real analysis - math.iitm.ac.in

Real Analysis∗

M.T.Nair

Contents

1 Set theoretic Preliminaries 3

2 Real Number System 5

3 Completeness of R 6

4 Metric spaces: Basic Concepts 9

4.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.2 Topological concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Completeness 19

6 Compactness 25

7 Connectedness 30

8 Continuity 32

8.1 Definition and characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

8.2 Relation with compactness and connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8.3 Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

9 A Few Important Theorems 38

9.1 Contraction mapping theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

9.2 Baire category theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

9.3 Arzela-Ascoli’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

9.4 Weierstrass approximation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

10 Power Series 44

∗The course MA5330: Real Analysis, July-Novemebr, 2016

1

11 Fourier Series 48

11.1 Trigonometric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

11.2 Trigonometric series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

12 Assignments 53

12.1 Assignment - I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

12.2 Assignment - II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

12.3 Assignment - III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

12.4 Assignment - IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

12.5 Assignment - Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2

1 Set theoretic Preliminaries

1. For sets A and B,A ⊆ B ⇐⇒ “x ∈ A⇒ x ∈ B”,

A ∪B := x : x ∈ A or x ∈ B,

A ∩B := x : x ∈ A and r x ∈ B,

2. For sets A1, A2, . . . , An,

n⋃i=1

Ai := A1 ∪ · ∪An := x : x ∈ Ai for some i = 1, . . . , n,

n⋂i=1

Ai := A1 ∩ · ∩An := x : x ∈ Ai for every i = 1, . . . , n,

3. Let Λ be a nonempty set. For each α ∈ Λ, let Aα be a set. Then

n⋃α∈Λ

Aα := x : x ∈ Aα for some α ∈ Λ,n⋂

α∈Λ

Aα := x : x ∈ Aα for every α ∈ Λ.

4. Let S be a nonempty set. Any subset of S × S is called a relation on S.

5. Let R be a relation on S. If (x, y) ∈ R, then we write xRy and say x is related to y.

6. Let R be a relation on S.

(a) R is said to be reflexive if xRx for every x ∈ S.

(b) R is said to be symmetric if for x, y ∈ S, xRy ⇒ yRx.

(c) R is said to be antisymmetric if for x, y ∈ S, xRy& yRx⇒ x = y.

(d) R is said to be transitive if for x, y, x ∈ S, xRy& yRz ⇒ xRz.

7. Let R be a relation on S.

(a) R is said to be an equivalence relation if it is reflexive, symmetric and transitive.

(b) R is said to be a partial order if it is reflexive, antisymmetric and transitive.

8. An equivalence relation is usually denoted by ∼, and a partial order is usually denoted by .

9. Given a partial order on S, and x, y ∈ S, we write x ≺ y if x ≺ y and x 6= y.

10. Let (S,) be a partially ordered set and A ⊆ S. Then A is called a totally ordered subset if forevery x, y ∈ A, either x y or y x.

11. Given an equivalence relation ∼ on S and x ∈ S, the set

[x] := y ∈ S : y ∼ x

is called the equivalence class of x.

12. Given an equivalence relation ∼ on S and x, y ∈ S,

[x] 6= [y] ⇐⇒ [x] ∩ [y] = ∅.

3

13. Given an equivalence relation ∼ on S, S =⋃x∈S

[x]. Thus, S is a disjoint union of equivalence

classes.

14. Given a partial order on S and A ⊆ S, and element x ∈ S,

(a) x is called an upper bound of A if a x for every a ∈ A.

(b) A is said to be bounded above if an upper bound exists for A.

(c) x is called a maximal element S if y ∈ S and x y implies y = x

(d) An element x ∈ S is called a least upper bound of A if x an upper bound of A and if y isany upper bound of A, then x y; equivalently, x an upper bound of A and if v ≺ x, thenthere exists w ∈ A such that v ≺ w.

15. Given a partial order on S and A ⊆ S, and element x ∈ S,

(a) x is called a lower bound of A if x a for every a ∈ A.

(b) A is said to be bounded below if a lower bound exists for A.

(c) x is called a minimal element S if y ∈ S and y x implies y = x

(d) An element x ∈ S is called a greatest lower bound of A if x a lower bound of A and if y isany lower bound of A, then y x; equivalently, x is a lower bound of A and and if x ≺ v,then there exists w ∈ A such that w ≺ v.

16. A relation ≺ on a set S is called an order relation if it is transitive and for every x, y ∈ S, oneand only one of the following true: x ≺ y, x = y, y ≺ x.

Exercise 1. Let (S,) be a partially ordered set and A ⊆ S.

1. Least upper bound of A, if exists, is unique.

2. Greatest lower bound of A, if exists, is unique.

3. If ≺ is an order relation on S, then defined by x y for either x ≺ y or x = y, is a partialorder.

♦

4

Example 2. Let us consider a few examples:

1. Let S = N× N and for (m,n), (p, q) in S, define

(m,n) ∼ (p, q) ⇐⇒ mq = np.

Then ∼ is an equivalence relation on S. In this case the equivalence class of (m,n) is denotedby m

n . The set of all equivalence classes is th set of all positive rational numbers.

2. Let n ∈ N. For a, b ∈ N, define a ∼ b ⇐⇒ a − b is a multiple of n. Then ∼ is an equivalencerelation on N.

3. Let G be a group and H be a subgroup of G. For a, b ∈ G, define a ∼ b ⇐⇒ a− b ∈ H. Then∼ is an equivalence relation on S.

4. For a nonempty set X let S = 2X . For A,B in S, define A B ⇐⇒ A ⊆ B. Then is apartial order on S.

5. For a, b ∈ R, define a b ⇐⇒ a ≤ b.Then is a partial order on R. In this case A := x ∈ R :x ≥ 0 has no upper bound, but 0 and every negarive number is a lower bound.

6. Let S = reiθ : 0 ≤ θ < 2π, 0 ≤ r ≤ 1. For z1 = r1eiθ1 , z2 = r2e

iθ2 in S, define

z1 z2 ⇐⇒ θ1 = θ2 and r1 ≤ r2.

Then is a partial order on S. In this case,

(a) for each θ, the set Aθ := reiθ : 0 ≤ r ≤ 1 is a totally ordered subset for which eiθ is anupper bounded.

(b) Every eiθ is a maximal element of S.

♦

2 Real Number System

We shall denote the set of real numbers by R.

1. An ordered field F is a field along with an order relation ≺ such that

∀x, y ∈ R, x ≺ y ⇒ x+ z ≺ y + z ∀z ∈ R,

∀x, y ∈ R, x ≺ y ⇒ xz ≺ yz ∀z ≺ 0.

Recall that if ≺ is an order relation, then defined by x y ⇐⇒ either x ≺ y or x = y is apartial order.

2. An ordered set is said to have least upper bound property if every every nonempty subset of itwhich is bounded above has the least upper bound.

3. A complete ordered field is an ordered field F such that every subset of F which is bounded abovehas least upper bound.

THEOREM 3. The field Q does not have least upper bound property.

5

Proof. Consider the set A = x ∈ Q : x2 < 2. Clearly A is nonempty and it is bounded above. Weshow that for every x ∈ A, there exists y ∈ A such that x < y.

Let x ∈ A with x > 0, and let y = 2x+2x+2 . Note that

x < y ⇐⇒ x(x+ 2) < 2x+ 2 ⇐⇒ x2 < 2 ⇐⇒ x ∈ A

and

2− y2 = 2− (2x+ 2)2

(x+ 2)2=

2(x+ 2)2 − (2x+ 2)2

(x+ 2)2=

2(2− x2)

(x+ 2)2

so that y2 < 2 ⇐⇒ x2 < 2 ⇐⇒ x ∈ A. Thus, we have proved that x ∈ A implies y ∈ A andx < y.

Remark 4. The proof of Theorem Th-lub-1 shows that if we define B = x ∈ Q : x2 > 2, then Bdoes not have a greatest lower bound in Q. ♦

Exercise 5. For x > 0 in R, let y = f(x) := 2x+2x+2 . Prove the following:

(i) y = x ⇐⇒ x2 = 2.

(ii) Let x1 = 1 and for n ∈ N, let xn+1 = f(xn). Then xn < xn+1 for all n ∈ N and xn →√

2. ♦

THEOREM 6. There exists a complete ordered field containing Q as a subfield, and any two completeordered fields are isomorphic.

Here are a few important consequences of the lub property of R.

THEOREM 7. (Archimedean property) If a, b ∈ R with a > 0, then there exists n ∈ N suchthat na > b.

Proof. Let a > 0, b > 0. Suppose there is no n ∈ N such that na > b. Then na ≤ b for all n ∈ N.That is, S := na : n ∈ N bounded above by b. Let β := supS. Then β − a < β so that there existsn ∈ N such that β − a < na. Thus, β < (n+ 1)a, which contradicts the definition of β.

THEOREM 8. (Denseness of Q) If a, b ∈ R with a < b, then there exists r ∈ Q such thata < r < b.

Proof. Let a, b ∈ R with a < b. Then b − a > 0. By the Archimedean property, there exists n ∈ Nsuch that n(b− a) > 1. Hence there exists m ∈ N such that na < m < nb. Thus, a < m

n < b.

3 Completeness of R

Definition 9. A sequence (xn) of real numbers is said to be convergent if there exists x ∈ R suchthat for every ε > 0, there exists n0 ∈ N such that |xn − x| < ε for all n ≥ n0, and in that case, wewrite

xn → x as n→∞ or xn → x or limn→∞

xn = x.

♦

1. Every convergent sequence is bounded: if xn → x, then there exists α > 0 such that |xn| ≤ αfor all n ∈ N.

6

Proof. Suppose xn → x. Let n0 ∈ N be such that |xn − x| ≤ 1 for all n ≥ n0. Then

|xn| ≤ |xn − x|+ |x| ≤ max1 + |x|, M, M := max|xk| : k = 1, . . . , n0.

2. Every convergent sequence is a Cauchy sequence, i.e., for every ε > 0, there exists n0 ∈ N suchthat |xn − xm| < ε for all n,m ≥ n0.

Proof. Suppose xn → x and ε > 0. Let n0 ∈ N be such that |xn−xm|〈ε for all n,m ≥ n0. Then

|xn − xm| ≤ |xn − x|+ |x− xm| ≤ 2ε ∀n,m ≥ n0.

THEOREM 10. Every Cauchy sequence in R converges.

As steps towards proving the above theorem, we prove some lemmas.

LEMMA 11. Suppose (xn) is a Cauchy sequence in R and S = xn : n ∈ N. If S is a finite set,then (xn) is eventually constant, and hence it converges.

Proof. Suppose S is a finite set, say S = α1, . . . , αk. Assume for a moment that (xn) is noteventually constant. Then, for any nN ∈ N there exists n,m ≥ N such that xn 6= xm. But, for suchn,m,

|xn − xm| ≥ min|αi − αj | : i, j = 1, . . . , k, i 6= j = d > 0.

This is not possible, because (xn) is a Cauchy sequence. Thus, either (xn) is eventually constant orelse S is an infinite set.

LEMMA 12. Suppose (xn) is a Cauchy sequence in R. Then S is bounded.

Proof. Let n0 ∈ N be such that |xn − xm| < 1 for all n,m ≥ n0. Hence,

|xn| ≤ |xn − xn0 |+ |xn0 | < 1 + |xn0 | ∀n ≥ n0.

Thus,|xn| ≤ max1 + |xn0 |,M+ |x|, M := max|xk| : k = 1, . . . , n0.

LEMMA 13. Suppose (xn) is a Cauchy sequence in R. If (xn) has a subsequence which convergesto some x ∈ R, then (xn) converges to x.

Proof. Suppose there exists a subsequence (xnk) of (x− n) such that xnk

→ x as k →∞. Let ε > 0be given, and let k0 ∈ N be such that |xnk

− x| < ε for all k ≥ k0. Since nk ≥ k for all k ∈ N, we have

|xk − x| ≤ |xk − xnk|+ |xnk

− x| ≤ |xk − xnk|+ ε ∀ k ≥ k0.

Also, let k1 be such that |xn − xm| < ε for all n,m ≥ k1. Then |xk − xnk| < ε for all k ≥ k1. Thus,

|xk − x| ≤ |xk − xnk|+ ε < 2ε ∀ k ≥ maxk0, k1.

7

LEMMA 14. Suppose (xn) is a Cauchy sequence in R. Then (xn) has a convergent subsequence.

Proof. Let S = xn : n ∈ N. If S is a finite set, then we know by Lemma 11 that (xn) itself converges.Therefore, assume that S is an infinite set. By Lemma 12, S is bounded.

Let S ⊆ I0 := [a, b]. Let c = (a+ b)/2. Since S is an infinite set, either S ∩ [a, c] is an infinite setor S ∩ [c, b] is an infinite set. Let

I1 := [a1, b1] =

[a, c] if S ∩ [a, c] is an infinite set[c, b] if S ∩ [a, c] is not an infinite set

.

Thus, S ∩ I1 is an infinite set. Next let c1 = (a1 + b1)/2. Then either S ∩ [a1, c1] infinite set orS ∩ [c1, b1] is an infinite set. Let

I2 := [a2, b2] =

[a1, c1] if S ∩ [a1, c1] is an infinite set[c1, b1] if S ∩ [a1, c1] is not an infinite set

.

Thus, S ∩ I2 is an infinite set. Continuing this process, we obtain closed intervals I1, I2, . . . such that

1. I1 ⊃ I2 ⊃ I3 ⊃ · · · , with E ∩ Ik is an infinite set for every k ∈ N,

2. `(Ik+1) = `(Ik)/2 for all k ∈ N so that `(In) = `(I0)/2n, n ∈ N.

Note also thata ≤ ak ≤ ak+1 ≤ bk+1 ≤ bk ≤ b ∀ k ∈ N,

Thus (an) is bounded above and (bn) is bounded below. Let

α := supnan, β := inf

nbn.

Then ak ≤ α ≤ β ≤ bk or all k ∈ N. Since

bk − ak = `(I0)/2k → 0 as k →∞,

it follows that α = β =: γ, say, and ak → γ and bk → γ.

Since each S ∩ [ak, bk] is an infinite set, xnk∈ S ∩ [ak, bk] can be chosen such that

nk ≤ nk+1 and xnk+16∈ xn1 , . . . , xnk

∀ k ∈ N.

Now, for ε > 0, let k0 ∈ N such that ak, bk ∈ (γ − ε, γ + ε) for all k ≥ k0. Then

xnk∈ [ak, bk] ⊂ (γ − ε, γ + ε) ∀ k ≥ k0.

Thus xnk→ γ as k →∞.

Proof of Theorem 10. Follows by Lemma 13 and Lemma 14.

A particular case the following result has already been used in proving Lemma 14

THEOREM 15. (Nested interval theorem) If (In) is a decreasing sequence of closed and boundedintervals, then ∩∞n=1In is nonempty. Further, if `(In)→ 0, then ∩∞n=1In is a singleton set.

8

Proof. Let In = [an, bn], n ∈ N. Since In ⊇ In+1 for all n ∈ N, an ≤ an+1 ≤ bn+1 ≤ bn for all n ∈ N.In particular, (an) is bounded above by b1 and (bn) is bounded below by a1. Let

a := sup an, b = inf bn.

Then an ≤ an+1 ≤ a ≤ b ≤ bn+1 ≤ bn for all n ∈ N (Why a ≤ b?). Thus, a, b ∈ ∩∞n=1In. Clearly, if`(In)→ 0, then bn − an → 0 so that a = b, and an → a, bn → b.

Exercise 16. If E ⊆ R is a bounded infinite set, then E contains a convergent sequence with distinctentries. ♦

Exercise 17. Let (xn) be a sequence in R. Prove the following:

(i) If (xn) is monotonically increasing and bounded above, and if x := supxn : n ∈ N, then (xn)converges to x.

(ii) If (xn) is a Cauchy sequence having a subsequence which converges to x, then (xn) alsoconverges to x.

(iii) if an → x, bn → x and if an ≤ xn ≤ bn for all n ∈ N, then xn → x.

(iv) If an ≤ bn for all n ∈ N and an → a and bn → b, then a ≤ b. ♦

Exercise 18. (i) Suppose (an) is such that |an+2−an+1| ≤ r|an+1−an| for all n ∈ N, where 0 < r < 1.Then (an) converges.

(ii) Suppose a1 = 1 = a2 and an+2 = an+1 + an and bn = an+1/an for all n ∈ N. Then (bn)converges to (1 +

√5)/2. ♦

4 Metric spaces: Basic Concepts

4.1 Definition and Examples

Definition 19. Let X be a non-empty set. A function d : X ×X → R is said to be a metric on Xif it satisfies the following properties:

(i) d(x, y) ≥ 0 ∀x, y ∈ X;

(ii) x, y ∈ X d(x, y) = 0 ⇒ x = y;

(iii) x, y ∈ X ⇒ d(x, y) = d(y, x);

(iii) x, y, z ∈ X ⇒ (d(x, y) ≤ d(x, z) + d(z, x).

A set together with a metric on it is called a metric space. ♦

Exercise 20. For every x, y, z is a metric space with metric d, |d(x, z)− d(y, z)| ≤ d(x, y)|. ♦

Example 21. (x, y) 7→ d(x, y) := |x− y| is a metric on R. ♦

Example 22. Let X be a set. For x, y ∈ X, let d(x, y) = 0. Then d is a metric on X, called theindiscrete metric. ♦

9

Note that, using Cauchy-Schwarz inequality,

‖x+ y‖2 =

n∑k=1

|xk + yk|2 =

n∑k=1

(|xk|2 + |yk|2 + 2|xkyk|

= ‖x‖2 + ‖y‖+ 2

n∑k=1

|xkyk| = ‖x‖2 + ‖y‖+ 2‖x‖ ‖y‖

= (‖x‖+ ‖y‖)2.

Thus, ‖x+ y‖ ≤ ‖x‖+ ‖y‖.

Example 27. For (x1, . . . , xn) and (y1, . . . , yn) in Kn, let

d(x, y) :=

(n∑k=1

|xk − yk|2) 1

2

.

The (Kn, d) is a metric space: First three conditions hold easily. To see the third condition, letz = (z1, . . . , zn) ∈ Kn. Then, by triangle inequality,

d(x, y) = ‖x− y‖ = ‖(x− z) + (z − y)‖ ≤ ‖x− z‖+ ‖z − y‖ = d(x, z) + d(z, y).

♦

Example 28. Let C[a, b] be the set of all real valued continuous functions defined on [a, b]. Recallthat every x ∈ C[a, b] is a bounded function and it attains a maximum at some point in [a, b]. Forx, y ∈ C[a, b], let

d(x, y) := maxa≤t≤b

|x(t)− y(t)|,

d1(x, y) :=

∫ b

a

|x(t)− y(t)|dt,

d2(x, y) :=

(∫ b

a

|x(t)− y(t)|2dt

) 12

.

Then d, d1, d2 are metrics on C[a, b]: It can be easily verified that d, d1 are metrics. The fact that d2

is also a metric will follow from the Triangle inequality proved below. ♦

For obtaining Triangle inequality, one may use Cauchy-Schwarz inequality.

THEOREM 29. (Cauchy-Schwarz inequality) For x, y in C[a, b],

∫ b

a

|x(t)y(t)| dt ≤

(∫ b

a

|x(t)|2dt

) 12(∫ b

a

|y(t)|2dt

) 12

.

Proof. Use the arguments as in Theorem 25.

THEOREM 30. (Triangle inequality) For x, y in C[a, b],(∫ b

a

|x(t) + y(t)|2 dt) 1

2 ≤(∫ b

a

|x(t)|2dt) 1

2

+(∫ b

a

|y(t)|2dt) 1

2

.

11

Proof. Use the arguments as in Theorem 26.

From the above theorem, we deduce:

Example 31. For a non-empty subset, let B(S) be the set of all real valued bounded functions definedon S. For x, y in B(S), let

d(x, y) := sup|x(s)− y(s)| : s ∈ S.

Then d is a metric on B(S) (Exercise).

Taking S = N, we see that B(N) is the set of all bounded sequences. This metric space B(N) isusually denoted by `∞. ♦

Example 32. Let

`1(N) := (an) ∈ `∞(N) :

∞∑k=1

|ak| converges,

`2(N) := (an) ∈ `∞(N) :

∞∑k=1

|ak|2 converges.

For x, y ∈ `1(N), let

d1(x, y) :=

∞∑k=1

|xn − yn|,

and for x, y ∈ `2(N), let

d2(x, y) :=

( ∞∑k=1

|xn − yn|2) 1

2

.

Then (`1(N), d1) and (`2(N), d2) are metric spaces:

(i) Let x = (x1, x2, . . .), y = (y1, y2, . . .) and z = (z1, z2, . . .) be in `1. We know that, for everyn ∈ N,

n∑k=1

|xk − yk| ≤n∑k=1

(|xk − zk|+ |zk − yk) ≤∞∑k=1

|xk − zk|+∞∑k=1

|zk − yk|.

Letting n→∞, we obtain d1(x, y) ≤ d1(x, z) + d1(z, y). Other axioms for d1(·, ·) to be a metric canbe easily verified.

(ii) Let x = (x1, x2, . . .), y = (y1, y2, . . .) and z = (z1, z2, . . .) be in `2. Since (x1, . . . , xn), (y1, . . . , yn)and (z1, . . . , zn) are in Rn, by the triangle inequality in Example 27, we have( n∑

k=1

|xk − yk|2) 1

2 ≤( n∑k=1

(|xk − zk|2) 1

2

+( n∑k=1

|zk − yk|2) 1

2

≤( ∞∑k=1

(|xk − zk|2) 1

2

+( ∞∑k=1

|zk − yk|2) 1

2

.

This is true for all n ∈ N. Consequently,( ∞∑k=1

|xk − yk|2) 1

2 ≤( ∞∑k=1

(|xk − zk|2) 1

2

+( ∞∑k=1

|zk − yk|2) 1

2

.

Thus, we have proved d2(x, y) ≤ d2(x, z) + d2(z, y). Other axioms for d2(·, ·) to be a metric can beeasily verified. ♦

12

Definition 33. Let X be a vector space over R. A map x 7→ ‖x‖ from X to the set of all non-negativereal numbers is called a norm on X if

1. for every x ∈ X, ‖x‖ = 0 ⇐⇒ x = 0,

2. ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X, and

3. ‖αx‖ = |α| ‖x‖ for all x ∈ X and α ∈ R.

A vector space together with a norm is called a normed linear space. ♦

The following result follows easily from the above definition.

THEOREM 34. If X is a vector space with a norm ‖ · ‖, then

d(x, y) := ‖x− y‖

defines a metric on X, called the metric induced by the norm.

Example 35. (i) On the vector space Rn

x := (x1, . . . , xn) 7→ ‖x‖1 :=

n∑i=1

|xi|,

x := (x1, . . . , xn) 7→ ‖x‖2 :=

(n∑i=1

|xi|2)1/2

,

x := (x1, . . . , xn) 7→ ‖x‖∞ := max|xi| : i = 1, . . . , n

are norms.

(ii) For x := (x1, x2, . . .) ∈ `1(N), let

‖x‖1 :=

∞∑i=1

|xi|.

Then `1(N) is a vector space and x 7→ ‖x‖1 is a norm on `1(N).

(iii) For x := (x1, x2, . . .) ∈ `∞(N), let

‖x‖1 :=

( ∞∑i=1

|xi|2)1/2

.

Then Then `2(N) is a vector space and x 7→ ‖x‖2 is a norm on `1(N).

(iv) For a nonempty set S, B(S) is a vector space and x 7→ ‖x‖ := sup|x(s)| : s ∈ S is a normon B(S). ♦

13

4.2 Topological concepts

Definition 36. A sequence (xn) in a metric space (X, d) is said to be convergent if there existsx ∈ X such that d(xn, x) → 0 as n → ∞, and in that case we say that (xn) converges to x, andwrite lim

n→∞xn = x or xn → x as n→∞, or simply xn → x.. ♦

• A sequence (xn) in a metric space (X, d) converges to x ∈ X iff for every ε > 0, there existsn0 ∈ N such that

d(xn, x) < ε ∀n ≥ n0.

Definition 37. Let (X, d) be a metric space. For x ∈ X and r > 0, the set

B(x, r) := y ∈ X : d(y, x) < r

is called an open ball with centre x and radius r > 0. ♦

• A sequence (xn) in a metric space (X, d) converges to x ∈ X iff for every ε > 0, there existsn0 ∈ N such that

xn ∈ B(x, ε) ∀n ≥ n0.

Exercise 38. Let (xn) be a sequence in a metric space X. If (xn) converges in X, then the limit isunique. That is, If xn → x and xn → y for some x, y ∈ X, then x = y. ♦

Exercise 39. Let X be a metric space, (xn) be a sequence in X and x ∈ X.

(i) xn → x iff every subsequence of (xn) converges to x.

(ii) If (xn) is a sequence, then xn → x iff (xn) has a subsequence which converges to x. ♦

Definition 40. Let (X, d) be a metric space, A ⊆ X and x ∈ X.

(i) x is called an interior point of A if there exists r > 0 such that B(x, r) ⊆ A.

(ii) x is called a closure point of A if for every r > 0, B(x, r) ∩A 6= ∅.

(iii) x is called a limit point of A if for every r > 0, B(x, r) ∩ (A \ x) 6= ∅.

(iv) x is called a boundary point of A if for every r > 0, B(x, r) ∩ A 6= ∅ and B(x, r) ∩ Ac 6= ∅.♦

Definition 41. Let (X, d) be a metric space and A ⊆ X.

(i) The set of all interior points of A is called the interior of A, and it is denoted by Ao or int(A)or intX(A).

(ii) The set of all closure points of A is called the closure of A, and it is denoted by A or cl(A)or clX(A).

(iii) The set of all limit points of A is denoted by A′.

(iv) A is called a perfect set if A = A′. ♦

Definition 42. Let (X, d) be a metric space and A ⊆ X.

(i) A is called an open set if every point in A is an interior point of A.

(ii) A is called a closed set if A contains all its limit points.

14

(iii) A together with all its closure points is called the closure of A, and it is denoted by A.

(iv) The set of all boundary points of A is called the boundary of A, and it is denoted by ∂(A).♦

THEOREM 43. Let (X, d) be a metric space and A ⊆ X. The following are equivalent:

(i) A is closed.

(ii) A contains all its closure points.

(iii) A contains all its limit points.

(iv) A contains all its boundary points.

(v) Ac is open.

Proof. The implications (i) ⇒ (ii)⇒ (iii) ⇒ (i), and (ii) ⇐⇒ (iv) follow from the definitions (Seeexercise below). Now, we prove (i)⇐⇒ (v):

Suppose Ac is not open. Then there exists x ∈ Ac such that for every r > 0, B(x, r) containssome point of A. Then x is a limit point of A - a contradiction to the fact that A contains all itslimit points. Thus (i) ⇒ (v). Conversely, suppose Ac is open, and x is a limit point of A. If x doesnot belong to A, then x ∈ Ac, and since Ac is open there exists r > 0 such that B(x, r) ⊆ Ac, i.e.,B(x, r)∩A = ∅ - a contradiction to the assumption that x is a limit point of A. Thus, (v)⇒ (1).

Exercise 44. Let (X, d) be a metric space and A ⊆ X.

(i) A = A ∪ ∂A = A ∪A′

(ii) A is open iff A = Ao.

(iii) A is closed iff A = A. ♦


(i) Ao is an open set.

(ii) Ao is the union of all open sets contained in A, that is,

Ao =⋃G ⊆ X : G open and G ⊆ A.

(ii) A is the intersection of all open sets containing A, that is,

A =⋂G ⊆ X : G open and G ⊆ A.

♦


(i) x ∈ A ⇐⇒ ∃(xn) in A such that xn → x.

(ii) x ∈ A′ ⇐⇒ ∀ r > 0, B(x, r) ∩A is an infinite set. ♦

Exercise 47. Every finite subset of a metric space is closed, and it does not have any limit point. ♦

Example 48. Let X = R with usual metric d(x, y) := |x − y|, x, y ∈ R. Let A = (0, 1]. Then wehave the following (verify):

15

(i) Ao = (0, 1) and A = [0, 1].

(ii) A is neither open nor closed.

(iii) ∂A = 0, 1. ♦

Example 49. Let X = R with usual metric d(x, y) := |x − y|, x, y ∈ R. Let A = Z. Then we havethe following (verify):

(i) Ao = ∅ and A = Z.

(ii) A is a closed set, not open.

(iii) ∂A = Z. ♦

Example 50. Let X be with discrete metric and A ⊆ X. Then we have the following (verify):

(i) Ao = A = A.

(ii) A is open and closed.

(iii) ∂A = ∅.

(iv) Every subset of X is open and closed.

(v) A sequence in X is a Cauchy sequence iff it is eventually constant. ♦

Example 51. Let X = R be with usual (Euclidian) metric and A = (0, 1)× 0. Then we have thefollowing (verify):

(i) Ao = ∅, A = [0, 1]× 0.

(ii) A is closed but not open.

(iii) ∂A = [0, 1]× 0. ♦

Exercise 52. Let (X, d) be a metric space.

(i) Every open ball in X is an open set.

(ii) Union of every arbitrary collection of open sets in X is open in X.

(iii) Finite intersection of opens sets in X is open in X.

(iv) Intersection of every arbitrary collection of closed sets in X is closed in X.

(v) Finite union of of closed sets in X is closed in X. ♦

THEOREM 53. If Y is a subset of a metric space X with metric d, then (x, y) 7→ d(x, y) defines ametric on Y .

Proof. Exercise.

Definition 54. The metric on Y as in the above theorem is called the metric induced by d. ♦

Let (X, d) be a metric space and Y ⊆ X. Let A ⊆ Y . Then, as per the above definition, A is openin Y (w.r.t. the induced metric) iff for each x ∈ A, there exists r > 0 such that

BY (x, r) := y ∈ Y : d(x, y) < r ⊆ A.

16

THEOREM 55. Let (X, d) be a metric space and Y ⊆ X. Let A ⊆ Y . Then A is open in Y iffthere exists an open set G in X such that A = G ∩ Y .

Proof. Let A be open in Y . Then, for every x ∈ A, there exists rx > 0 such that BY (x, rx) ⊆ A.Hence, A = ∪x∈ABY (x, rx). But, BY (x, rx) = BX(x, rx) ∩ Y . Hence,

A = ∪x∈ABY (x, rx) = (∪x∈ABX(x, rx)) ∩ Y.

Thus, taking G = ∪x∈ABX(x, rx), we obtain A = G ∩ Y .

Conversely, let A = G ∩ Y for some open set G in X. Let x ∈ A. Since B is open in X, thereexists rx > 0 such that BX(x, rx) ⊆ G. Hence, BY (x, rx) = BX(x, rx) ∩ Y ⊆ G ∩ Y = A.

Exercise 56. Let (X, d) be a metric space and Y ⊆ X. Let A ⊆ Y . Then the following are true.(i) If A is open in X, then A is open in Y .(ii), Converse of (i) holds if Y is open in X.(iii), Converse of (i) need not hold if Y is not open in X. ♦

THEOREM 57. Let (X, d) be a metric space and Y ⊆ X. Let B ⊆ Y . Then B is closed in Y iffthere exists a closed set F in X such that B = F ∩ Y .

Proof. Let B be open in Y . Then, Y \B is open in Y . Hence, there exists an open set G in X such thatY \B = G∩Y , that is, Y ∩Bc = G∩Y . This implies, (Y ∩Bc)c = (G∩Y )c, that is, Y c∪B = Gc∪Y c,so that Y ∩ (Y c ∪ B) = Y ∩ (Gc ∪ Y c), that is, (Y ∩ Y c) ∪ (Y ∩ B) = (Y ∩ Gc) ∪ (Y ∩ Y c), that is,B = Y ∩Gc. Take F = Gc.

Conversely, suppose F is a closed set in X such that B = F ∩ Y . Then,

Y \B = Y ∩Bc = Y ∩ (F ∩ Y )c = Y ∩ (F c ∪ Y c) = (Y ∩ F c) ∪ (Y ∩ Y c) = Y ∩ F c.

Taking G = F c, we see that Y \B is open in Y ; hence B is closed in Y .

Exercise 58. Let (X, d) be a metric space and Y ⊆ X. Let B ⊆ Y . Then the following are true.(i) If B is closed in X, then B is closed in Y .(ii), Converse holds if Y is closed in X.(iii), Converse of (i) need not hold if Y is not closed in X. ♦

Exercise 59. Let X = R with usual metric.

(i) If Y = Z, what are the open subsets of Y ?

(ii) If Y = 1, 12 ,

13 , . . ., then what are the open subsets of Y ?

(iii) If Y = 0, 1, 12 ,

13 , . . ., then what are the open subsets of Y ? Is 0 open in Y ? ♦

Exercise 60. Let X be a metric space and Y ⊆ X. If E ⊆ Y , then clY (E) ⊆ clX(E); and equalityneed not hold. ♦

Definition 61. Let X be a set and T be a family of subsets of X. Then T is called a topology onX if the following conditions are satisfied:

(i) X ∈ T , ∅ ∈ T .

(ii) Union of every sub-collection of members of T is a member of T .

(iii) Intersection of every finite sub-collection of members of T is a member of T .

The pair (X, T ) is called a topological space. Members of T are called open sets in X andcompliments of open sets are called closed sets. ♦

17

Example 62. Let X be a metric space. Then the family of all open sets w.r.t. the metric on X forma topology on X, and the concept of open sets and closed sets in the setting of metric space coincideswith the corresponding concepts in the setting of topological space. ♦

Definition 63. Let d and ρ be metrics on a set X. Then d and ρ are said to be equivalent if thereexists C1 > 0, C2 > 0 such that

C1d1(x, y) ≤ d2(x, y) ≤ C2d1(x, y) ∀x, y ∈ X.

♦

Exercise 64. Suppose d and ρ are equivalent metric on X. Then (xn) converges to x w.r.t. d iff (xn)converges to x w.r.t. ρ; ♦

Exercise 65. For x, y ∈ Rk, let d∞(x, y) := max|xk − yk| : k = 1, . . . , k. Then for every p with1 ≤ p < ∞, dp and d∞ are equivalent metrics on Rk. Further, for any p, r with 1 ≤ p ≤ ∞ and1 ≤ r ≤ ∞, the metrics dp and dρ on Rn are equivalent. ♦

Exercise 66. Let X = C[a, b]. Show that the metrics

d1(x, y) :=

∫ b

a

|x(t)− y(t)|dt, d∞(x, y) = supa≤t≤b

|x(t)− y(t)|

are not equivalent. ♦

Definition 67. Consider a metric space (X, d).

(i) A subset D of X is said to be dense in X if D = X.

(ii) X is said to be separable if it has a countable dense subset.

(iii) A set E ⊆ X is said to be nowhere dense if (A)0 = ∅. ♦

• A metric space (X, d) is separable iff there exists a countable set D such that for every x ∈ Xand for every ε > 0, there exists y ∈ D such that d(x, y) < ε.

Example 68. Let R be with usual metric.

(i) Q and Qc := R \Q are dense in R.

(ii) R is separable.

(iii) R \ Z is dense in R.

(iv) Z is nowhere dense in R. ♦

• Is Qc separable?

Yes. To see this consider the setD := r +

√2 : r ∈ Q.

Then D is a countable dense subset of Qc: Clearly, D is countable. To see that it is dense in Qc, letx ∈ Qc. Since Q is dense in R, for every ε > 0, there exists r ∈ Q such that |(x−

√2)− r| < ε. Hence,

|x− (r +√

2)| < ε. Thus, D is dense in Qc.

More generally, we have the following.

18

THEOREM 69. Every subset of a separable metric space is separable.

Proof. Let (X, d) be a separable metric space and Y be a subset of X. Let E := x1, x2, · · · be acountable dense subset of X. First we observe that, due to denseness of E in X, for each k ∈ N,X =

⋃∞n=1B(xn,

1k ). Let S := (k, n) ∈ N × N : B(xn,

1k ) ∩ Y 6= ∅. For such (k, n) ∈ S, let

yk,n ∈ B(xn,1k )∩ Y . Clearly, D := yk,n : k, n ∈ N is countable. We show that D is dense in Y . For

this let x ∈ Y . Then, for each k ∈ N , there exists n ∈ N such that d(x, xn) < 1. Hence,

d(x, yk,n) ≤ d(x, xn) + d(xn, yk,n) <1

k+

1

k=

2

k.

Now, let ε > 0 be given. Let k ∈ N be such that 2k < ε. Let n be as above. Then d(x, yk,n) < 2

k < ε.

Thus, we have shown that for every ε > 0, there exists y ∈ D such that d(x, y) < ε. Consequently,D is dense in Y .

5 Completeness

Definition 70. Let (X, d) be a metric space and (xn) be a sequence in X. We say that (xn) is aCauchy sequence if for every ε > 0, there exists N ∈ N such that d(xn, xm) < ε ∀n,m ≥ N. ♦

The following theorem is easy to prove.

THEOREM 71. If a Cauchy sequence (xn) in a metric space X has a subsequence which convergesto a point x ∈ X, then xn → x.

Definition 72. A metric space X is said to be a complete metric space if every Cauchy sequencein X converges to some point in X. ♦

Before giving some examples, let us observe the following result.

THEOREM 73. If (xn) and (yn) are Cauchy sequences in a metric space (X, d), then the sequence(d(xn, yn)) converges.

Proof. Note thatd(xn, yn) ≤ d(xn, xm) + d(xm, ym) + d(ym, yn)

so thatd(xn, yn)− d(xm, ym) ≤ d(xn, xm) + d(ym, yn).

Since (xn) and (yn) are Cauchy sequences, there exists N ∈ N such that d(xn, xm) < ε/2 andd(ym, yn) < ε/2 for all n,m ≥ N . Hence,

d(xn, yn)− d(xm, ym) ≤ d(xn, xm) + d(ym, yn) < ε ∀n,m ≥ N.

Similarly, d(xm, ym)− d(xn, yn) < ε for all n,m ≥ N. Hence,

|d(xn, yn)− d(xm, ym)| < ε ∀n,m ≥ N.

Thus, (d(xn, yn)) is a cauchy sequence of real numbers. Since R is complete, (d(xn, yn)) converges.

COROLLARY 74. If (xn) is a Cauchy sequence in a metric space (X, d), then for every x ∈ X,the sequence (d(xn, x)) converges.

19

Example 75. (i) R with d(x, y) = |x− y| is complete.

(ii) Q with (x, y) 7→ |x− y| is not complete.

(iii) R2 is complete with the metrics d1(x, y) := |x1 − y1| + |x2 − y2| and d∞(x, y) := max|x1 −y1|, |x2 − y2|.

(iv) (0, 1], [0, 1), (0, 1) with d(x, y) = |x− y| are not complete. ♦

Exercise 76. Let X be with discrete metric.

(i) Every Cauchy sequence in X is eventually constant.

(ii) X is complete. ♦

Exercise 77. Suppose d and ρ are equivalent metric on X, and (xn) is a sequence in X. Then (xn)is a Cauchy sequence w.r.t. d iff (xn) (xn) is a Cauchy sequence w.r.t. ρ. ♦

THEOREM 78. If d and ρ are equivalent metrics on a set X and if X is complete w.r.t. d, then itis complete w.r.t. ρ.

Proof. Suppose d and ρ be equivalent metrics on X. Let c1, c2 > 0 such that

c1d(x, y) ≤ ρ(x, y) ≤ c2d(x, y) ∀x, y ∈ X. (∗)

Suppose X is complete w.r.t. d. We show that X is complete w.r.t. ρ. For this, let (xn) be a Cauchysequence w.r.t. ρ. Then, by (∗), we see that (xn) be a Cauchy sequence w.r.t. d. Hence, there existsx ∈ X such that d(xn, x)→ 0. Hence, again by (∗), ρ(xn, x)→ 0..

Example 79. The metric space (Rk, d∞) is complete. Further, Rk with dp is complete for any p with1 ≤ p <∞. ♦

The above example is a a particular case of the following.

THEOREM 80. Let S be a nonempty set. Then B(S) is complete w.r.t. the sup-metric.

Proof. Let (xn) be a Cauchy sequence in B(S). For ε > 0, let N ∈ N be such that

supt∈S|xn(t)− xm(t)| = d∞(xn, xm) < ε ∀n,m ≥ N. (∗)

Hence, for each t ∈ S,|xn(t)− xm(t)| = d∞(xn, xm) < ε ∀n,m ≥ N

so that (xn(t)) is a Cauchy sequence in R. Since R is complete, limn→∞xn(t) exists for each t ∈ S.Let x(t) := lim

n→∞xn(t), t ∈ S. Note that

|xn(t)| ≤ |xn(t)− xn0(t)|+ |xn0

(t)| ≤ d∞(xn, xn0) + sup

t∈S|xn0

(t)| ≤ 1 + supt∈S|xN (t)| = M,

where n0 ∈ N is such that d∞(xn, xm) < 1 ∀n,m ≥ n0. Hence, |x(t)| = limn→∞

|xn(t)| ≤ M for all

t ∈ S so that x ∈ B(S). Now, from (∗),

|xn(t)− xm(t)| = limm→∞

|xn(t)− xm(t)| ≤ ε ∀n ≥ N.

This is true for all t ∈ S. Hence, d∞(xn, x) ≤ ε for all n ≥ N . Thus, we have shown that (xn)converges to x ∈ B(S).

20

Example 81. As special cases of the above theorem, we have the following:

(i) Taking S = 1, . . . , k, we have B(S) = Rk.

(ii) Taking S = N, we have B(S) = `∞, the set set of all bounded sequences of real numbers. ♦

Definition 82. A normed linear space which is complete w.r.t. the metric induced by the norm iscalled a Banach space. ♦

THEOREM 83.

THEOREM 84. The following hold.

(i) Every closed subset of a complete metric space is complete.

(ii) Every complete subset of a metric space is closed.

Proof. Let X be a metric space with metric d(·, ·), and let E ⊆ X.

(i) Suppose X is complete and E is closed. Let (xn) be a Cauchy sequence in E. Then (xn) is aCauchy sequence in E. Since X is complete, there exists x ∈ X such that xn → x. Since E is closed,x ∈ E = E.

(ii) Suppose E is a complete subset of X. Let x ∈ E. Then there exists a sequence (xn) in E suchthat xn → x. Hence, (xn) is a Cauchy sequence in E. Since E is complete, x ∈ E. Thus, E = E.

COROLLARY 85. If Y ⊆ X is not closed in X, then Y is not complete.

THEOREM 86. Let Ω be a closed and bounded subset of R and C(Ω) be the set of all real (orcomplex) valued continuous functions defined on Ω. Then C(Ω) is a closed subset of B(Ω) w.r.t. thesup-norm. In particular, C(Ω) is a complete metric space w.r.t. the sup-norm.

Proof. Let d(x, y) := supt∈Ω |x(t) − y(t)| for x, y ∈ B(Ω). Clearly C(Ω) ⊆ B(Ω). Let (xn) be asequence in C(Ω) such that d(xn, x) → 0 for some x ∈ B(Ω). To prove that x ∈ C(Ω). For this, lett0 ∈ Ω and let ε > 0 be given. Note that, for any t ∈ Ω,

|x(t)− x(t0)| ≤ |x(t)− xn(t)|+ |xn(t)− xn(t0)|+ |xn(t0)− x(t0)|.

Hence,|x(t)− x(t0)| ≤ d(x, xn) + |xn(t)− xn(t0)|+ d(xn, x). (∗)

Let N ∈ N be such that d(xn, x) < ε/3 for all n ≥ N , and let δ > 0 be such that |xn(t)−xn(t0)| < ε/3whenever t ∈ Ω and |t− t0| < δ. Hence, from (∗), we obtain

|x(t)− x(t0)| ≤ d(x, xN ) + |xN (t)− xN (t0)|+ d(xN , x)

< ε whenever t ∈ Ω, |t− t0| < δ.

Thus, x ∈ C(Ω).

Example 87. C[a, b] is not complete w.r.t the metric d1(x, y) :=∫ ba|x(t)− y(t)|dt:

Let a < c < b, and let xn(t) =

0, a ≤ t < c− 1n ,

1 + n(t− c), c− 1n ≤ t < c,

1, c ≤ t ≤ b,Also, let x(t) =

0, a ≤ t < c,1, c ≤ t ≤ b.

Then it can be seen that ∫ b

a

|xn(t)− x(t)|dt =

∫ c

c− 1n

|xn(t)− x(t)|dt =1

2n.

21

THEOREM 90. Let X be a normed linear space and let E be its closed unit ball. Then

1. E is closed in X,

2. E is complete ⇐⇒ X is complete.

Proof. (i) let (xn) be in E such that xn → x for some x ∈ X. Then

‖x‖ ≤ ‖x− xn‖+ ‖xn‖ ≤ ‖x− xn‖+ 1 ∀n ∈ N.

Since ‖x− xn‖ → 0, it follows that ‖x‖ ≤ 1. hence x ∈ E.

(ii) Suppose E is compete. Let (xn) be a Cauchy sequence in X. Then there exists M > 0 suchthat ‖xn‖ ≤M for all n ∈ N. Since xn/M ∈ E for all n ∈ N and since (xn/M) is a Cauchy sequence,by completeness of E, there exists u ∈ E such that xn/M → u. Hence, xn → x := Mu.

The reverse implication follows from (i) and Theorem 84.

THEOREM 91. Let (X, d) be a metric space. Then, for every Cauchy sequence (xn) in X, thereexists a metric space (Y, ρ) such that X ⊆ Y , d is the restriction of ρ and (xn) converges in Y .

Proof. Let (xn) is a Cauchy sequence in X. Let x := (xn), Y = X ∪ x and

ρ(x, y) :=

d(x, y), x, y ∈ X,0, x = y = x,limn→∞

d(x, xn), x ∈ X, y = x..

Then ρ is a metric on Y and ρ(xn, x) → 0 as n → ∞. [observe that for each n ∈ N, ρ(xn, x) =limm→∞ d(xn, xm) so that ρ(xn, x)→ 0 as n→∞.]

THEOREM 92. A metric space (X, d) is complete iff every sequence in X which converges in anysuper space of X also converges in X.

Proof. Suppose (X, d) is complete and (xn) is a sequence in X which converges to some x in a superspace (Y, ρ). Since X is closed in Y , x ∈ X.

Conversely, suppose that every sequence in X which converges in any super space of X alsoconverges in X. Let (xn) is a Cauchy sequence in X. Take (Y, ρ) and x as in Theorem 91. Thenρ(xn, x)→ 0. By hypothesis, (xn) converges in X.

COROLLARY 93. A metric space (X, d) is not complete iff there exists a sequence (xn) and a superspace Y of X such that (xn) converges in Y , but not in X.

THEOREM 94. (Completion) Let (X, d) be a metric space. Then there exists a complete metric(X, ρ) and an isometry f : X → X, i.e., ρ(f(x), f(y)) = d(x, y) for all x, y ∈ X, such that the rangeof F is dense in X. Further, if (Y, η) is any compete metric space and g : X → Y is an isometry withrange of g is dense in Y , then Y is isometric with X.

Sketch of the proof. Let X be the set of Cauchy sequences in X. recall that if (xn), (yn) are in X ,then (d(xn, yn)) converges. Define, (xn) ∼ (yn) iff d(xn, yn) → 0. Then, it can be see that ∼ is anequivalence relation on X . Let X be the set of all equivalence classes. For x = [(xn)], y = [yn)] in X,define

ϕ(x, y) := limn→∞

d(xn, yn).

23

It can be seen that ρ is well-defined on X × X and it is a metric on X. Now, define f : X → X by

f(x) = [(x, x, . . .)], x ∈ X.

Then the following can be verified (do it!):

1. ρ is a complete metric;

2. f is an isometry, i.e., ϕ(f(x), f(y) = d(x, y)∀x, y ∈ X;

3. range of f is dense in X;

4. if (Y, η) is any compete metric space and g : X → Y is an isometry with range of g is dense inY , then Y is isometric with X.

Definition 95. The metric space (X, ρ), or any complete metric space (Y, η) as in Theorem 94 iscalled the completion of (X, d). ♦

Following theorem describes one of the completions.

THEOREM 96. Let (X, d) be a metric space. Let x0 ∈ X. For each u ∈ X, let fx : X → R bedefined by

fu(x) := d(x, x0)− d(x, u), x ∈ X.

Then fu ∈ B(X) for every u ∈ X, and the map T : X → B(X) defined by

T (u) = fu, u ∈ X

is an isometry. In particular, R(T ) is a completion of X.

Proof. Note that for every u ∈ X,

fu(x) := d(x, x0)− d(x, u) ≤ d(u, x0) ∀x ∈ X.

Hence, fu ∈ B(X) for every u ∈ X. . Also, for u, v, x ∈ X,

fu(x)− fv(x) = [d(x, x0)− d(x, u)]− [d(x, x0)− d(x, v)] = d(x, v)− d(x, u) ≤ d(u, v).

Hence,d∞(fu, fv) := sup

x∈X|fu(x)− fv(x)| ≤ d(u, v).

Further,fu(u)− fv(u) = [d(u, x0)− d(u, u)]− [d(u, x0)− d(u, v)] = d(u, v).

Hence (why?)d∞(fu, fv) = d(u, v).

Remark 97. In the above theorem, the function fu is not only bounded, but it is also continuous.Thus, we have prove that every metric space Ω is isometric with a subset of Cb(Ω). ♦

24

Example 98. With respect to the usual metric,

(i) R is the completion of Q and Qc;

(ii) [0, 1] is the completion of (0, 1], [0, 1), (0, 1), [0, 1] ∩Q, [0, 1] ∩Qc. ♦

We shall see that

• C[a, b] with sup-metric is the completion of the set of all polynomial functions on [a, b].

6 Compactness

Let (X, d) be a metric space. Before introducing the concept of compactness, let us observe thefollowing property.

THEOREM 99. For every distinct points x, y ∈ X, there exists disjoint open sets and containing xand y, respectively.

Proof. Let x, y ∈ X with x 6= y. Let 0 < r < 12d(x, y). Then B(x, r) and B(y, r) are disjoint open sets

containing x and y, respectively: Clearly, x ∈ B(x, r) and y ∈ B(y, r). Further, if z ∈ B(x, r)∩B(y, r),then

d(x, y) ≤ d(x, z) + d(z, y) < r + r = 2r < d(x, y),

which is a contradiction.

• The property of of metric space stated in the above theorem is called the Hausdorff property.

• A general topological space need not have this property: For example, let X be a set withindiscrete topology T := ∅,X. If #(X) ≥ 2, then X is not Hausdorff.

Definition 100. A family Aα : α ∈ Λ of subsets of a set X is said to be a cover of a set E ⊆ X ifE ⊆

⋃α∈ΛAα. A cover Aα : α ∈ Λ of E is called an open cover if each Aα is an open set. ♦

Definition 101. A subset E of X said to be a compact subset of X or compact in X if for every opencover of E has a finite subcover; that is, for family Vα : α ∈ Λ of open sets in X with E ⊆

⋃α∈Λ Vα,

there exist α1, . . . , αn in Λ such that E ⊆⋃ni=1 Vαi

. ♦

Example 102. Let X = R with usual metric. Then the following sets are not compact. You mayfind an open cover in each case so that it does not have a finite subcover.

(i) E = (0, 1]: Take Vn = ( 1n , 2), n ∈ N.

(ii) E = [0, 1): Take Vn = (−1, nn+1 ), n ∈ N.

(iii) E = (0, 1): Take Vn = ( 1n ,

nn+1 ), n ∈ N.

(iv) E = [0,∞): Take Vn = (−1, n), n ∈ N. ♦

THEOREM 103. Compact subset of a metric space is closed.

Proof. Let E be a compact subset of X. It is enough to show that Ec := X \ E is open in X. So,let x ∈ Ec. Then for every y ∈ E, there exists disjoint open sets Uy and Vy containing x and yrespectively. Then Vy : y ∈ E is an open cover of E. Since E is compact, there exists y1, . . . , yn inE such that E ⊆ V :=

⋃ni=1 Vyi and x ∈ U :=

⋂ni=1 Uαi . Note that U is open and U ∩ V = ∅ so that

U ∩ E = ∅. Thus, U ⊆ Ec. This proves that Ec is open.

25

By this theorem we can assert that the sets (0, 1], [0, 1), (0, 1) are not compact in R, withoutactually constructing an open cover in each case so that it does not have a finite subcover.

THEOREM 104. Closed subsets of a compact metric space are compact.

Proof. Let X be a compact metric space and E be a closed subset of X. Let Vα : α ∈ Λ be anopen cover of E. Then Vα : α ∈ Λ ∪ Ec is an open cover of X. Since X is compact, there existα1, . . . , αn in Λ such that X = Ec∪ (

⋃ni=1 Vαi

). Then E = E∩X = E∩ (⋃ni=1 Vαi

) ⊆⋃ni=1 Vαi

.

THEOREM 105. Let E ⊆ Y ⊆ X. Then E compact in X iff E is compact in Y .

Proof. Suppose E compact in X, and let Vα : α ∈ Λ be a family of open sets in Y which covers E.Then there exist open sets Gα in X such that Vα = Gα ∩ Y, α ∈ Λ. Thus

E ⊆⋃α∈Λ

Vα =⋃α∈Λ

(Gα ∩ Y ) ⊆⋃α∈Λ

Gα.

Since E is compact in X, there exist α1, . . . , αn in Λ such that E ⊆⋃ni=1Gαi . Hence,

E ⊆n⋃i=1

(Y ∩Gαi) =

n⋃i=1

Vαi.

Conversely, suppose E is compact in Y , and let Gα : α ∈ Λ be a family of open sets in X whichcovers E. Then Y ∩Gα : α ∈ Λ is a family of open sets in Y which covers E. Since E is compactin Y , there exist α1, . . . , αn in Λ such that E ⊆

⋃ni=1(Y ∩Gαi

). Hence, E ⊆⋃ni=1Gαi

.

THEOREM 106. Every compact metric space is separable.

Proof. Let (X, d) be a compact metric space. For each n ∈ N, B(x, 1n ) : x ∈ X is an pen cover of

X. Then there exists x(n)1 , . . . , x

(n)kn

in X such that X =⋃kni=1B(x

(n)i , 1

n ). Consider the set

D :=⋃n∈Nx(n)

i : i = 1, . . . , kn.

We show that D is dense in X. For this, let x ∈ X and ε > 0 be given. Then, for each n ∈ N, there

exists in ∈ 1, . . . , kn such that x ∈ B(x(n)in, 1n ). Take n large enough such that 1

n < ε. Then we have

d(x, x(n)in

) < 1n < ε. Thus, the countable set D is dense in X.

Definition 107. A subset of a metric space is said to be bounded if it is contained in a ball. ♦

• A subset E of a metric space X is bounded if and only if

daim(E) := supd(x, y) : x, y ∈ E <∞

if and only if there exists M > 0 such that for every x ∈ X, d(x, y) ≤M for all y ∈ E.

• Every Cauchy sequence in a metric space is bounded. In particular, every convergent sequenceis bounded: To see this, let (xn) be a Cauchy sequence in a metric space X. Then there existsN ∈ N such that d(xn, xN ) < 1 for all n ≥ N . Therefore,

d(xn, xN ) < ρ := 1 + max1≤i≤N

d(xi, xN ) ∀n ∈ N,

i.e., xn ∈ B(xN , ρ) for all n ∈ N.

26

THEOREM 108. Every compact subset of a metric space is bounded.

Proof. Let E be a compact subset of a metric space. If E = ∅, then it is bounded. So, assume thatE 6= ∅. Let x0 ∈ E. Then E ⊆

⋃∞n=1B(x0, n); because, if y ∈ E, then there exists n ∈ N such

that d(y, x0) < n. Since E is compact, there exists N ∈ N such that E ⊆⋃Nn=1B(x0, n). Since

B(x0, n) ⊆ B(x0, N) for every n ≤ N , we have E ⊆ B(x0, N).

Proof. (Aliter) Note that, for each r > 0, B(x, r) : x ∈ E is an open cover of E. Then there existsx1, . . . , xk in E such that B(xi, r) : i = 1, . . . , k is also a cover of E. Note that (verify) for anyx ∈ X there exists jx ∈ 1, . . . , k such that d(x, xjx) < r. Hence

d(x, x1) ≤ d(x, xjx) + d(xjx , x1) ≤ r + max1≤j≤k

d(xj , x1) = ρ, say.

Thus, E ⊆ B(x, ρ) so that E is bounded.

By this theorem we can assert that the sets (0,∞), [0,∞), (−∞, 1], R, Z, N, Q are not compact inR, without actually constructing an open cover in each case so that it does not have a finite subcover.

• A closed and bounded set need be compact. For example, an infinite set with discrete metric isclosed and bounded, but not compact.

THEOREM 109. (Finite intersection property) Suppose Kα : α ∈ Λ is a family of com-pact sets in a metric space such that every finite subfamily of it has nonempty intersection. Then⋂α∈ΛKα 6= ∅.

Proof. Suppose⋂α∈ΛKα = ∅. Let α0 ∈ Λ. Then there exists α 6= α0 such that Kα0

∩ Kα = ∅.Hence, Kα0

⊆⋃α6=α0

Kcα. Since Kα0

is compact, there exists α1, . . . , αk different from α0 such that

Kα0⊆⋃ki=1K

cαi

. This implies that Kα0∩(⋂k

i=1Kcαi

)c= ∅, i.e., Kα0

∩(⋂k

i=1Kcαi

)∅. This is a

contradiction.

COROLLARY 110. Suppose (Kn) is a sequence of compact sets in a metric space such that K1 ⊇K2 ⊇ · · · . Then

⋂∞n=1Kn 6= ∅.

THEOREM 111. Every compact metric space is complete.

Proof. Let (X, d) be a compact metric space. Assume for a moment that it is not complete. Thenthere exists a Cauchy sequence (xn) which does not converge in X. This also implies that (xn) doesnot have any convergent subsequence1. Hence, for any given x ∈ X, there exists rx > 0 such thatB(x, rx) contains only a finite number of terms from (xn). Since B(x, rx) : x ∈ X is an open cover

of X and X is compact, there exist x(1), . . . , x(k) in X such that X =⋃ki=1B(x(i), ri), ri := rx(i) .

This implies that X contains only a finite number of terms from xn : n ∈ N - a contradiction.

Definition 112. A subset E of a metric space is said to be totally bounded if for every ε > 0,there exists a finite number points x1, . . . , xn in X such that E ⊆

⋃ni=1B(xi, ε). ♦

• Every totally bounded set is bounded (verify). In particular, R with usual metric is not totallybounded.

1Recall that if a Cauchy sequence has a convergent subsequence, then the sequence itself will converge.

27

• A bounded set need not be totally bounded: For example, if X is an infinite set with discretemetric, then every subset of it is bounded, but infinite subsets are not totally bounded (Why?)

• A set is totally bounded iff its closure is totally bounded.

Exercise 113. Every totally bounded metric space is separable.(Hint: Look at the proof of Theorem 106). ♦

THEOREM 114. Every bounded subset of R, with usual metric, is totally bounded (verify).

Proof. Let E be a bounded subset of R. Then E ⊆ [a, b] for some a, b ∈ R with a < b. Let

a0 = a, ai = a+ i(b−a)n , i = 1, . . . , n. Then [a, b] ⊆

⋃ni=1BR(ai,

2i(b−a)n ). Let

Λn := i ∈ 1, . . . , n : BR(ai,2i(b− a)

n) ∩ E 6= ∅.

Then [a, b] ⊆⋃i∈Λn

BR(ai,2i(b−a)

n ). For i ∈ Λn, let xi ∈ BR(ai,2i(b−a)

n ) ∩ E. Then

E ⊆⋃i∈Λn

BR(ai,4i(b− a)

n).

Now, given ε > 0, we may choose n such that 4i(b−a)n ) < ε. Then we have E ⊆

⋃i∈Λn

BR(xi, ε).

• Every subset of a totally bounded set is totally bounded.

THEOREM 115. Every compact subset of a metric space is totally bounded.

Proof. Let E be a compact subset of a metric space X. For ε > 0, B(x, ε) : x ∈ E is an open coverof E. Then there exists x1, . . . , xk in E such that B(xi, ε) : i = 1, . . . , k is also a cover of E.

THEOREM 116. A metric space is compact iff it is complete and totally bounded.

Proof. Let (X, d) be a metric space.

⇒): Suppose E is a compact compact subset of X. We have already seen that X is complete andtotally bounded.

⇐): Suppose X is complete and totally bounded. To show that it is compact. Assume for amoment that X is not compact. Then there exists an open cover G of X which does not have anyfinite subcover.

Let (rn) be a sequence of positive reals such that rn > rn+1 for all n ∈ N and rn → 0. Since X is

totally bounded, there exist x(1)1 , x

(1)2 . . . , x

(1)k1

such that X =⋃k1i=1B(x

(1)i , r1). Since X is not covered

by any finite sub-collection of G, atleast one of B(x(1)1 , r1), B(x

(1)2 , r1), . . . , B(x

(1)k1, r1) is not covered by

a finite number of members from G. W.l.g, assume that B(x(1)1 , r1) is not covered by a finite number

of members from G. Since B(x(1)1 , r1) is totally bounded, there exist x

(2)1 , x

(2)2 . . . , x

(2)k2

in B(x(1)1 , r1)

such that B(x(1)1 , r1) ⊆

⋃k2i=1B(x

(2)i , r2). Since B(x

(1)1 , r1) is not covered by any finite sub-collection

of G, atleast one of B(x(2)1 , r2), B(x

(2)2 , r2), . . . , B(x

(2)k2, r2) is not covered by any finite sub-collection

of G. W.l.g, assume that B(x(2)2 , r2) is not covered by any finite sub-collection of G. Continuing this,

we obtain a sequence (x(n)n ) in X such that for each n ∈ N, B(x

(n)n , rn) is not covered by any finite

sub-collection of G.

28

Note that, for each m ∈ N, x(n)n ∈ B(x

(m)m , rm) for all n ≥ m. Hence (x

(n)n ) is a Cauchy sequence

in X. Since X is complete, there exits x ∈ X such that x(n)n → x, and2 d(x

(m)m , x) ≤ rm. Let G ∈ G

be such that x ∈ G, and let ε > 0 be such that B(x, ε) ⊆ G. We show that B(x(m)m , rm) ⊆ B(x, ε) for

all large enough m. For this, let y ∈ B(x(m)m , rm). Then

d(y, x) ≤ d(y, x(m)m ) + d(x(m)

m , x) < 2rm.

Hence, taking m sufficiently large such that 2rm < ε, we obtain B(x(m)m , rm) ⊆ B(x, ε) ⊆ G. This

contradicts the fact that B(x(n)n , rn) is not covered by any finite sub-collection of G.

THEOREM 117. A subset E of a metric space is totally bounded iff every sequence in E has aCauchy subsequence.

Proof. Let (X, d) be a metric space and E ⊆ X.

⇒): Suppose E is totally bounded. Let (xn) be a sequence in E, and let ε > 0 be given. Since E is

totally bounded, there are x(1), x(2) . . . , x(k) in E such that E ⊆⋃ki=1B(x(i), ε2 ). Hence, there exists

` ∈ 1, . . . , k such that B(x(`), ε2 ) contains infinite number of terms of (xn), say xnj∈ B(x(`), ε2 ) for

j ∈ N. Thus, for every i, j ∈ N, d(xni, xnj

) ≤ d(xni, x(`)) + d(x(`), , xnj

) < ε. W.l.g we can assumethat (nj) is strictly increasing. Hence, we have a Cauchy subsequence (xnj ).

⇐): Assume for a moment that E is not totally bounded. Then there exists ε > 0 such that Ecannot be covered by a finite number of ε-balls. Let x0 ∈ E. Since E cannot be covered by B(x0, ε),there exists x1 ∈ E \ B(x0, ε), x2 ∈ E \ B(x0, ε) ∪ (B(x1, ε), x3 ∈ E \ B(x0, ε) ∪ B(x1, ε) ∪ B(x2, ε)etc. xn+1 ∈ E \

⋃ni=1B(xi, ε). From this it follows that d(xn, xm) ≥ ε for every distinct n 6= m in N.

Thus, the sequence (xn) does not have any Cauchy subsequence.

As a corollary we obtain the following:

THEOREM 118. A subset E of a metric space is compact iff every sequence in E has a subsequencewhich converges in E.

Proof. Let X be a metric space and E ⊆ X. Suppose E is compact in X. Let (xn) be a sequencein E. By Theorem 118, (xn) has a Cauchy subsequence. Since E is complete, this Cauchy sequenceconverges to a point in E.

Conversely, every sequence in E has a subsequence which converges in E. Then by Theorem118, E is totally bounded. Now, let (xn) be any Cauchy sequence in E. By assumption, (xn) hasa subsequence which converges to some point x ∈ E. Since, (xn) is a Cauchy sequence, it itselfconverges to x. Thus, E is complete as well. Therefore, E is complete and totally bounded, and hencecompact.

THEOREM 119. If d and ρ are equivalent metrics on a set X and if X is compact w.r.t. d, thenit is compact w.r.t. ρ.

Proof. Suppose X is compact w.r.t. d. Then by Theorem 116, it is complete and totally boundedw.r.t. d. Hence, by Theorem 78, it is complete w.r.t. ρ. Now, let c1, c2 > 0 such that

c1d(x, y) ≤ ρ(x, y) ≤ c2d(x, y) ∀x, y ∈ X. (∗)2If (an) in X such that an → a and if b ∈ X, then d(an, b) → d(a, b). Also, if d(an, b) < ε for all n ∈ N, then

d(a, b) ≤ ε.

29

Since X is totally bounded w.r.t. d, for every ε > 0, there exist x1, . . . , xk in X such that X =⋃ki=1Bd(xi, ε). But, by (∗), Bd(x, ε) ⊆ Bρ(x, c2ε) for every x ∈ X. Hence, X =

⋃ki=1Bρ(xi, c2ε).

Hence, X is totally bounded w.r.t. ρ. Thus, by Theorem 116, X is compact w.r.t. ρ.

Definition 120. A subset E os a metric space is called a relatively compact set if its closure iscompact. ♦

• If X is a complete metric space, then a subset of X is relatively compact iff it is totally bounded.

7 Connectedness

Definition 121. Let X be a metric space.

1. X is said to be disconnected if there exist nonempty sets A and B such that

X = A ∪B, A ∩B = ∅, A ∩ B = ∅.

2. X is said to be connected if it is not disconnected.

3. A subset E of X is disconnected if it is disconnected with respect to the induced metric.

4. A subset E of X is connected if it is connected with respect to the induced metric.

♦

THEOREM 122. A metric space X is disconnected iff there exist nonempty disjoint open sets Aand B such that X = A ∪B.

Proof. ⇒) : Suppose X is disconnected. Let A and B be nonempty sets such that X = A ∪ B,A ∩ B = ∅ and A ∩ B = ∅. We show that A and B are open: Let x ∈ A. Since A ∩ B = ∅,x 6∈ B. Hence there exists r > 0 such that B(x, r) ∩B = ∅. Thus, B(x, r) ⊆ Bc = A. Thus, we haveprove that every point of A is an interior point of A so that A is open. Similarly, using the fact thatA ∩B = ∅, it can be shown that B is open.

⇐) : Suppose there exists nonempty disjoint open sets A and B such that X = A ∪ B. ThenB = Ac is closed so that B = B and hence A ∩ B = A ∩ B = ∅. Similarly, using the fact that B isopen, we obtain A ∩B = A ∩B = ∅.

The reader may verify the following:

THEOREM 123. Let E be a subset of a metric space X. Then the following are equivalent.

(i) There exist nonempty sets A and B such that

E = A ∪B, A ∩B = ∅, A ∩ B = ∅,

(ii) E = V1 ∪ V2, where V1 and V2 are nonempty disjoint sets which are open in E.

(iii) There exist open sets G1 and G2 in X such that

E ⊆ G1 ∪G2, E ∩G1 6= ∅, E ∩G2 6= ∅, E ∩ (G1 ∩G2) = ∅.

30

Proof. (i) ⇒ (ii): By (i), A ⊆ (B)c and B ⊆ (A)c. Take V1 = E ∩ (B)c and V2 = E ∩ (A)c. ThenA ⊆ V1, B ⊆ V2, and E = V1 ∪ V2, since

E = E ∩ [A ∪B] ⊆ E ∩ [(B)c ∪ (A)c] = V1 ∪ V2 ⊆ E.

Thus, (ii) holds.

(ii) ⇒ (iii): Let V1 and V2 be as in (ii). Let V1 = E ∩ G1 and V2 = E ∩ G2, where G1 and G2

are open in X. Then E ∩ G1 = V1 and E ∩ G2 = V1 are nonempty, and E ⊆ G1 ∪ G2. Further,E ∩ (G1 ∩G2) = V1 ∩ V2 = ∅. This proves (iii).

(iii) ⇒ (i): Let G1 and G2 be as in (iii). Take A = E ∩G1 and B = E ∩G2. Then A and B arenonempty, and E = A ∪ B. Now, suppose that A ∩ B 6= ∅. Then there exists x ∈ B = E ∩ G2 andx ∈ A so that there exists r > 0 such that B(x, r) ⊆ G2 and B(x, r) ∩ A 6= ∅. Let y ∈ B(x, r) ∩ A.Hence, y ∈ B(x, r) ⊆ G2 and y ∈ A = E ∩ G1 so that y ∈ E ∩ G1 ∩ G2, which is a contradiction tothe hypothesis in (iii).

In view of Theorem 122 and Theorem 123,

THEOREM 124. A subset E of a metric space is disconnected iff any of the three conditions inTheorem 123 is satisfied.

What are connected subsets of R w.r.t the usual metric?

Recall that a subset E of R is an interval if for every x, y ∈ E and z ∈ R, x < z < y impliesz ∈ E.

THEOREM 125. A subset of R, which is not singleton, is connected iff it is an interval.

Proof. Let E ⊆ R. Suppose E is not an interval. Then, there exists x, y ∈ E and z ∈ R withx < z < y, but z 6∈ E. Then E ⊆ Az ∪ Bz, where Az := (−∞, z) and Bz := (z,∞) are disjoint opensets in X with x ∈ Az, y ∈ Bz. Hence E is not connected.

Conversely, suppose E is not connected. Then there exist nonempty sets A and B such thatE = A ∪ B, A ∩ B = ∅, A ∩ B = ∅. let x ∈ A and y ∈ B. W.l.g., assume that x < y. Letz = sup(A ∩ [x, y]). Then z ∈ A. Hence, z 6∈ B so that x ≤ z < y. Let us consider the two cases (i)z 6∈ A and (ii) z ∈ A.

(i) z 6∈ A implies x < z < y, and we are done.

(ii) if z ∈ A, then z 6∈ B so that there exists an open interval containing z which does not containany point from B. Hence, there exists z′ such that z < z′ < y. Note that z′ 6∈ A and z′ 6∈ B so thatz′ 6∈ E.

Thus we have proved that there exists α ∈ R \ E such that x < α < y, so that E is not aninterval.

Exercise 126. A metric space X is disconnected iff there exist disjoint nonempty closed sets A andB such that X = A ∪B. ♦

Exercise 127. A metric space X is connected iff there is no nonempty set which is both open andclosed in X. ♦

Definition 128. A metric space X is said to be totally disconnected if singleton sets are the onlynonempty connected sets. ♦

31

Example 129. The set Q is totally disconnected w.r.t. the usual metric. To see this suppose E isa nonempty subset of Q, which is not singleton. Let x, y ∈ E with x < y. Let z ∈ Qc such thatx < z < y. Then E = A ∪ B, where A := E ∩ (−∞, z) and B := E ∩ (z,∞) are disjoint nonemptyopen subsets of E. ♦

THEOREM 130. Let X be a metric space and E ⊆ X. If E is connected then E is connected; andthe converse is not true.

Proof. Suppose E is not connected. Then there exist disjoint nonempty open subsets V1, V2 in E suchthat E = V1 ∪ V2. Then V1 = E ∩G1, V2 = E ∩G2 for some open sets G1, G2 in X.

Let x ∈ V1 = E∩G1. Then there exists r > 0 such that B(x, r) ⊆ G1; also, B(x, r)∩E 6= ∅. Hence,E∩G1 6= ∅. Similarly, E∩G1 6= ∅. Thus, E = (E∩G1)∪(E∩G2). Note that U1 = E∩G1, U2 = E∩G2

are open sets in E with U1 6= ∅, U2 6= ∅ and U1∩U2 = (E∩G1)∩(E∩G2) ⊆ (E∩G1)∩(E∩G2) = ∅.Thus, we have proved that E connected implies E disconnected.

To see that the converse is not true, consider X = R and E = Q.

8 Continuity

8.1 Definition and characterizations

Definition 131. Let X be a metric space with metric dX and Y a metric space with metric dY . LetE ⊆ X and let f : E → Y be a function.

(i) f is said to be continuous at a point a ∈ E if for every ε > 0, there exists δ > 0 such that

x ∈ E, dX(x, a) < δ ⇒ dY (f(x), f(a)) < ε.

(ii) f is said to be continuous on E ⊆ X if it is continuous at every point in E. ♦

Observe that

• f is continuous at a point a ∈ E iff for every ε > 0, there exists δ > 0 such that x ∈ BE(a, δ)implies f(x) ∈ BY (f(a), ε).

Example 132. Let X = [0, 1] with usual metric. Let f(x) =

0, 0 ≤ x < 1

2 ,1, 1

2 ≤ x ≤ 1.Then f is continuous

at every point except at 12 . ♦

Example 133. Let X = [0, 1] with usual metric. Let f(x) =

0, x ∈ Q ∩ [0, 1],1, x ∈ Qc ∩ [0, 1].

Then f : X → R

is not continuous at any point. ♦

THEOREM 134. A function f : E → Y is continuous at a point a ∈ E iff for every sequence (xn)in E,

dX(xn, a)→ 0 ⇒ dY (f(xn), f(a))→ 0.

Proof. Suppose f : E → Y is continuous at a ∈ E. Let (xn) be in E such that dX(xn, a) → 0. Toshow that dY (f(xn), a)→ 0.

32

Let ε > 0 be given. Since f is continuous at a ∈ E, there exists δ > 0 such that x ∈ E,dX(x, a) < δ⇒ dY (f(x), f(a)) < ε. Since dX(xn, a)→ 0, there exists n0 ∈ N such that dX(xn, a) < δfor all n ≥ n0. Hence, dY (f(xn), f(a)) < ε for all n ≥ n0. Thus, dY (f(xn), f(a))→ 0.

Conversely, suppose that for every sequence (xn) in E, dX(xn, a) → 0⇒ dY (f(xn), f(a)) → 0.Suppose f is not continuous at a. Then there exists ε > 0 such that for any δ > 0 there is x ∈ E withdX(x, a) < δ, but dY (f(x), f(a)) ≥ ε. In particular, there exists ε > 0 such that for any n ∈ N thereis xn ∈ E with dX(xn, a) < 1

n , but dY (f(xn), f(a)) ≥ ε. Thus, we obtain a sequence (xn) in E suchthat dX(xn, a)→ 0, but dY (f(xn), a) 6→ 0. This is a contradiction to the hypothesis.

THEOREM 135. A function f : E → Y is continuous at a point a ∈ E iff for every open set V inY containing f(a), there exists an open set U in E containing a such that f(U) ⊆ V .

Proof. ⇒) : Suppose f : E → Y is continuous at a ∈ E. Let V be an open set in Y containingf(a). Then there exists ε > 0 such that BY (f(a), ε) ⊆ V . Since f is continuous at a, there existsδ > 0 such that x ∈ E and dX(a, x) < δ implies dY (f(x), f(a)) < ε. That is, x ∈ BE(a, δ) impliesf(x) ∈ dY (f(a), ε) ⊆ V . Thus, taking U := BE(a, δ), f(U) ⊆ V .

⇐) : Let ε > 0 be given. Consider the open ball V := BY (f(a), ε). Let U be an open set in Econtaining a such that f(U) ⊆ V . Let δ > 0 be such that BE(x, δ) ⊆ U . Then we have x ∈ BE(u, δ)implies f(x) ∈ V := BY (f(a), ε).

THEOREM 136. A function f : E → Y is continuous on E iff for every open set V in Y , f−1(V )is open in E.

Proof. ⇒) : Suppose f : E → Y is continuous on E. Let V be an open set in Y . Let x ∈ f−1(V ).Then f(x) ∈ V . Hence, there exists open set U in E containing x such that f(U) ⊆ V . Hence,U ⊆ f−1(V ). Thus, we have proved that f−1(V ) is open in E.

⇐) : Suppose f−1(V ) is open in E for every open set V in Y . Let a ∈ E and let ε > 0 be given.Consider the open ball V := BY (f(a), ε). Since f−1(V ) is open in E, there exists δ > 0 such thatBE(a, δ) ⊆ f−1(V ). That is, x ∈ BE(a, δ) implies f(x) ∈ V := BY (f(a), ε).

8.2 Relation with compactness and connectedness

THEOREM 137. Let f : X → Y be a continuous function and E ⊆ X.

(i) If E is compact in X, then f(E) is compact in Y .

(ii) If E is connected in X, then f(E) is connected in Y .

Proof. (i) Suppose E is compact in X. To show that f(E) is compact in Y . For this, let Vαα∈Λ

be an open cover of f(E). Then f−1(Vα)α∈Λ is a cover of E. Further, since f is continuous,for each α ∈ Λ, f−1(Vα) ∩ E = x ∈ E : f(x) ∈ Vα is open in E. Since E is compact, thereexist α1, . . . , αn in Λ such that E ⊆

⋃ni=1

(f−1(Vαi

) ∩ E). Hence, E ⊆

⋃ni=1 f

−1(Vαi), and hence3

f(E) ⊆⋃ni=1 f

−1(Vαi).

(ii) Suppose E is connected in X. Assume for a moment that f(E) is disconnected. Then, byTheorem 123, there exist disjoint open sets V1 and V2 in Y such that

f(E) ⊆ V1 ∪ V2, f(E) ∩ V1 6= ∅, f(E) ∩ V2 6= ∅.3using f(∪αAα) ⊆ ∪αf(Aα) and f(f−1(A)) ⊆ A.

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Since f is continuous, f−1(V1) and f−1(V2) are open sets such that

E ⊆ f−1(V1) ∪ f−1(V2), E ∩ f−1(V1) 6= ∅, E ∩ f−1(V2) 6= ∅.

Thus, again by Theorem 123, E is disconnected; a contradictions to the hypothesis.

Following are important corollaries of the above theorem.

THEOREM 138. (Existence of maxima and minima) Let E be a compact subset of a metricspace X and f : E → R be a continuous function. Then there exists u, v ∈ E such that

f(u) = supx∈E

f(x), f(v) = infx∈E

f(x).

Proof. Since f is continuous and E is a compact subset of X, f(E) is compact in R. In particular,f(E) is bounded above and bounded below. Let

α := sup f(E), β := inf f(E).

Since f(E) is compact, there exist (un) and (vn) in E such that f(un)→ α and f(vn)→ β as n→∞.Since E is compact, (un) and (vn) have subsequences, (un) and (vn) such that un → u and vn → vfor some u, v ∈ E. Continuity of f implies, f(un) → f(u) and f(vn) → f(v). Hence, f(u) = α andf(v) = β.

Alternate proof. Since E compact and f continuous, f(E) is compact, and hence it is closed andbounded. Hence, α, β ∈ R, and

α := sup f(E) ∈ f(E) = f(E) and β := inf f(E) ∈ f(E) = f(E).

Thus, there exists u, v ∈ E such that f(u) = α, f(v) = β.

THEOREM 139. (General intermediate value theorem) Let E be a connected subset of ametric space X and f : E → R be a continuous function. Let u, v ∈ E and γ ∈ R be such thatf(u) ≤ γ ≤ f(v). Then there exists x ∈ E such that f(x) = γ.

Proof. Since E is a connected subset of X, f(E) is connected in R. Since connected subsets ofR are either singleton sets or intervals, [f(u), f(v)] ⊆ f(E). Hence, there exists x ∈ E such thatf(x) = γ.

THEOREM 140. (Intermediate value theorem) Let I be an interval and let f : I → R be acontinuous function. Let u, v ∈ I and γ ∈ R be such that f(u) ≤ γ ≤ f(v). Let a := minu, v andy = maxu, v. Then there exists x ∈ [a, b] such that f(x) = γ.

Proof. Since E := [a, b] is a connected subset of R, f(E) is connected in R. Since f(u), f(v) ∈ f(E),and since connected subsets of R are either singleton sets or intervals, [f(a), f(b)] ⊆ f(E). Hence,there exists x ∈ E such that f(x) = γ.

THEOREM 141. A metric space X is disconnected iff there exists continuous surjective functionf : X → 0, 1, where 0, 1 is endowed with discrete metric.

34

Proof. Suppose X is disconnected, and let A and B be disjoint open sets A and B such that X = A∪B.Let f : X → 0, 1 be defined by f(x) = 0 for every x ∈ A and f(x) = 1 for every x ∈ B. Then f iscontinuous (why?) and onto.

Conversely, suppose there is a continuous surjective function f : X → 0, 1. Then A := f−1(0)and B := f−1(1) are nonempty disjoint closed sets such that X = A∪B, so that X is disconnected.

8.3 Uniform continuity

Recall that a function f : X → Y is continuous on E ⊆ X iff for every ε > 0 and for every a ∈ E,there exists δ > 0 such that x ∈ E, dX(x, a) < δ implies dY (f(x), f(a)) < ε. Note that, the δ abovemay depend not only on ε, but also on the point a.

Definition 142. Let X and Y be metric spaces. A function f : X → Y is said to be uniformlycontinuous on E ⊆ X if for every ε > 0, there exists δ > 0 such that

x, y ∈ E, dX(x, y) < δ ⇒ dY (f(x), f(y)) < ε.

♦

THEOREM 143. Let X and Y be metric spaces and f : X → Y . Then f is uniformly con-tinuous on E ⊆ X iff for every sequence (xn) and (yn) in E satisfying dX(xn, yn) → 0, we havedY (f(xn), f(yn))→ 0.

Proof. ⇒): Suppose f : X → Y is uniformly continuous on E ⊆ X and (xn) and (yn) are sequences inE such that dX(xn, yn)→ 0. Let ε > 0 be given and let δ > 0 be such that x, y ∈ E with dX(x, y) < δimplies dY (f(x), f(y)) < ε. Let N ∈ N be such that dX(xn, yn) < δ for all n ≥ N . Then, it followsthat dY (f(xn), f(yn)) < ε for all n ≥ N .

⇐): Suppose f is not uniformly continuous. Then there exists ε > 0 such that for every δ > 0,there exist x, y ∈ E (depending on δ) such that dX(x, y) < δ, but dY (f(x), f(y)) ≥ ε. In particular,for every n ∈ N, there exist xn, yn ∈ E such that dX(xn, yn) < 1

n , but dY (f(xn), f(yn)) ≥ ε. Thus,we have proved that, if f is not uniformly continuous, then there exist sequences (xn) and (yn) in Esatisfying dX(xn, yn)→ 0, but dY (f(xn), f(yn)) 6→ 0.

Example 144. Consider f : (0, 1]→ R defined by f(x) = 1x . This function is continuous on (0, 1], but

not uniformly continuous. To see this consider the sequences ( 1n ) and ( 1

n2 ). We see that | 1n −1n2 | → 0,

but |f( 1n )− f( 1

n2 )| = |n− n2| 6→ 0. ♦

THEOREM 145. Let X and Y be metric spaces and f : X → Y be continuous. If E ⊆ X iscompact, then f is uniformly continuous on E.

Proof. Let E be a compact subset of X. Let ε > 0 be given. Then for every u ∈ E, there exists δu > 0such that x ∈ E and dX(x, u) < δu implies dY (f(x), f(u)) < ε

2 . Since BX(u, δu2 ) : u ∈ E is an open

cover of E and since E is compact, there exist u1, . . . , uk in E such that E ⊆⋃ki=1BX(ui,

δi2 ), where

δi := δui . Now, let x, y ∈ E be such that

dX(x, y) < δ := minδi2

: i = 1, . . . , k.

Let i ∈ 1, . . . , k be such that dX(x, ui) <δi2 . Then we have

dX(y, ui) ≤ dX(y, x) + dX(x, ui) < δ +δi2< δi.

35

Hence,dY (f(x), f(y)) ≤ dY (f(x), f(ui)) + dY (f(ui), f(y)) < ε.

Alternate proof. Let E be compact. Suppose f is not uniformly continuous on E. Then, by Theorem143, there exist sequences (xn) and (yn) in E such that dX(xn, yn) → 0, but dY (f(xn), f(yn)) 6→ 0.Then, there exist ε > 0 such that dY (f(xn), f(yn)) ≥ ε for infinitely many n’s. Thus, (xn) and (yn)have subsequences (xn) and (yn), respectively, such that

dY (f(xn), f(yn)) ≥ ε ∀n ∈ N. (∗)

Since E is compact, (xn) and (yn) have convergent subsequences, say (xn) and (yn), respectively. Letxn → x and yn → y. Since dX(xn, yn)→ 0, it follows that x = y. Consequently, dY (f(xn), f(yn))→ 0.This is a contradiction to (∗).

Remark 146. Note that, in Example 144, X := (0, 1] is not compact. ♦

Remark 147. Theorem 145 is important in the context of Riemann integration, where to show thatevery continuous function f : [a, b] → R is Riemann integral, what we essentially use is the uniformcontinuity of f . ♦

Exercise 148. Let X and Y be metric spaces. If f : X → Y is uniformly continuous, then for everyCauchy sequence (xn) in X, the sequence (f(xn)) is Cauchy in Y . Give an example to show that theassumption of uniform continuity of f cannot be dropped. ♦

The proof of the following theorem is left as an exercise.

THEOREM 149. Let X and Y be metric spaces and f : X → Y be such that there exists κ > 0satisfying

dY (f(x), f(y)) ≤ κ dX(x, y) ∀x, y ∈ X.Then f is uniformly continuous on X.

Definition 150. Let X and Y be metric spaces. A function f : X → Y is said to be Lipschitzcontinuous if there exists κ > 0 such that

dY (f(x), f(y)) ≤ κ dX(x, y) ∀x, y ∈ X.

The number κ above is called a Lipschitz constant.

If f : X → X is a Lipschitz continuous function with Lipschitz constant κ < 1, then f is called acontraction. ♦

• Every Lipschitz continuous function is uniformly continuous.

THEOREM 151. Let (X, d) be a metric space. For x0 ∈ X, the map x 7→ d(x, x0) is Lipschitzcontinuous from X to R (with usual metric), with Lipschitz constant κ = 1.

Proof. Note that for x, y ∈ X,

d(x, x0)− d(y, x0) ≤ d(x, y), d(y, x0)− d(x, x0) ≤ d(x, y).

Hence,|d(x, x0)− d(y, x0)| ≤ d(x, y).

36

THEOREM 152. Let (X, d) be a metric space. For E ⊆ X and x ∈ X, let

d(x,E) := inf(d(x, y) : y ∈ E.

(i) For x ∈ X, d(x,E) = 0 iff x ∈ E.

(ii) The map x 7→ d(x,E) is Lipschitz continuous from X to R (with usual metric), with Lipschitzconstant κ = 1

Proof. (i) Let x ∈ X. Then d(x,E) := inf(d(x, y) : y ∈ E = 0 iff for every ε > 0, there exists y ∈ Esuch that d(x, y) < ε iff x ∈ E.

(ii) Let x, y ∈ E. Then for every u ∈ E,

d(x, u) ≤ d(x, y) + d(y, u).

Hence,d(x,E) ≤ d(x,E) := inf

v∈Ed(x, v) ≤ d(x, y) + d(y, u),

d(x,E) ≤ d(x, y) + infu∈E

d(y, u) = d(x, y) + infu∈E

d(y, u) = d(x, y) + d(y,E).

Thus, d(x,E) ≤ d(x, y) + d(y,E). Similarly, d(y,E) ≤ d(x, y) + d(x,E). Thus,

|d(x,E)− d(y,E)| ≤ d(x, y).

Exercise 153. Let X and Y be metric spaces and D be a dense subset of X. If f : D → Y isuniformly continuous and if Y is complete, then there exists a unique uniformly continuous functionϕ : X → Y such that ϕ|D = f .

Hint: Steps involved:

1. For x ∈ X, take a sequence (xn) in D such that xn → x.

2. Observe that f(xn) is a Cauchy sequence in Y .

3. Write F (x) = limn→∞

f(xn).

4. Observe that if (un) in D with un → x, then limn→∞

f(un) = limn→∞

f(xn).

5. Observe that F : X → Y is well-defined.

6. Observe that F (u) = f(u) for every u ∈ D.

7. Show that F is uniformly continuous.

♦

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9 A Few Important Theorems

9.1 Contraction mapping theorem

Definition 154. Let X be a nonempty set and let f : X → X. A point x0 ∈ X is said to be a fixedpoint of f if f(x0) = x0. ♦

Existence of a fixed point of a function is important in the context of solving equations. Forexample, F : X → X is a given function and y ∈ X. Suppose we want to solve the equation

F (x) = y.

This problem can be converted into a problem of finding fixed point for the function f : X → Xdefined by

f(x) = x+ F (x)− y.

THEOREM 155. Let X be a complete metric space and f : X → X be a contraction. Then f has aunique fixed point. In fact, if x0 ∈ X, and for n ∈ N, if xn = f(xn−1), then (xn) converges to a fixedpoint of f , and it is the only fixed point of f .

Proof. Note that, for n ∈ N,

d(xn, xn+1) = d(f(xn−1), f(xn)) ≤ κd(xn−1, xn).

Hence, it follows that d(xn, xn+1) ≤ κnd(x0, x1). Thus, for any n,m ∈ N with n > m,

d(xm, xn) ≤ d(xm, xm+1) + · · ·+ d(d(xn−1, xn)

≤ (κm + κm+1 + · · ·+ κn−1)d(x1, x0)

≤ κm(1 + κ+ · · ·+ κn−1−m)d(x1, x0)

≤ κm

1− κd(x1, x0).

Since 0 < κ < 1, it follows that (xn) is a Cauchy sequence in X. Since X is complete, there existsx ∈ X such that xn → x. Now, since

d(xn, f(xn)) = d(xn, xn+1) ∀n ∈ N.

Hence, taking limit as n → ∞, d(x, f(x)) = d(x, x) = 0. Thus, x is a fixed point of f . If u ∈ X isanother fixed pint of f . Then we get

0 ≤ d(x, u) = d(f(x), f(u)) ≤ κd(x, u).

Since 0 < κ < 1, the above inequality implies that x = u.

9.2 Baire category theorem

We know that intersection of two dense sets in a metric space need not be dense. But, intersection oftwo dense open sets will be dense.

THEOREM 156. Suppose A and B be dense opens sets in a metric space (X, d). Then A ∩ B isdense.

38

Proof. Let D = A ∩ B. Let x ∈ X and r > 0. To prove that B(x, r) ∩ D 6= ∅. Since A isdense, B(x, r) ∩ A 6= ∅. Since B(x, r) ∩ A is open, there exists x1 ∈ X and 0 < r1 < r such thatB(x1, r1) ⊆ B(x, r)∩A. Since B is dense, B(x1, r1)∩B 6= ∅. Since B(x1, r1)∩B is open, there existsx2 ∈ X and 0 < r2 < r1 such that B(x2, r2) ⊆ B(x1, r1) ∩B. Thus,

B(x2, r2) ⊆ B(x1, r1) ∩B ⊆ [B(x, r) ∩A] ∩B = B(x, r) ∩ (A ∩B).

In particular, B(x, r) ∩D 6= ∅.

in the above theorem, there is nothing special about two sets. It can be easily generalized to anyfinite number of sets. Thus, we have

THEOREM 157. Suppose A1, A2, . . . , An be dense opens sets in a metric space (XS, d). Then⋂nk=1Ak is dense.

Proof. Left as an exercise.

Can we generalize the above theorem into infinitely many sets?

THEOREM 158. (Baire category theorem) Suppose (X, d) is a complete metric space andA1, A2, . . . , are dense opens sets in X. Then

⋂∞n=1Ak is dense.

Proof. Let D =⋂∞n=1Ak. Let x ∈ X and r > 0. To prove that B(x, r) ∩D 6= ∅.

Since A1 is dense, B(x, r)∩A1 6= ∅. Since B(x, r)∩A1 is open, there exists x1 ∈ X and 0 < r1 < rsuch that B(x1, r1) ⊆ B(x, r) ∩ A1. Since A2 is dense, B(x1, r1) ∩ A2 6= ∅. Since B(x1, r1) ∩ A2 isopen, there exists x2 ∈ X and 0 < r2 < r1 such that B(x2, r2) ⊆ B(x1, r1) ∩ A2. Continuing this, weobtain x1, x2, . . . and r > r1 > r2 · · · such that

B(xn, rn) ⊆ B(xn−1, rn−1) ∩An ∀n ∈ N with x0 = x, r0 = r.

In particular,xn ∈ B(xn, rn) ⊆ B(xm, rm)∀n ≥ m.

Without loss of generality, we may assume that rn → 0 as n→∞ (e.g., we may take rn+1 < rn/2 forall n ∈ N). Hence, (xn) is a Cauchy sequence in X. Since X is complete there exists x ∈ X such thatxn → x.

Now, since d(xn, xm) < rm for all n ≥ m, letting n → ∞, d(x, xm) ≤ rm for all m ∈ N. In otherwords, x ∈ B(xm, rm). But,

B(xn, rn) ⊆ B(xn−1, rn−1) ∀n ∈ N

so thatx ∈ B(xm, rm) ⊆ B(xm−1, rm−1) ∩Am ⊆ B(x0, r0) ∩Am ∀m ∈ N.

Thus,

x ∈ B(x0, r0)

∞⋂m=1

Am.

In particular, B(x, r) ∩D 6= ∅.

Why the above theorem is called Baire category theorem?

39

Definition 159. A metric space X is said to be of first category if it can be represented as acountable union of nowhere dense sets. It is said to be of second category if it is not of firstcategory. ♦

LEMMA 160. Let A be a subset of a metric space. Then A is nowhere dense iff (A)c is dense.

THEOREM 161. A compete metric space is of second category.

Proof. Let (X, d) is a complete metric space. Suppose it is of first category. Then there exists nowheredense sets B1, B2, . . . such that X =

⋃∞n=1Bn. Then, X =

⋃∞n=1 Bn and hence,

⋂∞n=1(Bn)c = ∅.

But, since Bn is nowhere dense, (Bn)c is open and dense. Hence, by Theorem 158, we arrive at acontradiction.

9.3 Arzela-Ascoli’s theorem

We know that closed and bounded subset of a metric space need not be compact. For example, if X isan infinite set with discrete metric, then it is closed and bounded, but not compact. More generally,if E is a discrete subset of a metric space, then it cannot be compact unless it is a finite set. We mayrecall that a subset of metric space is called a discrete set if every singleton subset of it is an open set.For instance, if R is with usual metric and E = 1, 1

2 ,13 ,

14 . . ., then E is not compact. In Rn with

usual metric, a subset is compact iff it is closed and bounded. This result is called the Heine-Boreltheorem. In particular, the closed unit ball in Rn is compact. Let us consider a few more examples ofmetric spaces in which closed and bounded sets need not be compact. For 1 ≤ p ≤ ∞, let X = `p(N)with metric d defined by d(x, y) := ‖x− y‖p for x, y ∈ `p(N), where we used the notation

‖x‖p :=

( ∞∑i=1

|x(i)|p) 1

p

, 1 ≤ p <∞,

sup|x(i) : i ∈ N, p =∞.

For n ∈ N, let en(j) := δnj , j ∈ N. Then en belongs to the unit circle S := x ∈ `p(N) : ‖x‖p = 1.Note also that ‖en − em‖p = 1 for every n 6= m. Hence, (un) does not have a Cauchy subsequence.Thus, S is not compact.

In this section we consider a characterization of compact subsets of C(Ω) w.r.t. the sup-metricwhen Ω is a compact metric space. Essentially we would like to give a characterization for the totallybounded subsets of C(Ω), which is known as the Arzela-Ascoli’s theorem.

Clearly, if Ω is a finite set with k elements, then C(Ω) can be identified with Rk with sup-metric.Hence, in this case, a subset of C(Ω) is compact iff it is closed and bounded.

What about if Ω with infinite number of elements?

In this case, a closed and bounded set in C(Ω) need not be compact. Consider, for example,Ω = 0, 1, 1

2 ,13 , . . .. Then the unit circle S := x ∈ C(Ω) : ‖x‖∞ = 1 in C(Ω) is closed and bounded.

It is not compact. To see this, let un( 1j ) = δjn for j, n ∈ N and un(0) = 0. Then un ∈ S for all n ∈ N

and ‖un − um‖∞ = 1 for every n 6= m. Thus, (un) does not have a Cauchy sequence, and hence S isnot compact.

Also, consider the case Ω = [0, 1] and X = C(Ω) with sup-metric. Let (tn) be a sequence in (0, 1)such that t1 > t2 > · · · (for example tn = 1

n+1 , n ∈ N). Let un be the hat function centered at tn + 1,

40

that is,

un(t) :=

t− tn+2

tn+1 − tn+2, tn+2 ≤ t ≤ tn+1,

tn − ttn − tn+1

, tn+1 ≤ t ≤ tn,

0, elesewhere

Then, it can be easily seen that ‖un‖∞ = 1 for all n ∈ N and ‖un − um‖∞ = 1 for all n 6= m, so that(un) is a bounded sequence in X which does not have a convergent subsequence. Consequently, Theset u ∈ C[0, 1] : ‖u‖∞ = 1 is not compact.

The Arzela-Ascoli’s theorem gives a characterization of compact subsets of C(Ω) with sup-metricwhenever Ω is a compact metric space.

Definition 162. Let (Ω, δ) be a metric space and F be a family of functions from Ω to R. The familyF is said to be equicontinuous if for every ε > 0, there exists δ > 0 such that

x, y ∈ Ω, d(x, y) < δ ⇒ |f(x)− f(y)| < ε ∀ f ∈ F .

♦

Exercise 163. Let F be a family of functions from Ω to R. If F is equicontinuous, then every f ∈ Fis uniformly continuous. ♦

THEOREM 164. (Arzela-Ascoli’s theorem) Let (Ω, d) be a compact metric space, X := C(Ω)be with sup-metric and E ⊆ C(Ω). Then E is totally bounded iff it is bounded and equicontinuous.

Proof. ⇒): Assume that E is totally bounded. Clearly, E is bounded. For showing its equicontinuity,

let ε > 0 be given. Since E is totally bounded, there exist f1, . . . , fk in E such that E ⊆⋃ki=1BX(fi, ε).

Let f be any arbitrary element in E and let s, t ∈ Ω. Then there exists j ∈ 1, . . . , k such thatf ∈ BX(fj , ε). Thus,

|f(s)− f(t)| ≤ |f(s)− fj(s)|+ |fj(s)− fj(t)|+ |fj(t)− f(t)|≤ ‖f − fj‖∞ + |fj(s)− fj(t)|+ ‖f − fj‖∞≤ 2ε+ |fj(s)− fj(t)|.

Since fj is continuous and Ω is compact, it is uniformly continuous. Hence, there exists δj > 0 suchthat |fj(s) − fj(t)| < ε for every s, t ∈ Ω with d(s, t) < δj . Thus, s, t ∈ Ω with d(s, t) < δj implies|f(s)− f(t)| < 3ε for all f ∈ E.

⇐): Assume that E is bounded and equicontinuous. To show that it is totally bounded. For this,let ε > 0 be given. Let δ > 0be such that |f(t)− f(s)| < ε whenever s, t ∈ Ω and d(s, t) < δ. Since Ω

is totally bounded, there exist t1, . . . , tn ∈ Ω such that Ω =⋃ki=1BΩ(ti, δ). For each i ∈ 1, . . . , n,

let Ai := f(ti) : f ∈ E. Since E is bounded, there exists M > 0 such that ‖f‖∞ ≤ M for allf ∈ E. In particular, |f(ti)| ≤ M for all f ∈ E. Thus, Ai is bounded in R for every i = 1, . . . , n.

Since every bounded subset of R is totally bounded, there exist f(i)1 , . . . , f

(i)mi in E such that Ai ⊆⋃mi

j=1BR(f(i)j (ti), ε). Now, let f be an arbitrary element in E and t ∈ Ω. Let i ∈ 1, . . . , n be such

that d(t, ti) < δ and . Then we have

|f(t)− f (i)j (t)| ≤ |f(t)− f(ti)|+ |f(ti)− f (i)

j (ti)|+ |f (i)j (ti)− f (i)

j (t)|.

41

Hence from (∗),

nx(1− x) =

n∑k=0

(k − nx)2rk(x) ≥ (nδ)2∑k∈Bx

rk(x).

Thus, ∑k∈Bx

rk(x) ≤ nx(1− x)

n2δ2≤ 1

4nδ2.

Therefore,

|f(x)− pn(x)| ≤ ε+ 2‖f‖∞∑k∈Bx

rk(x) ≤ ε+2‖f‖∞4nδ2

.

Let N be such that 2‖f‖∞4nδ2 < ε for all n ≥ N . Then we obtain

|f(x)− pn(x)| ≤ 2ε ∀n ≥ N.

It remains to prove (∗):

n∑k=0

(k − nx)2rk(x) =

n∑k=0

(k2 − 2knx+ n2x2)rk(x)

=

n∑k=0

k2rk(x)−n∑k=0

2knxrk(x) +

n∑k=0

n2x2rk(x)

=

n∑k=0

k2rk(x)− 2nx

n∑k=0

k rk(x) + n2x2n∑k=0

rk(x)

=

n∑k=0

k2rk(x)− 2nx

n∑k=0

k rk(x) + n2x2.

Note that for any x, y ∈ R,

(x+ y)n =

n∑k=0

(n

k

)xkyn−k.

Differentiating w.r.t. x,

n(x+ y)n−1 =

n∑k=1

k

(n

k

)xk−1yn−k,

n(n− 1)(x+ y)n−2 =

n∑k=2

k(k − 1)

(n

k

)xk−2yn−k.

Evaluating the above at y = 1− x,

n =

n∑k=1

k

(n

k

)xk−1(1− x)n−k,

n(n− 1) =

n∑k=2

k(k − 1)

(n

k

)xk−2(1− x)n−k.

⇒

nx =

n∑k=1

k

(n

k

)xk(1− x)n−k =

n∑k=0

krk(x),

43

n(n− 1)x2 =

n∑k=0

k(k − 1)

(n

k

)xk(1− x)n−k

=

n∑k=0

k(k − 1)rk(x)

=

n∑k=0

k2rk(x)−n∑k=2

k rk(x)

=

n∑k=0

k2rk(x)− nx.

Hence,

n∑k=0

(k − nx)2rk(x) =

n∑k=0

k2rk(x)− 2nx

n∑k=0

k rk(x) + n2x2

= [n(n− 1)x2 + nx]− 2nx(nx) + n2x2

= nx(1− x).

10 Power Series

Let (an) be a sequence in R. Consider the power series∑∞n=0 anx

n.

THEOREM 167. (Abel’s theorem) Suppose the power series∑∞n=0 anx

n converges for somex0 6= 0. Then it converges absolutely at every x with |x| < |x0|.

Proof. Note that

|anxn| = |anxn0 |∣∣∣∣ xx0

∣∣∣∣n ∀n ∈ N.

If |x| < |x0|, then∑∞n=0

∣∣∣ xx0

∣∣∣n converges. Hence, by comparison test,∑∞n=0 |anxn| converges.

COROLLARY 168. Suppose the power series∑∞n=0 anx

n diverges at x1 6= 0. Then it diverges atevery x with |x| > |x1|.

Let

D := x ∈ R :

∞∑n=0

anxn converges,

and letR := sup|x| : x ∈ D.

Then we have the following:

1. If R = 0, then∑∞n=0 anx

n only at x = 0.

2. If 0 < R ≤ ∞, then∑∞n=0 anx

n at every x with |x| < R.

44

Definition 169. In the above,

(i) the set D is called the domain of convergence of∑∞n=0 anx

n,

(ii) R is called the radius of convergence of∑∞n=0 anx

n, and

(iii) if R > 0, then (−R,R) is called the interval of convergence of∑∞n=0 anx

n. ♦

We may recall the following:

THEOREM 170. (d’Alembert’s ratio test) Suppose bn 6= 0 for all n ≥ n0 for some n0 ∈ N such

that either

(∣∣∣∣bn+1

bn

∣∣∣∣) converges or diverges to ∞. Let ` := limn→∞

∣∣∣∣bn+1

bn

∣∣∣∣ .(i) If ` < 1, then

∑∞n=1 bn converges absolutely.

(ii) If ` > 1, then∑∞n=1 bn diverges.

THEOREM 171. (Cauchy’s root test) Let(bn) be a sequence in R such that either (|bn|1/n)converges or diverges to ∞. Let ` := lim

n→∞|bn|1/n.

(i) If ` < 1, then∑∞n=1 bn converges absolutely.

(ii) If ` > 1, then∑∞n=1 bn diverges.

The following two theorems are immediate consequences of Theorems 170 and 171.

THEOREM 172. Suppose an 6= 0 for all n ≥ n0 for some n0 ∈ N such that either

(∣∣∣∣an+1

an

∣∣∣∣)converges or diverges to ∞. Let ` := lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ . Then we have the following:

1. If ` = 0, then the power series∑∞n=0 anx

n converges for every x ∈ R.

2. If ` =∞, then the power series∑∞n=0 anx

n converges only at x = 0.

3. If 0 < ` <∞, then the power series∑∞n=0 anx

n converges for every x with |x| < 1/` and divergesfor every x with |x| > 1/`.

THEOREM 173. Let(an) be a sequence in R such that either (|an|1/n) converges or diverges to ∞.Let ` := lim

n→∞|an|1/n. Then we have the following:






n converges for every x with |x| < 1/` and divergesfor every x with |x| > 1/`.

There are situation in which the limits in the above theorems may not exist or if exists then thelimit can be 1.

Example 174. Let bn =

1, n even,2, n odd.

Then∣∣∣ bn+1

bn

∣∣∣ =

2, n even,1, n odd

so that(∣∣∣ bn+1

bn

∣∣∣) is bounded

but limn→∞

∣∣∣∣bn+1

bn

∣∣∣∣ does not exist. Note that |bn|1/n =

1, n even,21/n, n odd

so that limn→∞

|bn|1/n = 1.

Thus, Theorem 172 cannot be applied, whereas Theorem 173 can be applied. ♦

45

LEMMA 175. Suppose (bn) is a sequence of non-negative real numbers and α := lim supn

bn, i.e.,

α := limk→∞ supn≥k bn. Then we have the following:

(i) If α < β , then there exists N ∈ N such that bn ≤ β for all n ≥ N .

(ii) If α > β then there exists a subsequence (bkn) such that bkn ≥ β for all n ∈ N.

Proof. Let ck = supn≥k bn.

(i) Suppose α := limk→∞ supn≥k bn = limk→∞ ck < β. Then there exists k0 such that ck :=supn≥k bn < β for all k ≥ k0. Hence, bk < β for all k ≥ k0.

(ii) Suppose α := limk→∞ supn≥k bn = limk→∞ ck > β. Then Then there exists k0 such thatck := supn≥k bn > β for all k ≥ k0. Hence, for each k ≥ k0, there exists nk such that bnk

> β.

The following two theorem can be proved (Exercise) using the above lemma.

THEOREM 176. (Cauchy’s Root test) Let ` := lim supn|bn|1/n.

1. If ` < 1, then∑∞n=1 bn converges absolutely.

2. If ` > 1, then∑∞n=1 bn diverges.

The following theorem is an immediate consequence of Theorem 176.

THEOREM 177. Let ` := lim supn |an|1/n and R := 1/`. Then we have the following:






n converges for every x with |x| < R and divergesfor every x with |x| > R.

Using the convention 1/0 =∞ and 1/∞ = 0, R :=1

lim supn|an|1/n

is the radius of convergence of

the series∑∞n=0 anx

n.

Example 178. Let an =

1, n even,2n, n odd.

Then |an|1/n =

1, n even,2, n odd

so that limn→∞

|an|1/n does

not exist. Thus, Theorem 173 cannot be applied. But, lim supn|an|1/n = 2. Hence Theorem 177 can

be applied. ♦

THEOREM 179. Let R > 0 be the radius of convergence of∑∞n=0 anx

n. Let f(x) :=∑∞n=0 anx

n

for x ∈ (−R,R). Then for 0 < r < R,∑∞n=0 anx

n converges uniformly on [−r, r], and f is continuouson (−R,R).

Proof. Let 0 < r < ρ < R. Then for |x| ≤ r,

|anxn| = |aρn|∣∣∣∣xρ∣∣∣∣n ≤ |aρn| ∣∣∣∣ rρ

∣∣∣∣n ≤M ∣∣∣∣ rρ∣∣∣∣n ,

46

where M is a bound for (|anρn|). Since∣∣∣ rρ ∣∣∣ < 1, it follows that

∑∞n=0 |anxn| converges uniformly

on [−r, r]. Hence, f is continuous on [−r, r]. Now, let x0 ∈ (−R,R). Taking r > 0 such thatx0 ∈ [−r, r] ⊆ (−R,R), f is continuous at x0.

THEOREM 180. Let R > 0 be the radius of convergence of∑∞n=0 anx

n. Let f(x) :=∑∞n=0 anx

n

for x ∈ (−R,R). Then we have the following.

1. For [a, b] ⊆ (−R,R),∫ baf(x)dx =

∑∞n=0 an

bn+1−an+1

b−a

2. f is differentiable on (−R,R), and its derivative is given by f ′(x) =∑∞n=1 nanx

n−1 on (−R,R).

We make use of the following:

PROPOSITION 181. Let (fn) be a sequence of continuous functions on [a, b].

1. Suppose (fn) converges uniformly on [a, b] and f(x) := limn→∞ fn(x), x ∈ [a, b]. Then f is

continuous and∫ bafn(x)dx→

∫ baf(x)dx.

2. Suppose fn is differentiable for each n ∈ N with f ′n continuous on [a, b], (f ′n) converges uniformlyon [a, b] and (fn) converges at some point x0 ∈ [a, b]. Then (fn) converges uniformly on [a, b],the function f defined by f(x) := limn→∞ fn(x), x ∈ [a, b], is differentiable and f ′ = g, whereg(x) := limn→∞ f ′n(x), x ∈ [a, b].

Proof. 1. Since fn ∈ C[a, b] for all n ∈ N and (fn) converges uniformly on [a, b] and since C[a, b] is aclosed subset of B[a, b] with respect to ‖ · ‖∞, f ∈ C[a, b]. Hence,∣∣∣∣∣

∫ b

a

fn(x)dx−∫ b

a

f(x)dx

∣∣∣∣∣ ≤∫ b

a

|fn(x)− f(x)|dx.

Let ε > 0 be given. Since fn → f uniformly on [a, b], there exists N ∈ N such that |fn(x)− f(x)| < εfor all n ≥ N and for all x ∈ [a, b]. Hence,∣∣∣∣∣

∫ b

a

fn(x)dx−∫ b

a

f(x)dx

∣∣∣∣∣ ≤∫ b

a

|fn(x)− f(x)|dx < (b− a)ε ∀n ≥ N.

2. By the fundamental theorem of Riemann integration,

fn(x) = fn(x0)−∫ x

x0

f ′n(t)dt.

Let α := limn→∞ fn(x0). Then we have

fn(x) = fn(x0)−∫ x

x0

f ′n(t)dt→ α+

∫ x

x0

g(t)dt.

Let

f(x) := α+

∫ x

x0

g(t)dt, x ∈ [a, b].

Then, f(x0) = α and

fn(x)− f(x) = [fn(x0)− α] +

∫ x

x0

[f ′n(t)− g(t)]dt.

47

Let ε > 0 be given. Since f ′n → g uniformly on [a, b], g ∈ C[a, b] and there exists N1 ∈ N suchthat |f ′n(x) − g(x)| < ε for all n ≥ N and for all x ∈ [a, b]. Also, there exists N2 ∈ N such that|fn(x0)− α| < ε for all n ≥ N . Hence,

|fn(x)− f(x)| ≤ |fn(x0)− α|+∫ x

x0

|f ′n(t)− g(t)|dt < ε+ ε(b− a) ∀n ≥ N ∀x ∈ [a, b].

Hence fn → f uniformly on [a, b]. Also, since f(x) := α+∫ xx0g(t)dt for all x ∈ [a, b], f is differentiable

and f ′ = g.

Proof of Theorem 180. Let f(x) :=∑∞n=0 anx

n and fk(x) :=∑kn=0 anx

n for x ∈ (−R,R). Let0 < r < R such that [−r, r] ⊆ (−R,R). Then, by Theorem 179, fn → f uniformly on [−r, r].

1. We may take r such that [a, b] ⊆ [−r, r] ⊆ (−R,R). By Proposition 181,∫ b

a

f(x)dx = limk→∞

∫ b

a

(k∑

n=0

anxn

)dx = lim

k→∞

k∑n=0

∫ b

a

anxn = lim

k→∞

k∑n=0

anbk+1 − ak+1

k + 1.

2. We have

f ′k(x) =

k∑n=1

nanxn−1, x ∈ (−R,R).

Sincelim sup

n|nan|1/n = lim sup

nn1/n|an|1/n = lim sup

n|an|1/n,

R := 1/(lim supn |an|1/n

)is the radius of convergence of

∑∞n=1 nanx

n−1 as well. Hence,∑∞n=1 nanx

n−1

converges for all |x| < R and diverges for |x| > R. Let g(x) :=∑∞n=1 nanx

n−1, |x| < R. Then, usingthe arguments as in (1) above, f ′k → g uniformly on [−r, r], and by Proposition 181, f ′ = g on [−r, r].

If x0 ∈ (−R,R), we may take r such that x0 ∈ (−r, r) ⊆ [−r, r] ⊆ (−R,R) so that f ′(x0) = g(x0).This completes the proof.

11 Fourier Series

11.1 Trigonometric polynomials

Definition 182. A function of the form

f(x) := c0 +

k∑n=1

[an cosnx+ bn sinnx]

is called a trigonometric polynomial. ♦

Note that f is a 2π-periodic function, i.e., f(x+ 2π) = f(x) for all x ∈ R, and for m,n ∈ N,

• 1

π

∫ π

−πcosmx cosnxdx =

1

π

∫ π

−πsinmx sinnxdx =

1, m = n,0, m 6= n

• 1

π

∫ π

−πcosmx sinnxdx =

1, m = n,0, m 6= n

48

Hence,

1

2π

∫ π

−πf(x)dx = c0,

1

π

∫ π

−πf(x) cosmxdx = am,

1

π

∫ π

−πf(x) sinmxdx = bm.

Since

cosnx =einx + e−inx

2, sinnx =

einx − e−inx

2i,

we obtain

f(x) = c0 +

k∑n=1

[an cosnx+ bn sinnx]

= c0 +

k∑n=1

[an

(einx + e−inx

2

)+ bn

(einx − e−inx

2i

)]

=

k∑n=−k

cneinx,

where, for n = 1, 2, . . . , k,

cn =an2

+bn2i

=an − ibn

2=

1

2π

∫ π

−πf(x)[cosnx− i sinnx]dx =

1

2π

∫ π

−πf(x)e−nxdx,

c−n =an2− bn

2i=an + ibn

2=

1

2π

∫ π

−πf(x)[cosnx+ i sinnx]dx =

1

2π

∫ π

−πf(x)enxdx,

so that

cn =1

2π

∫ π

−πf(x)e−nxdx for n = ±1,±2, . . . ,±k.

THEOREM 183. A function defined on R is a trigonometric polynomial iff it is of the formk∑

n=−k

cneinx with cn ∈ C.

Proof. ⇒): This part we have already proved.

⇐): Exercise.

Also, observe that

1

2π

∫ π

−πenxe−mxdx =

1

2π

∫ π

−πe(n−m)xdx =

1, m = n,0, m 6= n

Let

ϕ0 =1√2π, ϕ2n−1 =

1√π

sinnx, ϕ2n(x) :=1√π

cosnx, n ∈ N,

and

en(x) :=1√2πeinx, n ∈ Z.

Then the sets ϕn : n ∈ N0 and en : n ∈ Z are orthonormal in the sense that

〈ϕn, ϕm〉 :=

∫ π

−πϕn(x)ϕm(x)dx =

1, m = n,0, m 6= n

〈en, em〉 :=

∫ π

−πen(x)em(x)dx =

1, m = n,0, m 6= n

49

11.2 Trigonometric series

Definition 184. A series of the form c0 +

∞∑n=1

[an cosnx + bn sinnx] with an, bn ∈ R is called a

trigonometric series. ♦

THEOREM 185. Every trigonometric series is of the form

∞∑n=−∞

cneinx. Conversely every series

of the form

∞∑n=−∞

cneinx is a trigonometric series.

Suppose the trigonometric series∑∞n=−∞ cne

inx converges at some point x ∈ R. Then it alsoconverges at x + 2π. Thus, it is enough to discuss its convergence in an interval of length 2π, inparticular in [−π, π].

THEOREM 186. Suppose cn ∈ C such that∑n∈Z |cn| converges. Then

∑∞n=−∞ cne

inx converges

uniformly on [−π, π], and in that case c0 = 12π

∫ π−π f(x)e−nxdx and

cn =1

2π

∫ π

−πf(x)e−nx for n = ±1,±2, . . . ,±k.

Definition 187. The trigonometric series∑∞n=−∞ cne

inx is called a Fourier series of a 2π-periodic

function f which is integrable on [−π, π] if c0 = 12π

∫ π−π f(x)e−nxdx and

cn =1

2π

∫ π

−πf(x)e−nx for n = ±1,±2, . . . ,±k,

and in that case we write

f(x) ∼∞∑

n=−∞cne

inx.

♦

THEOREM 188. Suppose f is square integrable on [−π, π] and f(x) ∼∑∞n=−∞ cne

inx. If fk(x) :=k∑

n=−k

cneinx, then ∫ π

−π|f(x)− fk(x)|2dx ≤

∫ π

−π|f(x)− g(x)|2dx

for every g of the form g(x) :=∑kn=−k dne

inx with dn ∈ C, n = 0,±1, · · · ,±n. Further,

k∑n=−k

|cn|2 ≤∫ π

−π|f(x)|2dx ∀ k ∈ N,

and consequently, cn → 0 as |n| → ∞.

Proof. For any square integrable functions ϕ,ψ on [−π, π], let us denote

〈ϕ,ψ〉 :=

∫ π

−πϕ(x)ψ(x)dx.

50

THEOREM 189. Let f be square integrable on [−π, π] and f(x) ∼∑∞n=−∞ cne

inx. Let x ∈ (−π, π).If there exists δ > 0 and M > 0 such that

|f(x+ t)− f(x) ≤M |t| ∀ t ∈ (−δ, δ),

then∑∞n=−∞ cne

inx converges to f(x).

Proof. First we observe that

fk(x) :=

k∑n=−k

cneinx

=

k∑n=−k

(1

2π

∫ π

−πf(t)e−inydy

)einx

=1

2π

∫ π

−πf(y)Dk(x− y)dy

=1

2π

∫ π

−πf(x− y)Dk(y)dy,

where

Dk(y) :=

k∑n=−k

einy,

called the Dirichlet kernel. Clearly,1

2π

∫ π

−πDk(y)dt = 1.

Further,

Dk(y) :=

k∑n=−k

einy =sin (k + 1

2 )y

sin y2

,

which can be seen by observing that

(eix − 1)Dk(t) = ei(k+1)t − e−ikt.

Thus,

fk(x)− f(x) =1

2π

∫ π

−π[f(x− y)− f(y)]Dk(y)dy =

1

2π

∫ π

−π[f(x− y)− f(y)]

sin (k + 12 )y

sin y2

dy

=1

2π

∫ π

−π

f(x− y)− f(y)

sin y2

sin (k +1

2)ydy.

Let g(y) := f(x−y)−f(y)sin y

2. Then,

fk(x)− f(x) =1

2π

∫ π

−πg(y)sin (k +

1

2)ydy =

1

2π

∫ π

−πg(y)[sin ky cos

y

2+ cos ky sin

y

2]dy

=1

2π

∫ π

−π

[g(y) cos

y

2] sin ky + [g(y) sin

y

2] cos ky

dy.

Since g(y) cos y2 and g(y) sin y2 are bounded sqyuare integrable functions, by Theorem ??,

1

2π

∫ π

−π[g(y) cos

y

2] sin kydy → 0,

1

2π

∫ π

−π[g(y) sin

y

2] cos kydy → 0

as k →∞.

52

12 Assignments

12.1 Assignment - I

Prove/verify/Justify the following statements:

1. Let (S,) be a partially ordered set and A ⊆ S.

(a) Least upper bound of A, if exists, is unique.

(b) Greatest lower bound of A, if exists, is unique.

(c) If ≺ is an order relation on S, then defined by x y for either x ≺ y or x = y, is apartial order.

2. (a) Let S = N×N and for (m,n), (p, q) in S, define (m,n) ∼ (p, q) ⇐⇒ mq = np. Then ∼ isan equivalence relation on S. In this case the equivalence class of (m,n) is denoted by m

n .The set of all equivalence classes is th set of all positive rational numbers.

(b) Let n ∈ Z. For a, b ∈ N, define a ∼ b ⇐⇒ a−b is a multiple of n. Then ∼ is an equivalencerelation on Z.

(c) Let G be a group and H be a subgroup of G. For a, b ∈ G, define a ∼ b ⇐⇒ a− b ∈ H.Then ∼ is an equivalence relation on S.

(d) Let V be a vector space and W be a subspace of V . For x, y ∈ V , define x ∼ y ⇐⇒x− y ∈W . Then ∼ is an equivalence relation on V .

(e) For a nonempty set X let S = 2X . For A,B in S, define A B ⇐⇒ A ⊆ B. Then is apartial order on S.

(f) Let S = reiθ : 0 ≤ θ < 2π, 0 ≤ r ≤ 1. For z1 = r1eiθ1 , z2 = r2e

iθ2 in S, definez1 z2 ⇐⇒ θ1 = θ2 and r1 ≤ r2. Then is a partial order on S. In this case,

i. for each θ, the set Aθ := reiθ : 0 ≤ r ≤ 1 is a totally ordered subset for which eiθ isan upper bounded.

ii. Every eiθ is a maximal element of S.

3. Let x ∈ C[a, b]. Then there exists t0 ∈ [a, b] such that supa≤t≤b

|x(t)| = |x(t0)|.

4. For x ∈ C[a, b], show that∫ ba|x(t)| dt = 0⇒x(t) = 0 for all t ∈ [a, b].

5. For a non-empty subset, let B(S) be the set of all real valued bounded functions defined on S.For x, y in B(S), let d(x, y) := sup|x(s)− y(s)| : s ∈ S. Then d is a metric on B(S). What isB(N)?

6. Let 1 < p <∞ and let q > 1 be such that 1p + 1

q = 1. For x, y in Rn,

(a)∑nk=1 |xkyk| ≤ (

∑nk=1 |xk|p)

1p (∑nk=1 |yk|q)

1q . (Holder’s inequality)

[You may assume the relation αβ ≤ αp

p + βq

q for every α ≥ 0, β ≥ 0.]

(b) (∑nk=1 |xk + yk|p)

1p ≤ (

∑nk=1 |xk|p)

1p + (

∑nk=1 |yk|p)

1p . (Minkowski’s inequality)

(c) dp(x, y) := (∑nk=1 |xk − yk|p)

1p defines a metric on Rn.

7. Let 1 < p <∞ and let q > 1 be such that 1p + 1

q = 1. For x, y in C[a, b],

53

(a)∫ ba|x(t)y(t)| dt ≤

(∫ ba|x(t)|pdt

) 1p(∫ b

a|y(t)|qdt

) 1q

. (Holder’s inequality)

[You may assume the relation αβ ≤ αp

p + βq

q for every α ≥ 0, β ≥ 0.]

(b)( ∫ b

a|x(t) + y(t)|p dt

) 1p ≤

( ∫ ba|x(t)|pdt

) 1p

+( ∫ b

a|y(t)|pdt

) 1p

. (Minkowski’s inequality)

(c) dp(x, y) :=( ∫ b

a|x(t)− y(t)|pdt

) 1p

defines a metric on C[a, b].

8. Two metrics d and ρ are said to be equivalent if there exist constants C1 > 0, C2 > 0 such thatC1d(x, y) ≤ ρ(x, y) ≤ C2d(x, y) ∀x, y ∈ X.

(a) For any two metrics d and ρ on X, define d ∼ ρ iff d and ρ are equivalent. Then ∼ is anequivalence relation on the set of all metrics on X.

(b) For x, y ∈ Rk, let d∞(x, y) := max|xk − yk| : k = 1, . . . , k. Then for every p with1 ≤ p <∞, dp and d∞ are equivalent metrics on Rk. Further, for any p, r with 1 ≤ p ≤ ∞and 1 ≤ r ≤ ∞, the metrics dp and dρ on Rn are equivalent.

(c) For x, y ∈ C[a, b], let d∞(x, y) := max|x(t) − y(t)| : a ≤ t ≤ b. Then for every p with1 ≤ p <∞, dp and d∞ are not equivalent metrics on C[a, b].

9. Suppose X is a set. A function ρ×X → [0,∞) is called a pseudo metric on X if

(i) ρ(x, x) = 0 ∀x ∈ X,

(ii) ρ(x, y) = ρ(y, x) ∀x, y ∈ X,(iii) ρ(x, y) ≤ ρ(x, z) + ρ(z, y) ∀x, y, z ∈ X.

(a) Let R[a, b] be the set of all real valued Riemann integrable functions defined on [a, b]. For

x, y ∈ C[a, b], let d(x, y) :=∫ ba|x(t)− y(t)|dt. Then d is a pseudo metric, but not a metric

on R[a, b].

(b) The function ρ defined by ρ((x1, x2), (y1, y2)) = |x1 − y1|, (x1, x2), (y1, y2) ∈ R2 is apseudo metric on R2, but not a metric.

(c) Let t0 ∈ [a, b]. Then ρ defined by ρ(x, y) = |x(t0)−y(t0)|, x, y ∈ C[a, b] is a pseudo metricon C[a, b], but not a metric.

10. Let ρ be a pseudo metric on X. For x, y ∈ X, define x ∼ y ⇐⇒ ρ(x, y) = 0.

(a) ∼ is an equivalence relation on X.

(b) X be the set of all equivalence classes. On X, define d([x], [y]) := ρ(x, y). Then d is ametric on X, called the metric induced by the pseudo metric ρ.

(c) Taking X = R[a, b] and ρ as in Problem 4(a), what are the equivalence classes of anx ∈ R[a, b]?

(d) Taking X = C[a, b] and ρ as in Problem 4(c), what are the equivalence classes of anx ∈ C[a, b]?

11. Let (X, d) be a metric space and A ⊆ X. The following are equivalent:

(i) A is closed.




(v) Ac is open.

(vi) For every (xn) in A, if xn → x for some x ∈ X, then x ∈ A.

54

12. Let (X, d) be a metric space, A ⊆ X.

(a) If x0 is a limit point of A, then every open ball containing x0 contains infinitely manypoints from A.

(b) If A is a finite set, then A does not have any limit point.

13. Let (X, d) be a metric space and A ⊆ X.

(i) A = A ∪ ∂A(ii) A = A ∪A′, where A′ is the set of all limit points of A.

(iii) A = ∩F : F is closed in X.(iv) A is open iff A = Ao.

(v) A is closed iff A = A.

14. If A is a subset of R and A is bounded above, then sup(A) ∈ A.

15. Let (X, d) be a metric space.




(iv) Intersection of every arbitrary collection of closed sets in X is closed in X.

(v) Finite union of of closed sets in X is closed in X.

16. Let (X, d) be a metric space and (xn) be a sequence in X. Then xn → x iff for every r > 0,there exists N ∈ N such that xn ∈ B(x, r) for all n ≥ N .

17. Let X be with discrete metric.

(i) Every subset of X is open and closed.

(ii) A sequence in X is convergent iff it is eventually constant.

(iii) A sequence in X is a Cauchy sequence iff it is eventually constant.

18. Let X be with discrete metric, (xn) be a sequence in X and x ∈ X.

(i) xn → x iff every subsequence of (xn) converges to x.

(ii) If (xn) is a sequence, then xn → x iff (xn) has a subsequence which converges to x.

19. (a) Every closed subset of a complete metric space is complete.

(b) Every complete subset of a metric space is closed.

20. Suppose X1, X2 are metric spaces with metrics d1, d2, respectively, and X1 ⊆ X2 and d2|X1×X1=

d1. If X1 is not closed in X2, then X1 is not complete.

21. (a) For a nonempty set S, the metric space B(S) w.r.t. the sup-metric is complete.

(b) C[a, b] is a closed subset of B[a, b].

(c) C[a, b] with sup-metric is complete.

22. If d and ρ are equivalent metrics on a set X, and if d is complete, then ρ is also complete.

23. For 1 ≤ p <∞, (C[a, b], dp) is not complete.

24. Let (X, d) be a metric space and Y ⊆ X. let ρ := d|Y×Y. Then a set A ⊆ Y is open iff there

exists an open set G in X such that A = G ∩ Y .

55

12.2 Assignment - II



(i) A = A ∪ ∂A = A ∪A′

(ii) A is open iff A = Ao.

(iii) A is closed iff A = A.



(ii) A is a closed set.


(i) Ao is the largest open set (w.r.t. the partial order ⊆) contained in A, i.e., if G is an open setsuch that G ⊆ A, then G ⊆ Ao.(ii) A is the smallest closed set (w.r.t. the partial order ⊆) containing A, i.e., if F is a closedset such that A ⊆ F , then A ⊆ F .


(i) Ao is the union of all open sets contained in A, that is,

Ao =⋃G ⊆ X : G open and G ⊆ A.

(ii) A is the intersection of all open sets containing A, that is,

A =⋂G ⊆ X : G open and G ⊆ A.

5. Let (X, d) be a metric space and A and B are subsets of X.

(i) A ⊆ B⇒ A ⊆ B.

(ii) A ⊆ B⇒A0 ⊆ Bo.(iii) If A ⊆ B, then it is not necessary that ∂A ⊆ ∂B.



(ii) x ∈ A′ ⇐⇒ ∀ r > 0, B(x, r) ∩A is an infinite set.

7. Let A ⊆ R.

(i) If A is bounded above, then sup(A) ∈ A.

(ii) If A is bounded below, then inf(A) ∈ A.

8. Every finite subset of a metric space is closed, and it does not have any limit point.

9. Let X = R with usual metric d(x, y) := |x− y|, x, y ∈ R. Let A = (0, 1]. Then

(i) Ao = (0, 1) and A = [0, 1].


(iii) ∂A = 0, 1.

56

10. Let X = R with usual metric d(x, y) := |x− y|, x, y ∈ R. Let A = Z. Then

(i) Ao = ∅ and A = Z.


(iii) ∂A = Z.

11. Let X = R with usual metric d(x, y) := |x− y|, x, y ∈ R. Let A = 1n : n ∈ N. Then

(i) Ao = ∅ and A = A ∪ 0.(ii) A is neither closed nor open.

(iii) ∂A = 0.

12. Let X be with discrete metric and A ⊆ X. Then we have the following (verify):

(i) Ao = A = A.


(iii) ∂A = ∅.

13. Let X = R be with usual (Euclidian) metric and A = (0, 1)× 0. Then we have the following(verify):

(i) Ao = ∅, A = [0, 1]× 0.(ii) A is closed but not open.

(iii) ∂A = [0, 1]× 0.

14. Let X = (0, 1] with usual metric d(x, y) := |x − y|, x, y ∈ (0, 1]. Let A = (0, 12 ] and B = ( 1

2 , 1].Then

(i) A is closed, but not open in X

(ii) B is open, but not closed, in X.

(iii) ∂A = 12 = ∂B.

(iv) Ao = (0, 12 ) and B0 = ( 1

2 , 1]

(v) A = (0, 12 ] and B = [ 1

2 , 1]

15. Let (X, d) be a metric space and Y ⊆ X. Let A ⊆ Y . Then A is open in Y iff there exists anopen set G in X such that A = G ∩ Y . Let X = R with usual metric.

(i) If Y = Z, what are the open subsets of Y ?

(ii) If Y = 1, 12 ,

13 , . . ., then what are the open subsets of Y ?

(iii) If Y = 0, 1, 12 ,

13 , . . ., then what are the open subsets of Y ? Is 0 open in Y ?

16. C[a, b] is not complete w.r.t the metric dp(x, y) :=(∫ b

a|x(t)− y(t)|pdt

) 1p

for any p ∈ [1,∞).

12.3 Assignment - III


1. Every closed and bounded subset of Rk has a limit point.

2. If Ω is a closed and bounded subset of Rk, then C(Ω) ⊆ B(Ω).

57

3. Let Ω is a metric space, and Cb(Ω) be the set of all continuous bounded functions defined on Ω.Then

(a) Cb(Ω) is a closed subset of B(Ω),

(b) Cb(Ω) is complete w.r.t. the sup-metric.

4. The set R \Q of rational numbers is separable w.r.t.the usual metric.

5. A closed set is nowhere4 dense iff its compliment is dense.

6. Every infinite subset of a compact set K has a limit point in K.

7. If E ⊆ Rk and if every infinite subset of E has a limit point in E, then E is compact.

8. If Every infinite subset of a metric space has a limit point, then the metric space separable.

9. Every compact subset of a metric space is separable.

10. Every bounded infinite subset of Rk has a limit point. (This result is known as Bolzano-Weierstrass theorem)

11. Let E be a subset of a metric space X. Then E is disconnected iff there exist disjoint open setsA and B in X such that E ⊆ A ∪B, E ∩A 6= ∅, E ∩B 6= ∅.

12. If E1 and E2 are connected subsets of a metric space such that E1 ∩ E2 6= ∅, then E1 ∪ E2 isalso connected.

13. Closure of a connected set is connected.

14. Interior of a connected set need not be connected.

15. Every nonempty open subset of R is a countable disjoint union of open intervals.

16. If E is a nonempty connected subset of a metric space, and if E is not singleton, then E is anuncountable set.

17. Let f : [a, b]→ R be a continuous function such that for every α ∈ R, either f−1(α) is singletonor #(f−1(α)) ≤ 2. Then, there exists β ∈ R such that f−1(β) is singleton. [ Hint: β is eitherthe maximum value or the minimum value of f .]

18. Let X and Y be metric spaces. If f : X → Y is uniformly continuous, then for every Cauchysequence (xn) in X, the sequence (f(xn)) is Cauchy in Y . Give an example to show that theassumption of uniform continuity of f cannot be dropped.

12.4 Assignment - IV


1. Let (fn) be a sequence of functions defined on an interval I and f is also a function defined onI. Then the following are true:

(a) If there exists a sequence (αn) of positive real numbers such that αn → 0 as n → ∞ and|fn(x)− f(x)| ≤ αn ∀n ∈ N, ∀x ∈ I, then fn → f uniformly.

4A subset of a metric space is nowhere dense if interior of its closure is empty.

58

(b) If fn → f uniformly, then for every sequence (xn) in I, |fn(xn)− f(xn)| → 0 as n→∞.

2. For n ∈ N, let fn(x) = xn, x ∈ I := (0, 1). Then

(a) (fn) converges pointwise to the zero function on I, but not uniformly,

(b) (fn) converges uniformly on (0, a) for any a with 0 < a < 1.

3. For n ∈ N, let fn(x) = nx1+n2x2 , x ∈ I := [0, 1]. Then (fn) converges pointwise to the zero

function on I, but not uniformly.

4. For n ∈ N, let fn(x) = 2nx1+n4x2 , x ∈ I := [0, 1]. Then (fn) converges to the zero function

uniformly.

5. Give an example of a sequence (fn) such that

(a) each fn has continuous derivative on an interval I and

(b) (fn) converges uniformly to a differentiable function f on I,

but f ′n 6→ f pointwise.

6. The series∑∞n=1

cosnxn2 converges uniformly on R.

7. Let R > 0 be the radius of convergence of the power series∑∞n=0 anx

n and let f(x) :=∑∞n=0 anx

n, −R < x < R. Then

(a)∑∞n=0 anx

n converges uniformly on [−r, r] for any r with 0 < r < R,

(b) for any k ∈ N,∑∞n=k

n!(n−k)!akx

n−k converges uniformly on [−r, r] for any r with 0 < r < R,

(c) for any k ∈ N, f (k)(0) exists and ak = f(k)(0)k! .

8. Let R1 > 0 and R2 be the radii of convergence of the power series∑∞n=0 anx

n and∑∞n=0 bnx

n,respectively. If

∑∞n=0 anx

n =∑∞n=0 bnx

n for all x in a neibourhood of 0, then an = bn for alln ∈ N ∪ 0.

9. From the relation 11+x =

∑∞n=0(−1)nxn for |x| < 1, derive log(1 + x) =

∑∞n=0(−1)n x

n+1

n+1 for|x| < 1.

10. From the relation 11+x2 =

∑∞n=0(−1)nx2n for |x| < 1, derive π

4 =∑∞n=0(−1)n 1

2n+1 .

12.5 Assignment - Final


1. Let (S,) be a partially ordered set and A ⊆ S. Then

(a) Least upper bound of A, if exists, is unique.

(b) Greatest lower bound of A, if exists, is unique.

(c) If ≺ is an order relation on S, then defined by x y for either x ≺ y or x = y, is apartial order.

2. (a) Let n ∈ Z. For a, b ∈ N, define a ∼ b ⇐⇒ a−b is a multiple of n. Then ∼ is an equivalencerelation on Z.

59

(b) Let G be a group and H be a subgroup of G. For a, b ∈ G, define a ∼ b ⇐⇒ a− b ∈ H.Then ∼ is an equivalence relation on S.

(c) Let V be a vector space and W be a subspace of V . For x, y ∈ V , define x ∼ y ⇐⇒x− y ∈W . Then ∼ is an equivalence relation on V .

(d) For a nonempty set X let S = 2X . For A,B in S, define A B ⇐⇒ A ⊆ B. Then is apartial order on S.

(e) Let S = reiθ : 0 ≤ θ < 2π, 0 ≤ r ≤ 1. For z1 = r1eiθ1 , z2 = r2e

iθ2 in S, definez1 z2 ⇐⇒ θ1 = θ2 and r1 ≤ r2. Then is a partial order on S. In this case,

i. for each θ, the set Aθ := reiθ : 0 ≤ r ≤ 1 is a totally ordered subset for which eiθ isan upper bounded.

ii. Every eiθ is a maximal element of S.

3. (a) For x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn, let

d∞(x, y) := max|xk − yk| : k = 1, . . . , n,

d1(x, y) =

n∑k=1

|xk − yk|,

d2(x, y) =( n∑k=1

|xk − yk|2)1/2

.

Then d∞, d1, d2 are metrics on Rn; and any two of them are equivalent.

(b) For x, y ∈ C[a, b], let

d∞(x, y) := sup|x(t)− y(t)| : t ∈ [a, b],

d1(x, y) =

∫ b

a

|x(t)− y(t)|dt,

d2(x, y) =(∫ b

a

|x(t)− y(t)|2dt)1/2.

Then d∞, d1, d2 are metrics on C[a, b], and they are not equivalent.

(c) For a non-empty subset, let B(S) be the set of all real valued bounded functions definedon S. For x, y in B(S), let d(x, y) := sup|x(s) − y(s)| : s ∈ S. Then d is a metric onB(S). What is B(N)?

4. Suppose X is a set. A function ρ : X ×X → [0,∞) is called a pseudo metric on X if

(i) ρ(x, x) = 0 ∀x ∈ X,

(ii) ρ(x, y) = ρ(y, x) ∀x, y ∈ X,(iii) ρ(x, y) ≤ ρ(x, z) + ρ(z, y) ∀x, y, z ∈ X.

(a) Let R[a, b] be the set of all real valued Riemann integrable functions defined on [a, b]. For

x, y ∈ C[a, b], let ρ(x, y) :=∫ ba|x(t)− y(t)|dt. Then ρ is a pseudo metric, but not a metric

on R[a, b].

(b) The function ρ defined by ρ((x1, x2), (y1, y2)) = |x1 − y1|, (x1, x2), (y1, y2) ∈ R2 is apseudo metric on R2, but not a metric.

(c) Let t0 ∈ [a, b]. Then ρ defined by ρ(x, y) = |x(t0)−y(t0)|, x, y ∈ C[a, b] is a pseudo metricon C[a, b], but not a metric.

60

5. Let ρ be a pseudo metric on X. For x, y ∈ X, define x ∼ y ⇐⇒ ρ(x, y) = 0.

(a) ∼ is an equivalence relation on X.

(b) Let X be the set of all equivalence classes. On X, define d([x], [y]) := ρ(x, y). Then d is ametric on X, called the metric induced by the pseudo metric ρ.

(c) Taking X = R[a, b] and ρ as in Problem 4(a), what are the equivalence classes of anx ∈ R[a, b]?

(d) Taking X = C[a, b] and ρ as in Problem 4(c), what are the equivalence classes of anx ∈ C[a, b]?

6. Let (X, d) be a metric space and A ⊆ X. The following are equivalent:

(i) A is closed.




(v) Ac is open.

(vi) For every (xn) in A, if xn → x for some x ∈ X, then x ∈ A.


(i) A = A ∪ ∂A(ii) A = A ∪A′, where A′ is the set of all limit points of A.

(iii) A = ∩F : F is closed in X.(iv) A is open iff A = Ao.

(v) A is closed iff A = A.

8. Let X = R with usual metric d(x, y) := |x− y|, x, y ∈ R.

(a) Let A = (0, 1]. Then

(i) Ao = (0, 1) and A = [0, 1].


(iii) ∂A = 0, 1.(b) Let A = Z. Then

(i) Ao = ∅ and A = Z.


(iii) ∂A = Z.

(c) Let A = 1n : n ∈ N. Then

(i) Ao = ∅ and A = A ∪ 0.(ii) A is neither closed nor open.

(iii) ∂A = 0.

9. Let X be with discrete metric and A ⊆ X. Then we have the following (verify):

(i) Ao = A = A.


(iii) ∂A = ∅.

61

10. Let X = R be with usual (Euclidian) metric and A = (0, 1)× 0. Then we have the following(verify):

(i) Ao = ∅, A = [0, 1]× 0.(ii) A is closed but not open.

(iii) ∂A = [0, 1]× 0.

11. Let X = (0, 1] with usual metric d(x, y) := |x − y|, x, y ∈ (0, 1]. Let A = (0, 12 ] and B = ( 1

2 , 1].Then

(i) A is closed, but not open in X

(ii) B is open, but not closed, in X.

(iii) ∂A = 12 = ∂B.

(iv) Ao = (0, 12 ) and B0 = ( 1

2 , 1]

(v) A = (0, 12 ] and B = [ 1

2 , 1]

12. Let (X, d) be a metric space, A ⊆ X.

(a) If x0 is a limit point of A, then every open ball containing x0 contains infinitely manypoints from A.

(b) If A is a finite set, then A does not have any limit point.

13. If A is a subset of R and A is bounded above, then sup(A) ∈ A.

14. Let (X, d) be a metric space.




(iv) Inersection of every arbitrary collection of closed sets in X is closed in X.

(v) Finite union of of closed sets in X is closed in X.

15. Let (X, d) be a metric space and (xn) be a sequene in X. Then xn → x iff for every r > 0, thereexists N ∈ N such that xn ∈ B(x, r) for all n ≥ N .

16. Let X be with discrete metric.

(i) Every subset of X is open and closed.

(ii) A sequence in X is convergent iff it is eventually constant.

(iii) A sequence in X is a Cauchy sequence iff it is eventually constant.

17. Let X be a metric space, (xn) be a sequence in X and x ∈ X.

(i) xn → x iff every subsequenece of (xn) converges to x.

(ii) xn → x iff every subsequence of (xn) has a subsequence which converges to x.

18. Let (X, d) be a metric space and Y ⊆ X. Then a set A ⊆ Y is open in Y (w.r.t. the inducedmetric on Y ) iff there exists an open set G in X such that A = G ∩ Y .

19. Let X be a metric space and A and Y be such that A ⊆ Y ⊆ X. Then

(a) Suppose Y is closed in X. Then A is closed in Y iff A is closed in X.

62

(b) Suppose Y is open in X. Then A is open in Y iff A is open in X.

(c) Result in (19a) need not hold if Y is not closed in X.

(d) Result in (19b) need not hold if Y is not open in X.



(ii) A is a closed set.


(i) Ao is the largest open set (w.r.t. the partial order ⊆) contained in A, i.e., if G is an open setsuch that G ⊆ A, then G ⊆ Ao.(ii) A is the smallest closed set (w.r.t. the partial order ⊆) containing A, i.e., if F is a closedset such that A ⊆ F , then A ⊆ F .


(i) Ao is the union of all open sets contained in A, i.e., Ao =⋃G ⊆ X : G open and G ⊆ A.

(ii) A is the intersection of all open sets containing A, i.e., A =⋂G ⊆ X : G open and G ⊆ A.

23. Let (X, d) be a metric space and A and B are subsets of X.

(i) A ⊆ B⇒ A ⊆ B.

(ii) A ⊆ B⇒A0 ⊆ Bo.(iii) If A ⊆ B, then it is not necessary that ∂A ⊆ ∂B.



(ii) x ∈ A′ ⇐⇒ ∀ r > 0, B(x, r) ∩A is an infinite set.

25. (a) Every closed subset of a complete metric space is complete.

(b) Every complete subset of a metric space is closed.

26. Let X be a metric space and Y ⊆ X. If Y is not closed in X, then Y is not complete (w.r.t.the induced metric).

27. (a) For a nonempty set S, the metric space B(S) w.r.t. the sup-metric is complete.

(b) C[a, b] is a closed subset of B[a, b].

(c) C[a, b] with sup-metric is complete.

28. If d and ρ are equivalent metrics on a set X, and if d is complete, then ρ is also complete.

29. Then metric space (C[a, b], dp) for p ∈ 1, 2 is not complete.

30. Let X be a complete metric space and (Fn) is a sequence of closed nonempty subsets of X suchthat Fn ⊇ Fn+1 for all n ∈ N and diam(Fn) → 0 as n → ∞. Then

⋂∞n=1 Fn is nonempty and

singleton.

31. Let A ⊆ R.

(i) If A is bounded above, then sup(A) ∈ A.

(ii) If A is bounded below, then inf(A) ∈ A.

32. Every finite subset of a metric space is closed, and it does not have any limit point.

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33. Every closed and bounded subset of Rk has a limit point.

34. If Ω is a closed and bounded subset of Rk, then C(Ω) ⊆ B(Ω).

35. Let Ω be a metric space, and Cb(Ω) be the set of all continuous bounded functions defined onΩ. Then

(a) Cb(Ω) is a closed subset of B(Ω),

(b) Cb(Ω) is complete w.r.t. the sup-metric.

(c) Ω compact implies C(Ω) = Cb(Ω).

36. If Ω is a compact metric space, then for every f ∈ C(Ω), there exists u, v ∈ Ω such thatf(u) = sup

x∈Ωf(x) and f(v) = inf

x∈Ωf(x).

37. Let X and Y metric spaces and f : X → Y is a continuous bijective function. If X is compact,then f−1 : Y → X is continuous.

38. Let E be a bounded subset of R (w.r.t. the usual metric). If E is not closed, then

(a) there exists a continuous function f on E such that f(E) is not bounded.

(b) there exists a continuous bounded function f on E such that f does not attain maximumon E.

[Hint: Take x0 ∈ E \ E and f1(x) := 1/(x− x0), f2(x) = 1/[1 + (x− x0)2].]

39. The set R \Q of rational numbers is separable5 w.r.t.the usual metric.

40. A closed set is nowhere6 dense iff its compliment is dense.

41. Every infinite subset of a compact set K has a limit point in K.

42. If E ⊆ Rk and if every infinite subset of E has a limit point in E, then E is compact.

43. If Every infinite subset of a metric space has a limit point, then the metric space separable.

44. Every compact subset of a metric space is separable.

45. Every bounded infinite subset of Rk has a limit point. (This result is known as Bolzano-Weierstrass theorem)

46. Let X be a metric space and E ⊆ X. Then the following are equivalent:

(a) There exist disjoint open sets A and B in X such that E ⊆ A∪B, E∩A 6= ∅, E∩B 6= ∅.(b) E is disconnected w.r.t. the induced metric on E.

(c) There exist nonempty sets A and B such that E = A ∪B, A ∩B = ∅, A ∩ B = ∅.

47. Let E be a subset of a metric space X. Then E is disconnected iff there exist disjoint open setsA and B in X such that E ⊆ A ∪B, E ∩A 6= ∅, E ∩B 6= ∅.

48. If E1 and E2 are connected subsets of a metric space such that E1 ∩ E2 6= ∅, then E1 ∪ E2 isalso connected.

49. Closure of a connected set is connected.

5A metric space is separable if its closure is the whole space.6A subset of a metric space is nowhere dense if interior of its closure is empty.

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50. Interior of a connected set need not be connected.

51. Every nonempty open subset of R is a countable disjoint union of open intervals.

52. If E is a nonempty connected subset of a metric space, and if E is not singleton, then E is anuncountable set.

53. Let f : [a, b]→ R be a continuous function such that for every α ∈ R, either f−1(α) is singletonor #(f−1(α)) ≤ 2. Then, there exists β ∈ R such that f−1(β) is singleton. [ Hint: β is eitherthe maximum value or the minimum value of f .]

54. Let X and Y be metric spaces. If f : X → Y is uniformly continuous, then for every Cauchysequence (xn) in X, the sequence (f(xn)) is Cauchy in Y . Give an example to show that theassumption of uniform continuity of f cannot be dropped.

55. Let (fn) be a sequence of functions defined on an interval I and f is also a function defined onI. Then the following are true:

(a) If there exists a sequence (αn) of positive real numbers such that αn → 0 as n → ∞ and|fn(x)− f(x)| ≤ αn ∀n ∈ N, ∀x ∈ I, then fn → f uniformly.

(b) If fn → f uniformly, then for every sequence (xn) in I, |fn(xn)− f(xn)| → 0 as n→∞.

56. For n ∈ N, let fn(x) = xn, x ∈ I := (0, 1). Then

(a) (fn) converges pointwise to the zero function on I, but not uniformly,

(b) (fn) converges uniformly on (0, a) for any a with 0 < a < 1.

57. For n ∈ N, let fn(x) = nx1+n2x2 , x ∈ I := [0, 1]. Then (fn) converges pointwise to the zero

function on I, but not uniformly.

58. For n ∈ N, let fn(x) = 2nx1+n4x2 , x ∈ I := [0, 1]. Then (fn) converges to the zero function

uniformly.

59. (Dini’s theorem): Let (fn) be a sequence of continuous (real valued) functions defined on acompact metric space Ω such that fn ≥ fn+1 for all n ∈ N and fn → f pointwise on Ω. Thenfn → f uniformly on Ω.

(Hint: Take gn = fn − f and for ε > 0, define En := x ∈ Ω : gn(x) ≥ ε; observe that Ens arecompact and En ⊇ En+1 for all n ∈ N; use nested theorem.)

60. Let (fn) be a sequence of continuous (real valued) functions defined on a compact metric spaceΩ and f : Ω → R. Suppose fn → f pointwise on Ω. If (fn) is equicontinuous, then fn → funiformly on Ω.

61. Give an example of a sequence (fn) such that

(a) each fn has continuous derivative on an interval I and

(b) (fn) converges uniformly to a differentiable function f on I,

but f ′n 6→ f pointwise.

62. The series∑∞n=1

cosnxn2 converges uniformly on R.

63. Let R > 0 be the radius of convergence of the power series∑∞n=0 anx

n and let f(x) :=∑∞n=0 anx

n, −R < x < R. Then

(a)∑∞n=0 anx

n converges uniformly on [−r, r] for any r with 0 < r < R,

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(b) for any k ∈ N,∑∞n=k

n!(n−k)!akx

n−k converges uniformly on [−r, r] for any r with 0 < r < R,

(c) for any k ∈ N, f (k)(0) exists and ak = f(k)(0)k! .

64. Let R1 > 0 and R2 be the radii of convergence of the power series∑∞n=0 anx

n and∑∞n=0 bnx

n,respectively. If

∑∞n=0 anx

n =∑∞n=0 bnx

n for all x in a neibourhood of 0, then an = bn for alln ∈ N ∪ 0.

65. From the relation 11+x =

∑∞n=0(−1)nxn for |x| < 1, derive log(1 + x) =

∑∞n=0(−1)n x

n+1

n+1 for|x| < 1.

66. From the relation 11+x2 =

∑∞n=0(−1)nx2n for |x| < 1, derive π

4 =∑∞n=0(−1)n 1

2n+1 .

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