Is the identity map on A. Suppose that
a 5A 5 in D, Let x e A V Then there exists x E A such
13
that iT Cxl = x . Since x c A, g9(x) = f (x ). But
x 6 X Therefore,
fV (x ) = f o f (x ) = f (x ) = x.
Since x c A1 ,
f Cx g = g (xX
Hence,
Since g (x) = x and xX E A,
Therefore,
g (x ) = g 9 gX(x
Hence, (A , g ,D) is an inverse system.
Let a<1 in D and x 6 A . Then there exists x in A
such that = Tr(x). Then,
Tr (x) = x F _ A
Since x X, f (x ) = x. But
g9j(x f) = if(x x =a
Hence,f is onto.
(b) Let i be the a-th projection function of HA
restricted to A , Let x 6 A. Then for each a and 1 in D
such that cc 1, g1 jx ) = x(a. Hence, f (x =x . There-
fore x e X. Since x c A0, x HIA . Hence, x c X nB.
Therefore, A -c XO B,
Let x 6 B nX, . Suppose that x A. Then for each
y 6 A, there exists A E D such that x y . Therefore,y y y
14
there exist disjoint sets U and V, , open in X. , suchy y y
that x EU and y Vx . Now, each of 7T (U )y y y y y y
and f ~(V 1 is open in A and their intersection isx . x 00
y y
empty. Let
{ I (V )jy 6 A}.
Then W covers A. There exists a finite subset V of *
covering A, Let
7 = { J (V ))IA }
where M is a finite subset of D. There exists a D such
that acya for all a e M. For each A M,
g IuX)ng (Vx) = e.
Recall that U is the open set of X that is disjoint from
V and contains x . If z e Aa, there is z e A such that
S(z) = z. Since z A, z ' (V) 7. Hence, zX VA*
Since z 6 X, z a E g aj,(VX). Therefore {g x 1 (Vx)IA M
covers A , But x )g (Vx) for any A M. Therefore,
x a Aa, But if xg G Aa then x { HA . This is a contra-
diction to x E B. Hence x F A. Therefore A X_ qB cA.
Let x e A, Then x FXs and for each aE : D, x eA .
Therefore, x E Xq B, Hence AcX OB.
Let x F BnX. Since x c B, xc A for each a.
Since x 6 XCO,, for a: , f5x) =fx . But x 6 A . Hence,
g a(x = f a(x )=x a .
15
Therefore, x F A, Thus AX qBGA,. Hence,
A0 = X O B = A.
(c) Let x E X,0 \ A. Suppose that for each a sD,
x A . Then x c TA .nX. By part b, x 6 A. This con-
tradicts x e XQO\A. Therefore, there exists A E D such that
x A.-Hence, xX E XX\AX. Therefore, x C T[A 1 (Xx\ AX).
Therefore, x 6 U{ '(Xa\Aa)ja D}. Hence,
Xw , A E U (6 1(Xa\Aa)a E D).
Let x 6 U{rj-(Xa\Aa)Ia e D}. Then there exists
A c D such that x 6 r 1 (XX\AX). Hence, x AX. There-
fore, 7Xj1 (xX) OA = 4. Since x 6 Tr-j(x,), x A. Hence,
x F X \A. Thus,
X,\ A = LTr a'(Xa\Aa)I|ac D} .
Theorem 1.9: Let A and B be compact subsets of X,,
C = AqnB, C = a(C), and h = C Then
C = lim(Ca~,h ,D) = C..
Proof: By Theorem 1.8, part a, (ca, h ,aD) is an in-
verse system. Let 7a (A) = Aa and F a(B) = B . for each
a F D. For a:" in D, let g, = f and j = f W B
Then, by Theorem 1.8, part b, A0 = lim(A, ,g,D) = A and
50= lim(Ba VD) = B. Let x S Co.. Then for each a,
x C. Therefore, xa F Acx , Baand for a:5 in D,
hO(x ) = f C(x$) = fWA BCx) = Xa
Therefore, x c A00 and x Boo. Consequently ~c B,. Let
x c AqB,. Then for @ in D,
gPCx)( = Cxx) =x .
But
g(x ) = f A (x ) = Xa
and
j (x ) = ff3 B(X ) = x
Hence,
X f O A qB C x f3 tICjxf)
Therefore, x 6 C . Now, A OB = C . Since A = A and
Bo = B, C0 = AnB = C. Hence,
C = lim(C ,h ,D) = C0 .4- a O 0
Theorem 1.10: If E is cofinal in D, then (X ,f ,E)
is an inverse system and X = lim(X ,f ,D) is homeomorphic00 + Qa t3
to Y = lim(X ,f ,E).
Proof: For each a t in E, f 3 is continuous and f
is the identity map on X If a56 5 in E, then a56 t3
in D. Hence, f = f o f1 . Therefore, (X ,f ,E) is
an inverse system. Let 'ir be the a-th projection function
of II X restricted to X, and r be the a-th projectionacD
function of II X restricted to Y . Define a relationactE C0
g;X + Y by g(x) = t where for all A E, f (t) = (x).
It is apparent that g:X +÷Y,, for if x X and a:5 in E,
then f (t ) f Cx ) = x = t . Hence, t c Y . If
g(x) = w and g(x) = t, then for all a cE, i (w) =7r (x)= (t).T o=nin
Therefore, w = z. Hence, g is a function.
16
17
Suppose that x y in X . Then there exists X s D
such that xX y . Since E is cofinal in D, there exists
6 E such that A 56. Hence xX yX. Therefore,
Hence, g(x} ) g(yL. Therefore g is one to one.
Let y c Y If a 6 D, there exists a'c E such that
a5 at. Let x 6 II X such that for each a sCDcD a
71 a x) = f aO(7r I(Y))
If a' and a" are elements of E such that ac5 o' and c 5oa",
then there exists ac" in E such that at <ct"' and a" l<a"'".
Then,
But
fct 0 f y = f ? (i n(y))Ot Ot a Ot a a a
and
a a O 1 t,, =' f tf YOtI ?,(Y))
Therefore,
fc 7T aGi t(~Y)) f= tOt 'o Ct (Y
Let a f in D. Then,
HF(X)n=f(Y)
Hence,
f ((x))OTf (f (To (
f f O'=t y f ,(-f ,(y)= T ( x)
18
Hence, x 6 X. and
g(x) = y.
Therefore, g is onto.
Let x e X0 and g(x) = y in Y. Let U be a basic open
neighborhood of y in Y. Then U = Ta '(U ) where U is open
in Xa and a c E. Now a (y) = Tra (x) . Therefore, U is open
in X and a (x) e U. Hence, x 6 '7r -(U ) which is open
in X, Let Tra '(U ) = Ut . If s E U1, (g(s)) a U . There-
fore, g(s) T rij(CU 0 ). Hence g(U t ).U. Consequently, g
is continuous.
Let y e Y . Then g'(y) = x 6 X. Let U = r ~1(U )
be a basic neighborhood of x in X00. Then, 'a (x) c Ua. But,
by the way in which an inverse element was found in proving
g onto, Tr (x) =f a( ,(y)) where caz5a t and a' E. There-
fore, 7a (y)F- f 1 a(U ). Let
f ~1(U ) = U '.
Hence, y ' U ). Since a' c E, a , 1(U 0) is open
in YO. Lets 6 7 '(U,) . Then g~(s) c X0 and
I (s a) C U ,. Hence,
f TECs(0)) = I(g-'(s)) E US a lota t a
Therefore, g-'(s) 7 ~t(U ). Now, g~(F ,71 (U ,)) c 1(U )
Hence, g~1 is continuous. Therefore, g is a homeoporphism.
Hence, Y. is homeomorphic to X,.
Definition 1.3; A continuous function. f, from Y onto
Z, is monotone if and only if for each p E Z, f~ 1 (p) is
connected.
19
Lemma l,3; A continuous function f from Y onto Z,
where Y is compact and Hausdorff and Z is Hausdorff, is
monotone if and only if for each connected set C in Z,
f (C) is connected in Y.
Proof: Suppose that f is monotone. Let C be a set
connected in Z. Suppose that f'1(C) is not connected in Y.
Then, f~'(C) = AUB where A( B = p, A $ / B, and A and B
are closed in f '(C)., Let p c C. Suppose that
f -,(p) n A $/f~-,(p) n B.
Then,
f~(p) = (f(p)nA) U(f-1 (p) nB).
But,since A and B are closed in f1'(C), A = H f 1(C) and
B = Knf-'(C) where H and K are closed in Y. Therefore,
f 1 (p)qnA = f 1 (p)qnH,
fc(p)qnB = f~1(p)flK
and each is closed in f~1(p). Hence,
f'(p) = (f~l(p)nH) u (f-1 (p)qnK).
But (f~'(p)qH) and(f' 1(p)q K) are disjoint and nonempty.
Hence, f '(p) is not connected. This is a contradiction to
the supposition that f is monotone. Therefore, f~1(p)c A
or f 1 (p)cB.
Let
M = {p E CIf~I(p)c A}
and
N = {p CJf 1 Cp)_cB}.
20
Then MN =, C = MU N and M # # N. Suppose that M is
not closed in C, Then, there exists p E C\M such that,
if U is open in C and p c U then UN M X$ Since Z is
Hausdorff, for each y c Z such that y p, there exist
disjoint sets Uy and Vy, open in Z such that p 6 Uy and
y F V . Since f is continuous, f'(Vy) is open for each y.
Therefore,
{ ={ff'(Vy)jy # p}
covers Y\f_'(p). Since p M, f~'(p)c B. Suppose that
y c f~ 1 (pffiH. Then, y E f'1(C)q nH. Hence, y E A. But
y t f~(p) c B, Therefore, A nB 4. This is a contra-
diction to AnB = $, Hence f~'(p)qH = 4. Therefore,
HE Y\f'(p}. Hence, ( covers H. Since H is closed, it is
compact. Therefore, there exists a finite subcollection
of V covering H. Let
= {f(V Y)Ii in}
nbe this finite cover. Furthermore, let U = n U
i=l Yi
Then, p 6 U and U is open in Z. Hence, Un C is open in C
and contains p. Let W = US C. If W #M X$, then there is
q c Wq M. But q c M. Therefore, f~'(q) _A. Since q C W,
nq E r U , it follows that f~ 1(q).Gf" 1(U ) for each i.
I=1 i
Hence,
nf W7I(q)gn f-il(U )
21
But,n
( q f 1(U )) A = d.1=1
This contradicts the fact that f-1 (q)._A. Therefore,
W m = p. Hence, M is closed in C. Similarly, N is closed
in C. Therefore, C is not connected. This contradicts
the supposition that C is connected. Hence, f'1(C) is
connected in Y.
Suppose that f' CC) is connected in Y for each C
connected in Z. Since {p} is connected in Z for each p,
f '(p) is connected in Y. Hence, f is monotone.
Lemma 1.4; Let X= lim(X ,f ,D) where each X is
compact and Hausdorff. If A B = ( and each of A and B
are closed in X, then there exists 6 e D such that
76(A) nTr 6(B) = (.
Proof: Let x e A. Then, for each y c B there exists
A e D such that x # y . Since X is Hausdorff, there exist
disjoint sets U and V , open in XV, such that x e U
and y EV . Let
?'= {7T 1(V )|y cB}.
Then, ?r covers B. Since B is closed in X , B is compact.
Therefore, there exists a finite subcollection ofrcovering
B. Let
r{T[- 1(V ) 1 F}
where F is a finite subset of D, be this finite cover.
22
There exists a E D such that at y for each y E F. Let
H = U{fj' ()|X F}
and
K = U{fcxj' (UJ)|X c F}.
Then each of H and K is open in X and H nK =a a a a aHowever, x a K and 7,r (B) H . Hence, for each x c A,
there is 3 c D such that there exists M and N disjointx
and open in X, such that x F M and 7 (B)cN . LetSx 13 -x
T{1(M )|x E A}.
Then Z covers A. Therefore, there exists a finite sub-
collection 9, of 4 covering A. Then, there exists 6 e D
such that, if a c D and 1(m ) e 9, then 6 !a. Leta x
U = q{f ~1(1 )|r ~1(M ) e #u{ , ( x a x '
Then U is open in X6 and 76(B). U. Let
V = U{f6 (N )|r '(7 ) E}.
Then, V is open in XI and Tr6(A) cV. If s E U, then
f 6(s) c NY for each N such that ia ~1 (Mx ) c . There-
fore, f6y(S) 4 m for each M such that 1( G.
Hence,
f 6U1(f (s))f '(M ) =6f oSa a x
for each NM such that 'r '(M ) 9. Therefore, s 4; V.
23
Similarly, if e V, U. Hence, Un V =. But,
7T (B} U and 71 (A) 1&V . Therefore,
T6(A)rn7T 6 (B) = $.
Theorem 1.11: Let (X ,f ,D) be an inverse system of
compact Hausdorff spaces such that each f is monotone.
Then, each i-r is monotone.
Proof: Suppose that there exists a s D such that ff
is not monotone. Then, there exists x c X such that
7 aI(x ) = A UB where each of A and B is closed in 7T (-'(x )and A(B = 4 and A 7 4 7 B., Since A and B are closed in
7 aI(x ), they are closed in X, and IIX . By Lemma 1.4,
there exists 6 6 D such that 74(A) Rr)7 6 (B) = $. Since X6is Hausdorff, there exists disjoint sets H6 and K, open
in X6., such that 6 (A) cH6 and ff6 (B) .K6. There exists
a F D such that 6 ca and ac a. Let
H=f 1(H6
and
K = f ~(K
Then, HnK = and each of H and K is open in X . If s c A,
sa a ( f 6 (H6 ).If s c B,
s E f ~1(s )cf 1(Ka 6 6 a (K6
If pa 6 fa U1(x ) then there is p e X such that 7a (p) =p .
Now,
p = f (p ) = f (x ) = xS Ga a aot, a
24
Therefore, p E x ). It then follows that either p 6 A
or p E B. Hence, p H or p c K. But, this means that
f !(,x ) is separated by the nonempty, disjoint, open
sets H and K. This contradicts the supposition that f., is
monotone. Hence, 71 must be monotone,
Definition 1,4: A space Y is locally connected if for
each y c Y and each open set U containing y, there exists
an open connected set V such that y e VZU.
Definition 1.5: A space Y is semi-locally connected
if for each y E Y and each open set U containing y, there
exists an open set V such that y e V.GU and Y\V has only
finitely many components.
For the remaining theorems of Chapter 1, it is assumed
that each f is monotone.
The following theorem returns to the examination of
properties passed to X by the factor spaces.
Theorem 1.12: Let (X ,f ,D) be an inverse system of
compact Hausdorff spac es . If for all and ain D, f is
mono t one and X6 is locally c onne ted (semi-loally c connected ,
then Xead is locally connected (semi-locally connected).
Proof: Suppose that for all f and a in D such that
a, f is monotone and X is locally connected. Let
x e6 X and U be open in X00 such that x c U. Then, there
exists a basic neighborhood of x, 7Tr 1 (U ), such that
Tr (U ) _CU. Since X is locally connected and x E UX,
there exists V , connected and open in X such that
25
xx 6 VXU , Since Tis monotone, 1Tr1(V,) is connected.
Butx 7T ('V )_qj ~(Ux)SU. Hence, X is locally
connected.,
Suppose that for all ( and a in D such that f3 a,
f is monotone and X is semi-locally connected. Let
y c U where U is open in X. Then, there exists a basic
neighborhood, ra 1(U ), of y such that Ra '(U.).9u. There-
fore, ya c U and Ua is open in Xa. Since Xa is semi-
locally connected, there exists V open in X such thata a
ya a CUa and Xa\U has only finitely many components.
Let C2,,C2 , ...,Cn denote these components. Now,
y E 7T '(V ) CU.
Suppose that there exists more than finitely many components
of X0\ 7TaI(V a'). Then, there exists C, a component of
X,\ 'ij(Va,), such that C X TFa '(Ci) for any i. Since
Tr a (Ci) is connected for each i, if CC'rr~'(C.), for some
j, then
C = T '(C.).
But C f (C ) for any i. It then follows that CP 'c.)(for any i. However, 7Tr (C) is connected in X and
Ta (C) SX \V . Hence, 7T (C) EC for some j. But,
a, '~, ,a, a JCC gT OT (ra(C))-SIa,'(C *
Therefore , C C1f (C.). Consequently,
C = a ~'(C.).
This contradicts C a '(C.). Hence, X is semi-locally
connected,
26
The following example illustrates that Theorem 1.12
will not be true if each f is not monotone.
Example 1.3: For each n F Z+ let Xn be the unit
circle in the complex plain. Give Xn the relative topology.
An element of Xn e 1, will be denoted, in degree, by 0.
Let fn+1 n; Xn + Xn be defined by
fn+l n(e) = 20
and
f (0) = 0,
Then each f n+1 nis continuous. For m >n, let f :Xm + Xn
be defined by
fmn e) = fn+l n 0 fn+2 n+l o fm m-1(8).
Then (Xn fmn',Z+) is an inverse sequence. Hence, an inverse
system, Let
X = lim(X ,f ,z ).+ n mn +
Suppose that X, is locally connected. Let U be an
open, connected, nonempty subset of Xn not containing 00.
Then,
U = {a EX n < t < S}
where 00-<5 and 6:53600. Let
H = {y n+
and
K=(YcXr+ 1 if- + 1800 < y < + 1800}.
Then H and K are disjoint regions and 01 H UK. If 0 U,
then < e<a .,But fn 1(e) = + 1800}.n+1 n 2 2'
27
Hence, < '< -and + 18O< + 18o0<p6 + 180' . Therefore,2 7 f2 2
fn+1 n (e) L;HU K.
Consequently, fn+ n '(U).SHUK, Furthermore, fn~l nn+l n
meets both H and K. If a E H U K, then a E H or a K.
But if r F_ H, then (< $.< Therefore, 3< 2 a< 6. Hence,
<f n+i nCa<6, Consequently, fn+l n(a) c U. Also, if
a P K, then + 1800< a< + 180, It then follows that
2( + 1800)< 2a< 2( + 1800). Hence, < 2a < 6. Therefore,
fn+l n(a) e U. It has now been shown that
fn+l n(U) = H UK.
By induction, f (n 1-(U) is the disjoint union of 2n-l
mutually disjoint regions. Further, if X c U, fn 1 (x)
meets each region. Let z 6 X . If p c Tn (7~1(z1)), then
7n (P)r r1 I(z1 ) A q. Therefore, there exists s 6 X such
that s c Tn -(P)nq 1 '(z 1 ). Now sn = p and s = z .
Hence, fn 1 (P) = z1 . Therefore, p c fn 1 '(z1 ). It then
follows that 7 n r 17(z 1 )) Cfn 1 ~1(z1 ). Let p c fn 1 ~1 (z1 ).
Let s eX0 such that sn=np. Since s = fn n (p) = z ,
s F Tr 1 (z1). Consequently, in n(s) r ( ~i 1(z )). But
sn = p. Hence, p 6 n- (.I7(z1 )). Therefore,
nl 1 1(z1)) = n 1
Now, U can be chosen such that 1800 U. Then,
7F " (1800).s7 1 1(U).
Since X, is locally connected for each z c 711(1 8 0 0 ), there
28
exists V an open connected subset of X such that
z 6 V z-Tj(U, Then,
?r= {VzIz & 1-1 (1800)}
is an open cover of 7 (1801). Since 7 1(1800) is closed
in X., it is compact. Therefore, there is a finite sub-
collection 2,of V/, covering Tr1 (18 00). Let
'{V ,V2 ''''Vm .
There exists n e Z+ such that 2n-1>m. Now, fn ~1(U)
is the union of 2n-1 disjoint regions and fn 1 180)
meets each region. For each i, 7T (V.) is connected.
Therefore, n n(V.) is a subset of one of the regions.
Since there are more than m regions, there exists at least
one region R, such that n(Vn ) OR = for each i. But
m7-1(1800) c U V.
i=l
Therefore,
n T1 (18oo)) OR = q.
However,
Sn OT (1800)) = fn 1 (18 00 )
and f n 11(1800) meets each region. Consequently, it meets
R. Therefore, there exists some j such that n (V.) OR .
This contradicts the fact that n (V1) OR = for each i.
Hence, X is not locally connected.
Theorem 1.13: Let (X ,f ,D) be an inverse system
of compact Hausdorff spaces. Let C be a closed subset of
X, and a and b be distinct points of Xc,\C. For a c D, let
29
C =. (C). If, for a E D, X \C = A UB where A and Ba c. a a a a aare disjoint regions containing a and b respectively,
a athen there exists disjoint regions A and B in X containing
a and b respectively, such that X\ C = A UB.
Proof: By Theorem 1.8,
X\C = U{7r ~'(X\C )Ia s D}.Sa a a
But Xa\C = A UB . Hence,a a a a
X\C = U{7rj (A UB )fa FD}.
Since
7T 1'(A UBa) = Tr ~ (A )UT ~-(B ),a a a a aX C = U{Tr~'(A ) U 71(B )1a 6sD
= { UT ~1(A )}U{ U 7 <~1(B )}.
acD a a asD a
Let
M, = U 7r 1 (A )
and
N = U ~(B ).a.:D
Then, a c M and b e N.
Let x E m'N. Then there exists A and 6 in D such
that x e XA ) and x e f6~ 1(B ) There is a c D such
that A 5a and 6:5a. It then follows that x P Aand
x6 c B 6 . Suppose that zX6 A and f~'1(z X). A. Then,
there exists z c X such that z f A and f (z ) = z .
Hence, z a B or z a C . If z r C , there exists
z e C such that P (z) = z . Since z e C, z c X . There-
fore,
f(z ) = z = 7A (z).
30
Hence, z EC , But z EC and z A cA Consequently,
z e AnC . However, AA/C = . Therefore, zo C .
It must follow that z E B . Since each f is monotone,
f ~(z ) is connected, Thus, f '(z )CB But, since
A is connected and z e AA, f ~'(A A).9B , However,
a O f '1(A ) Hence, a c B .. This contradicts the
hypothesis. Therefore, f 1(Ax)cA . Similarly,
f 1(B )_cB Since xx AA and x6 B6, x cEf x )_cAand x e f 6
1 (B6 )-B., Hence, A flB = . This contra-
dicts the hypothesis, Therefore, MqnN =
Suppose that M = HUK where HOK = 4, H X X # K and
each of H and K are open inM. Then, ae H or a cK.
Suppose that a 6 H, Suppose also that for some a c D,
y C 7r '(A ), Now, ra -(A )cH UK and a -'(A ) is con-
nected. If y K, i- 1(A )_cK. However, a 6 'rr '(A )and a K, Therefore, for each a D, r -'(A ) OH.
Hence, K = $. This contradicts the supposition that H and
K disconnect M. Consequently, M is connected. If a K
then, by a similar argument, H = $. The supposition that
M is not connected is again denied. Hence, M is connected.
Clearly, by substituting N for M, N is also connected.
Now, X\C = MUN where M and N are disjoint regions con-
taining a and b, respectively.
It is noteworthy to point out that if D is countable
and each X is a separable metric space, then X is a
separable metric space,
31
Definition 1.6; Any space homeomorphic to [01] is
an arc.
Definition 1,7: A connected space is irreducibly
connected between two points if no proper connected subset
contains both points.
The following two theorems are beyond the scope of
this paper and will be assumed. Theorem A has been proven
by Wilder (5) and Theorem B by Whyburn (4).
Theorem A: A space Y is irreducibly connected
between a and b if and only if for y c Y\{ab}, Y\y = A UB
where A and B are disjoint regions such that a e A, b s B.
Theorem B: A separable metric continuum Y is an arc
if and only if Y is irreducibly connected between some two
points.
The following two lemmas and Theorem 1.15 are proven
in preparation for showing that if Xa is an arc for each
az, then X, is an arc.
Lemma 1.5: If a continuum X is irreducibly connected
between a and b and X is irreducibly connected between
c and d, then
{a,b} = {c,d}.
Proof; Suppose that a X y # b. Suppose that X is
irreducibly connected between a and y. Then X\b = H UK
where a E H, y K, i(K = $ and each of H and K is a
region. But, since X is irreducibly connected between
a and b, X\y = MUN, where M and N are disjoint regions
32
containing a and b, respectively. There exists r K
such that r y, If not, {y} is both open and closed,
contradicting the supposition that X is connected. Since
r A y, r E M or r s N, If r s M, then r c mOK. Hence,
MUK is open and connected. Therefore,
X = (MUK) UN,
But (MUK) ON A $. If not, X is not connected. Since
M nN = 4, K N A $. Hence, K UN is connected and open.
But, (K UN) UH = X. Therefore, (K UN) OH A $. Otherwise,
X is not connected. However, H nK = $. Consequently,
N H A $. It follows that NUH is connected and open.
Since y E N and y 6 H, N UH is a proper subset of X. But,
a e N and b c H, This contradicts X being irreducibly
connected between a and b. Therefore, X is not irreducibly
connected between a and y. A similar argument shows that
X is not irreducibly connected between b and y. Hence, if
{a,b}fl{c,d} A $,
then,
{a,b} = {c,d}.
Suppose that {a,b} O{cd} = $. Then a A c A b and
a A d A b. By the previous argument, X is not irreducibly
connected between a and c or between a and d. By Definition
1.7, there must be a proper connected subset H containing
a and c, There must also be a proper connected subset K
containing a and d, Now, if X 7 H UK then H UK is a proper
connected subset of X. But C F H UK and d E H UK. There-
33
fore, X is not irreducibly connected between c and d. This
contradicts the hypothesis. Hence, X = H UK. However,
b e X. Consequently, b 6 H or b c K. But, a c HqK. It
then follows that X is not irreducibly connected between
a and b. This also contradicts the hypothesis. As a
result,
{ab} = {c,d}.
Therefore, Lemma 1.5 guarantees that if a continuum
X is irreducibly connected between two points they are
unique.
Lemma 1.6: If f is a monotone function from a con-
tinuum X onto a continuum Y and X is irreducibly connected
between a and b, then if Y is irreducibly connected between
c and d,
{f(a),f(b)} ='{c,d}.
Proof: Suppose that f(a) = f(b). Then, f~'(f(a))
is connected and contains both a and b. Hence, f~'(f(a)) =X.
Therefore, Y = f(a). Consequently, c = d and Y is not
irreducibly connected between c and d. Consequently,
f(a) # f(b), Suppose that Y is not irreducibly connected
between f(a) and f(b). Then there exists a proper connected
subset U of Y such that f(a) U and f(b) c U. Since f is
monotone, f-'(U) is connected in X. But a and b are both
elements of f'CU). Hence, f1'(U) = X. However, U is a
proper subset of Y. Therefore, there exists x E Y such
that x ( U. Consequently, f~1(x) qf~'(U) = p. But
f '(U) = X. Hence, f '(x) = 4. This contradicts f being
34
onto. Therefore, Y is irreducibly connected between f(a)
and f(bY.. By Lemma l,5, {f(a),f(b)} = {c,d}.
Theorem 1.14; If, for each a e D, Xa is a continuum
irreducibly connected between two points, then X, is a
continuum irreducibly connected between two points.
Proof: By Theorem 1.5, X, is a continuum. Let a e D.
Let
E = 'sO Da0}.
Then E is cofinal in D. Hence, by Theorem 1.10,
YCO= lim(Xa ff E)
is homeomorphic to X. For A E, let TX be the X-th
projection function of II X restricted to Y. LetY,,beAXE
irreducibly connected between a. and b.. By Lemma 1.6,
for each C E, there exists as and b such that X is
irreducibly connected between as and b where f (a$ ) = a
and f (b ) = b . If A 5 in E, let X be irreducibly
connected between a and b,. Then
f,,(a) = a .
If not, f (a) = b (aX and b are unique). Therefore,
fa(a ) = f o f (a,) = f(bX) = b .
But,
f (aa) =aa
This contradicts a b . Similarly,
f (b ) = b .
For 6 E, let H = 1T (a ) and K% = 6 (b6). Each of
35
H and K6 is closed, compact, connected, non-empty and
Hausdorff. Let
V = {H616 e E}
and V = {K6 |6 cE}.
Let H and H be elements of t(. There exists a E such
that <cy and , o. If s c HV, then s= a and
s = f (a ) = a .
Hence, s c H . Similarly, s H . Therefore, H ._cH nH
By Lemma 1.1.,1Q/ cp. Similarly, rn? 4. Let a c nwand b e r7. For each 6 e , (a) = a6 and T6(b) = b6
Let t c Y,\{a,b}. Then, there exists r E such that
a t b . Let
F = {6 c EfnT6}.
Then F is cofinal in E. Then Z = lim(X ,f ,F) is homeo-00 + a B
morphic to Y . Let f:X+-* Y, and g;Y÷+ Z, denote these
homeomorphisms. For each A F,
X\(g(t)) = MUN
where M and N are disjoint regions containing (g(a))
and (g(b))X respectively. Recall that from Theorem 1.10,
(g(a)) = a . By Theorem 1.13, Z\ g(t) = MUN where M and
N are disjoint regions containing g(a) and g(b), respec-
tively. Therefore,
Y 0\ t = g 1 (M) U g'(N)Hence, Y. is irreducible between a and b. Consequently,
X, is irreducible between some two points.
36
Theorem 1.15: If D is countable and for each a s D,
X is an arc, then X is an arc.
Proof: By Theorem B, for each a E D, X is irreducibly
connected between some two points. By Theorem 1.14, X, is
a continuum irreducibly connected between some two points.
As was previously noted, X, is a separable metric space.
Therefore, by Theorem B, X, is an arc.
Definition 1.8: Any space homeomorphic to S', the
unit circle in the plane, is a simple closed curve.
Definition 1.9: A space Y has property Q if it is
connected, non-degenerate and for every pair of distinct
points x and y, Y\{x,y} = A UB where A and B are nonempty,
disjoint regions.
This paper will assume the following theorem by
Whyburn (4).
Theorem C: A separable metric continuum Y is a simple
closed curve if and only if Y has property Q.
Theorem 1.16: If, for each acED, X is a continuum
having property Q, then X, is a continuum having property Q.
Proof: By Theorem 1.5, X is a continuum. Let
x ; y e X. Then there exists a c D such that x y . Let
E = {X c Dja <x}.
Since X has property Q for all c D, X \ {x ,y a} = A UBwhere A and B are disjoint nonempty regions of X . Let
6 e E. Then x6 . Hence, X\{x' 6Y6} = H UK where H
and K are disjoint nonempty regions of X6. Let t c f 1(A ).6 Ba'S
37
If t = x, then f (t) = x . Hence, x e A . But x A
Therefore, t x, Similarly, t y . Hence,
f a(A) c HUK.
In a like manner,
f ~(B )gH UK.
Suppose that f6a~'(Au ).H and f,,(A ) 1K. Then
f~ '(A ) meets both H and K. But, f6 is assumed to be
monotone. Hence f 1(A ) is connected. Therefore,
f '(A ) cannot meet both H and K. Consequently,
f 1(A ) cH
or
f '(A) _cK.
Similarly,
f 6-(B )cH
or
f ~1B )cK.
For 6 c E, denote the region containing f 1(A ) as A
and the region containing f6 1(B ) as B .
Now, E is cofinal in D. Hence.
Y = lim(X ,f ,E)C* + tc
is homeomorphic to X . Let 7a be the a-th projection
function of II X restricted to Y . Let g be the homeo-
morphism from X onto Y, as described in Theorem 1.10.
Then Cg(x)) = x and (g(y)) = y for all f c E.
Let t Y0\ {g(x),g(y)}. Then, there exists A E
such that (g(x)) = x t y = ( Therefore,
38
t A or tX B .Hence, t s X'(A ) or t e i~1(BX).
Consequently,
t 6C U Tr (A ))U( U T'(B ))cE FE
Let
A =U '(A
and
B = U EI1(B ).
Then
Y\00 {g(x),g(y)} cA UB.
Let t AUB, If t j Y .\{g(x),g(y)}, then t = g(x) or
t = g(y). Suppose that t = g(x), then t = x for all
6 E. Hence, t A and t B for any 6 E. There-
fore, t 4 AUB. Similarly, if t = g(y), t 4 AUB. This
contradicts t c A UB. Hence,
t F Y \{g(x),g(y)}.
Consequently,
A UB = Y \ {g(x),g(y)}.
Since A is nonempty, there exists v a A . Now each f
is monotone. Therefore each f is onto. Hence, i is
onto for each a c E. Therefore, it follows that there
exists v e Y such that i (v) = v . Since v A A ,
f '(v )CA for each E. Hence,
v O W ~'(A ).aE:E a 06
Therefore, A is connected. Similarly, B is connected.
Since A and B are open in X for each a c E, T ~'(A ) and
39
7T (3 are open in Y' Q Consequently, A and B are open
in Y0. Suppose that w E AQnB, Then w E F (A6 ) and
w 6 7 (B ) for some 6 and in E, Hence, w6 . A and
w e B . There exists A E such that a SA and cy X5.
Therefore, wA A and w B .B This contradicts the
fact that A ( B = $ for all E .E, Hence, AnB = p.
Therefore,
Y0\{g(x),g(y)} = AUB,
where A and B are disjoint regions of YC. Hence, Y.0 has
property Q. As a result, XC, has property Q.
An obvious result of Theorem 1.16 is that if D is
countable and if for each a sD, X is a simple closed
curve, then X0, is a simple closed curve.
CHAPTER BIBLIOGRAPHY
1. Capel, C. E., "Inverse Limit Spaces," Duke MathematicalJournal, 21 (1954), 233-245.
2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1968.
3. Whyburn, Gordon T., "Analytic Topology," AmericanMathematical Society Colloquium Publications,2 8-(1942)0.
4. Wilder, Raymond L., "Topology of Manifolds," AmericanMathematical Society Colloquium Publications,32 (1949).
40
CHAPTER II
MAPPINGS BETWEEN INVERSE LIMITS
Before attempting to characterize compact metric spaces,
it will be helpful to investigate the following situations.
Let (X ,f ,D) and (Y ,g ,D) be inverse systems
where for each a in D, f and g are continuous. By
a mapping Q:(.X ,f ,D) + (Ya,g ,D) is meant a collection
of functions {Qa} such that Q :X -+ Y and for a:5
in D,
Q o f = g o Q .Qa 0 fa = a 0
Define a relation Q :X + Y by
Q,({xaaFD9 = {Q Q(x aOD*
For H X , let -T be the ot-th projection function restrictedUED
to X., and for H Y., let TF be the a-th projection functionaE:D
restricted to Y00. Let x 6 X0 such that x = {x }D Then,
Q00(x) = {QU (xa)}acD
Let a 5in D. Now,
T(jQ0 0(x)) =QU~(xU).
But
g(QSox$)) = that(x
So that
9 CQ ( ))= Q,(x9,.
41
42
Therefore,
7F( Q" 'x)) = a(--" Qc(x)))
Hence, Q,,x) c Y . It then follows that Q (X ) gy.
Let x = y in X . Then, for each a s D, x = y
Since Q is a function, Q (x ) = Q (y ). Therefore,
{Q Cx )} = {Q~ (y)}D.{Q a IacD {Qa a aE:D
Consequently,
Q (x) = Q(y).
Now Q is a function,
To say that Q has a certain property will mean that
Q has that property for each a c D. By placing the proper
restrictions on Q, Q. can be forced to possess certain
properties, as is shown by the following theorems.
Theorem 2.1: Let each of (X3,fD) and (Y,g ,D)
be inverse systems and Q be a mapping from (X ,f ,D)
to (Y ,g D). Then, if Q is continuous (one to one and
onto) (a homeomorphism), then Q, is continuous (one to one
and onto) (a homeomorphism) .
Proof: Suppose that Q is continuous. Let s E X C
and Qo(s) = w. Let U be open in Y, such that w c U. Then,
there exists a basic neighborhood, if (U9, of w such
that w F V 1 (U 6 ) CU. Now, w6 e U and
Q(s) ='{Q (s )} D= {wa}D.
Since Q6 is continuous and Q(s = w6 E U., there exists
V, open in X, such that Q(V)cU6 and s6 6'V Also,
s e 'r'(V I which is open in X . Let p (.V
43
Then, p, C V6. Hence, Q(P9 16. But Q0 (p) c Y and
a (Q a (Pa acDSince Q6(p1 s Up, Qp) e j1(U )~U. Consequently,
Q00 (7T (V6)) cU. As a result, Q. is continuous.
Suppose that Q is one to one and onto. Also, suppose
that Q xl = Q (yl. Then,
{Qa Cx )}ac = {Qa(ya)}aDa a aD a a asD'Hence, Q (x ) = Q (y ) for all a. Since each Q is one
to one, x =y for all a. Therefore, x = y. It then
follows that Q is one to one. Let t c Y . Then for each
a, ta c Y a. Each Q is one to one and onto. Consequently,
for each a, there is a unique p in X such that Q (p) =t
Let p be the element in TIX such that the a-th projection
of p is pa for each a. Let as5 in D and suppose that
f (p) p.' Since % is one to one, Q (f (p )) Q(p ).
Hence, Q (f (p t . However,
QtC a (pag~a= 0 Q (p ) 9g t )=taThis contradicts the previous statement. Therefore,
f (p) =p
Hence, p c X, But
QO(P) = Q (p )}aeD = t.
Consequently, Q is onto.
Suppose that Q is a homeomorphism. Then, Q is con-
tinuous, one to one and onto. By the first two arguments,
Q. is continuous, one to one and onto. Let y c Y, and
QI 'Cy) = x. Let U be open in X00 such that x csU. Then,
44
there exists a in D such that U is open in X and
x C 7 T (U since Q is a homeomorphism, Q (U ) = V
is open in Y . Therefore, 7~-(V ) is open in Y.. Since
x 6 ff _1U ), x a U . Consequently, Q (x ) c V However,Q x cx T -1(ccx 1
Qaa x ) a Hence, y ' (V ). Let w c T ~a(V ).
Then w 6 V , As a result,
Q -'(w ) Q ~(V ) = U
It then follows that Q ~'(w) E: '(U ). From this it
is observed that Q ~ is continuous. Hence, Q is a homeo-
morphism.
Definition 2.1: If f is a function from a space X to
X, x e X and n c Z+, then fn(x) means f composed with it-
self n times acting on x.
Definition 2.2: A function f from X to X is periodic
if there exists some positive integer n such that, for each
x C X, fn(x) = x. The least such n is the period of f.
If (X, d) is a metric space then f is called almost periodic
if for each c > o, there is a positive integer n such that
for all x c X, d(x,fn(x))< c.
Theorem 2.2: Let ((Xnd ),f ,Z ) be an inverse se-nn mn +
quence of metric spaces and
Q:((Xnjd ),f Z+)÷+ ((xn,dn3m, Z+
be continuous. Then, if Q is periodic, Q is almost per-
iodic.
Proof: It will be helpful to recall that, if for
each n c Z+, CXndn) is a metric space, then a metric d
45
00
for R (X d ) ia given byn=l n, n
d (x ,y )d(x,y} = 2 nn 1
n=1 2[1 + dn n nCxy)]
where
00
and
t- 00
y ={yn In n=1For each i c Z+, let Q have periodicity n1 and let
c > o. Then, there exists k Z+ such that Z<C.p=k+l 2P
Let m be the product of the n Is for 1:55k.1
and let
Let z c X
d (Qm (z ),z ) = sp pp p p
Thenco
000d(Q (z),z) =
p=l
k
p=l
sp
2[ + s]
po
p=k+1
s-.p
2 [1 + SP]
sp
L 2 + s ]p
For p:k, let
... -* nk= rp.
Then Qmp
n rP P. Hence, Qm(z ) = z. Therefore, forp p p p
p ! k, sp = 0. Consequently,
dCQm (Czi,zI =00
p=k+l 2P[1 + s ]p
00
p-k+l 2p
As a result, Q, is almost periodic .
n + 2 +..-n ~ - np+1
The following example illustrates that requiring each
Qn to be periodic will not force Q to be periodic.
Example 2.1: For each n c Z+, let Xn be the unit
circle in the plane. Give each Xn the relative metric
topology, dn. An element, e o, of Xn will be denoted by
0, expressed in degrees. Define Q :X + X byn n n
Q (e) =+ 3600n n
Clearly, each Qn is continuous and of period n. Define
fnn :Xn Xn to be the identity map and fn+l n:X n+l - Xn
by
f =6 n+l()n+l n n
Obviously, each fn+l n is continuous and onto. For m> n in
Z+, let f :X X be f a f 0mn m n n+ln n+2 n+l0..'fm m-l'
Then fmn is continuous and ((Xn3,dn). fmnZ+) is an inverse
sequence of metric spaces and Qn is onto for each n. Also,
fmn 0 Q M(e) = f ( + 360
n+l n+2 m + 360n n+ lm-l n
m 3600=-0 + .n n
However,
Q f (e) = Q (n+l n+2 mn mn n n n1 n-1
= Qn(m( m) = + 360.T=n n n
Therefore,
mn om = n ofmn
Now, suppose that Q, is of period s. Then, Qs(x) = x
for each x E X, and Q(x) ='{Q5 (x )}neZ+ If P> s, then00n~n nZ.I ,te
there exists y c X such that Q (y) / y. Let x e 7r 1 (y).p p Pp
Then.,
QSc)={Q s(x ) =X.
Qx) ={nxn neZ+
Therefore, Q (y) = y. This is a contradiction to Q (y) # y.p , p
Hence, Q. is not of period s.
Theorem 2.3; Let ((Xndn ) f Z+) be an inverse se-
quence and Q:((Xn,dp, fm,Z+) + ( (Xndn),f ,Z+) be con-
tinuous. If Qn is almost periodic and Xn is compact for
each n, then Q, is almost periodic.
Proof: Let a >o. Then, there exists b such that
o <b minimum {l,a}. There exists k E:Z+ such that k> b
00
and p= +1 < . Let 0< 6< Since each Xn is compact,p=k+l 2P -b
each f is uniformly continuous. Therefore, for each i
such that i <k, there exists 6. such that if U is set of
Xk of diameter less than 6i, the diameter of fki(U) is
less than 6. There exists c such that
0< c< minimum {6, 1 1<1 i k} .
But Qk is almost periodic. Consequently, there exists m
such that for x Xk, dk(Q m(.x),x)< 2-. If x E Xk, then
there is an open set UX of diameter less than c such that
{x,Q (x)}}SU. But fkiCUx) has diameter less than 6.
48
Ience, d (fki x),fki (Q Cx)) <6. Therefore, dk(Qi(f W),
fki(x)) <6. Let y 6cX and d (Qm(y ),y ) = s . Then,PkPpp p
dCQm(y),y s=P5 -00 5p(m3+
p=1 2P[ 1 + s I p=k+l 2 [1 + s ]p p
But for p5k, s 6< kb Hence,
b
s < _ kbk(l-l)k
Consequently, s [l + ] , so that s < + sp(b). Now,
s < [l + s ]. Thus, 1+sP < b. Therefore,pkp l+ s P ip
k s k k1Di() I 51- _=1
p=l 2P[1 + s p=1 2p 2 k 2 k 2pp
As a result,
b 00 5 00
d(Qm b++p+ -l<b +b<a2 p=k+l 2P[1 + s ] p=k+l 2P 2 2
p
It is then evident from this that Q is almost periodic.
Theorem 2.4; Let (X , f,D) and (y , g ,D) be inverse
systems and Q:C(X a ,fD) + (Y , g ,D) be continuous. If
Q is monotone and X is compact and Hausdorff for each a,
then Q00 is monotone.
Proof: Let y 6 YO, such that Q00 1(y) X p. For each
a, let Q '(y ) = A . Since Q is monotone, A is a con-
tinuum. Hence, .II A is a continuum. For ac 35 in D,acD
let J,, = f$aIAK Then j a:A + A . For if x c A,
49
Q (xI = y 5. Hence,
g o0Q (W = g(y)=y.
However,
0oQa(x) = Q of (x)=ya.
Therefore,
f (x) 6 Q -1(y ) = A.
Now, (A ,j ,D) is an inverse system and A = rlim(A, j ,D)
is a continuum.
Lets 6 A. Then, for as in D, j a(s ) = s.
Therefore, f (s = a. Consequently, s c X . Since
A SHA , A iHA aX0. Let s e HA F)X. Then, s F X. As
a result, for a:50 in D, f (s ) = s . But s HA . Hence,
s c A So that j (s ) = s . Therefore, s c AM. Now,
A = X FUTA.00 0o a
Let s c A0. Then, s c X. and for all a e D, s e A
Consequently,
Q (s ) = y .
As a result, Q,(s) = y. It then follows that A SQ ~1 (y).
Let s E QW~ 1(y). Then s c X and, for o c D, s a Q ~1(y ).
Hence, s A . Therefore, s c HA , so that s E HA qX = A
Hence, A00 = Q00 1(y). Since A0 is a continuum, Q0 ~1(y) is
connected. It then follows that Q is monotone.
Theorem 2.5: Let
(aXndn), fmn, z+and
(CYn ,can nZ
50
be inverse sequences and
Q;C(Xn, nJfmn,3Z+) * n n), gmn,3Z+)
be continuous and open. Suppose that if gn+i n n+i1 n
and Q (x y, then
nn+1 n+1 n+1 xn
Then Q, is open.
Proof: Let 7 n be the n-th projection function of
R Xn restricted to X. and Tn be the n-th projectionn nZ+
function of R Yn restricted to Y,. Let U = n ~ 1(U )nZn+ n
where Un is. open in Xn. If x e U, then xn 6 Un. Hence,
n n) n(U)n Since Q,(x) = {Qn(x)}n=l'
Q(x) E 7n (Qn(Un))
Therefore,
Q, (U)_nIn(Un)).
Let x c 7n'(Qn(U) )). Then, xn n)Q.(U Hence, there
exists kn E -U such that Qn(kn) x n. But x c Y,. Conse-
quently, xm m for each m and g (Xm ) = Xn for all n 5m.
By the hypothesis, since Q (k ) = x and g n(xn+l) = Xnin n n n1nn1 n
Qn+ii'(xn+)mfn+i n(kn) *34p.
Let kn+1 be an element of this intersection. By induction,
for each m c Z such that m > n, there exists k c X such
that Qm(km=xm andf(m m-1(k ) = kM-1. Let k cllXn such
that for n!E i in Z+, 7Ti(k) = k and for n< i,
7(k) = f .(k ).7i ni n
51
Then k E X and Q Ckl ='{Q (k )} = x. Since k c Un n n=l n nk E 7 n (U)n, Therefore,
Q(k) = x c Q(U),
Hance, ' 7T CQn (U ))ZQ(U), It follows thatn n n"
Q (U) = ~7 Q (U )).n n n
Consequently, QcjiU) is open in Y . Now, Q, is open.
The following example proves that requiring each Qn
to be open is not sufficient to guarantee that Q will be
open.
Example 2.2: Consider Example 1.3. Let H be
{fnl~'({o0 ,18O})|n scZ+ and l< n}.
For each n c Z+, let Xn and y n be the unit circle with H
deleted. Give Xn and Yn the relative topology from the
plane. An element of Xn or Yn will be represented the
same as in Example 1.3. Let f+ nXn++ X be defined byn+l n' n+l n
fn+li n( e = 26. For m >n, let fmn;Xm - Xn be defined as
fn+ n n+2 n+0 fm M-1 Then (Xn ,f Z+) is an
inverse sequence. Define gn+1 nYn+l + Yn by gn+ is
4e if 00< 0 < 900, 3600 - 2e if 900< 0< 2700 and 20 if
270' < e<3601 . Then gn+1 n is continuous. For m> n let
mn m n begn+n1 n n+2 n+l 0 m M-1. Then
gm is continuous. Now (Yn +) is an inverse sequence.
Define Qn.;Xn n nY by Qn() is 3600 -aif 0 0< < 1800 and
0 if 18G 0 < 0 <3600. Then each Qn is continuous and open
for each n.
52
If Q < 0 < 900 then 0 < 20 < 1800 and 2700 < 3600 - 0< 3600.
Therefore,
S(n+ e() =Qn(20) = 360Q 26
and
gn+1 n 0n+i(0) = n+1 n(36 00 0)
2(3600 - e) = 3600- 20.
If 900 < 0< 1800, then 1800 < 26 < 3600 and 1800< 3600 - < 2700.
The refore,
n fn+i n(0) = Qn(20) = 20
and
n+1 n 0n+j 0( = ) n+1 n(3600 )
- 360o - 2(360 0a) = 20.
If '1800 < 0 <2700, then 0 < 20 < 1800. Hence,
Qn nofn+ n(0) = Qn(20) = 360o - 20
and
gn+1 n *0 n+i 0 ) = gn+1 n(8) = 3600 - 20.
If 270 < 0 <3600, then 18o <26< 3600. Therefore,
Qn *fn+i n(0) =n (20) = 20
and
gn+1 n 0 Qn+1 0 = n+ 1 n(e) = 20.
Now Q is a mapping.
Let 7r be the i-th projection of fXn restricted to
X and ~ii be the i-th projection of HYn restricted to Y,.
Let V = Tr iCU) where U is open in X1 and 14Qo c U. Then,
x = (14o0 ,700,350,...) c V.
Q0(x) = (2200,2900,3250,-...) F Q (V).
53
Let QCx) = y. For each open set H of Y such that y c H,
there existsHi open in Y such that y 6 F '(H ) SH.
Since y C ' 1i 1(Hi)-, yi c Hi. But y1 is such that
180, < y. < 3600. Therefore, there is eO6 Y +1 such that
0 < 6 <900 and 4e = yp. Hence, gi+l i(e) = y . But
0G < 6< 9Q0 . Hence, 001<-< 90O. Therefore,
g+2 1+6
By induction,
z = (2200,2900,...,yi,4,+,...) y
But z, = y . Hence, z c fi (Hi) CH. Now z Q00 (v).
Therefore, Hq(Y \ Q (V)) 4p for each H. Hence Q00(V) is not
open in Y , Therefore, Q is not open.
If each of (X ,f ,D) and (Y,g ,D) is an inverse
system of compact Hausdorff spaces and Q is a continuous
mapping from (X fD) to (Y ,g, then Q will be
closed. Example 2.3 illustrates that requiring Q to be
closed will not guarantee that Q is closed.
Example 2.3: For each n c Z+, let Xn = Z+. Give
X the discrete topology. Define f :Xn+ X to ben n+l n' n+l nthe identity function. For m>n, let f :X + X be
mn m n
fn+l n 4 n+2 n+l0 '0' f m m-1. Then (Xnf mn Z+) is
an inverse sequence. For each n s Z+, let
n p{ l1pin}.
Give Y the discrete topology also. Define gn+ nn+l
1 1 1 1 1 1 1by gn+1 n(-) if f and if -<, For m> n, let
54
gmn:Xm + Xn beg n+1 n n+2 n+1 a 0" mo m-1. Then
gm is continuous and CY n mn,Z+) is an inverse sequence.
For each n, let Q ;X -* Y be defined by Q (z) = ifn n n n zz n and 1 if z >n, Then each Q is continuous and closed.nn
Let t X+ such that t < n+l. Then, t <n and 1 1andn+l scn -t
1 1n+l <:. Hence,
Qn f (t) = Q (t) 1n n+1 n n t
and
n+1 n Q0n+Qit) = gn n =
If t X n+1 and t = n+lthen,
Q f (t) Q (t)
n n+1 n n n'
Now,
gn+ n Q1t)1n+1 n+1 nt '
If t C Xn+1 and t >n+l, then t > n. Therefore,
Q o f (t) = Q (t) =1n n+1 n n n
But
n+1 n n+l n+l n n+ 1 '
Now Q is a mapping.
An element N of X is described by N =(n,n,n,..),
1 1 11 1and Q(-N) = (1,1,,..1, ,,,,...). Now
1 1 1y = 17,]P...* ) 6 Y .
A basic open set U of y is T ~'(U ) where c U . Butn n n n
N = (n,n,n,. .) X,, and Q (n,n,n,...)= (1 1 ., 1 n
where - is the n--th coordinate. Therefore Q (N) c 'n 1(U 'nny Xs n n
Qonaequently, Q,.XO) is not closed. HenceQ. is not closed.
55
Definition 2.3: A space X is totally disconnected
provided it has no nondegenerate components.
Definition 2.4; A function f from a space X to a
space Y is called light if for each y 6 Y, f~'(y) is to-
tally disconnected.
Theorem 2,6: Let (X ,f ,,D) and (Y ,g ,D) be inverse
systems and Q be a continuous mapping from (X ,f ,D) to
(Ya', D). If Q is light then Q, is light.
Proof: Let ff and TF be the respective projection
functions of X, and Y. Suppose that there exists t 6cY,
such that QW_1(t) is not totally disconnected. Then there
exists H, a nondegenerate connected subset of Q '~(t). For
each u, 7T(H) is connected. If x 6 H, then x c f (H).
But, Q (x ) = ta. Consequently, xa Qa~1(t ). Since Q
is light, Q a1(t ) contains no nondegenerate connected
subsets. Hence, 7a (H) must be a single point. Let
7T(H) = k .
If x and y are elements of H such that x # y, then, for
some a, xa y a. As a result, 7a (H) is not a single point.
This is a contradiction to 7F (H) = ka, a single point.
It then follows that Q is light.
Although Theorem 2.7 breaks the tradition of investi-
gating Q 0,, it is interesting and will be useful later.
Theorem 2.7: Let (X ,f ,D) be an inverse system. If
X is totally disconnected for each a, then X00 is totally
disconnected.
56
Proof; Suppose that there exists H, a nondegenerate
connected subset of X,,,. Then, TVCH) is connected for each
a. Hence, (TH} must be a one point set. But, H is non-
degenerate., Consequently, there exists x and y in H such
that x A y. Therefore, there exists c D such that
x ; y. As a result, {xVy }.r (H). This is a contra-
diction to 7 (H) being a single point. It then follows
that X is totally disconnected.
Definition 2.5: A function f from a space X to a
space Y is compact if for each compact subset K of Y,
f- (K) is compact.
Theorem 2.8: Let (X ,f ,D) and (Y ,g ,D) be inverse
systems of Hausdorff spaces. If Q:(X ,f ,D) + (Y ,g ,D)
is continuous and compact, then Q, is compact.
Proof: Let K be a compact subset of Y. Let T
and TV be the respective projection functions of X and
Y . Since TV is continuous for each a., v (K), denoted by
K , is compact. However, Q is compact for each a. There-
fore, each Q a1(K ) is a compact subset of X . Hence,
R (Q1(K)) is a compact subset of IX . Let t c Q0 ~1(K).
Then, there exists w 6 K such that Q0 0(t) = w. For each a,
Qa(t ) = w aEK .
Consequently, t 6 Q '(K ). As a result, t H(Q ~1(K )).S a a a a
It then follows that Q"'(K).S (Q '(K )). Since K iso K a a
compact and Y is Hausdorff, K is closed in Y . Therefore,
57
Q '(K) is closed., But, Q~(K) SIICQ '(K )) which isca a
compact. Hence, Q ~'(K) is compact. It is then evident
that Q, is compact.
Now that it is understood how certain properties can
be built into Q, it will be helpful, in characterizing
compact metric spaces, to examine the following seemingly
unrelated material.
Definition 2.6: If x and p are elements of a space
X, then x and p are separate din X if there exist open
disjoint sets H and K such that x c H, y c K and X = H UK.
Definition 2.7: For an element p in a space X,
Cp = {x X such that x and p cannot be separated in X}.
is called the quasi-component of p.
Theorem 2.9: If p c X, then Cp = flX where
Y = {Up e U and U is both open and closed in X}.
Proof: Let x E C . If U c ( and x U, then U isp
open, X\U is open, p c U and x c X\U. Since X = U U(X\U),
U and X\U separate x and p. This contradicts x c CP.Hence, x c U. Therefore, Cp .g nK.
Let x c n('. If x Cp, then there exist disjoint
sets W and V, open and closed in X, such that p c W, x e V
and X = W UV, Now, p e W and W is open and closed in X.
Hence, W F . Since x V, x W. Therefore, x n
This contradicts the supposition that x F . Conse-
quently, x c CP . It then follows that fl(CP, . As a result,
p pp
58
It is now revealed in Theorem 2.10 that in a compact
Hausdorff space componenets and quasi-components are iden-
tical.
Theorem 2.10: If p is an element of X, a compact
Hausdorff space, then Cp is a component.
Proof: Let z 6 X\Cp. Then, there exists disjoint
sets U and V, open in X, such that X = U UV, z c V and
p e U. Suppose that w 6 Vnc . Then p F U and w c V.
Hence, w CP . Therefore, V(n Cp = . Now, z V._X CPConsequently, X\Cp is open. As a result, Cp is closed in
X. It then follows that Cp is compact.
Suppose that CP is not connected. Then, there exists
H and K, disjoint, closed, nonempty subsets of Cp_, such
that Cp = H UK. Since H and K are closed in Cp, H and K
are both compact, Therefore, there exist disjoint sets
U and V, open in X, such that HcU and KLcV. But X\(UUV)
is closed and compact. If x E X\(U UV), then x C CP.
Hence, there exist disjoint sets U and V , open and closedx x
in X, such that
X = UxUV ,
p 6 Vx and x e U . Also, Cc CV . Now, ( = {UxIx s X\(UUV)}
covers X\(UUV). Consequently, there is a finite subcol-
lection of 'U covering X\(UUv). Let
W* ={Ul,U2 , ...Un
be this finite cover, M = V Then, C c M. Sincei=l
59
each Vi is open and closed in X, M is open and closed in X.
nLet N = U U It then follows that X\(U UV) cN and N is
1=1
open and closed in X. Clearly, mON = . If z c M,
z 4 N. Consequently, z X\(U UV). As a result, z s (U UV).
Therefore, Mc (U UV). Since C cM, C c(MOU) U(MNOV).
Each of M and U is open in X. Consequently, M OU is open
in X. Suppose that MNU is not closed in X. Then, there
exists z E X\(MOU) such that if z S H and H is open in X,
then H O(r ON) $, But V is open and U OV = . Hence,
V O(MOU) = $. Therefore, z V. If z U, then z U Uv.
As a result, z 6 X\(UUV). It then follows that z c N.
However, N is open and NONm = $. Consequently, NO(mnOU) =.
This contradicts the statement that if H is open and z c H
then HO(mNOU) # . Therefore, z N. It then follows
that z e U.
If z E M, then z c MOU. However, z E X\(mOU).
Hence, z 4 M. Hence, z c X\M which is open in X. But
(x\M) ON = q. Therefore, (X\m) O(mOU) = 4. This is
also a contradiction to the previous statements about z.
It now follows that mnOU is closed in X. Now
x = (mOU)U (x\ (mnOU)).
Hence, CP;mnOU or Cpgcx\(mu). If CpCmnOU, then
Cp SU. Since uOv = 4, Cp Ov = $-. However, Cp = H UK
and K CV. Consequently, K = 4. This contradicts the
supposition that K $. Therefore, C cX\(m nu). Hence,
60
C n (rm r =ui =. But Cp SM. Thus, C U = $. Since HCU,P P p
H -. This. contradicts H / $p It now follows that C isp
connected..
If C is a connected subset of X and CP is a proper
subset of C, then there exists x c C such that x Cp.Since x Cp, there exists disjoint sets UX and VX, open
and closed in X, such that x c U , p c V and
X = UK ,UVx x
But Ux and Vx separate C. Therefore, C is not connected.
This contradicts the supposition that C is connected.
Consequently, Cp is a component.
Theorem 2.11: A compact Hausdorff space X is totally
disconnected if and only if for distinct points x and y
in X, there is a set U, open and closed in X, such that
x e U and y 4 U.
Proof: Let x and y be distinct points of a totally
disconnected, compact Hausdorff space X. Since X is com-
pact and Hausdorff, CX is a component. But, since X is
totally disconnected, C = {x}. Now Cx = n (where
Y(= {U cXIx c U and U is open and closed in X}.
Since y C , there exists U c ( such that y U. Hence,
x F U, y U and U is open and closed in X.
Let X be a compact Hausdorff space such that for
distinct points x and y in X, there exists a set U, open
and closed in X, such that x c U and y U. Suppose that
X is not totally disconnected. Then, there exists H, a
61
nondegenerate component of X. Let x and y be distinct
elements of H, By supposition, there exists an open and
closed subset U of X such that x E U and y U. Hence,
y E X\U and X\U is open in X. However, U H is open in H
and (X\U) (nH is open in H. Consequently,
H = (UrnH) U ((X\ U) oH),
x c UqH, y e (X\U)nH and (U H) n((x\u)nH) = $. There-
fore, H is not connected. This contradicts the statement
that H is a component. Hence, X is totally disconnected.
Theorem 2.12: If (X,d) is a compact metric space
and V is finite open cover of X, then there exists c> o
such that if V is an open set of X having diameter less
than o, then V is a subset of some element ofV.
Proof: Let V ( = {Ul,U 2 , ... ,Un } be a finite open cover
of X. Suppose that for each c > o, there exists Vc, an
open set of diameter less than c, such that V0c-U1 for
1l5in. For each m E Z+, let xm Vm where V is an open
set of diameter less than 1. Then the sequence {x }M n n=l
has a cluster point x. Since x c X, x 6 U for some i.
There exists S(x,6), a spherical open neighborhood of x
of diamter less than 6, such that S(x,6)._U . However,
there is m c Z such that < . Since {x } clusters+ m T n n=l
at x, there exists p e Z such that p > m and xF S (x,).
Let y c Vp . Then d y,x):5d(y,xp) + d(xp,x). But,
6d( y,x )< and d(xx) < Conse quently,p p p 6 6 6
dCy, x) < + < + = <zCS
62
As a result, y E SCx,6), Hence, V S(x Thisp 1)CU1 Ti
contradicts the supposition that VC ui for any i. There-
fore, the theorem is proven.
Definition 2.8: Let 74 and V be covers of X, then 2
refines V, written as ?(< V, means that for each A E
there exists B F V such that AcB.
The following theorem establishes a connection between
inverse systems and theorems 2.9 through 2.12.
Theorem 2J13: If (X,d) is a totally disconnected,
compact, metric space, then X is homeomorphic to the in-
verse limit of an inverse sequence of finite spaces.
Proof: Let U be open in X and x 6 U. Since X\U
is closed in X, it is compact. For each y 6 X\U, there
exists Vy open and closed in X, such that y e Vy and
x V. Then ? ='{V |y E X\U} covers X\U. Hence, there
exists a finite subcollection of ?r covering X\U. Let
W={V V,23*''V}n
be this finite cover. Now, UW is open and closed in X
and does not contain x. Let z E X\ Uk. Then, z U2.
Therefore, z X\U. Consequently, z F U. Hence, X\U2 CU.
As a result, x 6 X\ U9c U. Now, X has a base of open and
closed sets.
Let n e Z+ Then, for each x c X, there exists Hxsuch that x 6 Hx and H is open and closed in X and has
a diameter less than 11. Therefore, W( = {HxIx Xcovers X.2n
63
Since X is compact, there exists a finite subcollection
$ of K covering X. Let
= {H1 ,H2 ,..j,Hm).
Let U = H3,n-l
U2 = H2\ H,..,Un = Hn\ U H
Continue this process as long as the U1 s remain nonempty.
Let
S= {UIlim}
be this collection of nonempty U is. For each i,
i-l i-1H\ UH = (X\ UH )rnH .
p=lp p=lp
Consequently, U is both open and closed in X. Let x c X.
Then, x E Hi for some i. Let j be the least such i. Now
j-1x 6 H.\ UH = U..
i p=j p J
Therefore, tW covers X. Suppose that 1-5j m and 1 5 m
and j Z. Either j <k or k < j. Suppose that j< k.
Then,k-1
U = H\ U Hp=l P
So, if x E U., x 42 H, Hence,
j-1
x4H.\ UH =U..J p=lp J
j --l -1If x e U , since j < , xE UH H. As a result, x 6 H\ UH
p=l p=lP
Hence, x 42 U." Therefore u% RnU, = $. Similarly, if k <
64
U qU = $. Then t4 is a cover of mutually disjoint, closed
1and open seta each having diameter less than - By2 n
Theorem 2.12, there exists 6 > 0 such that if V is a subset
of X of diameter less than 6, then V CU 1 ?{. Let A be
less than the minimum of {6 , .- By the previous con-2n+l
struction, there exists a finite open cover of X by mutually
disjoint, open and closed sets of diameter less than A.
Let ?r be this cover. Then, 7f< V. By induction, for each
n c Z+, there exists n, a cover of X by mutually disjoint,
open and closed aets of diameter less than such that2 n
4n+l< nn Let {? iil be a sequence of such covers. For
each n, let Yn be {'nI with the discrete topology. Define
n+l n n+l n by
fn+l n (U) = V
where V is the unique element of Yn such that USV. Then,
each f n+1 nis a continuous function. Let f :Xn - Xn be
the identity map and for m >n,
f f of o .fmn n+ln 0 n+2 n+l 0 '' m m-l'
Then, (Yn3fm, Z+) is an inverse sequence. Let
Y = lim(Y ,f Z)S n + mn,+and y c Y0 where y = '{U} such that U s ?cI. Define
F:YO X by000
000
F(y) = FC{U } ) = ui=i s
Since f n+ n(CU n+1 =U n,3U n+1 U n. But X is compact and
65
each U1 is closed. Therefore, U1 is compact. By Lemma 1.1,
00
q u1=1
If x and y are elements of n U1 such that x # y, theni=1
there is n cEZ such that -<d(xy). If x 6 Un, y Un*
00 00 00
Hence, if x c q U,, y n U1 . Therefore, q U1 is aI=1 1=1 1=1
single point. Consequently, if F(y) = p and F(y) = q,
then p = q. As a result, F is a function.
Suppose that F(X) = F(y) where x ='{U 1 }O and
y = {V }0 0 . If U1 V. for some I, then U i V = Hence,
00 00
1=1 1=1
This contradicts FCx) = F(y). Consequently, U1 = V. for
each i. Therefore, x = y. It then follows that F is one
to one.
Let x 6 X. Then, for each n, there exists Un n
such that x c U andUn is iniquein Y . Let y iYn n n n
such that Pn(y) = Un for each n. Suppose that for some m,
f m+ m(U)m+1 )7Um. Then
mm+ m(Um+ )lUm
But x E Um+l. Therefore, x fm+1 fm(U m+1). Since x 6 UM3
m+1 m(Um+) num X '
This is a contradiction to fm+1 M(Um+ 1 )Um = . Conse-
quently,
m+l m(Um+ ) =um0
66
As a result, y , YO.. But00
E~y) = 0 U. = x
1=1Hence, F is onto.
Let t ='{w } e Y and Z be an open set of X such
that F(t) e Z, Then, there exists a basic open neighborhood
U = S(F(t),6), the spherical neighborhood about F(t) of
radius 6, such that F(t) c U.SZ. Let
H = {y = {U } 1 e Y.1there exists n ycZ+ such that U nCU}.
00
Since t = {w } and F(t) = { fl 1W}LU, by Lemma 1.2,i=1
there exists m e Z+ such that wmLU. Therefore, t c H.
Hence, H / p. Let y = {U H. Then, for some j,00
U U. But F(y) = q U. U.. As a result, F(y) c U.i=i'
Consequently, F(H)5U. Let
z = {V } O H.
Then for some n, V CU. For each i, {V.} is open in Y..n T
Hence, 7Tr '({Vn}) is open in Y and contains z. If
n n { u={s}00 - I(r{V
ii-l n n1then {s n{V}, Therefore, s = V. But V U. Sincen . 1sn n n :U icsn =Vn' nU, It then follows that s c H. Now, H is
open, t E H and F(H)cUcZ. Consequently, F is continuous.
Now, F is a continuous function from a compact space onto
a Hausdorff space. Hence, F is a homeomorphism. Therefore,
X is homeomorphic to Y00 the inverse limit of an inverse
sequence of finite spaces.
67
Definition 2.9: A space X is perfect if each element
of X is a limit point of X.
Theorem 2,14; If U is a nonempty open subset of X,
a compact, Hausdorff, totally disconnected, perfect space
and if n is a positive integer, then U is the union of
exactly n open, mutually disjoint, nonempty sets.
Proof; Since U is nonempty, there exists x E X such
that x c U. Since X is perfect, x is a limit point of X.
Therefore, there exists y 6 U such that y X x. Since X is
compact, Hausdorff and totally disconnected by Theorem 2.11,
there exists a set V., open and closed in X, such that x c V
and y V. Now, x e VqCU which is open in X and
y 6 (X\V) qU which is also open in X. Let V nU = U1 and
((x\v) qU) = U2 . Then U = U1 UU 2 , U1 qU 2 = , and
U 1 X U2, By induction, for any n c Z+, U is the union
of n disjoint, open, nonempty sets.
Theorem 2.15: Any two compact, perfect, totally dis-
connected, metric spaces are homeomorphic.
Proof: Let X and Y satisfy the above hypothesis. For
a set H, let tHl denote the cardinality of H. For each
n E Z+, let "9 and V be - finite covers of X and Y, re-+n n
spectively, of mutually disjoint, closed and open sets such
that each element of "h and each element of 7/ has diametern
less than and %+l and n+1 n, By Theorem 2.14
can be made the same as IVnI for each n. For each
68
n c Z let X be'{0n} with the discrete topology and Y+ 1n nbe {V} with the discrete topology. Let
XO = lim(Xn fZ+)
and
Y = lim(Y ,g ,Z+ n mn +
be the same as in Theorem 2.13. Then, X is homeomorphic
to X, and Y is homeomorphic to Y,. Define Q :X - Y by
Qn(U ) = Vi,
for each n. Then, each Qn is a homeomorphism. Now,
n+1n Qn+l(Un+l) = + n(Vn+l) = Vnand
n n+ln( n+l n(Un n'
Therefore, by Theorem 2.1, Q :X-+ Y is a homeomorphism.
Hence, X is homeomorphic to Y.
Theorem_2.16: Every compact metric space is the con-
tinuous image of the cantor set C.
Proof: Let (X,d) be a compact metric space. Let
' = {A,1 1 A2 1 ,.*,An 11
be a cover of X by closed sets of diameter less than 2.
For each i such that 1515n1 , let Ki be a finite cover
1of A 11by closed sets of diameter less than. Since
A is closed in X, it can be assumed that each elementn
of Ki is a closed subset of Ai l:Let ?2 = U K be denotedi=l
by
2 2 ,A2 2, ,An22
69
Then 2 . and, if x Ai, there exists A 2 : 2 such
that x e A. j2A1 . By continuing this process a sequence
{2U } of finite covers of X by closed sets can be ob-
1tained, An element of 14 has diameter less than -. Eachn 2
Vn has the properties that n+ <4 n and if x 6 A iCVn
then there exists A. n+ n+1such that x E Aj nlA in'
For each m, let
2m = {Am,AA2m,...An m Km
For each i such that 1:i5 n , let
B(1,) = {(x,i,l)fx E A }
Let K be open in B(1,1) if and only if {xj (xi,l) E K} is
open in Au. Let n
H=1nb p B(i nly.i~l
Define a set V to be open in H1 if and only if Vn B(j,l)
is open in B(j,l) for each j.
For each Aj2 such that A .CS:A , letA. j2~ p1
B(j,p,2) = {(x,jp,2)Ix c Aj2
Letn1
H2 = U (UB(jip,2)|Aj2CAp ).p=l
Define open sets in the same manner as for B(il) and H1 .
For each AP 3 such that A, 3 -cA.j2 cA 1p, let
BC,j,p,3) = {(x,Qj,p,3)Ix E A }.
Letn
H3= U CUB , jp,3)jAP 3 -!A j2SAp1)'p=l
70
Define open sets in the same manner as for B(j,2) and H2.
By continuing this process, a sequence of sets {H i} i is
obtained. Now,n
H ,,CUBi,n)jA EA. . ..Hn n, U*CUBjj n -J JH n-1 .nl
i=1 n n-1
The method used to define open sets of Hn guarantees
that each B(inin-13,...,iin) is both open and closed in
Hn* It is also apparant that the B's are disjoint. Since
Hn is the union of the Bts and the B's are open, Hn is the
union of the open sets defined on Hn. Let V { ='{U }acDbe
a collection of open sets of Hn indexed by D. Let
K = B(jinin-1 ,'''j 1 ,n).
Since each U is open in Hn'
M = {x|(x,j ,jn-,'''i 1 ,n) s K(OU }
is open in A. . Therefore UM is open in A. . ButEnn asD Jnn
U Mc a ={x|(xjj n- 1 '.,1,n) 6 KnO(U L4)}.
Hence Uk is open in K. Therefore UV is open in H .
Let U and U 2 be open in Hn. Then for i c {1,2},
Mi = {x|(x'jn jn 1''', 1 ,n) P U K}
is open in A. . Therefore M 1Nm2 is open in A .But
MN m2 = {XI ('xj n l,n) 6 (U qU 2 )rK}.
Hence, U 1U 2 is open in K. Therefore, UJn U2 is open in
Hn, By induction, this can be extended to the intersection
of any finite number of open sets. Now the open sets of
Hn form a topology for Hn
71
For each n, let Q H + X be defined bynfn
QnC(-xnn-lj''l,*,jln)). = x
Clearly Qn is a function from Hn onto X. Let
y = (x,j n' 1n-l''j'1,n)
be an element of Hn and V be an open set of X containing
x. Since V is open in X, K = VFnA. is open in A.Jnn jnn
Let
U {(',jn'jn-l''j 1 ,n)|w e K}
and
Then, U c T.
open in T.
open subset
then V K.
quently, Qn
Define
T = B(jn'$n-l''' ,n).
Since {y(yj ,jn-l''063 1 ,n) c UI = K, U is
Since the B's are disjoint in Hn and U is an
of T, U is open in Hn. Clearly, y c U. If
h = (v,jnjn-l,,'',j'1 ,n) c U
Hence, v 6 V. Therefore Qn(U).V. Conse-
is continuous.
f *H n+ H byn+l n n+l n
S(+ , n n+1 .n'''ln+l))
Let f be the identity function. Clearly, f is ann n+l n
function. Let
t = (xjn n-l ',,j 1 ,n) Hn'
Then x e A. . There exists A. c n such that3n Jn+ln+l n+l
x c A. cA. . Thenin+1 n ann
S= (x jn+1lUn'''',j 1 ,n+1) e Hn+1
72
and fn+ n(s) = t. Hence f n+1 nis onto. Let
n ln1n+l1jn +'1,n+l) c Hn+1
and V be an open set of Hn containing fn+1 n(y). Let
N = B( jn'0n-l n'., n).
Then fn+i n y) e N, Since V is open in Hn3
M = {z (z,j n'in-l .j 1 ,n) E VON}
is open in A.n Since A. CA. , K = A.l mJ
nan+ln+1 J an+ln+l
is open in A j n+lAlso, x c K. Let
U =f{(w,jn+1,jn '...,jln+l)Iw e K}.
Now UL N and U is open in N. Hence U is open in Hn+i*'If
S = (t~jn+in ,'.'.,j 1,n+l) s U,
then t E K. Hence, t c A.in+n+1NM. Therefore, t c V.
Hence, fn+l n (s )c V. Now, fn+i n(U)L V. Since y s U
and fn+n) = x V, f n+1 n is continuous. For m> n,
let f :H + H be defined bymn m n
fmn ( f)1=0fn+ n +2 n+1 ' ' mm-
Then f is continuous and onto. Therefore, (Hn ,f Z+)
is an inverse sequence.
For each n in Z+, let Xn = X and in+1 be the identity
function from Xn+l onto Xn. For m> n, i:X + Xn be
in+1 n 0 1n+2 n+1 ..*.'0im m-l. Then (Xn mn ', Z+) is an
inverse sequence. Further, n;QHn X is continuous andn n n
onto. If
n+ 1 -jn'.''* 1,n+l) Hn+1
73
then
n0f n+inCY) QnC vjjn'in-,'' ,n)) = v.
But
in+l + in+l(v) = V.
Therefore, by Theorem 2.1, Q00:H0 -+ X is continuous.
Let w = (x,x,,..) c X . Then x A1 1for some i.
Hence (x,11 ,l) cH1 . There exists A1 22 F2 such that
x F A1 22 A*. Hence (x,12,11,2) c H2 and
f2 1 x,12 'I1 ,2) = (x,11 ,1).
By induction, there is an element s of H such that the00
n-th projection of s is (xin,1in-l.'. I 1 ,n). Hence,
Q (s) = w.
Therefore, Q is onto. Since X is homeomorphic to X,
X is the continuous image of H.
To complete the theorem it will be sufficient to show
that H is the continuous image of H X C and that H X C
is homeomorphic to C. Let
s = (x,jn ,jn-l '' '''d i,n)
and
t = (y'i ,i n-l-''''-' 1,n)
be elements of Hn such that s # t. If there is p such that
1 p n and j / iP, then s and t are in different B's.
Hence, there exists disjoint open sets
B(inI3n-l ''''',n) = T
74
and
S= B Q n 'j n.,- 3 '' ,.n)
containing t and s, respectively. If jP = iP for all pthen x # y and a and t are in T. Therefore, x and y are
elements of A ,Since x # y, there exist open disjoint
sets M and N of A. such that x e M and y e N. Let
J = {Cv n'nn-l'..,j,n)|v M}
and
I ={(w,j njn- ',.',i,n)|w N}.
Then c J and t6 I andJr)I=p. By the method of de-
fining I and J, each of I and J is open in T. Further,
J cT and I CT. Hence, I and J are open in H . Therefore,n
Hn is Hausdorff.
Let 7W = {VU},D be an open cover of Hn. Let
R = B(jnin-l''..,j,n)
be one of the disjoint collections of Hn' For each a c D,
V qR must be open in R. Therefore,
M Hn{(xlj '...,J ,n) E V mR}
is open in A . If y A , then s = (y,j ,...,jln) c R.
Therefore, there exists 6 6 D such that s 6 V Hence,
s 6 MN6R. Now, {MaRca D} covers A. . Since A. is
closed in X, A is compact, Hence, there exists a finitejfn . hreeitnafnt
subset F-of D, such that {M a e F} covers A. . Let
nz= (vj' in e R.
75
Then v e A 4. Therefore cM- for some a c F. Hence,
z e V . Therefore {V }aEF covers R. Hence, Hn is compact.
Now, H is compact and Hausdorff for each n. Therefore,
11H, is compact and Hausdorff. Hence, HO, is compact.
Let w A y be elements of X0,. Then there exists an n
in Z+ such that 7Tnr(w) A 7rn(y). Let
in(w) = Wn =(xn'j nn-l''...,jln)
and
=Tnn = (vn in -lW''' i1,n).
If for some p such that 1 p!5n, ip 7 j , then
Wn , BCjn .u n-l ''',n) = W
and
yn cB(in in-,'..,ii,n) = T
and W T. Therefore, WnT = . Hence, r (W)1n
closed in X., contains w and does not contain y.
for all p, then x A v. There exists m 6 Z+ such
1 d(.x V)1< d2x ) Therefore, xc Am e t( and x c A22mm n
where i . Further, A. nA.m lm1m jmm
is open and
If j =i1p p
that
. evj mmm
= $. Then
,ff.m(w) = (x, jm 1 ,.0.j 1,m)
and
=rm( Cvim im..-1' i 1,m).
Since j im, by the above argument there exists an openmm
and closed set of X. containing w and not y. Therefore by
Theorem 2.10, X, is totally disconnected.
76
Since X0 is compact and Hausdorff, it is regular. If
X is second countable then by Urysohnts metrization theorem,
X, is metrizable, Now, if H has a countable base for each00 n
n, then TtHn will be second countable. Hence, X will be
second countable and metrizable, Since X is a compact
metric space, X is second countable. Let 7 = {V}i be
a countable base for X. For each BKjnVn'..'''iln), let
M n' =l''xj n'dn-l''' ,il,n)|x s VqA }.
Then,
MiQn.n-l''...,jln)
is an open subset of
BQjn'$n1-1 ' ' ' 1 ,n)
and for each B there are only countably many such sets. Let
B(jnn-''..',jin) = R
and U be open in R. Then
.{Xj (x~,n'n-l''n,.1,j,n) c U nR} = P
is open in Ajn, But 7/ covers X. Hence, there is a sub-
collection C of 7 such that (U3)nA. = P. LetJnn
= {Vp1,VP
2
Then, Mp Q nn .,j 1 ,n) c U. If
t = (w,jn' n-l''.,j 1 ,n) E U,
then w c P. Therefore, there is V P -6 such that w c V .
jn
77
{M (nj'V1n) }0 is a base for 3(j njn-n1 '''.jl,n).{PiQn Jn-I's ~nl
As a result, Hn is second countable. Therefore, H, is
metrizable.
Let K = HOO X C. An element of K can be written as
(h,c) where h s H0 and c c C. Since the projection function
from K to H00 is continuous and onto, H00 is the continuous
image of K.,
Since each of H00 and C is a compact metric space,
K is a compact metric space. If (h1 ,c1 ) # (h2 ,c2 ) in K,
then h1 $ h2 or c1 c2 . If h1 X h2 , then since H. is
totally disconnected, there exists U open and closed in H00
such that h 1 U and h2 U. Consequently, (h1 ,c1 ) 6 U X C
and (h2 ,c2 ) is not. A similar argument holds for the
assumption that c1 c2 . As a result, K is totally dis-
connected. Let (hic) K and U be a basic open set of K
containing (h,c). Then U = U1 X U 2 where U1 is open in
H and U2 is open in C. Since C is perfect, there exists
c E U2 such that c c1 . Therefore, (h,c1 ) e U and
(h,c) (h,c1 ). It then follows that K is perfect. Now,
by Theorem 2.15, K is homeomorphic to C. Since H00 is the
continuous image of K and X is the continuous image of H0,,
X is the continuous image of C.
CHAPTER BIBLIOGRAPHY
1. Gentry, K. R., "Some Properties of the Induced Map,Fundamenta Mathematicae, LXIV(1969), 55-59.
2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.
78
CHAPTER III
AN INVERSE LIMIT CHARACTERIZATION OF
COMPACT HAUSDORFF SPACES
To characterize compact Hausdorff spaces, it will be
necessary to establish the proper background.
Definition 3.,1: Let En be euclidean n-space. Then,
,n 1 2 na point ai in E is expressed by ai (a ,a ,j.., ).Let
{aO,al,...,ar} be r + 1 linearly independent points in
En. Define [acial,...,ar I to be
1 2 n i{(x ,x ,...,x )Ix
r r= 2a 1.a,5ii5n, and 2 D .
j=0 j=1 0
= 1 and for each j, O .l1}.
Then, [a0 ,a1 ,.. .,ar] is called an r-simplex and each a.
r .is called a vertex. If x is such that x = x.a. where
j=0 J
r+l for each j, then x is called the barycentre of
[a 0 ,a,...,ar-]
To simplify notation, whenever x = (x 1 x2 ,...,xn) is
r .expressed by x = X.a., for 1 i<n, the superscript i
j=o c su
will be deleted if no confusion results.
79
8o
Definition 3,2: If a1 ,ai ,...,a are s + 1 pointso 1 s
of {aoal...,ar}, then [ai ,ai ,...,a ] is called ano 1 5
s-face of [a0 ,a1 ,.., arI.
Definition 3.3: Let K be a finite collection of
simplices in En auch that every face of a simplex of K is
in K and the intersection of any two simplices of K is
either empty or a face of each. Then K is called a complex
and UK is a polyhedron.
It is important to notice that each simplex of K is
closed and compact. As a consequence, UK is compact.
Theorem 3.1: Let Xn be an n-simplex in K and
A(Xn) be the barycentre of Xn. Let n0 > n1 > ... >ni be
a finite descending sequence of positive integers and
n0 n n. n.X ,X ,...,X be such that each X 0 is an n.-simplexUO a 1 i a1i
nk n n0 nwhere X is a face of X kl. Then, [A(X ),...,A(X )]
ak ak-1 a 0 Ot
is a simplex.
Proof: For notational purposes, let
n. 0 1 n.X X =[a ,a .. ,a
n a aa jaj
andn.
ACX 3) = Ba. n.
3 J
Suppose that'{BnnBn ,n1..,n } is not linearly independent.
81
Since X is a face of X , the following arrangement3 ~j-1l
can be made, Let
B be the barycentre of [a0 ,a1 , ... ,an]
Bn be the barycentre of [a0,a1 ,...,a]
Bn be the barycenre of [a 0 ,a1 ,... ,an .ni
Since the barycentres are not linearly independent, there
exists B such thatnp
B =cB +c +...+C Bn On ln 1nl n-p 0 1 n - n -
+ c B +.. + Bnp+l np+1 i
where all the c 's are not all 0. Therefore,13n c0 n0 c n
-n 1 a = s + ... + n- asn+ln0+1nn1 + +12Ss=s=0
+ p+l np+lC n
p+l -On 1+1 s
As a result,
c n0c n n0 =Da + .*. + pa1
n0,+10an +1 s in+ ias= p -l 5=0 p s=0
n n+ p+ L p+l c i i
+ j aF+,,.. + IFa.np+l+1s=0 n1 s=0 S
82
c.For o 5j <i and j p, let K. = Let K
S n.+l p n+13 P
Then, n n
Q= K Za + .,. + K a + .. + K Das=0 Es=0 S=0
Hence,
i i0= E K )a0 + ( EK )a +
m=0 m m=0 ml11-1
+ ( E K )a + ( mK )am=0 m n1 m=0 m n1+l
+ ( K K n)an+2 ,..+ E Km)am=0 m g2m=0 m ni1y
0an1+1+ K0an+2 O+n + K0an0
However, {a 0 ,a1 ,..n.,anI } is linearly independent. Therefore
jeach coefficient must be 0. mLKet c. E K . Now, c = K .S m=0m
If Kp = KO, then K ~ 0. This is a contradiction~p 0, 0 n +1p
to each coefficient being 0. As a result, Kp # KO'
Therefore, K0 = 0. Hence c0 = 0. But,
0 = c1 = K + K = 0 + K .
If n =n , K = 1 -0. This contradicts each coeffi-p 1 1 n 1+1X
cient being 0. Therefore, n / n1 ,Hence, K = 0. Con-
sequently, .= , Thus, c1 = 0. By induction, n nn +1 p
83
for any i such that 0 5i5n This contradicts the assump-
tion that {Bn 13 n ,'. n.} was not linearly independent.0 1 in
Hence {B n ,'.'Bn } is linearly independent. Now,
n n[ACX)0 ,...,ACX )] is a simplex.
0i
Theorem 3.2: Let K be a complex and Xn be an n-simplex
of K., Let n 0>n1 > ... > ni be a finite descending sequence
n n nof positive integers and X ,X ,...,X be such that each
a0
n. nk nklX is an n.-simplex where X is a face of X . Let
a k ak-l
nB be the barycentre of X k. Then [B ,B ,...,B ] is
a simplex. If K is the collection of all such simplices
formed from X n for each Xn in K, then K is a complex.
Proof: Since K is a complex, it contains only
finitely many simplices. Each simplex of K gives rise
to only finitely many simplices of K .6Consequently,
K has, only finitely many simplices.
Let Xni = [B0 ,B,**.,Bn ] be a simplex in K where
B is the barycentre of [bo,b,...,bn]0 n01
B1 is the barycentre of (b0 ,b1 ,...,bbn ]3
Bn is the barycentre of [b,b ,.. .,b ].
84
Let [B ,B ,.. ,, ]be a face ofX . ThenUa alajn
Baois the barycentre of [b 0,b1 ,...,ba]
is the barycentre of [b 0 ,b1 ,.. .,b al
B is
However, since
the barycentre of [b 0 , b,...,b ].
for each i, [b0,b ,..,b3] is a face of
[b.b ,*w .,bno], it is a simplex of K. Therefore,
[B a 0 3Ba 1'04.,B a ]jI c K101
The following lemma will help to show that if the
intersection of elements of K is nonempty, then the in-
tersection is a face of each.
Lemma 3.l: Let X = [Bo,B1 , ...,B ] be an element of
K where
Let S =
B0 is the barycentre of [b 0,b, .. ,bm 0
B is the barycentre of [b 0 , b1 ,.3. ,bIm
Bp is the barycentre of [b 0 ,b,...,bm ]p
[a0,ai,...,an] be a face of [b0,b1 ,...,bm ] such
that XOS 7k. Then XqS is a face of X.
85
Suppose that there does not exist i such that o 5i:p
and B e S. Then for each i such that o 5i 5p, there
exists b such that o j< m and b {aOa , .. ,an}.
Let x E X n~S, Then,
x X + 1 B 1 + ++ . . pBp
=0a0 + 61a 1+...+na n'
Since each Bi is the barycentre of [b 0,b, ...,bm],
A m0
o+s=0 pm +s=0
0a0 + ... + nnaX.
For each j such that o:j!p, let K. -=. Then,
pX Z= K.)b0 +. +( K.)bj=0 JO j0 p
p--1 p-i+ ( ZK)b +1 + . (EK.)bm
j=0 p j=0 Jmp-1
+ .,.,, + K 0bm 1 + . . . + K 0bm 0
6a0 + . .+. + 6a . +K0 0 n n
However, there is b. such that b. a for any i such that
o i :n. Hence, the coefficient of b . is 0. Since the
coefficient of b. is K.,
i j=0 0
+ Kp =0.
86
Therefore,
0 1 p
This contradicts the statement that if x c X then the
sum of the coefficients must be 1. Consequently, there
exists i such that oSi p and B S.
Let {B ,B ,.,.,B } be the collection of B 'sa0 al at
such that o:5i.<p and Bi c S. Let x s Xq S. Since x c S,
x ao + 61a 1 + ... + 6a nn*
Since x X,
x = A0 B0 + 1B1 + ... +AB
For each i such that o -<i !p, let K =m.iTheni m.+P Thn
p px = ( K)b0 + ... + ( K)b
m=O m=O p
p-1 p-1+(rK )b + + K)+ ... bm=m m +1 m mM=O p M=0 p-
+.,. + K0 b n1 +1+ + K0bn0
Suppose that o 5 i:5p and Bi i' B for any j such that
o < j !5t and 0. Let s be the largest such i. Then
there exists br {b 0 ,...,bm } such that br {aO,...,an}9
Now, for each j s, ar c {b 0 ,.*..bm} . The coefficients
sof b is DK. = 0, Hence, U = A1 = . = A 0r j=0s
87
By assumption, 7 0. Therefore, x [EBa,.0..,Ba].
Clearly, if x E [E3B,*a .' B at], x X s, Therefore,
0 1t0Ba ,Ba
which is a face of X,
Now to complete the proof of Theorem 3.2, let X and
T be simplices of K such that XnT # . Let
Xs = [BQ,B...,Bp]
and
Xt = [HOH .. .Ht
be the smallest faces of X and T respectively such that
X qTcXs and xrnTg;Xt. Let
B0 be the barycentre of [b 0 ,b1 ,.. .,bn0=P
B be the barycentre ofD[b,3 , b.n.1.,bn
Bp be the barycentre of [b 0 , b1 ,.. .,bn ] and
p
H0 be the barycentre of [h0,h,. . .,hm0 =L
H1 be the barycentre of [h 0,h,.. .hm]1
Ht be the barycentre of [h 0 ,h1 ,...,hm t
Now, let
[beb , .. ] (Minh 0 ,h1 , ...,hm ] = [a0 ,a1 ,...a ] = R.0 0p
88
The XOTZCR and R is, a face of P and L. By Lemma 3.1,
X QR is a face of X S But XQ T.GXs DR, hence, xs Q R = X sAlso, Xt R is a face of X and XQTCX R. Hence,t -tXt nR = Xtq. It can be assumed that R is the smallest
face of P and L containing X nT, If a1 c{asa ,. .a },
then a. c {b,b ,...bn00
} n{hh ,3...,5h }. For each j0
such that o j p, there exists x e Xs such that
x = AB$B + X11 + .. + AB. + ... + A B
and A. 0. For each j, let K. .n0 n1 n+
x = K(L b ) + K (Lb ) + ... +
S=0 s=0
Then,
n
K ( L bp S=0 s
p= ZL K )b
s=0 s
p+.Z+ K s)b ns=0 np
p-1+ ( Ky)b n+1
s=0 np
p-1+ .,.+s )b n
s0 sn0+ .,. +K~b.
+ ... +Kb0 1
Since A. 0, the coefficients of b ,b ,.. .,bn.b0' bj3*3bn.3
zero. But, x=60a0 + ... Therefore,
{bobl,..,b } c {a .,ap,...ap
Consequently,
As a result,
{b 0 ,b ,..,bn }_{a0 ,a1 ,...,a }.
{ b Q b j ,.) *. b n } = ' { a ,a . . . ,a0p }
are non
89
By a similar argument,
{hohihh,.,hm } = {aoa ,,,,a{h hm0 0 ' ' 'p
Therefore, B0 = H0 .
If Bi A H for some j such that b <j t, then either;
Case 1; there exists b c {b 0 ,b1 ,. ..,nb } such that
b {h ,h. .. ,hI}
or
Case 2: there exists h e {h 0,h1 ,...,hM } such that
h {bo,b ,..,,b }.
Suppose that B1 A H1 . Let x E [B,B ,...,Bp] where
x = 0 B0 + . B and A1 7 0. If x e [HoHi,...,Ht],
x 0 n0n = - b sn0 S=0 s
+
60 sM0=m----- h +
Let K = and L. =i n +1 1 m.1+1A
x np.. +-- Z b
np+1s S=Os
6t mt.. + TE h
mt+1s=0s
Therefore,
px = ( D K )b0 +...
m=0
+ .,, + K0hn+1
tCZLm h0 +,,m=0
+ ., + Lhm+1
p+ ( Zi K )b
m=0 m np
+ *.. + K0bn0
t+ E 2 L )hm
m=0 mint
+ , O
p-1+ ( Z K )b
m=0 m np+1
t.-1+ C Z Lm)hm
m=0 t
then,
90
IIf Case 1 is true, then the coefficient of b is Z) K
m=0 m
for 1. However, since b {{h0 ,h1 ,...,hm }, the co-
efficient of b must be Lo. Therefore,
LO = K0 + K1 + ,,, + K1 ,
PIf Lo - Ko, then K =E Lm for some p 21, Hence,
m=0
Lo= L0 + L + ... + Lp + K +... + K.
As a result, K1 = 0. It must follow that A = 0. This
is contrary to the supposition that A 1 j 0. Therefore,
Ko = L, But, Lo = K0 + K + ... + K. Hence, K = 0.
It then follows that A = 0. This contradicts A 0.1 1Hence, Case 1 is not true. Therefore, Case 2 must be true.
iNow, the coefficient of h is Lm for some i'l: and it is
m=0
also Kg . If Ko / Lo, then Lo = Ko + K + ... + K. for
some j e l. But, Ko = Lo + L + ... + Li for some i1.
Consequently,
Ko= K0 + K + . +..+K. + L + + L .
Therefore, K1 = 0. Hence, A = 0. This contradicts
A1 l 0. As a result, Ko = Lo. It then follows that since
Ko = LO + L1+ .o + L , L1 = 0. Now, 61 = 0.
Suppose that if B # Hr for 1l5 r5Z, then Ko =LO and
L = L2 =..' = L =0.
Suppose that B1 H +1, Then, if Case 1 is true, the
91
I Icoefficient of b is Ek , for 1, and it is also Z Ls=0 s=0
for some j such that 0< j< k + 1. Therefore, K1 = 0.
Hence, A1 = Q. But, X13 0. Consequently, Case 2 must
be true. As a result, the coefficient of h is K0 and also
iFL for some i Z! + 1, Then,s=0
K = L 0+ L1+... + L +1 + ... + L .
Since K0 = L0 , L+l = 0. Consequently, 6%+l = 0. Hence,
by induction, if B1 Hp, then 6p = 0. However, on the
tcontrary, F 61 = 1. Therefore, 60 = 1. Since 6 0 =
i=0 0
A = 0. This contradicts A1 $0. It then follows that
x P [Ho, .. ,Ht]. Consequently, B1 is not a necessary
vertex of [B0QB 1 ,. ..,Bp]. But, by the supposition, B1 is
necessary. Therefore, B = H for some j such that O<j t.
By a similar argument, H = B for some i such that 0< i 5p.
Suppose that B = H. and j >1. Then
{b0,b ,1...,b n 1} $ {h0,h ,...,)h }
which is a proper subset of {hh1,...,hm }. But
{h0 ,h1 ,..., hm } = {b0 ,b1 ,...,bn }
Therefore, {b 0 ,b1 ,...,b } is a proper subset of {b 0,b1,...,b }.
However, 1 1. This is a contradiction. Hence, H = B
There exists, b & {b0 , ., .,b n} such that b {b 0 ,...,bn .
92
Since B=H1, b E{h0,''.,hm} and b {h0 ,...,h '
Therefore, if x X Xt andx t
x =XA B + . .+ X B =6 H + .0.0 + 6tH'0. Q +PApB 0 0 t t3
then A0 =60*1
Suppose that for each i< j, B = Hi and if x 6 X nxtsuch that
x=AB+ . AB 6H + ... + 6H0p p 0 O t t3
then A = Suppose that B7 H. Let x csX
such that x = A B + .. + AB and A. 0. Then,0 0 p p
px = ( ZKm)b0 + ... + ( ZK)b
m=0 m=0 p
p-1+(K)b + +Kb
m=0 m n+1 0 n +1M- p 1
+,.+ K 0bn 0a
If x E Xt, then x = 60H0 + ... + 6t H. Hence,
t tx = CL )h
L= m)0 m= mZ m t
t"-1+ (2Lm)h + ... + L0h
M0m mt+I m+1
+ 0.w. + L 0h m 0
If Case 1
some i1 4j
1is true, then the coefficient of b is nD K for
and Z Lm for some s <4. Hence,m=0
0+1K+ ... + K + ... + K = L +
93
Since s <j, K = 0. Therefore, A. = 0. But A. # 0.
Consequently, Case 2 must be true. Hence, the coefficient
i 3of b is t-Lm for some i j and Z Km for some s< j.
m=m m=O
Therefore,
KO + . +Ks =L0 +... + L + ... +L
Since s <j, L = 0. Hence, 6 = 0. By a similar argument,
if B. / H. , then 6 = 0.3 j +l' +1Suppose that if B. H. for 0 s< i then 6 = 0.
3 3+s j+5Suppose that B. :Hj+. If Case 1 is true, then the co-
kefficient of b is Z K for some k >j and r L for some
p=OP p= 0 P
k <j +i. Hence, K. = 0. Therefore, A. = 0. But A. 0.
Then, Case 2 must be true. Consequently the coefficient of
kb must be L for some k tj + i and 2 K for Z< j + i.
s=0 s=0
Therefore,
L0 + L1+... + L. + + LS 1j+i kK +K 1 + ... + K
Hence, L = 0. As a result, 6 = 0. Therefore, by
induction, for all sk j, 6 = 0. Hence, 6 + . + 6 = 1.
Therefore, A0 + A1 + ... + = 1. Hence A. =0. This
is a contradiction to A # 0. Therefore, x [Ho, H1 ,...,Ht].
Consequently, B is not a necessary vertex of [Bo,B1 ,...,B ].
This is a contradiction. Therefore, B. = Hk for some kQ.
By a similar argument, H. = B for some k. The same3 k
94
argument that proved H = B1 will hold for H. BJ Je
Therefore,
[BOB 'EEp] = [HO,Hi, ...,Ht].
It now follows that K is a complex.
Now, K1 is called the barycentric subdivision of K.
It is noteworthy to point out that if K is the barycentric
subdivision of K, then UK = UK.
Theorem 3.3: If T = [AO,A**..,Ap] is a simplex in
En where d is the usual metric on En, then the diameter
of T is the maximum {d(A.,A.)I0 i p,o j :5p}.
Proof: Let
D = maximum {d(Ai,A )o5isp and Oj p}.
Let x and y be in T. Let A be a vertex farthest from y
and d(y,Ai) = 6. Then A. 6 S(y,6), the spherical neighbor-
hood around y of radius 6, for each j. Consequently,
TcS(y,6). If x T, d(x,y) d(y,A1 ). Let A. be vertex
farthest from A and X = d(Ai,A .). Then Tc S(AiA).i -3s
Therefore, d(y,A) 5d(A1 ,A3) :D. Consequently,
d (,y) :5 d (y, A 1) 5 d(A ,A )< D.
Hence, the theorem is proven.
Theorem 3.4: If K = [A0,A , *..,Am] is a simplex of
diameter D in En, K1 is the barycentric subdivision of K
and D is the diameter of K then D !5 D.1' lm+l
Proof: Let d be the usual metric on En. By Theorem
3.3, D1 is. the maximum of the distances between the vertices
95
of K 1, Let D= d(B.,B.) where i sj, E1 is the barycentre
of [AQ,A1 ,...,A1 ] and B. is the barycentre of [A05,A ,...,A].
Then, B A and B = A for 1 <r <n. Therefore,
n i
d(B.,B ) = ( (k ZAr i :Ax) 2)2r=1 i+lp=Q P 3 +1p=p
n1C 2 (( -) Ar 2 A2)r=1 + + p=0 P + 1 p=I+lP
n'_r__ 1 r 2 2
r=1 (+1)(j+1) p jp=+
n1= 2 7' ( (U) A r_ 1 Ar)2 2
n1. p jr=1 j+1 1+=p=0 p=I+1
r=l j+1 1+1p= Ap=+In= i11=:(Li('E Ar Ar 2
r= +1 +1 OP p ilp
ar1 +p=0 p=+1n 1 1
However, (+)=1, () (j-1) and AIs a vertex of K
for each 1 sj. Consequently, x = (x 2 ,..,n) where-
xI =l2 Ar, and y = (y 2yy )Ewherey =k Ar
p=r = 3 p=I+1
are elements of K, Hence,
n 1(C , Ar 1 Ar 2 ) 2 2D.
r=1 +1_ 0 ~ =1+1
96
Therefore, dCB ,B ) .5j-'D. But, j-i 5j and j 5m. As a
result, d lBB}<-D <m+1,D
Dieflhition 3.4. If K is a complex in En, then the
mesh of K is the maximum of the diameters of simplices of K.
Theorem 3.5: If{K }00 is a sequence of complexesi 1=1
such that for each i, K is the barycentric subdivisioni+lof Ki and M is the mesh of Ki, then{Mi}+0 as i+oo.
00Proof: Suppose that {K i}j 1is a sequence of complexes
satisfying the hypothesis and that the mesh of K. is M .1 i.
Let k = [AOA1 ,. .,Am] be a simplex of K of diameter M1
such that k has a maximum number of vertices. Then, if
k2 is a simplex of K2 , by Theorem 3.4, the diameter of
km M Hence, M2M ml M Furthermore, k has at
most m vertices. If k3 is a simplex of K3 , then the dia-
meter of k is less than or equal to m M Hence,3 m+l 2 H
M3 mE M2 By induction, M m:l(_ n-lM Since N is a
constant and ( 0n+s as n+oo, {M1}I 1-*0 as i+o.
This paper will assume without proof that if N is a
finite indexing set, then I= N I where I is theaoN 6?a
closed unit interval for each a, is a polyhedron.
Theorem 3.6; If N is a finite indexing set, F is a
closed subset of IN and U is an open subset of IN such that
FE Uc IN, then there exists a polyhedron P such that
Fc POc P c U.
97
NProof: Since EU and Iis a compact metric, there
exists an open subset V of IN such that F cVcVLU. Now,
since IN ia assumed to be a polyhedron, there exists a
complex H such that IN =UH. Let {H}L1 be a sequence
of complexes such that H = H and H.+ is the barycentric
subdivision of H . For each i, let Mi denote the mesh of
NNH, Since V SU, VU(I \U) = $ and each of 7 and IN\U is
closed in IN. Let (IN\U) = W. There exists 6> 0 such
that if x e V and y c W, d(x,y) 6, where d is the usual
Nmetric on IN. There exists j e Z+ such that M4. <6. Let
R = {S|S is a simplex of H. and S n7 }
Then R covers V. Let
K = {sjs is a face of some S c R}.
Now, K is complex, UK is a polyhedron and V. UK. Suppose
that x e UK and x U. Then x c W and there exists s e K
such that x e s. Since s c K, s is a face of some S c R.
Therefore, SqV v 4 and x E S. Let y F VM S. Then, since
y E 7 and x E W, d(x,y) 6. However, x E S, y c S and
S E R. Since S 6 R, S is a simplex of H.. Therefore, the
diameter of S is less than 6. Hence, d(x,y)< 6. This
contradicts the statement that d(x,y) 6. Consequently,
x e U. As a result, UK.9U. Now, F1V. V U K SU. It
now follows that F c UKOC U K.SU.
Theorem 3.7; If Y is a compact Hausdorff space, then
Y is homeomorphic to the inverse limit of an inverse system
of polyhedra.
98
Proof; Let Y be a compact Hausdorff space. Then Y
is homeomorphic to a subspace of H I , denoted by IAcA cA
where I is the closed unit interval and A is some indexing
set. Denote the homeomorphic image of Y in IA as X. Then,
AX is a closed compact subset of IA. Let
13 ={B|B is a nonempty finite subset of A}.
Now, for B and D in/s, B 3 D if and only if IB -riD
For each B e '1, let fAB1 IA÷B be defined by
f A({x a} :) = {xa } pBfAB caoacA a cwB'
Then, fAB is continuous and onto for each B cS and fAB
Bis a closed subset of I By Theorem 3.6, there exists P,
a polyhedron, such that fAB(X) CPO cP CB. Let
9 = {P IP is a polyhedron such that for some B c 1,
Bf AB (X) CP CPCI }
Let &be an indexing set for . If A e.6, and B and B
are elements of /3 and
fAB(-Q. ciB
and
fAB(X) _P _oP _IB
then B = B . If not, IB 1 B p. Therefore, for each
A & ,, there is a unique B c 8 such that fABXCPOC CIB
Let
E C= {(AB)(A,B) x
and fAB(X) O LIB
99
Then, E is an indexing set for P. Let (XB) c E, be
denoted by ,X 2
If C-B-CA, where C and R are elements of /3, then
define C +f Iby
fBC U a aE:B = a aeC'
Then each fBC is continuous. If x = {x a} A 1A, then
fAC(X) = {xaaC
However, fAB(ax) = aEB and f(BCx a IaB = xa aC. Hence,
fAC =fBC 0AB
Direct E in the following manner. Let rB and 6D be
elements of E. Then, B and D are unique in /3 such that
FAB * B WB cIB
and
f (X) gP0 O p LAD 6 D DcID
Now, define B 6 D if and only if B cD and fDB(6D B
Suppose that AD e E. Then DOCD and
D D D
Therefore, XD XD
Suppose that X V 5% and B 56 D. Then,
fAF(X) CPF FcIF
f AB (X) g P -P 01fABB SB
and
AD 6D 6D
100
Now, since FOCB and BSCD, F!D. AlsPo, D )D and
fBF CPO P. Consequently, fDF(P ) cP . It thenB y D ~ F4
follow-s that
The remaining property, necessary to show that "t<"
is a direction for E, is that of a common follower for two
elements of E. The following lemma will make this easy
and be helpful later.
Lemma 3.2: If 0B is in E and D is in / such that
B _cD, then fAD(X fDB B(P )
Proof: Suppose that x c fAD(X) and x f DB' B
Now, fDB (x) EfDB(fAD(X)). Therefore, fDBx) fAB(X).
But, x f DB B). Hence, fDB(x).P B Consequently,
fDB (x) fAB(X). This is a contradiction to fDB(x ) fAB .As a result, x fD B)( . It then follows that
fAD fDB( B
Now, to complete the proof that " " is a direction
for E, let XF and B be elements of E. There exists D c 1
such that B9 D and FcZD. By Lemma 3.2,
DB DE
which is open in ID. By Theorem 3.6, there exists a poly-
hedron P such that
f D(X)BP P BfI(PO )qf ( O CDAD DB B DF XF
101
Now, P e 9 and must be indexed by some 6 in E. SinceDP I , M =D, ThereforeD E.B ,fB6D E BtD B(BP
f (P)c P0 Fc D and BED. Therefore, XF 53 D and
B :6D. Consequently, " " is a direction for E.
Let PB<D in E. Then BcD. If B = D, let
f B PD -*PB be the inclusion map. If B is a proper
subset of D, let
f P + P .bef DBI .D B D B Dp
Now, (PsB f 6 ,E) is an inverse system.
Let x = {xa} A E X. For P DP
f (X) PO c cFAD D f cf D ADD D 6Dt D
AD~ ~ a c DD
Let x be the element in II P such that for each 6D D6E,FSE F
the 6D-th
If B 6D
P6D x
projection of x , P (x), is fAD ({X aA)
in E, then (xB )=XfAB{xa} A) and
fAD ix} . Consequently,
DB % Cx )) = (f (x} )AD a DaDB D caA
Since fAD( a{a} A D
and
102
f (f'({1x} ))6 D B' AD- a ac A
= f DB (fAD a XTatA)
AB QA6A)x )-B
Therefore, x s X = lim(P6 f ,E).00 +66IE3 )
Let g be a relation from X to X00 defined by g(x) X
and =D DX6Suppose that x = y in X where x = {x }
and y = {y }a Then, g(x) = x is such that
Tr1 (x ) =f ({x } )7 B AB aaEA
and g(y) = y such that
Y
BAB)=
a aA
for each 1B in E. Since x = y, x = y for each a cA.
Hence,
fAB({Xa}EA) AB aaAfor each B F-8. Therefore, 7TB B(X) = (y ) for each B'
t IConsequently, x = y . As a result, g is a function.
Suppose that x A y in X. Then there exists a c A
such that x aXy. Let B c 1 such that a e B. Then,
fAAB B) There exists XB c E such thatff (X) CAOThereCB
fABE#X) B- AB
B BBConsequently., . A 3 x ) X 7 X(y ).Therefore,, g(x) 74.g(y).
Hence, g is one to one.
Let x e X0 and 1B e E. Suppose that 7rT13 (x) fAB(X).
Then there exist disjoint open sets U and V of IB such that
103
fA(X)cU and 7rBCx) e V. Hence, fB(xX) unpo which isAB ABB
open in IB. Consequently, there is S B c E such that
fAB( X) c ~ycP , cUnp ,B B B
No w, TrB) B.However, BoB and fBB(PVB B P PB * B B B3? B
Therefore, B:. B. Since x X. and B:5 B'
Tr (x) = f I (Tr'B(x)).
Since B = B, f BOB is the inclusion map. Consequently,
f ' (7T'(x)) = Tr ' (x).
f OB B B W BTherefore,
Tr (x) = r (x) P .
Hence, )rX e P cU rP B.Now, T B(x) c U andB B B
7T rB(x) 6 V. This contradicts the statement that Uq- = p.
Therefore, if x E X, Tr B (X) ,fAB(X) for each B F E.
Suppose that there does not exist y c X such that g(y) = x.
Then for each y c X, there exists r3B E such that
fAB (y) B (x). Therefore, there exists Uy SB and Vy 0BBopen and disjoint in I such that f AB(y) Uy and
Tr (x) E V .Hence,y B
( =.{fAB 1(Uy)|y c X} covers X.
104
There exists a finite subcollection H of ?( covering X.
Let
H ='{fAD-(uD)1 6D D
where 9 is a finite subset of E. There exists M _1 such
that B cM for each B s ' By Lemma 3.2, fAN(X)cf ~1(PO )y ema 32, fAM (X-. fMB BB
for each .B . Let R = n fMBXP0 ) which is open in
I Then, fAMCX).CR. There exists am E such that
fAM(X) cP 0 c:P CR.AM m- m'
Consequently, fMB am B and B cM for each B e
Hence, B5m for each B c7. As a result,
7T r (x) = f GB(7Tra M(x))770B MOB MN
for each B c7. Now, since 7rM (x)cfAM(X), there exists
z c X such that fAM(z) = 7TrM(x). But z fAD(U6 D)for
some 6D 69 .- Therefore, fAD(z) U6 . However,
(6D) = 6D M
fM6 D fAM(z)) = fMD 0 fAM(z)
fAD(z).
Consequently, UD nv6D . This is a contradiction to
U 8 V = . As a result, there is y c X such that g(y) =x.HD D
Henc e, g is onto .
105
Suppose that y 6 X and that U is open in X., such that
g(iy) c U. Then there exists a basic open neighborhood K
such that g(y) K.U., Let K = 7r 4 (U6 ) NowD D
Tr, (g(y)) 6 U6 and fAD is continuous. Therefore, there
exists V open in X such that y e V and fAD (V) ;;U 6 ID f
z E V, fAD(z) E U6 .*Therefore, g(z) c K. Consequently,
g is continuous. Since X is compact Hausdorff, XO is
Hausdorff and g is a one to one continuous function from
X onto X,, g is a homeomorphism. Now, Y is homeomorphic
to X,, the inverse limit of an inverse system of polyhedra.
CHAPTER BIBLIOGRAPHY
1. Nagata, Jun-iti, Modern General Topology, Amsterdam,North Holland Publishing Company, 1968.
2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.
106
BIBLIOGRAPHY
Capel, C. E., "Inverse Limit Spaces," Duke MathematicalJournal, 21 (1954), 233-245.
Gentry, K. R., "Some Properties of the Induced Map,"Fundamenta Mathematicae, LXIV (1969), 55-59.
Nagata, Jun-iti, Modern General Topology, Amsterdam,North Holland Publishing Company, 1968.
Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.
Whyburn, Gordon T., "Analytic Topology," American Mathe-matical Society Colloquium Publications, 23 (1942).
Wilder, Raymond L., "Topology of Manifolds," AmericanMathematical Society Colloquium Publications,32 (19' 9) .
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