Metric Spaces

Lecture Notes, Exercises and Sample Examination Paper, Fall 2015

M.van den Berg

School of MathematicsUniversity of BristolBS8 1TW Bristol, UKmamvdb@bristol.ac.uk

1 Definition of a metric space.

1

Let X be a set, X 6= . Elements of X will be called points.

Definition 1.1 (a) A map d : X X R is called a metric on X if

(M1) ((x, y) X X) [0 d(x, y)

This proves (M1). (M2) and (M3) follow immediately from Definition 1.1. To prove (M4) we letz `2. Let n N be arbitrary and let ai = xi zi, bi = zi yi. Since{

ni=1

(ai + bi)2

}1/2{

ni=1

a2i

}1/2+

{ni=1

b2i

}1/2()

we have that {ni=1

(xi yi)2}1/2

d(x, z) + d(z, y).

We obtain (M4) by letting n.To prove () we note that this equivalent to

ni=1

aibi {

ni=1

a2i

}1/2{ ni=1

b2i

}1/2.

The above follows directly from the fact that for Rni=1

(ai + bi)2 = A2 + 2H+B 0,

where A =ni=1 a

2i , B =

ni=1 b

2i , H =

ni=1 aibi. Since the quadratic in is non-negative we

have that H2 AB. 2

Theorem 1.7 Let X be a non-empty set, and let B(X) be the set of all bounded real-valuedfunctions from X to R. Define d : B(X)2 R by

d(f, g) = supxX|f(x) g(x)|.

Then d is a (supremum) metric on B(X).

Proof. Since f, g B(X) there exist constants c1, c2 such that |f(x)| c1, |g(x)| c2 on X.Then 0 |f(x) g(x)| |f(x)|+ |g(x)| c1 + c2. Hence

0 supxX|f(x) g(x)| c1 + c2.

This proves (M1). (M2) and (M3) follow immediately from Definition 1.1. To prove (M4) we leth B(X). Then there exist c3 such that |h(x)| c3 on X. Hence

|f(x) g(x)| |f(x) h(x)|+ |h(x) g(x)| supxX |f(x) h(x)|+ supxX |h(x) g(x)| = d(f, h) + d(h, g),

andsupxX|f(x) g(x)| d(f, h) + d(h, g).

2

Definition 1.8 Let (X, d) be a metric space and let Y be a non-empty subset ofX. The restrictionof d : X X R to d : Y Y R induces a metric on Y .

Definition 1.9 (a) Let X be a vector space over R (or C). A mapping

: X R

2

is called a norm if

(N1) (x X) [0 x

2 Convergence in metric spaces.

Definition 2.1 Let (X, d) be a metric space. A sequence of points x1, x2, . . . (shorthand (xn))inX is convergent if there is an element x X such that

limn d(xn, x) = 0.

We write limnxn = x or xn x.

Example 2.2 Let (RN , d) be the standard Euclidean space. Thenx(n) = (x

(n)1 , . . . , x

(n)N ) x = (x1, . . . , xN ) if and only if x(n)j xj for all 1 j N .

Proof.

d((x(n)1 , . . . , x

(n)N ), (x1, . . . , xN )) =

Nj=1

(x(n)j xj)2

1/2

Nj=1

|x(n)j xj |.

Also for j = 1, . . . , N

|x(n)j xj | d(x(n), x).2

Definition 2.3 Let X be a set and let (Y, d) be a metric space. Let (fn), fn : X Y , be asequence of functions. Then (fn) converges pointwise if there exists a function f : X Y suchthat

(x X) [ limn d(fn(x), f(x)) = 0].

(fn) converges uniformly if there exists a function f : X Y such that

limn supxX

d(fn(x), f(x)) = 0,

or( > 0) (N N) [(n N) = sup

xXd(fn(x), f(x)) <

].

Remark 2.4 Uniform convergence implies pointwise convergence. The converse need not be true.Convergence of a sequence of (fn), fn B(X), n = 1, 2, with the supremum metric is nothingbut uniform convergence.

Example 2.5 Let fn : (0, 1) R, fn(x) = xn. Then fn f pointwise, where f(x) = 0, 0 < x 0 there exists N N such that n N, x [a, b] implies|fn(x) f(x)| < . Hence for any partition P of [a, b], and for n N

U(P, f) U(P, fn) + (b a), L(P, f) L(P, fn) (b a).

5

Since fN is Riemann integrable, there exists a partition P of [a, b] such that U(P , fN )L(P , fN ) 0 was arbitrary, f is Riemann integrable. Hence for n N b

a

f(x)dx = inf{U(P, f) : P a partition of [a, b]} inf{U(P, fn) : P a partition of [a, b]}+ (b a)

=

ba

fn(x)dx+ (b a).

Similarly, for n N ba

f(x)dx ba

fn(x)dx (b a).

Combining these two inequalities we have for n N

| ba

f(x)dx ba

fn(x)dx| 2(b a).

2

Remark 2.12 Let fn : [0,) R, fn(x) = 1nex/n. Then (fn) converges uniformly to the 0function on [0,). However,

0fn(x)dx = 1.

Remark 2.13 If fn f pointwise on [a, b], and fn is Riemann integrable for all n N. Then fneed not to be Riemann integrable. For example, let {q1, q2 . . . } be an enumeration of the rationalsin [0, 1], fn(x) =

{1, x {q1, . . . , qn},0, otherwise.

Then fn is Riemann integrable with 10fn(x)dx =

0, fn f pointwise with f(x) ={

1, x Q [0, 1]0, x [0, 1] Q.

But U(P, f) = 1, L(P, f) = 0, and f is not Riemann integrable.

Corollary 2.14 Ifn=1 fn converges uniformly to s on [a, b], and if each fn is Riemann inte-

grable on [a, b]. Then s is integrable, and bas(x)dx =

n=1

bafn(x)dx.

Proof. Put sn = f1 + + fn. Then bas(x)dx = limn

basn(x)dx

= limnnk=1

bafk(x)dx =

k=1

bafk(x)dx. 2

Theorem 2.15 Let (fn) be a sequence of real-valued differentiable functions on [a, b]. Supposethat (fn(x)) converges for at least one x [a, b], and that f n converges uniformly on [a, b]. Then(i) fn converges uniformly on [a, b],(ii) f = limn fn is differentiable on [a, b], and for all x [a, b], f (x) = limn f n(x).

Proof. Let limn f n = g, and let > 0. Since (fn) converges uniformly on [a, b], there exists

N N such that supaxb |f n(x) g(x)| < for n N . Hence supaxb |f n(x) f m(x)| 2 form,n N . If 1, 2 are points of [a, b] then, by the mean value theorem for fn fm

fn(1) fm(1) (fn(2) fm(2)) = (1 2).{f n() f m()}where [1, 2]. So for all m,n N ,

(1) |fn(1) fm(1) (fn(2) fm(2))| 2|1 2|.

6

If x0 is a point for which (fn(x0)) converges, then there is an N N such that for m,n N

(2) |fn(x0) fm(x0)| .Let x be any point of [a, b]. By (1), (2) with 1 = x, 2 = x0 we have for m,n max{N,N },

(3) |fn(x) fm(x)| (1 + 2|x x0|) (1 + 2(b a)).Hence limm fm(x) exists for all x [a, b]. Taking the limit m in (3) we see that (fn)converges uniformly on [a, b]. Let a < c < b, and let h 6= 0 be such that c+ h [a, b].By (1) with 1 = c+ h, 2 = c we have for m,n Nfn(c+ h) fn(c)h fm(c+ h) fm(c)h

2.Since fm(c+ h) fm(c) f(c+ h) f(c) as m, the above inequality shows that for n Nfn(c+ h) fn(c)h f(c+ h) f(c)h

2.By the definition of the derivative f N (c) there exists on > 0 such that

(0 < |h| ) =(fN (c+ h) fN (c)h f N (c)

) .Then for 0 < |h| f(c+ h) f(c)h g(c)

f(c+ h) f(c)h fN (c+ h) fN (c)h

+

fN (c+ h) fN (c)h f N (c) + |f N (c) g(c)| 2+ + .

Hence f (c) exists and f (c) = limn f n(c). 2

Corollary 2.16 Let un : [a, b] R be such that

(i) (x0 [a, b])( n=1

un(x0) is convergent

).

(ii) (n N) un is differentiable.

(iii) The series

n=1

un(x) is uniformly convergent on [a, b].

Then the series n=1

un(x)

converges uniformly to s(x) and

s(x) =n=1

un(x).

Theorem 2.17 If xn x in (X, d) and xn y in (X, d) then x = y.Proof.

0 d(x, y) d(x, xn) + d(xn, y) 0 = x = y.2

7

Theorem 2.18 Let xn x and yn y in (X, d). Then d(xn, yn) d(x, y).

Proof.0 |d(xn, yn) d(x, y)| d(x, xn) + d(yn, y) 0.

2

Definition 2.19 Let (X, d) and (X, e) be metric spaces. The metrics d and e are equivalent if

d(xn, x) 0 e(xn, x) 0.

Remark 2.20 If there are c1, c2 > 0 such that

x, y X c1d(x, y) e(x, y) c2d(x, y)then d and e are equivalent.

Example 2.21 On R2

d((x1, y1), (x2, y2)) =

(x1 x2)2 + (y1 y2)2

ande((x1, y1), (x2, y2)) = |x1 x2|+ |y1 y2|

are equivalent.

Proof.12e((x1, y1), (x2, y2)) d((x1, y1), (x2, y2)) e((x1, y1), (x2, y2)).

2

3 Limit points, open sets, closed sets

Let (X, d) be a metric space, x0 X, r > 0.

Definition 3.1 The set B(x0; r) = {x X : d(x, x0) < r} is the open ball with center x0 andradius r. B(x0; r) = {x X : d(x, x0) r} is the closed ball with center x0 and radius r.

Example 3.2 Let R1 have its usual metric, a R1. Then B(a; r) = (a r, a+ r),B(a; r) = [a r, a+ r]. Similarly for Rn, a Rn, B(a; r) = {(x1, . . . , xn) Rn :

ni=1(xi ai)2 0there is an x E such that 0 < d(x, c) < , i.e. (B(c; )\{c}) E 6= .

Theorem 3.5 Let c be a limit point of E. Then for every > 0 the ball B(c; ) contains infinitelymany points of E. Moreover, there exists a sequence (xn) of points in E such that d(xn, c) ispositive, and strictly decreasing to 0.

Proof. For n N, (B(c; 1n )\{c}) E 3 xn. Then 0 < d(xn, c) < 1n (). Hence xn c. The setof distinct points {x1, x2, . . . } is infinite. Otherwise min{d(x1, c), . . . , d(xn, c)} is bounded awayfrom 0, contradicting (). 2

8

Example 3.6 Let E be a finite set in a metric space (X, d). Then E does not have any limitpoints.

Proof. Since E is finite B(c; 1) contains finitely many points of E. Then c is not a limit point ofE by Theorem 3.5. 2

Example 3.7 (i) Let R have the Euclidean metric. Then any c R is a limit point of Q.Proof. For any > 0, (c , c) will contain a point of Q. Hence (B(c; )\{c}) Q 6= . 2

(ii) Let E ={

1, 12 ,13 , . . .

}. Then 0 is the only limit point of E.

Proof. B(0; ) 3 1n for n N, n > 1 . If x E then x = 1n for some n N. ThenB(1n ;

1n 1n+1

)\{x} does not contain points of E. If x < 0, then B (x;x) E = . If

0 < x < 1, x / E then B(x; ) E = for = min{ 1n x, x 1n+1}, where n is such that1

n+1 < x 1, then B (x;x 1) E = . So 0 is the only limit point of E. 2

Example 3.8 A discrete metric space does not have any limit points.

Definition 3.9 A point c of E is an isolated point of E if c is not a limit point of E.

Example 3.10 A discrete metric space consists of isolated points. The subset [0, 1) of R doesnot have isolated points. The set of limit points of [0, 1) is the set [0, 1].

Definition 3.11 Given a set E X. Then c is an interior point of E if there is an > 0 suchthat B(c; ) E. A set E X is open if every point of E is an interior point of E.

Example 3.12 Let (X, d) be a metric space. Then and X are open sets.

Example 3.13 The open ball B(a; r) is open.

Proof. Let x B(a; r). Then d(x, a) < r. Let = r d(x, a). Then > 0. Let y B(x; ).Then d(x, y) < and d(a, y) d(a, x) + d(x, y) < d(a, x) + = d(a, x) + r d(x, a) = r. Hencey B(a; r). Since y B(x; ) was arbitrary, we conclude B(x; ) B(a; r). Hence x is an interiorpoint, and B(a; r) consists of interior points. 2

Theorem 3.14 Let {A : T} be a collection of open sets in (X, d). ThenT

A

is open.

Proof. Suppose a T A. Then there exists 0 T such that a A0 . Since A0 is openwe have

r > 0 : B(a; r) A0 T

A.

2

Theorem 3.15 Let {Ai : i = 1, 2, . . . , N} be a finite collection of open sets in (X, d). ThenNk=1

Ak

is open.

9

Proof. Suppose a Nk=1Ak. Then ( k N)[(1 k N) a Ak]. Since Ak is open, wehave

(k) (1 k N) (rk > 0) : B(a; rk) Ak.Let r = min{r1, . . . , rN}, Then r > 0, and

(k N)[(1 k N) B(a; r) =Nk=1

B(a; rk) B(a; rk) Ak].

2

Example 3.16 Let Ak = ( 1k , 1k ) R1, k N, where R1 has the usual metric.Then

k=1

Ak = {0},

which is not open.

Definition 3.17 A set A is closed in (X, d) if X\A is an open set in (X, d).

Example 3.18 X and are closed sets.

Theorem 3.19 A set A in a metric space (X, d) is closed if and only if it contains all its limitpoints.

Proof. Suppose A is closed. Let x X\A. Since X\A is open there exists r > 0 such thatB(x; r) A = . Hence x is not a limit point of A. Suppose A contains all its limit points. Letx X\A. Then x is neither a limit point of A nor a point of A. Then there exists > 0 such thatB(x; ) A = . Then X\A is open and A is closed.

Theorem 3.20 The intersection of any collection of closed sets is closed. The union of a finitecollection of closed sets is closed.

Proof. Let {A : T} be a collection of closed sets. Then {Ac : T} is a collection ofopen sets. By Theorem 3.14

T A

c is open. By Definition 3.17

(T A

c

)cis closed. By De

Morgans law we have that(

T Ac

)c=T A

cc =

T A is closed.

Let {Ai : i = 1, 2, . . . , N} be a collection of closed sets. Then {Aci : i = 1, 2, . . . , N} is acollection of open sets. By Theorem 3.15

Ni=1A

ci is open. By Definition 3.17

(Ni=1A

ci

)cis

closed. By De Morgans law(N

i=1Aci

)c=Ni=1A

cci =

Ni=1Ai is closed.

Example 3.21 (i) The closed ball B(a; r), a X, r > 0 is a closed set.Let x X\B(a; r). Then d(x, a) > r. Let = d(x, a) r and y B(x; ). Then y / B(a; r)since d(a, y) d(a, x) d(x, y) > d(a, x) = r.

(ii) A finite set in a metric space is closed.

(iii) [0, 1] is a closed subset of R1.

Definition 3.22 Let E be any subset of X, in a metric space (X, d).

(i) The interior of E, denoted by int(E), is the set of interior points of E.

(ii) The set of limit points of E is denoted by E.

(iii) The closure of E, denoted by E, is the set E E.

10

(iv) The boundary of E, denoted by b(E) is the set E\int(E).

Theorem 3.23 For any set E in a metric space (X, d)

(i) int(E) is open,

(ii) E is closed,

(iii) b(E) is closed,

(iv) E is closed.

Proof.

(i) If int(E) = then it is open. If int(E) 6= , let x int(E). Then there exists a > 0 suchthat B(x; ) E. Let y B(x; ). Then B(y; d(x, y)) B(x; ) E. Hence y int(E).Hence B(x; ) int(E). Then int(E) is a union of open balls, and hence is open. Note that xis an interior point of int(E). So int(E) int(int(E)) int(E). Hence int(E) = int(int(E)).

(ii) If E = X. Then E is closed. If X\E 6= , let x X\E. Then x / E, x / E. Hence thereexists a > 0 such that B(x; )E = . We will show that B(x; )E = . Suppose to thecontrary and let x0 B(x; )E. Then d(x, x0) > 0, and B(x0; d(x, x0))E 6= sincex0 is a limit point of E. Then B(x; )E 6= for all > 0 since B(x0; d(x, x0)) B(x; ).Contradiction. Hence B(x; ) E = for some > 0, and B(x; ) E = . Then X\E isopen and E is closed.

(iii) b(E) = E\int(E) = E (X\int(E)). Since X\int(E) is closed, b(E) is intersection of twoclosed sets.

(iv) Let x be a limit point of E. Then for any > 0, B(x; )\{x} contains a point x0 of E.Let > 0 be a such that B(x0; ) B(x; )\{x}. Since B(x0; )\{x0} contains a point ofE,B(x; )\{x} contains a point of E. Then x E and E contains its limit points. HenceE is closed.

2

Theorem 3.24 For any set E in a metric space (X, d),

b(E) = {c X : for all > 0, B(c; ) E 6= and B(c; ) (X\E) 6= }.Proof. Suppose x b(E). Then x E\int(E). Then x is not an interior point of E, and so forall > 0, B(x; ) (X\E) 6= . Also x E or x E. Hence B(x; ) E 6= for all > 0. Hencex {c X : for all > 0, B(c; ) E 6= , B(c; ) (X\E) = }.

Suppose x {c X : for all > 0, B(c; ) E 6= , B(c; ) (X\E) 6= }. Then x 6 int(E)and, x E or x E. Hence x / int(E), x E. 2

Theorem 3.25 Let C[a, b] be the set of continuous real valued functions. Then C[a, b] is a closedsubset of the metric space (B[a, b], d) with the supremum metric.

Proof. Let f C[a, b], and let (fn) be a sequence in C[a, b] such that fn f in B[a, b]. Itsuffices to show that f C[a, b]. Fix t0 [a, b], > 0. Let N N be such that d(fN , f) < /3.Since fN is continuous we have

> 0 t (t0 , t0 + ) [a, b] : |fN (t) fN (t0)| < /3.Therefore for the same t

|f(t) f(t0)| |f(t) fN (t)|+ |fN (t) fN (t0)|+ |fN (t0) f(t0)| /3 + /3 + /3 = .Hence f C[a, b]. 2

11

Theorem 3.26 (i) If A B, and B is closed then A B.(ii) If C D and C is open then C int(D).

Proof. (i) If A is closed then A = A, and there is nothing to prove. Otherwise, assume thatA \ A 6= . Let x A \ A. Suppose that x 6 B. Then x is an interior point of Bc since Bc isopen. Therefore x is an interior point of Ac since Bc Ac. This is a contradiction since x is alimit point of A.

(ii) Let x C. Since C is open there exits > 0 such that B(x; ) C. Hence B(x; ) D,so that x int(D). 2

Theorem 3.27 Let (X, d) be a metric space, x X and r > 0. Then B(x; r) B(x; r). Theinclusion is not necessarily valid in the other direction. However, in Rn with the usual metricB(x; r) = B(x; r)

Proof. B(x; r) B(x; r). By 3.21 B(x; r) is closed. By 3.26(i) B(x; r) B(x; r). To see that theinclusion is not necessarily valid in the other direction we let d be the discrete metric on a set Xwith at least two elements say x0 and x1. Then B(x0; 1) = {x0} and B(x0; 1) = {x0} since thereare no limit points in a discrete metric space. But B(x0; 1) = X and X 6= {x0}. Let Rn have theusual metric, and let x Rn, y Rn with d(x, y) = r. It suffices to show that y B(x; r). Let (0, r), x = (x1, . . . , xn), y = (y1, . . . , yn) and x = ((1 2r )y1 + 2rx1, . . . , (1 2r )yn + 2rxn).Then d(x, y) = 2rd(x, y) =

2 . Hence x 6= y. Moreover d(x, x) = (1 2r )d(x, y) = r 2 < r. Hence

x B(x; r) and y is a limit point of B(x; r). 2

4 Functions in metric spaces

Definition 4.1 Let (X, ) and (Y, ) be metric spaces. Let A X, x0 be a limit point of A, i.e.x0 A, and y0 Y . Let f : A Y . Then y0 is the limit of f at x0 if

( > 0) ( > 0) (x A)[(0 < (x, x0) < ) = ((f(x), y0) < )].

Notation: limxx0

f(x) = y0 orf(x) y0 as x x0.

Theorem 4.2 Let (X, ) and (Y, ) be metric spaces. Let A X, x0 be a limit point of A, i.e.x0 A, and y0 Y . Let f : A Y . Then y0 = limxx0 f(x) if and only if for every sequence(xn) in A \ {x0}

((xn, x0) 0) = ((f(xn), y0) 0).

Proof. The proof is similar to the corresponding proof for real functions and shows that theCauchy and Heine definitions of a limit are equivalent.

Theorem 4.3 Let f(x) y1 as x x0 and f(x) y2 as x x0. Then y1 = y2.

Proof. (y1, y2) (y1, f(x)) + (f(x), y2) 0 as x x0. Hence y1 = y2. 2

Example 4.4 Let f : R2\{(0, 0)} R be defined by f(x, y) = x2y

x4 + y2.

Then lim(x,y)(0,0) f(x, y) does not exist. By Theorem 4.2 it is sufficient to find a sequence((xn, yn))nN with (xn, yn) (0, 0) such that (f(xn, yn))nN does not converge. Let (xn, yn) =(

1

n,

(1)nn2

). Then f(xn, yn) = (1)n/2.

12

Example 4.5 Let f : R2\{(0, 0)} R be defined by f(x, y) = x2y2

x4 + y2.

Since |x2y| 12 (x4 + y2) and |f(x, y)| |y|2

we have that d((x, y), (0, 0)) < 2 implies |f(x, y)| 0 there is a > 0 such that (x, x0) < implies (f(x), f(x0)) < (orf(B(x0; )) B(f(x0); )). If f is continuous at all points of a set then f is continuous on thatset.

Example 4.7 Let (X, d) be a metric space, let a X, and f : X R, f(x) = d(a, x). Then f iscontinuous.

Proof. Let x X be arbitrary. By Proposition 1.11 |f(x)f(y)| = |d(a, x)d(a, y)| d(x, y) < for y B(x; ). 2

Theorem 4.8 Let (X, ), (Y, ) be metric spaces. Then f : X Y is continuous if and only iffor every open set G in Y, f1(G) is open in X.

Proof. Suppose f is continuous and G is open in Y . If f1(G) = then it is open. Iff1(G) 6= let x0 f1(G). Then f(x0) G and since G is open there is an > 0 such thatB(f(x0); ) G. By continuity of f there exists > 0 such that f(B(x0; )) B(f(x0); ). ThenB(x0; ) f1(B(f(x0); )) f1(G). Then x0 is an interior point of f1(G), and f1(G) isopen.

Suppose that the open set condition holds. Let x X. For any > 0, B(f(x); ) is open in(Y, ). By hypothesis f1(B(f(x); )) is open in (X, ). Since x f1(B(f(x); )) there is a > 0such that B(x; ) f1(B(f(x); )). Then f(B(x; )) B(f(x); ), and so f is continuous at x.2

Theorem 4.9 Let (X, ), (Y, ) be metric spaces. Then f : X Y is continuous if and only ifthe inverse of every closed set C in Y is closed in X.

Proof. Suppose the closed set condition holds. Then C closed implies f1(C) closed. Let G beopen. Then Gc is closed, and f1(Gc) = (f1(G))c is closed. Then f1(G) is open. Then f iscontinuous by 4.8.

Suppose f is continuous, and C is closed in Y . Then Cc is open, and by 4.8 f1(Cc) =(f1(C))c is open. Then f1(C) is closed. 2

5 Complete metric spaces

Definition 5.1 A sequence (xn) of points in a metric space (X, d) is a Cauchy sequence if

( > 0) (N N) (m,n N) [(m,n N) (d(xn, xm) < )].

Lemma 5.2 Let (xn) be a Cauchy sequence in a metric space (X, d). Then (xn) is bounded.

Proof. Let = 1 in Definition 5.1. Then there exists N N such that m,n N impliesd(xm, xn) < 1. Hence for m N, d(xm, xN ) < 1, and so for all m,

d(xm, xN ) max{d(x1, xN ), . . . , d(xN1, xN ), 1}.

2

13

Theorem 5.3 A convergent sequence in a metric space is a Cauchy sequence.

Proof. Let xn x in (X, d). Then( > 0) (N N) (n N) [(n N) (d(xn, x) < /2)].

Let > 0 and let N be as above. Let m,n N be such that m,n N . Thend(xn, xm) d(xn, x) + d(xm, x) < /2 + /2 = .

Hence (xn) is a Cauchy sequence. 2

Remark 5.4 The converse of Theorem 5.3 is false in general. Let (Q, d) have the standard metricd(x, y) = |x y|. Let xn = 10nsn, where sn = min{s N : s2 > 2.102n}. We will show that (xn)is a Cauchy sequence in (Q, d) which does not converge. Since s2n > 2.102n (sn 1)2 we havethat (sn1)2 4.102n. Hence sn 2.10n+1 3.10n, and s2n 2.102n+2sn1 2.102n+6.10n.Then 2 < x2n 2 + 6.10n. It follows that limn x2n = 2, and so (x2n) is a Cauchy sequence.Since |xn xm| |xn xm|.|xn + xm| |x2n x2m| 6.max{10n, 10m} we have that (xn) isalso a Cauchy sequence. Suppose limn xn = x, x Q. Then limn x2n = x2. But x2 6= 2 forx Q. Contradiction.

Definition 5.5 A subset E of a metric space (X, d) is complete if for any Cauchy sequence ofpoints (xn) in E there exists x E such that xn x.

Example 5.6 The set of real numbers R with the usual metric d(x, y) = |x y| is a completemetric space. (See Lecture notes in Further Topics in Analysis).

Theorem 5.7 The Euclidean space RN is complete.

Proof. Let (xn) be a Cauchy sequence in RN . For a fixed n, the point xn is an N -tuplexn = (x

(1)n , x

(2)n , . . . , x

(N)n ). Then for every i = 1, 2, . . . , N , (x

(i)n ) is a Cauchy sequence of real

numbers. This follows from the inequality

|x(i)n x(i)m | {

Ni=1

(x(i)n x(i)m )2}1/2

,

which is valid for every i = 1, 2, . . . , N . By the completeness of R with the usual metric we concludethat, for every i = 1, 2, . . . , N , there exists x(i) such that x

(i)n x(i). Let x = (x(1), x(2), . . . , x(N)).

Then x RN , and the inequality{Ni=1

(x(i)n x(i))2}1/2

Ni=1

|x(i)n x(i)|

implies that xn x. 2

Theorem 5.8 Let (X, d) be a metric space, and let E be a subset of X.

(i) If E is complete then E is closed.

(ii) If X is complete, and E is closed, then E is complete.

Proof. (i) Let x X be a limit point of E. Then there exists a sequence (xn) of points in E suchthat xn x. Then (xn) is a Cauchy sequence, by Theorem 5.3. Since E is complete, xn y forsome y E. By uniqueness of the limit, x = y. Hence x E.

(ii) Let (xn) be a Cauchy sequence in E. Since X is complete, xn x for some x X. SinceE is closed, x E. 2

14

Definition 5.9 Let (X, d) be a metric space. A subset E of X is called bounded if there existsan open ball that contains E.

Remark 5.10 A subset E of a metric space (X, d) is bounded if and only if

diam(E) := supx,yE

d(x, y) 0 there exists N N such that m,n N implies(m, n) . Hence m,n N implies for x X, (m(x), n(x)) . Taking the limit n(see again Theorem 2.18) we obtain that m N, x X implies (m(x), (x)) . HencesupxX (m(x), (x)) for m N. Hence m N implies (m, ) . So m . 2

Corollary 5.12 (C[a, b], d) is a complete metric space.

Proof. Since R with the usual metric space is complete, B[a, b] with the supremum metric d, iscomplete by Theorem 5.11. By Theorem 3.25 C[a, b] is a closed subset of B[a, b]. Then C[a, b] iscomplete by Theorem 5.8. 2

Definition 5.13 A normed space which is complete as a metric space is a Banach space.

Remark 5.14 R, RN , B(X,Y ) with Y a complete normed space are examples of Banach spaces.C[a, b] with norm f g = supx[a,b] |f(x) g(x)| is a Banach space.

Example 5.15 Let C[0, 1] be the continuous real-valued functions on [0, 1]. Define d(f, g) = 10|f(t) g(t)|dt. Then (C[0, 1], d) is an incomplete metric space.

Proof. Define for n = 1, 2 . . .

fn(t) =

0, 0 t 12 ,

2n(t 12 ), 12 t 12 + 12n ,1, 12 +

12n t 1.

15

Then (fn) is Cauchy since d(fn, fm) = 10|fn(t) fm(t)|dt = 14 | 1n 1m |. Hence d(fn, fm) for

n,m 1 Suppose fm f . Then there exists N such that m N implies

d(f, fm) =

10

|f(t) fm(t)|dt

=

12

0

|f(t)|dt+ 1

2+12m

12

|f(t) fm(t)|dt+ 1

12+

12m

|1 f(t)|dt < .

We conclude that f(t) = 0, 0 t 12 , f(t) = 1, 12 < t 1. Hence f is not continuous at 12 .Contradiction. Hence C[0, 1] with this metric is not complete. 2

Theorem 5.16 A metric space (X, d) is complete if and only if for every sequence (Vn) of non-empty closed subsets of X

{(n N) [Vn+1 Vn]} [diam(Vn) 0][ n=1

Vn 6= ].

Proof. Suppose X is complete, and let (Vn) be a sequence of non-empty closed sets withVn Vn+1,diam(Vn) 0. Let > 0. There exists N N such that diam(VN ) < . Letxn Vn, n = 1, 2, . . . . Then for any m,n N, xm, xn VN and d(xm, xn) diam(VN ) < .Hence (xn) is a Cauchy sequence. Since X is complete xn x for some x X. Since xn Vmfor all n m,x Vm = Vm. Since m was arbitrary x

m=1 Vm, and

m=1 Vm 6= . Note that

{x} = m=1 Vm. (If y m=1 Vm and y 6= x, then diam(Vm) d(x, y) > 0).Suppose X is not complete. Then there exists a Cauchy sequence (xn) which does not converge.

Let Vn = {xi : i n}. Then Vn does not have any limit points (If x is a limit point of Vn,then there exists a subsequence of (xn) which converges. Since (xn) is Cauchy, (xn) converges.Contradiction). Hence Vn is closed, and Vn Vn+1. Given any > 0 there exists N such thatm,n N implies d(xm, xn) < . Hence diam(VN ) and so diam(Vn) 0. Suppose x Vn forall n. Then xn x since d(x, xn) diam(Vn) 0. Contradiction. So

n=1 Vn = . 2

Definition 5.17 A subset A of a metric space (X, d) is said to be dense if A = X. I.e. if

(x X) ( > 0) (y A) [d(x, y) < ].

Remark 5.18 A is not dense in X if

(x X) ( > 0) (y A) [d(x, y) ].

Definition 5.19 A metric space (X, d) is said to be separable if it contains a countable densesubset.

Example 5.20 (i) R with the standard metric is separable as Q is countable and dense in R.(ii) RN with the standard metric is separable as QN is countable and dense in RN .(iii) The sequence space l1 with the usual metric is separable.

Definition 5.21 Two metric spaces (X1, d1) and (X2, d2) are isometric if there exists a bijectionT : X1 X2 such that

(x, y X1)[d2(Tx, Ty) = d1(x, y)

].

Theorem 5.22 Let (X, d) be a metric space. Then there exists a complete metric space (X, d)

which has a subset W which is isometric with X and which is dense in X. If X is any completemetric space having a subset W isometric with X and dense in X then X and X are isometric.

16

The proof is in four steps: (a) the construction of a metric space (X, d), (b) the construction

of an isometry T from X onto W where W = X, (c) the proof of completeness of (X, d), (d) theproof of the uniqueness except for isometries.

Proof. (a) Let (xn) and (xn) be Cauchy sequences in (X, d). We define (xn) to be equivalent

to (xn) written (xn) (xn) if limn d(xn, xn) = 0. Then defines an equivalent relation.((xn) (xn) since d(xn, xn) = 0, (xn) (yn) limn d(xn, yn) = limn d(yn, xn) =0 (yn) (xn), (xn) (yn) and (yn) (zn) implies limn d(xn, zn) limn(d(xn, yn) +d(yn, zn)) = 0 and so (xn) (zn)). Let X be the set of all equivalence classes x, y, . . . of Cauchysequences. We write (xn) x if (xn) is a member of x. Define d : X X R by d(x, y) =limn d(xn, yn) where (xn) x, (yn) y. This limit exists since |d(xn, yn) d(xm, ym)| d(xn, xm) + d(yn, ym), which goes to 0 as n,m . Hence (d(xn, yn)) is a Cauchy sequence inthe complete metric space R. Moreover this limit is independent of the choice of representatives:Let (xn) (xn) and (yn) (yn). Then |d(xn, yn) d(xn, yn)| d(xn, xn) + d(yn, yn) 0. Hencelimn d(xn, yn) = limn d(xn, y

n) = d(x, y). To complete part (a) we show that d is a metric

on X. (M1) 0 d(x, y) < since 0 limn d(xn, yn) < for Cauchy sequences (xn) and(yn), (M2) d(x, y) = 0 limn d(xn, yn) = 0 for representatives (xn) x, (yn) y (xn) (yn) x = y, (M3) d(x, y) = limn d(xn, yn) = limn d(yn, xn) = d(y, x), (M4) Let(zn) z. Then d(x, y) = limn d(xn, yn) limn(d(xn, zn)+d(zn, yn)) = limn d(xn, zn)+limn d(zn, yn) = d(x, z) + d(z, y) since each of the above limits exists.(b) With each b X we associate the class b X which contains the constant Cauchy sequence(b, b, b . . . ). This defines a map T : X X. T is an isometry since d(b, c) = limn d(b, c) =d(b, c), where c is the class containing (c, c, . . . ). Put W = T (X). Then T is onto W . T is also

one-to-one. (Suppose b = c. Then d(b, c) = 0 implies d(b, c) = 0. Hence b = c). So X and W are

isometric. We will show that W is dense in X. Let x X be arbitrary, and let (xn) x. Forevery > 0 there exists N N such that n N implies d(xn, xN ) < 2 . Let (xN , xN , . . . ) xN .Then xN W . Moreover d(xN , x) = limn d(xN , xn) 2 < . Hence W is dense in X.(c) Let (xn), be a Cauchy sequence in (X, d). Since W is dense in X we have that for every xn Xthere exists zn W such that d(xn, zn) < 1n . By the triangle inequality d(zn, zm) d(zn, xn) +d(xn, xm) + d(xm, zm) 1n + + 1m for all m,n N and some N N. Hence (zm) is a Cauchysequence in W . Since T : X W is an isometry the sequence (zm), zm = T1zm is a Cauchysequence in X. Let x X be the equivalence class containing (zm). We will show that xn xin (X, d). Since d(xn, x) d(xn, zn) + d(zn, x) < 1n + d(zn, x), zn contains the Cauchy sequence(zn, zn, . . . ), and x contains the Cauchy sequence (zm) we have that d(zn, x) = limm d(zn, zm).So d(xn, x) 1n + < 2 for n sufficiently large.(d)We prove that X and X are isometric by constructing a bijective isometry U : X X. LetT : X W and T1 : X W be bijective isometries. Then TT11 : W W is a bijectiveisometry. Suppose x X\W . Then there exists a sequence zn W such that d(x, zn) < 1n . Then(zn) is a Cauchy sequence in X. Define zn = TT

11 zn, n = 1, 2, . . . .. Then (zn) is a Cauchy

sequence in W since TT11 is an isometry. Since X is complete zn x for some x X. Thisdefines a map U : X X by putting x = Ux. We will show that U is an isometry. Let x, y X.Then, by the triangle inequality,

|d(Ux, Uy) d(x, y)| |d(Ux, Uy) d(xn, yn)|+ |d(xn, yn) d(x, y)| |d(Ux, Uy) d(xn, yn)|+ |d(xn, yn) d(xn, yn)|+ d(xn, x) + d(yn, y) d(Ux, xn) + d(Uy, yn) + d(xn, x) + d(yn, y) 0

since Ux = limn xn, Uy = limn yn, xn x, yn y and d(xn, yn) = d(xn, yn). Hence Uis an isometry. But isometrics are injective (one-to-one). Finally we show that U is onto. Let

x X\W . Then there exists a sequence (zn) W such that zn x. Then zn = T1T1zn givesa Cauchy sequence (zn). Then xn x and Ux = x. 2

17

6 The contraction mapping theorem and applications

Let (X, d) be a metric space. Let T : X X be a function (mapping) from X into itself.

Definition 6.1 A point x X is a fixed point of the mapping T, ifTx = x.

Definition 6.2 A mapping T : X X is a contraction (or contraction mapping) if

( [0, 1)) (x, y X)[d(Tx, Ty) d(x, y)

].

Theorem 6.3 Let (X, d) be a complete metric space, and let T : X X be a contraction. ThenT has a unique fixed point.

Proof. Let x0 X. Define xn+1 = Txn, n = 0, 1, 2, . . . Thend(xn+1, xn) = d(Txn, Txn1) d(xn, xn1) = d(Txn1, Txn2)

2d(xn1, xn2) = nd(x1, x0).Let m n. Then

d(xn, xm) d(xn, xn+1) + d(xn+1, xn+2) + + d(xm1, xm) nd(x1, x0) + n+1d(x1, x0) + + m1d(x1, x0) n1d(x1, x0).(1)

It follows from (1) that (xn) is a Cauchy sequence. Since (X, d) is complete, there exists x X

such that xn x. Taking the limit m in (1) we obtain

(2) d(xn, x)

n

1 d(x1, x0).

By the triangle inequality and (2)

(3) d(Tx, x) d(Tx, Txm) + d(Txm, x) d(x, xm) + d(xm+1, x) 2m+1

1 d(x1, x0).

Taking the limit m in (3) we obtain that d(Tx, x) = 0. Hence Tx = x, and x is a fixedpoint of T . Suppose that z X is also a fixed point of T , i.e. Tz = z. Then

d(x, z) = d(Tx, T z) d(x, z).Since 0 < 1 it follows that d(x, z) = 0, and hence x = z. 2

Example 6.4 Let X = [0, 1] with the standard metric. Since R is complete, and [0, 1] is closed,[0, 1] is complete. Let T : X X be defined by

Tx = cosx.

Then TX = [cos 1, 1]. Also, by the Mean Value Theorem we have

|Tx Ty| = | cosx cos y| max[0,1]

| sin ||x y| = | sin(1)| |x y|.

Since | sin(1)| < 1 we conclude that T is a contraction mapping. Iteration of an arbitrary value in[0, 1] gives x = 0.739085133 . . . .

18

Example 6.5 Let X = C[0, 1] with the supremum (maximum) metric. Let T : X X be definedby

(Tf)(t) = 1 t0

f()d.

The T is a contraction mapping since,

d(Tf, Tg) = maxt[0,1]

t0

(g() f())d maxt[0,1]

t0

|g() f()|d

10

|g() f()|d 10

d(f, g)d =1

2d(f, g).

Note that Tf C[0, 1]. Since C[0, 1] is complete the equation Tf = f has a unique solution,by Theorem 6.3. Since t (Tf)(t) is differentiable we have that t f(t) is differentiable, andf (t) =

d

dt(Tf)(t) = tf(t). Since f(0) = 1 we have f(t) = et2/2 on [0, 1]. It can be shown that

this solution can be extended to X = C(R).

Example 6.6 Let K : [a, b] [a, b] R be a continuous function, and let : [a, b] R becontinuous. Consider the integral equation (x) =

baK(x, y)(y)dy + (x), C[a, b] where

is a constant (Fredholm integral equation of the second kind). We shall prove that for ||sufficiently small, there exists a unique C[a, b] which solves the equation. In Chapter 7 weshall prove that continuous functions on closed, bounded subsets of Rm are bounded: i.e. thereexists M such that |K(x, y)| M . We will also show that continuous functions on closed, boundedsubsets of Rm are uniformly continuous i.e. for all > 0 there exists > 0 such that |x1x2| < implies |(x1) (x2)| and |K(x1, y) K(x2, y)| < . Also for C[a, b], there exists Nsuch that |(y)| N for all a y b. Define T on C[a, b] by

(T)(x) =

ba

K(x, y)(y)dy + (x), a x b.

Then |T(x1) T(x2)| || ba

(K(x1, y) K(x2, y))(y)dy + |(x1) (x2)| ||N(b

a) + for |x1 x2| . Hence T C[a, b]. For any 1, 2 C[a, b] we have d(T1, T2) =maxaxb |T1(x) T2(x)| ||

ab

|K(x, y)| |1(y) 2(y)|dy ||M(b a)d(1, 2). HenceT is a contraction map on a complete metric space for || < (M(b a))1 and T = hasa unique solution. Then (x) =

baK(x, y)(y)dy + (x) has a unique solution in C[a, b] for

|| < (M(b a))1.

Example 6.7 Consider the integral equation (x) =

11

(x y)(y)dy+ 1, C[1, 1]. Thenba = 2, and |xy| 2 on [1, 1] [1, 1]. Hence there exists a unique solution for || < 14 . Wesee that the solution is of the form (x) = c+ dx. Hence c+ dx =

11

(x y)(c+ dy)dy + 1 =

1 2d3

+ 2cx. So c = 1 2d3, d = 2c with solution c =

3

3 + 42, d =

6

3 + 42. We see that

this is also the unique solution for || 14.

Theorem 6.8 Let S be the rectangle [x0 a, x0 + a] [y0 b, y0 + b]. Suppose that f : S R iscontinuous, and |f(x, y)| B for (x, y) S, and there exists a constant A such that

|f(x1, y1) f(x1, y2)| A |y1 y2|

19

for all (x1, y1) S, (x1, y2) S (Lipschitz condition). Let [0, a] be such that < min{1A ,

bB

}.

Then the differential equationdy

dx= f(x, y), subject to the initial condition y(x0) = y0 has a

unique solution in the interval [x0 , x0 + ].

Proof. Let M = [x0, x0 +], N = [y0 b, y0 + b]. Let C(M,N) be the collection of continuousfunctions from M into N with metric d(1, 2) = sup

x0tx0+|1(t)2(t)|. Since N is complete,

(C(M,N), d) is a complete metric space. Define T on C(M,N) by

T(x) = y0 +

xx0

f(t, (t))dt.

Then T is differentiable with (T)(x) = f(x, (x)), and hence T is continuous. Moreover,

|(T)(x) y0| = | xx0

f(t, (t))dt| xx0

|f(t, (t))|dt B|x x0| B < b.

Hence T C(M,N). By the Lipschitz condition

d(T1, T2) = maxxM

| xx0

f(t, 1(t))dt xx0

f((t, 2(t))dt|

maxxM

xx0

|f(t, 1(t)) f(t, 2(t))|dt maxxM

xx0

Ad(1, 2)dt

= Ad(1, 2).So T : C(M,N) C(M,N) is a contraction map with = A < 1. Hence T has a unique fixedpoint C(M,N) satisfying

T = i.e. (x) = y0 +

xx0

f(t, (t))dt.

We see that (x0) = y0. Since the right hand side is differentiable, is differentiable, and (x) =

f(x, (x)). Hence is a solution ofdy

dx= f(x, y) with y(x0) = y0. On the other hand any solution

of this o.d.e. with this initial condition is a fixed point since (Ty)(x) = y0 +

xx0

f(t, y(t))dt =

y0 +

xx0

dy

dtdt = y0 + y(x) y(x0) = y(x). 2

Example 6.9 Consider the equationdy

dx= x2 + y2, y(0) = 0. Choose b > 0 in Theorem 6.8.

Then |f(x, y1) f(x, y2))| = |y21 y22 | = |y1 + y2||y1 y2| 2b|y1 y2| since y1, y2 [b, b]. SoA = 2b. Choose S = [a, a] [b, b]. Then |f(x, y)| b2 + a2 on S. Hence B = b2 + a2. Choose < min

{1

2b,

b

b2 + a2

}, [0, a]. For a = b = 1

2we obtain that there exists a unique solution

on (21/2, 21/2). Approximations can be obtained by iterations: let y1 : (21/2, 21/2) (21/2, 21/2) be y1(0) = 0. ThenTy1(x) =

x0

f(t, y1(t))dt =

x0

f(t, 0)dt =x3

3, Ty2(x) =

x0

f

(t,t3

3

)dt =

x0

(t2 +

t6

9

)dt

=x3

3+x7

63, . . .

Remark 6.10 It is possible to prove local existence and uniqueness for equations of the typey(k)(x) = f(x, y, y(1) . . . , y(k1)) under suitable Lipschitz conditions on f , and under suitableinitial conditions.

20

Theorem 6.11 (Implicit function Theorem) Let (x0, y0) be a point of an open set G in R2,and suppose that f : G R satisfies

(i) f(x0, y0) = 0,

(ii) f is continuous on G,

(iii)f

yexists in G and is continuous at (x0, y0), and

f

y(x0, y0) 6= 0.

Then there exists a rectangle MN = [x0, x0+] [y0, y0+], and a continuous function : M N such that y = (x) is the only solution lying in M N of the equation f(x, y) = 0.

Proof. Put q =

(f

y(x0, y0)

)1. Since

f

yis continuous at (x0, y0), there is a rectangle

[x0 , x0 + ] [y0 , y0 + ] G on which1 q fy (x, y)

< 12 . Using conditions (i), (ii) wecan find a positive number such that for x M = [x0, x0 +], we have |qf(x, y0)|

2.

Since N is complete, C(M,N) with the usual metric is complete. Define on C(M,N) by()(x) = (x)qf(x, (x)), x M . Then is continuous. We first show that maps C(M,N)into itself. Let C(M,N). Then for x M , we have ()(x) y0 = (x) qf(x, (x)) y0 =(x) qf(x, (x)) (y0 qf(x, y0)) qf(x, y0) = ((x) y0)

(1 q fy (x, u)

) qf(x, y0) by the

Mean Value Theorem applied to the second variable: u (y0, (x)). Hence

|()(x) y0| |(x) y0| |1 q fy

(x, u)|+ |qf(x, y0)| 12

+

2= .

To prove that is a contraction map we let 1, 2 C(M,N), and let x M . Then(1)(x) (2)(x)

= (1(x) qf(x, 1(x))) (2(x) qf(x, 2(x)) = (1(x) 2(x))(

1 q fy

(x, v)

)for some v between 1(x) and 2(x). Then (1,2) 12(1, 2). Hence by the contractionmapping theorem has a unique fixed point. Hence there exists a unique : M N, C(M,N) such that () = i.e f(x, (x)) = 0, x M . 2

Example 6.12 Let f : R2 R, f(x, y) = cos(xy) + x y 1, (x0, y0) = (0, 0). Then q =(fy (0, 0)

)1= 1. Furthermore

1 q fy (x, y) = |x sin(xy)| |x|. Hence the hypotheses of the

theorem are satisfied with = = 12 and = . If we choose 0 = 0 then 1(x) = x, 2(x) =x+ f(x, x) = x+ cos(x2) 1 = x x42 +O(x6).

7 Compact sets

Definition 7.1 Let (X, d) be a metric space. A subset E of X (which may be X itself) is compactif every sequence in E has a subsequence which converges to a point in E. If X is compact then(X, d) is a compact metric space.

Example 7.2 A finite set in a metric space is compact.

Proof. If E = then E is compact. Suppose E = {p1, . . . , pk}. If (xn) is a sequence of points inE then at least one point pi appears infinitely often in (xn). 2

21

Theorem 7.3 The subset [a, b] of R with the usual metric is compact.

Proof. First we show that every sequence in R has a monotone subsequence. Let (xn) bea sequence in R. We call xp a terrace point if xn xp for all n p. If there are infinitelymany terrace points then let x1 be the first, x2 be the second, . . . . Then (xk) is a decreasingsubsequence of (xn). If there are finitely many terrace points (perhaps none), let 1 be a naturalnumber such that no xn for n 1 is a terrace point. There is a 2 > 1 such that x2 > x1 .There is a 3 > 2 such that x3 > x2 . Continuing this way we obtain an increasing subsequenceof (xn).

Let (xn) be a sequence of points in [a, b]. By the above there exists a monotone subsequence(xnk) of (xn). Since xn [a, b] for all n we have that xnk [a, b] for all k. Then (xnk) is boundedand monotone. Let x = limk xnk . Then either xnk = x for all k sufficiently large or x is a limitpoint. Since [a, b] is closed, x [a, b]. 2

Theorem 7.4 Let E = [a1, b1] [an, bn] Rn with the usual metric. Then E is compact.Proof. Let (xm) be a sequence of points in E, xm = (1m, . . . , nm). Then (1m)m1 is a sequencein [a, b]. By 7.3(1m)m1 has a convergent subsequence (1mi)i1. Let 1 = limi 1mi . Then(2mi)i1 is a sequence in [a2, b2] and has a convergent subsequence with limit 2. By repeatingthis process altogether n times we end up with a sequence (xk)k1 of points in E such thatlimi ji = j , j = 1, . . . , n. Let x = (1, . . . , n). Then

d(x, xi) =

nj=1

|j ji |2

1/2

nj=1

|j ji | 0

as i. Hence limi xi = x. Then xi = x for all i sufficiently large, or x is a limit point ofE. Since E is closed, x E. 2

Theorem 7.5 A compact set in a metric space is complete.

Proof. Let E be compact set in (X, d), and let (xn) be a Cauchy sequence in E. Then (xn) has asubsequence (xni) which converges to a point x E. Hence d(xn, x) d(xn, xni) + d(xni , x) 0as n, i. Hence xn x, x E. 2

Theorem 7.6 Let E be a set in a metric space (X, d).

(i) If E is compact, then E is bounded and closed.

(ii) If X is compact and E is closed then E is compact.

Proof. (i) Since E is compact, E is complete by 7.5. Then E is closed by 5.8 (i). Suppose E isunbounded. Then there exists a sequence (xn) in E such that d(x1, xn) n1, n N. Then (xn)does not have a convergent subsequence since d(xni , x) d(xni , x1)d(x1, x) ni1d(x1, x) as i(ii) Let (xn) be any sequence in E. Since (xn) is a sequence in X, X compact, there exists asubsequence (xni) with a limit x X. Since (xni) is a convergent sequence in E, and since E isclosed, x E. 2

Corollary 7.7 A set in Rn with the usual metric is compact if and only if it is bounded andclosed.

Proof. Suppose E is bounded and closed. Then E [N,N ] [N,N ] for some N > 0.Then E is a closed subset of the compact set [N,N ] [N,N ]. Then E is compact by 7.6(ii). The converse follows from 7.6 (i). 2

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Theorem 7.8 A set E in a metric space is compact if and only if every infinite subset of E hasat least one limit point in E.

Proof. Suppose E is compact. If A E and A is infinite then A contains a subset {x1, x2, . . . }of distinct points. Since E is compact (xn) has a subsequence (xni) which converges to a pointx E. Then x is a limit point of {xn1 , xn2 , . . . } and hence of A. Suppose every infinite subsetof E has at least one limit point in E. Let (xn) be a sequence in E. The set of points {x1, . . . }may be finite or infinite. If {x1, . . . } is finite then at least one of its points occurs infinitely manytimes in the sequence (xn). Then (xn) has a convergent subsequence, converging to a point in E.If {x1, . . . } is infinite then it has a limit point, and this is the limit of some subsequence of (xn).2

Theorem 7.9 (BolzanoWeierstrass theorem). A bounded, infinite set in Rn has at leastone limit point.

Proof. A bounded set lies in the compact set [N,N ] [N,N ] for some N > 0, and thetheorem follows by 7.8. 2

Theorem 7.10 Let (X, d), (Y, e) be a metric spaces. If f : X Y is continuous, and X iscompact then f(X) is compact.

Proof. Let (yn) be a sequence in f(X). Then there exists a sequence (xn) in X such that yn =f(xn). Since X is compact, there exists a convergent subsequence (xnk), with xnk x, x X.Since f is continuous , f(xnk) f(x). Hence ynk f(x) and f(x) f(X). 2

Corollary 7.11 If the domain of a continuous function is compact then the range is bounded andclosed.

Proof. Theorem 7.10 and Theorem 7.6 (i).

Corollary 7.12 If the domain of a real-valued continuous function is compact, then the functionis bounded and attains its least upper bound and its greatest lower bound.

Proof. Since the range is bounded and closed, the range contains its supremum and infimum. 2

Theorem 7.13 Let f : X Y be a continuous bijection, where X is compact. Then f1 : Y Xis continuous.

Proof. Let F be a closed set in X. Since X is compact, F is compact by 7.6 (ii). Then f(F )is compact by 7.10. Hence f(F ) is closed by 7.6 (i). Then (f1)1(F ) is closed, and f1 iscontinuous by 4.9. 2

Corollary 7.14 Let f : [a, b] R be strictly increasing and continuous. Then f1 : [f(a), f(b)][a, b] is strictly monotone and continuous.

Proof. f is onto [f(a), f(b)], and since f is one-to-one f1 : [f(a), f(b)] [a, b] exists. Since[a, b] is compact, f1 is continuous by 7.13. 2

Definition 7.15 Let (X, d), (Y, e) be a metric spaces. f : X Y is uniformly continuous if forany > 0 there exists a > 0 such that x, y X, d(x, y) < implies e(f(x), f(y)) < :

( > 0)( > 0)(x X)(y X)[(d(x, y) < e(f(x), f(y)) < ]

Remark 7.16 f is not uniformly continuous if

( > 0)( > 0)(x X)(y X)[(d(x, y) < ) (e(f(x), f(y)) )]

23

Example 7.17 f : (0, 1] R, f(x) = 1x is continuous but not uniformly continuous since thereexist sequences (xn), (xn) in (0, 1] such that |xn xn| 0, |f(xn) f(xn)| 1. (Take xn =1n , xn =

12n ). Similarly f : R R, f(x) = x2 is continuous but not uniformly continuous since

xn = n, xn = n+1n gives |f(xn) f(xn)| 1.

Example 7.18 Let f : R R be differentiable with bounded derivative. Then f is uniformlycontinuous.

Proof. By the Mean Value Theorem f(y) = f(x) + (y x)f (u) for some x u y. Since f isbounded, |f (u)| K for some K, and all u. Hence |f(y) f(x)| K|x y|. Choose = K in7.15. 2

Example 7.19 f : [0,) R, f(x) = x1/3 is uniformly continuous. This follows from theinequality |x1/3 y1/3| |x y|1/3. The latter is proved by considering the case x y 0 andtaking cubed powers.

Theorem 7.20 Let f : X Y be continuous, with X compact. Then f is uniformly continuous.

Proof. Suppose f is not uniformly continuous. Then there are sequences (xn), (yn) of points inX such that d(xn, yn) 1n and e(f(xn), f(yn)) , for some > 0. Since X is compact (xn)has a convergent subsequence (xnk) with limit . Then d(ynk , ) d(ynk , xnk) + d(xnk , ) 1nk

+ d(xnk , ) 0. Hence ynk . Since f is continuous, f(xnk) f(), f(ynk) f().Hence e(f(xnk), f(ynk)) 0. Contradiction, since e(f(xnk), f(ynk)) . 2

Definition 7.21 A collection A of sets A is a covering of a set E if E AAA. The covering isfinite if A has a finite number of members only. When E, and the members of A are in a metricspace, then the covering is open if all members of A are open. If A,A1 are both coverings of Eand A1 A then A1 is a subcovering of E.

Example 7.22 The interval (0,1) is covered by A = {(x2, x) : 0 < x < 1}. Note that A is aninfinite cover of (0,1) with no finite subcover. Suppose A1 = {(x2, x) : x {x1, . . . , xn}}. Thenthe points min1in x2i , and max1in xi are not in AA1A.

Lemma 7.23 Let E be a compact set in a metric space (X, d) and let > 0. Then there exists afinite number of points x1, . . . , xp in E such that E pi=1B(x; ).

Proof. Let x1 E be arbitrary. If E1 = E\B(x1; ) 6= let x2 E1. Generally if it has beenpossible to choose x1, . . . , xk and if Ek = E\ ki=1 B(xi; ) 6= , let xk+1 Ek. Suppose that noEk is empty. Then there exists an infinite sequence (xn) in E such that when n > m, xn Emi.e. d(xn, xm) . Then (xn) does not have a convergent subsequence (since d(xnk , xnl) fork > l). Hence Ek = for some positive integer k = p. 2

Lemma 7.24 Let E be a compact set in a metric space (X, d) and let G be an open covering ofE. Then there exists a positive number such that for every x E, the open ball B(x;) iscontained in some member of G. is the Lebesgue number of E for the cover G.

Proof. Suppose the lemma is false. Then for every n N there exists xn E such that B(xn; 1n )is not contained in any G of G. Since E is compact (xn) has a subsequence (xnk) which converges toa point x E. Let G be a member of G which contains x. Since G is open, there is a > 0 suchthat B(x; 2) G. For all k sufficiently large d(xnk , x) . Hence there is an integer r suchthat d(xr, x

) and r > 1 . If x B(xr; 1r ) then d(x, x) d(x, xr) + d(xr, x) < 1r + < 2.Hence x B(x; 2) and B(xr; 1r ) B(x; 2) G. Contradiction, since the B(xn; 1n ) werechosen to be open balls not lying in any member G of G. 2

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Theorem 7.25 A set in a metric space is compact if and only if every open covering contains afinite subcovering.

Proof. Let E be a compact set in a metric space (X, d) and let G be an open cover of E.By Lemma 7.24 there exists > 0 such that for every x E,B(x;) is contained in somemember G of G. Having chosen , we can find, by Lemma 7.23 points x1, . . . , xp such thatE pn=1B(xn;). For each n = 1, . . . , p there is a member Gn of G such that B(xn;) Gn.Then E pn=1B(xn;) pn=1Gn, and {G1, . . . , Gn} is a finite subcover of E.

Let E be a non-compact set in (X, d). By 7.8 E contains an infinite subset D which has nolimit points in E. So for each c D there exists an open ball Gc = B(c; rc) such that GcD = {c}.Since (D\D) E = we have that Dc D Ec = X and Dc D E. So G = (D)c is such thatG D E. Hence G = {G} {Gc : c D} is an open covering of E. But the only member ofG which contains a given point c D is Gc. Hence every subcovering must contain the infinitecollection {Gc : c D}, and so G contains no finite subcovering. 2

Theorem 7.26 If E is a collection of compact subsets of a metric space such that the intersectionof every finite subcollection of sets in E is non-empty, then EE is compact and non-empty.

Proof. Since each set E in E is closed, EEE is a closed subset of the compact set E0, where E0is any member of E . Then EEE is compact by 7.6 (ii). Suppose EEE = . Then X = EEEcby de Morgans law. Since E is closed, Ec is open, and {Ec : E E} is an open covering of X andhence of E0. Since E0 is compact, {Ec : Ec E} has a finite subcovering say X\E1, . . . , X\En(by 7.25). Then E0 (X\E1) (X\En). By de Morgans law X\E0 E1 En. HenceE0(X\E0) E0 En. But the left hand side is empty, and the right hand side is non-emptyby assumption. Contradiction. 2

Definition 7.27 A subset M , which may be X itself, in a metric space (X, d) is totally boundedif for any > 0 there exists a finite set {x1, . . . , xp} of points in X such that M pi=1B(xi; ).If X is totally bounded then (X, d) is a totally bounded metric space.

Example 7.28 Compact metric spaces are totally bounded metric spaces by Lemma 7.23.

Lemma 7.29 Let M be a set in a totally bounded metric space (X, d). Then (M,d) is a totallybounded metric space.

Proof. Let > 0 be arbitrary. Then there exists a finite set {x1, . . . , xp} X such that X pi=1B(xi; 2 ). We discard all balls in this union which do not contain any points of M and denotethis set again by {x1, . . . , xp}. Let yi B(xi; 2 ) M for i = 1, . . . , p. Then B(xi; 2 ) B(yi; )and M pi=1B(yi; ). Hence (M,d) is a totally bounded metric space. 2

Remark 7.30 If M is a totally bounded subset in a metric space (X, d) then any subset of M istotally bounded.

Lemma 7.31 Let M be a totally bounded set in a metric space (X, d). Then M is totally bounded.

Proof. Since M is totally bounded for any > 0 there exists a finite set {x1, . . . , xp} X suchthat M pi=1B(xi; 2 ). If M = M then there is nothing to prove. If y M\M then thereexists x M with d(x, y) < /2. Hence there exists xi {x1, . . . , xp} such that d(x, xi) < 2 , andd(xi, y) d(xi, x) + d(x, y) < . Hence y pi=1B(xi; ).So M pi=1B(xi; ). 2

Remark 7.32 Totally bounded sets are bounded.

25

Proof. Let M be a totally bounded set in a metric space (X, d). Then for = 1 there exists{x1, . . . , xp} X such that M pi=1B(xi; 1). Then diam (M) 2 + max{d(xi, xj) : i =1, . . . , p, j = 1, . . . , p}. 2

Theorem 7.33 Any bounded subset of Rn with the usual metric is totally bounded.

Proof. Let M be bounded. Then M is bounded (diam (M) = diam (M)) and closed. Then Mis compact by Corollary 7.7, and totally bounded by Lemma 7.23. Then M is totally bounded byRemark 7.30. 2

Example 7.34 Let C[0, 1] have the usual (supremum) metric d. Then (C[0, 1], d) is complete,not totally bounded, and hence not compact.

Proof. Let fn : C[0, 1] R be given by fn(x) = max{1 2(n + 1)2|x 1n |, 0}, n = 1, 2, . . . .Then the {f1, f2, . . . } have disjoint support i.e. d(fn, fm) = 1 for m 6= n. Hence there do not existg1, . . . , gp in C[0, 1] such that {f1, f2, . . . } pi=1B(gi; 12 ), and so C[0, 1] is not totally bounded.2

Lemma 7.35 Let (X, d) be a totally bounded metric space. Then any sequence (xn) in X has asubsequence which is Cauchy.

Proof. Since X is totally bounded it is covered by a finite number of open balls with radius12 . At least one of the balls must contain xn for infinitely many values of n. Hence there is asubsequence, say (xn,1), of (xn) such that all xn,1 belong to a single open ball with radius

12 .

Hence d(xn,1, xm,1) < 1 for all m,n. Suppose inductively that for i = 1, 2, . . . , k 1 there exists asubsequence (xn,i) of (xn) such that d(xn,i, xm,i) 0)( > 0)( f =)( x [0, 1])[|x a| < = |f(x) f(a)| < ].

A subset = C[0, 1] is equicontinuous if it is equicontinuous at every a [0, 1].

Example 7.39 The family = = {f1, f2, } in the proof of Example 7.34 is equicontinuous atany a (0, 1], and is not equicontinuous at 0.Proof. Let a > 0 and let > 0. Let Na = min{n N : 1n + 12(n+1)2 a2}. For n Na,we have that fn(x) = 0, a/2 x 1. Hence {fNa , fNa+1, } is equicontinuous at a. Each of

26

{f1, f2, , fNa1} is continuous at a. Since this family is finite this family is also equicontinuous.Then = is equicontinuous at a. To see that = is not equicontinuous at 0 we note that fn(0) =0, fn(

1n ) = 1. 2

Theorem 7.40 If = C[0, 1] with the supremum metric d is equicontinuous then = C[0, 1] isuniformly equicontinuous i.e.

( > 0)( > 0)( f =)( x [0, 1])( y [0, 1])[|x y| < = |f(x) f(y)| < ].

Proof. Suppose the Theorem is false. Then

( > 0)( > 0)( f =)( x [0, 1])( y [0, 1])[|x y| < |f(x) f(y)| ].

Let = 1/n, n N. Then there exist sequences (fn) =, and xn [0, 1], yn [0, 1] such that|xn yn| 1/n and |fn(xn) fn(yn)| . Since [0, 1] is compact there exists a subsequence(xnk) of (xn) such that xnk a for some a [0, 1]. Then ynk a too. Since (fnk) is asequence in = we have that |fnk(xnk)fnk(a)| 0 and |fnk(ynk)fnk(a)| 0. This contradicts|fnk(xnk) fnk(ynk)| .

Theorem 7.41 If = is totally bounded in C[0, 1] with the supremum metric then = is uniformlyequicontinuous.

Proof. Let > 0. By Lemma 7.29 there exist {f1, . . . , fp} = such that = pi=1B(fi; 3 ).Since each fi is uniformly continuous there exists i such that x [0, 1], y [0, 1] and |x y| < iimplies |fi(x) fi(y)| < 3 . Let = min{1, . . . , p}, and let f =. There exists i {1, . . . , p}such that d(f, fi) 0 be arbitrary. Since = is uniformly equicontinuous there exists > 0 suchthat for any f = and all x, y [0, 1] with |x y| < we have that |f(x) f(y)| < 3 . Wepartition [0, 1] into n disjoint intervals Ii = [(i 1), i), i = 1, . . . n 1, and In = [(n 1), 1],where n = min{m N : m 1}. Since = is bounded there exists N such that for all f =,d(f, 0) N . Hence for any f =, f([0, 1]) [N,N ]. We partition [N,N ] into ` intervalsJj = [N + (j 1) 3 ,N + j3 ], j = 1, . . . , ` 1, and J` = [N + (` 1) 3 , N ], where ` = min{k N : k 6N}. Consider the collection of all maps : {1, 2, . . . , n} {1, . . . , `} (There are `nsuch maps). For each we define g : [0, 1] [N,N ] by letting g take the constant valueN + ((i) 1) 3 on Ii = 1, . . . , n. Each g is a step function, constant on each Ii. We will showthat for any f = there exists g with d(f, g) < . Let f = be fixed. Define by putting foreach i, (i) = j where j is such that f((i 1)) Jj (there is at least one such j, and at mosttwo). For a given x in [0, 1], x Ii for a unique i, and g(x) = N + ((i) 1) 3 . Hence

|f(x) g(x)| |f(x) f((i 1))|+ |f((i 1)) (N + ((i) 1)3

)| 23.

Hence taking the supremum over all x Ii and all i = 1, . . . , n we obtain d(f, g) < . 2

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8 Connected sets

Definition 8.1 A set E in a metric space (X, d) is connected if there do not exist open sets G1, G2such that G1 E 6= , G2 E 6= , G1 G2 E,G1 G2 E = . If X is connected then (X, d)is a connected metric space.

Example 8.2 Let (X, d) be a metric space. Then is connected, {a} is connected for any a X,and any finite set with at least two points is not connected.

Proof. If E = then G1 E = G2 E = for all open sets G1, G2. If E = {a} then G1 E 6= and G2 E 6= implies a G1, a G2. Hence G1 G2 E 6= . Let E = {x1, . . . , xn}, and let = min{d(x1, xi) : i = 2, . . . , n}. Let G1 = B(x1; 3 ), G2 = ni=2B(xi; 3 ). Then G1, G2 are openand G1 E 3 x1, G2 E 3 x2, G1 G2 E and G1 G2 = . 2

Example 8.3 Let R have the usual metric. Then Q and R\Q are not connected.

Proof. Let G1 = (,

2), G2 = (

2,), E = Q. Then G1, G2 are open and satisfy allremaining four requirements in Definition 8.1. Similarly if E = R\Q then G1 = (, 0), G2 =(0,) satisfy all requirements in Definition 8.1. 2

Theorem 8.4 A metric space (X, d) is connected if and only if X and are the only sets whichare both open and closed.

Proof. Suppose X is connected. Let G be both open and closed. Then G and X\G are open, andG (X\G) X,G (X\G) X = . Since X is connected either G X = or (X\G) X = (or both). Hence G = or G = X. Suppose X is not connected. Then there are open sets G,Hsuch that GH = , GH = X,GX 6= , H X 6= . Hence G 6= , H 6= and G = Hc. SinceH is open Hc is closed, and so G is both open and closed. 2

Theorem 8.5 A necessary and sufficient for a non-empty set in R with the usual metric to beconnected is that it is an interval.

Proof. Let E be a non-empty set which is not an interval. Then there exist numbers a, q, b suchthat a, b E and q / E with a < q < b. Then G1 = (, q), G2 = (q,) are open, intersect Ein a and b respectively, and G1 G2 = R\{q} E. Hence E is not connected.

Let I be an interval, and letG1, G2 be open sets which intersect I and are such thatG1G2I =. Let a G1I, b G2I. Then a 6= b and we may suppose a < b. Then [a, b] is a subinterval of I.Let u = sup{G1[a, b]}. Since a, b are interior points of G1, G2 respectively and G1G2[a, b] = it follows that a < u < b. Let > 0 be such that a < u < u+ < b. Then (u, u+ ) containsno points of G1 and (u , u] contains at least one point of G1. Hence (u , u + ) cannot bea subset of either G1 or G2. It follows that u / G1, u / G2 since G1, G2 are open and was anypositive number such that a < u < u+ < b. Hence G1 G2 63 u, and so G1 G2 6 I. 2

Theorem 8.6 A non-empty open set in Rn with the usual metric is connected if and only if anytwo of its points may be joined by a polygon lying entirely in the set. The polygon may be chosenso that its edges are parallel to the coordinate axes.

Proof. Let G be open and connected. Let c G and let G1 be the set of points which can bejoined to c by a polygon in G. Let G2 = G\G1 so that G1 G2 = , G1 G2 = G. SupposeG2 6= . We will show that G1, G2 are open. Let x G1 be arbitrary. Then B(x; ) G for some > 0. Since x can be joined to c by a polygon in G, all points in B(x; ) can be joined to c by apolygon in G. Hence G1 is open. Let y G2. Then B(y; ) G for some > 0. Suppose B(y; )would contain a point of G1. Then y could be connected by a line segment to that point and toc. Contradiction, and B(y; ) G2. Hence G2 is open. Moreover G1 G 3 c and G2 G 3 y.

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This contradicts the connectedness of G. Hence G2 = and G1 = G. The polygon could be onewith its edges parallel to the coordinate axes for a point in an open ball of Rn may be joined tothe center by n edges parallel to the axes.

Suppose G is open and any two points of G may be joined by a polygon in G. Let G1, G2be non-empty, open sets which intersect G such that G1 G2 G = . We have to show thatG1 G2 6 G. Take a point c1 G1 and c2 G2. There is a polygonal path in G which joins c1and c2. This path must have an edge with an endpoint p G1, and an endpoint q / G1. If q / G2then G1 G2 6 G, and G1 G2 6= G. Suppose q G2. The edge with endpoints p and q may berepresented by the vector equation x(t) = p + (q p)t, 0 t 1. Let u = sup{t : x(t) G1}.Since p, q are interior points of G1, G2 respectively and G1 G2 {x(t) : 0 t 1} = it followsthat 0 < u < 1. Let > 0 be such that 0 < u < u + < 1. Then {x(t) : u < t u}contains at least one point of G1, and {x(t) : u < t < u + } contains no points of G1. Then{x(t) : u < t < u + } cannot be a subset of either G1 or G2. Since G1, G2 are open,x(u) 6 G1, x(u) / G2. Then G1 G2 63 x(u), and G1 G2 6= G. 2

Corollary 8.7 Rn with the usual metric is connected.

Definition 8.8 Let (X, d) be a metric space, and let E be a non-empty subset of X. C is acomponent of E if (i) C E, (ii) C is connected, and (iii) C D E and D connected impliesD = C.

Lemma 8.9 Let E be a collection of connected sets in a metric space, EEE 6= , then EEEis connected.

Proof. Suppose H = EEE is not connected. Then there exist open sets G1, G2 such thatG1 H 6= , G2 H 6= , G1 G2 H = , G1 G2 H. Let a EEE. Since a H, we mayassume a G1. Let E E be arbitrary . Then a E H so that G1 E 6= , G1 G2 E =, G1 G2 E. Since E is connected, G2 E = . Hence G2 H = since E E was arbitrary.Contradiction. Hence H = EEE is connected. 2

Theorem 8.10 Any set in a metric space has a unique decomposition into components.

Proof. Let E be a set in a metric space. For any x E, denote by Cx the union of all theconnected subsets of E which contain x. Since {x} is connected, Cx 6= . By Lemma 8.9 Cx isconnected. If C is a connected subset of E which contains Cx then, by definition of Cx, Cx C,and so Cx is a component. Any component of E is a set of the form Cx. 2

Corollary 8.11 An open set in Rn is the union of countably many disjoint, open, connected sets.

Proof. Let G be an open set in Rn, and let C be a component of G. If x C then Cx = C. SinceG is open there exists a > 0 such that B(x; ) G. Since B(x; ) is connected, B(x; ) Cx = C.Hence C is open, and connected. Since components are disjoint, G is the union of disjoint opensets, each of which contains distinct rational points i.e. the union is countable. 2

Corollary 8.12 Any open set in R with the usual metric is the union of countably many openintervals.

Theorem 8.13 If the domain of a continuous function is connected then so is the range.

Proof. Let (X, d), (Y, e) be metric spaces, and let f : X Y be continuous. If f(X) isnot connected, then Y contains open subsets G1, G2 which intersect f(X), and are such thatG1 G2 f(X) = , G1 G2 f(X). Since f is continuous f1(G1), f1(G2) are open. Theseare also non-empty since G1 f(X) 6= , G2 f(X) 6= . Moreover, f1(G1) f1(G2) = , andf1(G1) f1(G2) X. Hence X is not connected. 2

29

Corollary 8.14 If the domain of a real-valued, continuous function is connected, then the rangeis an interval.

Lemma 8.15 Let A be a connected set in a metric space (X, d), and let x be a limit point of A.Then A {x} is connected.Proof. Suppose that x / A and that A {x} is not connected. Then there exist open setsG1, G2 such that G1 (A {x}) 6= , G2 (A {x}) 6= (1), G1 G2 (A {x}) (2),G1 G2 (A {x}) = (3). Suppose x G1. Since G1 is open and x is a limit point of A,G1A 6= . Since x G1 we have by (3) that x / G2. By (1) G2A 6= , and by (2) G1G2 A.Hence A is not connected. Contradiction. 2

Theorem 8.16 Let A be connected. Then A is connected.

Proof. IfA = A thenA is connected. SupposeA\A 6= . For any x A\A we define Ex = A{x}.Then {Ex : x A\A} is by Lemma 8.15 a family of connected sets whose intersection contains A,and hence is non-empty. Then xA\AEx = A is connected by Lemma 8.9 2

Definition 8.17 A path in a metric space (X, d) is a continuous function : [a, b] X. Asubset S of X is path connected if for any x1, x2, S there exists a path : [a, b] S with(a) = x1, (b) = x2.

Example 8.18 Open, connected sets in Rn with the usual metric are path connected.

Theorem 8.19 Let f : S X be continuous, where S is path connected in a metric space (X, d).Then f(S) is path connected.

Proof. Let x1, x2 f(S). Then there exist s1, s2 in S such that x1 = f(s1), x2 = f(s2). SinceS is path connected, there exists a path : [a, b] S with (a) = s1, (b) = s2. Since f, arecontinuous f : [a, b] f(S) is continuous, and hence is a path from x1 to x2 in f(S). 2

Theorem 8.20 Path connected sets in a metric space are connected.

Proof. Let K be path connected. If K = then K is connected. Suppose K 6= . Letp K. Then for each a K there exists a path a : [0, 1] K from p to a. Furthermorea a([0, 1]) K. Hence K = aKa([0, 1]). But a([0, 1]) is connected by Theorem 8.13.Moreover, aKa([0, 1]) 3 p and is non-empty. Hence K is connected by Lemma 8.9. 2

Remark 8.21 Connected sets in a metric space are not necessarily path connected. Let An ={(x, y) R2 : x 0, y = xn}. Then An is path connected, and hence connected. Since n=1An 3(0, 0), A = n=1An is connected. Let p = (1, 0). Then p is a limit point of n=1An, andso n=1An {p} is connected by Lemma 8.15. Suppose n=1An {p} is path connected. Let : [0, 1] n=1An {p} be a path from (0, 0) to p. Let t = inf{t : (t) = p}. By continuity(t) = 1. Without loss of generality we may take t = 1. (Otherwise we rescale the t variable.)Let t1 < t2 < < 1 be a sequence such that d((tm), p) < 1/m. Let (mk) be a subsequenceof (m) such that tmk Anmk . Then d((tmk), p) < 1mk . There exists a sequence (smk), withtmk < smk < tmk+1 such that (smk) = 0. This contradicts the continuity of at 1. 2

Theorem 8.22 Let (X, d) be a metric space. Suppose that H is dense in K, and H is connected.Then K is connected.

Proof. Let x K\H. Since K H, x is a limit point of H. By Lemma 8.15 H {x} isconnected. For each x K\H, H {x} H. By Lemma 8.9 K = xK\HH {x} is connected.2

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9 Topological Spaces

Definition 9.1 A topological space is a non-empty set X together with a family T of subsetsof X which satisfies (i) X T , T , (ii) Any union of elements of T belongs to T , (iii) IfT1 T , T2 T then T1 T2 T . We say that T is a topology on X. The elements of T arecalled T - open or open sets.

Example 9.2 (i) Let X be a non-empty set and let T = {X, }. Then T is the indiscretetopology on X.

(ii) Let X be a non-emptyset and let T be the collection of all subsets of X. Then T is thediscrete topology on X.

(iii) Let d : X X [0,) be a metric on X. Let T = {G X : G is open}. Then T is thetopology on X induced by the metric d.

(iv) Let X = {a, b, c, d, e}, T1 = {X, , {a}, {c, d}, {a, c, d}, {b, c, d, e}}. Then T1 is a topology onX.

(v) Let X = {a, b, c, d, e}, T2 = {X, , {a}, {c, d}, {a, c, d}, {b, c, d, }}. Then T2 is a topology onX.

(vi) Let X = {a, b, c, d, e}, T3 = {X, , {a}, {c, d}, {a, c, d}, {a, b, d, e}}. Then T3 is not a topologyon X.

Theorem 9.3 Let X be a non-empty set and let {Ti : i I} be a family of topologies on X, whereI is an index set. Then {Ti : i I} is a topology on X.

Proof. X and are in Ti, i I. Hence X {Ti : i I}, and {Ti : i I}. Any collection ofelements in {Ti : i I} belongs to Ti. Since Ti is a topology the union of this collection belongsto Ti. Since i I was arbitrary the union of this collection belongs to {Ti : i I}. If T1, T2 aresets in {Ti : i I}. Then T1 T2 T Ti. Since Ti is a topology this intersection belongs toTi. Since i I was arbitrary this of this intersection belongs to {Ti : i I}. 2

Definition 9.4 Let X be a non-empty set and let A be a collection of subsets. Then

{T is a topology on X, T A}

is the smallest topology on X which contains the collection A. Note that this topology is notempty since the discrete topology on X contains A.

Remark 9.5 Let X = {a, b, c}, T1 = {X, , {a}}, T2 = {X, , {b}}. Then T1T2 = {X, , {a}, {b}}is not a topology on X.

Definition 9.6 Let (X, T ) be a topological space. A point p is a limit point of a set A X ifevery open set G in T containing p contains a point of A different from p. I.e.

(G T )[(G 3 p) ((G A) \ {p} 6= )].

The set of limit points of A is denoted by A. A subset C is closed if X \ C T .

Example 9.7 (i) Let X = {a, b, c, d, e}, and let T1 be the topology defined in Example 9.2.Let A = {a, b, c}. Then A = {b, d, e}.

(ii) Let (X, T ) be an indiscrete topological space, and let A be any subset of X. Then A = ifA = , A = X \ {p} if A = {p}, A = X if A contains at least two points.

31

(iii) X and are both open and closed.(iv) Any subset of a discrete topological space is both open and closed.

(v) The set {b} T1 in Example 9.2 is neither open nor closed.

Theorem 9.8 Let (X, T ) be a topological space. The intersection of any number of closed sets isclosed. The union of any two closed sets is closed.

Proof. These assertions follow by De Morgans law as in the proof of Theorem 3.20. 2

Theorem 9.9 Let (X, T ) be a topological space, and let A be a subset of X. Then A is closed ifand only if A contains its limit points.

Proof. Suppose A is closed. Let p Ac. Then Ac is open and contains p. But A Ac = and so p is not a limit point of A. Suppose A A. Let p Ac. Then p is neither a limitpoint of A nor a point of A. Hence there exists an open set Gp which contains p and whichis such that (Gp \ {p}) A = . Since p / A we have Gp A = . Hence Gp Ac. HenceAc = pAc{p} = pAcGp. This is a union of open sets, and so Ac is open. 2

Theorem 9.10 Let (X, T ) be a topological space, and let A,B be subsets of X. If A B and Bis closed then A B. Moreover A is a closed.Proof. Suppose p Bc. Then p is neither a point of B nor a limit point of B. Hence there existsan open set Gp which contains p and which is such that Gp B = . Hence Gp A = . So p isnot a limit point of A. So A = AA B. Next suppose that p Ac. Then p is neither a point ofA nor a limit point of A. Hence there exists an open set Gp such that p Gp and A (Gp \ {p}).Since p / A, AGp = . Suppose that A Gp 6= and contains the point x. Then x 6= p. So Gpis an open set which contains the limit point x of A. Then (Gp \ {x}) A 6= . This gives thatGp A 6= , which is a contradiction.

Theorem 9.11 Let (X, T ) be a topological space, and let A be a subset of X. ThenA = {F :FA,F closed}F.

Proof. The set A is closed and contains A, and so appears in the intersection in the aboveright-hand side. Hence {F :FA,F closed}F A. We also have that A {F :FA,F closed}F. Theright-hand side is an intersection of closed sets and hence is closed. By the first part of Theorem9.10, A {F :FA,F closed}F.

10 Exercises.

1. Let X be a non-empty set, and suppose : X X R satisfies 0 (x, y) < for allx, y X, (x, y) = 0 if and only if x = y, and (x, y) (x, z) + (y, z) for all x, y, z X.Prove that is a metric on X.

2. Let (X, d) be a metric space, let 0 < 1 and let e(x, y) = d(x, y), x X, y X. Provethat e is a metric on X.

3. Prove that if (X, ) is a normed space then X is a metric space with metricx, y X d(x, y) = x y.

32

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

Prove that d(x0, xn) ni=1

d(xi1, xi) for all x0, x1, . . . , xn X.

5. Let (X1, d1), (X2, d2) be metric spaces. Let X = X1 X2.Define d : X2 R by

(x1, x2), (y1, y2) X

d((x1, x2), (y1, y2)) =d21(x1, y1) + d

22(x2, y2).

Verify the axioms of the metric space.

6. Let B[0, 1] be the set of bounded functions on [0, 1] equipped with the supremum metric.Let (fn) be the sequence in B[0, 1] defined by

fn(x) =

{1 nx , 0 x 1n0 , 1n x 1.

(i) Compute d(fn, fm), where n,m N, n m.(ii) Prove that (fn) does not converge in {B[0, 1], d}.

7. Let C[0, 1] be the set of continuous real valued functions on [0, 1]. Define for f, g C[0, 1],

d(f, g) =

{ 10

|f(x) g(x)|2dx}1/2

.

(i) Prove that d is a metric on C[0, 1].

(ii) Let e denote the supremum metric on C[0, 1]. Prove that fn f in {C[0, 1], e} impliesfn f in {C[0, 1], d}.

8. Let S be the space of sequences x = (x1, x2 . . . ) of real numbers such thatxn converges

absolutely. For x, y S put(x, y) =

n=1

|xn yn|.

Prove that is a metric on S.

9. Let X be the set of all bounded sequences of real numbers: every element in X can bewritten as x = (1, 2, . . . ) and there exists cx R such that |j | cx for all j. Defined(x, y) = supjN |j j |, where x = (j) = (1, 2, . . . ) X, y = (j) = (1, 2, . . . ) X.Prove that d is a metric on X.

10. Find int(E), E, b(E) and E in each of the following subsets of R2 with the euclidean metric:

(i) E = {(x, y) : x2 + y2 < 5}(ii) E = {(x, y) : 1 x2 + y2 < 5}

(iii) E = Q {0}(iv) E = Q (R\Q)

33

(v) E =

{1

n: n N

} Z.

11. Let (X, d) be a metric space, A closed subset, x 6 A. Prove that

d(x,A) := inf{d(x, y) : y A} > 0.

12. Prove that in Rn with the Euclidean metric any collection of disjoint open sets is at mostcountable. Is this true for an arbitrary metric space?

13. (i) Prove that E is closed if and only if b(E) E.(ii) Prove that b(F ) = if and only if F is open and closed.

14. (i) Prove that int(E) is the union of all open sets contained in E (and so is the largestopen set contained in E).

(ii) Prove that E is the intersection of all closed sets containing E (and so is the smallestclosed set containing E).

15. (i) Prove that E1 Em = E1 Em but that for an infinite collection of setsEi, i I

iIEi

iI

Ei.

(ii) Prove thatiI

Ei iI

Ei. Construct an example to show that equality need not hold

even for a finite collection.

16. (i) Prove that X\E = int (X\E).(ii) Prove that int(E) b(E) int(X\E) = X.

17. Let B[0, 1] have the supremum metric d, and let f B[0, 1] be given by f(t) = t2, 0 t 1.(i) Describe the open ball B(f ; 1).

(ii) Describe the closed ball B(f ; 1).

(iii) Find the least value of r such that B(f ; r) contains g, g(t) = t(1 t).(iv) Let C[0, 1] B[0, 1] be the set of continuous functions from [0, 1] into R. Find

int(C[0, 1]) and b(C[0, 1]).

18. A set A in a metric space is bounded if the diameter of A is finite: i.e.

diam(A) = sup{d(a1, a2) : a1 A, a2 A}

(iv) Give an example of bounded set E and F such that diam (EF ) > diam(E)+diam(F ).

19. Let (xi) be a sequence of distinct elements in a metric space, and suppose that xi x. Letf be a one-to-one map of the set of xis into itself. Prove that f(xi) x.

20. Let (X, d) be a metric space. A set D is said to be dense in E if D E D. Show that ifC is dense in D, and D is dense in E, then C is dense in E.

21. Let B[a, b] have the supremum metric d. Let E = {f B[a, b] : f(x) > 0, a x b},F = {f C[a, b] : f(x) > 0, a x b}.

(i) Describe int(E), E, and b(E).

(ii) Describe int(F ), F , and b(F ).

22. Let f : (0,) (0,) [0,) be given by

f((x, y)) =xy

x+ y.

Find all , such that lim(x,y)(0,0)

f((x, y)) exists.

23. Let (X, ), (Y, ) be metric spaces. Let f, g : X Y be continuous. Prove that the setB = {x X : f(x) = g(x)}

is a closed subset of X.

24. Let (X, ) and (X,) be metric spaces, and let I : (X, ) (X,) be the identity functiongiven by I(x) = x, x X.

(i) Prove that I is a bijection.

(ii) Prove that both I and I1 are continuous if and only if and are equivalent metrics.

(iii) Give an example to show that when and are not equivalent I may be continuouswithout I1 being continuous.

25. Let A be a non-empty set in a metric space (X, d). Define f : X R by f(x) = inf{d(a, x) :a A}. Prove that f is continuous.

26. Let (X, ), (Y, ) and (Z, ) be metric spaces, and let f : X Y , g : Y Z be continuousfunctions. Prove that g f : X Z is continuous.

27. Let (X, ) and (Y, ) be metric spaces and let f and g be continuous functions from X intoY which are equal on a dense subset of X. Prove that f = g on all of X.

28. Let (X, ) and (Y, ) be metric spaces. Prove that f : X Y is continuous if and only iff(H) f(H) for all sets H in X.

35

29. Which of the following subsets of R2 with the Euclidean metric are complete:(i) {(x, y) : x2 + y2 < 2}, (ii) Z Z, (iii) {(x, y) : x > 0, y 1x} ,(iv) {(x, y) : 0 < x < 1, y = 0}?

30. Define d : R R R by d(x, y) = |arctanx arctany|.(i) Prove that (R, d) is a metric space.(ii) Prove that the sequence (xn) with xn = n is a Cauchy sequence in (R, d) which does

not converge.

31. (i) Prove that the intersection of any collection of complete subsets of a metric space iscomplete.

(ii) Prove that the union of a finite number of complete subsets of a metric space is complete.

32. Let X be the set of all ordered n-tuples x = (x1, . . . , xn) of real numbers, and define d :X X R by d(x, y) = max

j|xj yj |, where y = (y1, . . . , yn).

(i) Prove that (X, d) is a metric space.

(ii) Prove that X is complete.

33. Let (X, d) and (Y, e) be complete metric spaces, and let M = X Y .Define f : M M R by f((x1, y1), (x2, y2)) = d(x1, x2) + e(y1, y2).

(i) Prove that (M,f) is a metric space.

(ii) Prove that M is complete.

34. Define : R R R by (x, y) = x1 + |x| y1 + |y|

.(i) Prove that (R, ) is a metric space.(ii) Prove that R is not complete.

(iii) Prove that is equivalent to the usual metric on R.(iv) The metrics , on a set X are equivalent. Are (X, ) and (X,) necessarily both

complete or both not complete?

35. (i) Prove that if every closed ball of a metric space (X, d) is complete then X is complete.

(ii) Prove that if every countable closed subset of a metric space (X, d) is complete then Xis complete.

36. Let T : [1,) [1,) be defined by

Tx = x+1

x.

Is T a contraction? Does it have a fixed point?

36

37. Let T : [1, 1] R be given by Tx = 1 cx2, where R has the usual metric.

(i) Find all values of c such that T is a contraction mapping from [1, 1] into [1, 1].(ii) Let c = 13 and x0 = 0. Compute the iterates T

jx0 for j = 1, 2, 3 and find limj

T jx0.

38. Let x be the positive real root of 2 sinx = x.

(i) Find a closed interval X in R containing x such that T : X X given by Tx = 2 sinxis a contraction mapping.

(ii) Use the Banach fixed point theorem and a calculator to find an approximation for x

within 104.

(iii) Suppose x0 =pi

2. Find an integer N such that |Tnx0 x| < 1012 for n > N .

39. Consider the integral equation (x) =

11xy(y)dy + x2, 1 x 1.

(i) Prove that there exists a unique solution C[1, 1] (the real valued continuousfunctions on [1, 1]) for || < 12 .

(ii) Find the solution of the integral equation for || < 12 . What happens if = 32?

40. Consider the differential equationdy

dx= x+ y2 with initial condition y(0) = 0.

(i) Prove that there exists a unique solution on [, ] for any (0, 22/3).(ii) Find the first few terms of the Taylor expansion of the solution about 0.

41. Let C[0, 1] have the supremum metric d, and let T : C[0, 1] C[0, 1] be defined by

Tf(t) =

t0

f()d, 0 t 1, f C[0, 1].

Prove that T is not a contraction mapping but that T 2 is a contraction mapping.

42. Let (X, d) be a complete metric space, and let Ti, i = 1, 2, . . . and T be contraction mappingson X with fixed points xi, i = 1, 2 . . . and x respectively. Prove that if Ti converges uniformlyto T then limi xi = x.

43. Let (X, d) be a complete metric space, and let T : X X. Suppose T 2 = T T is acontraction mapping. Prove that T has a unique fixed point p in X, and that for anyx X, limj T jx = p.

44. Let A1, . . . , An be compact sets in a metric space (X, d).Prove that A1 An is compact.

45. Let E be a compact set in a metric space (X, d). Let diam(E) denote the diameter of E.

(i) Prove that there exist points x, y E such that diam(E) = d(x, y).

37

(ii) Give an example to show that these points are not necessarily unique.

46. Which of the following subsets of R2 with the usual metric are compact?

(i) {(x, y) R2 : x = 0},(ii) {(x, y) R2 : x2 + y2 = 1},

(iii) {(x, y) Z2 : x2 + y2 < 25},(iv) {(x, y) R2 : x y},(v) {(x, y) R2 : x2 + y2 = 1, x Q}.(vi) {(x, y) R2 : y = sin 1x , 0 < x 1} {(x, y) R2 : 1 y 1, x = 0}.

47. Let T be a compact metric space with metric d, and suppose that f : T T is a continuousmap such that for every x T , f(x) 6= x. Prove that there exists an > 0 such thatd(f(x), x) for all x T . (Hint: consider g : T R given by g(x) = d(f(x), x)).

48. Determine whether f : A R is uniformly continuous on A, when(i) A = (0, 1), f(x) = x, (ii) A = (0, 1), f(x) = 1/(1 x),(iii) A = R, f(x) = cosx, (iv) A = (0, 1), f(x) = sin(1/x),(v) A = [0,), f(x) = x, (vi) A = [1,), f(x) = 1/x.

49. Let f : [0,) R be continuous, and suppose that limx f(x) exists. Prove that f isuniformly continuous.

50. The functions f, g : C[0, 1] C[0, 1] with the supremum metric are given by f()(x) = x0(t)dt, g()(x) =

x0(t)2dt ( C[0, 1], x [0, 1]). Show that both functions are con-

tinuous on C[0, 1]. Are they uniformly continuous?

51. Let B[0, 1] be the set of bounded real valued functions on [0, 1] with the usual metric

d(f, g) = sup0x1

|f(x) g(x)|. Let fn : [0, 1] R be given by fn(x) = 1 if x = 1n

, fn(x) = 0

if x 6= 1n

.

(i) Prove that fn B[0, 1].(ii) Prove that (fn) does not have a convergent subsequence.

(iii) Let B[0, 1] be defined by (x) = 0, 0 x 1. Prove that B(; 1) is closed,bounded, and complete but not compact.

52. Give an example of a metric space (X, d) with disjoint, closed subsets A, B such thatinf{d(x, y) : x A, y B} = 0.

53. Let (X, d) be a metric space with disjoint compact subsets A, B. Prove that there exista A, b B such that inf{d(x, y) : x A, y B} = d(a, b), and conclude that d(a, b) > 0.

38

54. A real-valued map f on a metric space (X, d) is lower semi-continuous if for every a R, f1(a,) is open in X. Prove that if X is compact and f : X R is lower semi-continuous then f is bounded below and attains its lower bound on X.

55. Which of the following subsets of R2 with the usual metric are connected?

(i) B((0, 0); 1) B((2, 0); 1),(ii) B((0, 0); 1) B((2, 0); 1),

(iii) B((0, 0); 1) B((2, 0); 1),(iv) the set of all points with at least one coordinate in Q.

56. Prove that the components of a closed subset of a metric space are closed.

57. Let f : (0,) R2 be given by f(x) = (x, sin 1x ).(i) Prove that f is continuous.

(ii) Prove that f(0,) is connected.(iii) Prove that f(0,) {(x, y) R2 : x = 0, 1 y 1} is connected.

58. Prove that the intersection of two connected sets in R is connected. Show by an examplethat this is false in R2.

59. Let Q be the rationals with metric d(q1, q2) = |q1 q2|. Prove that the only non-emptyconnected sets are those consisting of single points.

60. Show by means of an example that the components of an open set are not necessarily open.

61. Give an example of closed connected subsets Cn R2 such that Cn Cn+1 for alln = 1, 2, . . ., but

n=1

Cn is not connected.

62. Prove that C[0, 1] with the supremum metric is connected.

63. Let S = {(x1, x2) R2 : x21 + x22 = 1}, and let f : S R be a continuous map. Showthat there exists a point x = (x1, x2) S such that f(x) = f(x), where x = (x1,x2).(Hint: consider g : S R given by g(x) = f(x) f(x).)

64. Which of the following sequences (fn) converge uniformly on [0, 1]?

(i) fn(x) =x

1 + nx, (ii) fn(x) = nxe

nx2 , (iii) fn(x) = n12x(1 x)n,

(iv) fn(x) = nx(1 x2)n2 , (v) fn(x) = xn

1 + xn, (vi) fn(x) = n

xxn cos(nx).

39

65. The set X consists of the real functions on [a, b] such that on [a, b] exists and iscontinuous, and |(x)| M , |(x)| M . For n = 1, 2, . . . the functions fn : X Rare defined by fn() =

ba

(x) sin(nx)dx, X. Prove that the sequence (fn) convergesuniformly on X.

66. The functions fn on [0, 1] are given by

fn(x) =nx

1 + n2xp(p > 0).

(i) For what values of p does the sequence converge uniformly to its limit f?

(ii) Examine whether

10

fn(x)dx 10

f(x)dx for p = 2, p = 4.

67. The functions fn on [1, 1] are given by

fn(x) =x

1 + n2x2.

(i) Show that (fn) converges uniformly and that the limit function f is differentiable.

(ii) Show that the relation f (x) = limn f

n(x) does not hold for all x in [1, 1].

68. Let (fn) be a sequence of functions from [a, b] into R. Prove that if (fn) converges pointwiseon [a, b] to a continuous function f , and that if each fn is monotonically increasing, then theconvergence is uniform.

69. Let C[0, 1] and let fn(x) = xn(x), 0 x 1. Prove that (fn) converges uniformly on[0, 1] if and only if (1) = 0.

70. Let fn(x) = (1 + xn)1/n, 0 x 2. Prove that the sequence (fn) of differentiable functions

converges uniformly to a limit function which is not differentiable at x = 1.

71. Let fn(x) = (sin(npix))n, 0 x 1.

(i) Examine whether limn fn exists .

(ii) Examine whether limn 10fn(x)dx exists.

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11 Sample Examination Paper

UNIVERSITY OF BRISTOL

Examination for the Degrees of B.Sc., B.A. and M.Sci. (Level 2)

METRIC SPACES

MATH 20006(Paper Code MATH-20006)

January 2014, 2 hours 30 minutes

This paper contains five questionsThe FOUR best answers will be used for assessment.

Calculators are not permitted in this examination.

On this examination,