Mathematical Analysis: A Concise Introduction 9811221634, 9789811221637

Table of contents :
Preface
Contents
Chapter 1 - Metric Spaces and Limits for Sequences
1.1 Metric Spaces
1.2 Sequences and Limits
1.3 Sets in Metric Spaces
1.4 Properties of Metric Spaces
Chapter 2 - Functions on Metric Spaces
2.1 Continuity
2.2 Properties of Continuous Functions
Chapter 3 - Differentiaion
3.1 Derivatives
3.2 Fréchet Differentiability
3.3 Inverse Function and Implicit Function Theorems
3.4 Higher Order Derivatives
Chapter 4 - Riemann Integrals
4.1 Definition of Integrals
4.2 Properties of Integrals
4.3 Further Theorems
Chapter 5 - Uniform Convergence
5.1 Observations and Examples
5.2 Uniform Convergence
5.3 The Metric of Uniform Convergence
5.4 Limit Theorems
Chapter 6 - Series
6.1 Series of Numbers
6.2 Further Tests for Convergence
6.3 Series of Functions
6.4 More Convergence Criteria
6.5 Power Series
6.6 Fourier Series
Appendix

Citation preview

MATHEMATICAL ANALYSIS A Concise Introduction

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Jiongmin Yong University of Central Florida, USA

World Scientific NEW JERSEY • LONDON • SINGAPORE • BEIJING • SHANGHAI • HONGKONG • TAIPEI • CHENNAI

.

TOKYO

.

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Control Number: 2020055934

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. /

r

MATHEMATICAL ANALYSIS A Concise Introduction Copyright © 2021 by World Scientific Publishing Co. Pte. Ltd.

All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

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ISBN 978 981-122-163-7 ( hardcover) ISBN 978-981-122 164-4 (ebook for institutions) ISBN 978 981 122 165 1 (ebook for individuals)

For any available supplementary material, please visit https://www.worldscientific.eom/worldscibooks/10. l 142/11859#t=suppl

In the Memory of My Mother

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Preface

Mathematical analysis serves as a common foundation for many research areas of pure and applied mathematics. It is also an important and powerful tool used in many other science fields, including physics, chemistry, biology, engineering, finance, economics, to mention just a few. On the other hand , it is a desire that after having taken a sequence of courses on basic calculus and linear algebra, one could spend a reasonable length of time ( ideally, say, one semester ) , to build an advanced base of analysis sufficient for getting into various research fields other than analysis itself , and /or stepping into more advanced level (s) of analysis courses (such as real analysis, complex analysis , differential equations, functional analysis, stochastic analysis, etc. ). To meet such a need , I wrote this concise introduction on mathematical analysis for a course that can be completed within one semester. From the author’s viewpoint , mathematical analysis should present a theory for the functions between the best regular ones ( meaning analytic functions which are treated in complex variables / analysis) and the worst regular ones ( meaning measurable functions which are treated in real analysis ). More precisely, this course discusses functions with various continuity, differentiability, Riemann integrability, and relevant topics of limit , series, and uniform convergence. For the convergence and continuity of functions, the linear ( or algebraic ) structure of the underlying space is irrelevant . Therefore, we consider functions and sequences in general metric spaces. As a preparation for that , some basic properties of metric spaces are discussed , such as completeness, compactness, connectedness and separability, etc. For differentiability and Riemann integrability of functions, the linear structure of the underlying space is necessary. Therefore, we discuss the problems of differentiation and integration in the m-dimensional Euclidean space. Vll

vm

Mathematical Analysis

— A Concise Introduction

Series is an independent topic. Partially, it is an application of the previous chapters if one views it from the limit of partial sums. On the other hand , the study of the series convergence has lots of its own feature. Power series and Fourier series make this part of theory very rich . Also, the theory of series can even touch one of the most challenging open questions in mathematical research: the Riemann hypothesis. Real number system is traditionally included in the course of mathematical analysis. Due to the time limit of the course, we decided to put it in a self-contained appendix at the end of the book so that the interested readers could still be able to read it separately, and it becomes optional for those readers who are only interested in the main body of mathematical analysis other than the real number system.

Finally, problems listed at the end of sections are complementary to the material covered in the main body of the book , and those with asterisks might be a little difficult. Readers could skip them at the first reading.

Jiongmin Yong Orlando , USA July, 2020

Contents

Preface 1.

Metric Spaces and Limits for Sequences

1

Metric Spaces Sequences and Limits Sets in Metric Spaces Properties of Metric Spaces

1 10 21 31

1

2 3 4 2.

45

Continuity Properties of Continuous Functions Compactness preserving and uniform continuity 2.1 2.2 Connectedness preserving

Differentiation 1 2 3 4

4.

...

Functions on Metric Spaces

1 2

3.

Vll

61

Derivatives Frechet Differentiability Inverse Function and Implicit Function Theorems . Higher Order Derivatives

...

61 76 80 89

95

Riemann Integrals

1 2 3

.

45 52 53 58

Definition of Integrals Properties of Integrals Further Theorems . .

95 107 112

IX

Mathematical Analysis

x

5.

A Concise Introduction

121

Uniform Convergence 1 2 3 4

6.

—

Observations and Examples . . . . Uniform Convergence The Metric of Uniform Convergence Limit Theorems

121 123 132 140 151

Series 1 2 3 4 5

6

Series of Numbers Further Tests for Convergence Series of Functions More Convergence Criteria . . Power Series A general consideration 5.1 Exponential and logarithmic functions 5.2 The sine and cosine functions 5.3 Fourier Series Inner product 6.1 polynomial approximation Trigonometric 6.2 Fourier series 6.3

151

. 157 167 175 179 179 185 187 192 193 196 . 198

207

Appendix The Real Number System 1

2

3

4

Natural Numbers Peano axioms 1.1 Addition and multiplication 1.2 Orders 1.3

207

. 207

...

Integers and Rationals Integers 2.1 Rational numbers 2.2 Real Numbers Sequences of rational numbers . . 3.1 A construction of real numbers . . 3.2 Further properties of real numbers 3.3 Real exponentiation 3.4 Complex Numbers

.

209 214 218 218 226 236 236 237 243 252 256

Bibliography

257

Index

259

'


R as follows: Example 1.4 (i) Let X

*

do ( x , y ) =

° 1

x = y, x y.

^

It is easy to see that do ( , •) is a metric which is called the discrete metric on X . Such a metric can always be defined on any non-empty set X . Hence, for any non-empty set X , one can always define a metric so that it becomes a metric space. *

1See

the Appendix for a theory of the real number system .

Metric Spaces and Limits for Sequences

3

Example 1.5. Let ( X , d ) be a metric space and Y Cl. Then the restriction d |Y x Y of d on Y x Y is clearly a metric on Y; ( Y, d|y x y ) is called a subspace of ( X, d ) ; and d | y x y is called the metric on Y induced by d . For convenience, hereafter , we simply denote ( Y, d) instead of ( Y, d|y x y ) . Under the standard metric , N = {1, 2, 3, • • • } ( the set of all natural numbers) , Z ( the set of all integers) , Q ( the set of all rational numbers ) , and interval [a , b] are all subspaces of R.

To have more interesting examples, for any integer m

1, we define

= { x = (x1, x2, • • • , xm ) | xk G R , 1 /c m}. = ( x1, #2 , - - , xm ) , t/ = (2/19 2/ 2 , • • , y m ) £ Rm and Rm

For any x introduce

*

*

x + y = (x1 + 2/1, x 2 -f y2 ,

- - - , xm + t/

m

),

Ax = ( Ax1, Ax 2 ,

A 6 R , we

- - - , Axm ).

These operations are called addition and scalar multiplication, respectively, for which the following eight axioms are satisfied:

Vx, y G Rm x + ?/ = y 4- x, ( x + y ) + z = x + ( y 4- z ) Vx , ?/ , 2 G R m Vx G Rm x+0 = 0+x = x Vx Rm ® + ( -x ) = 0 , Vx , y G Mm A G R , A ( x + y ) = Ax + Ay , ( A 4- p ) x = Ax f /ix , Vx G Mm , A , fi G R , Vx G Rm lx = x . A (/zx ) = ( Ay) x, Vx G Rm A , fi G R. '

With the addition and scalar multiplication , Rm is called a linear space ( or a vector space ) . Each x G Rm is called a vector. The addition and scalar multiplication are called algebraic operations on Rm . Any properties of Rm involving addition and scalar multiplication are referred to as algebraic properties of Rm .

Now , for any x

= (x1, x 2 , • • • , xm ) G Rm , we define 771

II*IIP =

2.

( EVr)' =

p e [i , oo )

k 1

max

^^

l /c ra

P = oo ,

( 1.2 )

Mathematical Analysis

4

— A Concise Introduction

which is called the £ p- norm. We refer to ( Rm • ||p ) as a normed linear space. The following proposition is concerned with the fp- norm. Proposition 1.6 . L e t p e [ l , oo] . Then the above-defined £ p-norm satisfies the following:

( i) ( positivity ) ||x ||p x = 0.

for

^0

all x £ Rm , and \ \ x \\ p

( ii ) ( positive homogeneity ) || Ax ||p = |A|||x || p ,

for

only if

all x £ Rm and A £ R .

||x ||p + \ \ y \\ p ,

( iii) ( triangle inequality ) ||x -F y \\ p

= 0 if and

for

all x , y £ Rm .

From the definition of fp- norm , we easily see that (i) — ( ii) are true. To prove ( iii ) , we first look at the cases p = 1, oo. One has m

m

m

H= lxfcl + H=

F + yk \

x + vh = k =1

k 1

k 1

= ll ^ ll 1 + IMIi

.

and

x + y || oo

max |x* + {/fc|

=

max |xfe| + max |j/fc| = 11 x 11 «, + \\ y\\oo

-

Now , we look at the case p = 2. If y = 0 , the proof is trivial. Thus, let x, y £ R with 2/ 7^ 0. For any A £ R , one has m

0

m

=£

\\ x - X y\\ l

(*fc - A / ) 2

= H II

1 2

- 2A

fc = l

^

x V + A2||y|| .

^

fc = l

Taking

A

m

=

5

we obtain 0

0

||x||2

n

0

1

+

~

m

AI ( £ *•*)

0

which leads to m

Y= fc

^ l

xkyk

< ||x|| 2 ||y ||2

V x, y

e Rm .

(1.3)

5

Metric Spaces and Limits for Sequences

.

This is called the Cauchy- Schwarz inequality Then it follows that m

II* + y||!=

£ =

m

= £ ( 1*k 2 + 2 xkyk + |yfe|2 )

(** +

|

k =1

k l m

= IMI 2 + 2 £ ®V+ I|y|l 2 < Ha; l 2 + 2 lla; ll 2 ||y ||2 + ||2/||2 = (||a;||2 +||2/||2 ) /c = l

2

-

Consequently, (iii) holds for p = 2.

For general p £ (1, 2 ) U ( 2, 00 ) , the situation is a little more subtle. We need the following results which are interesting by themselves. Proposition 1.7. Let p , q £ ( 1 , 00 ) such that

^^

= 1.

( i) ( Young’s Inequality ) For any a , b £ (0, 00 ) , the following holds: ap b« . (1.4 ) ab — -b .

^

p

—q

( ii) ( Holder’s Inequality ) For any x , y £ Rm , the following holds: m

*V £ =

IMIpIlyllg .

(1.5)

k 1

(iii) ( Minkowski’s Inequality ) For any x , y

\\ x + y \\ p Proof , ( i) Let

^ I ^ IIP + IIPIIP

£

Rm , the following holds:

( 1.6 )

*

./ x rp 1 fir ) = + - - r, r £ [0, 00 ) . p 9 We try to find the minimum of / ( •) . To this end , 2 setting /

—

f' ( r ) we get r

= 1.

= rp

_1

Clearly, by the first derivative test , one has f (r )

This is equivalent to

/ (1) =

^^ +

rp

— 1 = 0,

1 f9 P

2 We

- 1 = 0,

Vr

£

( 0 , 00 ) .

r.

assume that single variable calculus is known to the readers.

(1.7)

6

— A Concise Introduction

Mathematical Analysis

Taking r = ab 1

i ~

p

, one has r p 1 h-= p q

=r Consequently, i

-

ab = ab 1 ~ p b

! P

/ apb 1 ~ p


.

V ( xk ) , ( yk ) e i p ,

p

p e [1, oo ) ,

k =1

and

Poo ( ( xk ) , ( yk ) ) = sup |xfe - ykI

V( xfe ) , ( / ) s

^

k l

-

^.

(p £ [l , oo] ) . In fact , again , properties ( a) Then pp ( , •) is a metric on and ( b) in the definition of the metric ( Definition 1.3) hold for pp ( , ) in f p . Now for the triangle inequality, we let N 1 and from the Minkowski’s inequality, one has

- -

^

( E i*

fc

fc = l

+

^^ P

)

( E i** r) + ( E I »* IP ) lp fc = l

o o , one can rigorously show that llx + y \\ p li ^ llp i II / IIP ?

Letting N

^ Hence, pp ( - , ) is a metric on •

“"

lp.

•

k =1

^

5Since some results on series will be involved , those who axe not familiar with this could skip this example until they have read Chapter 6.

9

Metric Spaces and Limits for Sequences

We point out that unlike Rm , the spaces £ p are different for different p G [1, oo ) , and all pp are not strongly equivalent.

To conclude this section , let us further look at the case p = 2. On Rm , we introduce a binary operation , called an inner product: m

( x , y ) = x T y = Ylxky

Vx, ?/ G Rm .

k=l

v

With such an inner product , Rm is called an inner product space. Clearly, ( x, y ) ( x , y ) satisfies the following conditions:

( x, y) = ( y, x ) Vx , y G Rm ( X x + y y , z ) = X ( x , z ) + /i ( y , z ) , (x , x) = 0 ( z , z ) 0,

VA , y e R , x , y , z £ Rm x = 0.

( l. n )

Further , one sees that

||x||2

= y/ ( x , x )

Vx G Rm .

Therefore, we say that || • H 2 is the norm induced by the inner product ( We call || • ||2 the Euclidean norm.

*

, ). *

Exercises 1

1. Prove the de Morgan’s Law: Let { A\ | A R. Then

G

A} be a family of sets in

( n ^)

( iMA = n A x xeA

A?

u ^-

=

A6A

A A

AGA

2. Let d\ and d 2 be two metrics on the non-empty set X . Then for any Ai , A 2 > 0, Aidi + A 2 2 is also a metric on X .

^

3. Define

d( x , y )

=1

x - y \\ p + \\ x - y \\ P

Vx , y

Rm

where || • ||p is defined by (1.2 ) . Prove or disprove that d ( , •) is a metric on Rm . *

-

4. Let d{ , •) be defined on R m as follows: d( x, y ) K

yk , = , , ^ Vx = ( x , - - - , xm , — (

= { k | xk

1

k

1 2

) y

•

y

•

1

, m} ,

, • • , ym ) G Rm

10

Mathematical Analysis

— A Concise Introduction

where \ A\ stands for the number of elements in A. Then d ( • , •) is a metric on Rm .

5. Let X be a non-empty set and let p : X x X -» [0 , oo ) be a map satisfying the following: p( x , y ) 0 , V x , y e X ; x = y, p( x , y ) = 0 p( x , y ) p ( z , z ) + p( z , y ) Vrr , y , z e X . > Define Mx , y e X . d ( x , y ) = m a x { p( x , y ) , p { y , x ) } Then d ( • , •) is a metric on X .

^

6. Let ( X, d ) be a metric space and / : X -» X be a one-to-one map. Define x, y X . d( x , y ) = d ( f ( x ) , f ( y ) ) Then d ( - , •) is a metric on X.

7*. Let 2

and

be defined in Example 1.9. Then £ l C

^ , and 2

^

Sequences and Limits

We introduce the following definition. Definition 2.1. Let {xn }n

^ i be a

sequence in a metric space ( X , d ) .

( i) The sequence is said to be bounded if there exists an x G X and a constant M > 0 such that d ( xn , x ) M , n 1. ( ii) The sequence is said to be Cauchy if for any e > 0, there exists an N 1 such that d { X n i %m ) £ » n, m N . (iii) The sequence is said to be convergent to x G X with respect to d if for any £ > 0, there exists aniV l such that n N. d ( xn , x ) < e , In this case, x is called a Zirait of the sequence {xn}n i The above is also

F

^

^

^

denoted by the following: lim d ( xn , x )

n —toc

^-

= 0.

When the metric d is clear from the context , we simply denote the above by lim xn = x .

—

n > oo

The sequence is said to be divergent if it is not convergent .

M

Metric Spaces and Limits for Sequences

11

From the above definition , we see that a sequence { xn }n i is bounded , Cauchy, convergent , or divergent , if and only if so is the sequence { xn }n k for any fixed k 1. In other words , the above-mentioned properties remain if we drop any first finitely many terms of the sequence. Likewise, by adding finitely many terms at the beginning of the sequence, the above- mentioned properties will not be changed. Hence, from now on , we will simply denote the sequence by { xn } instead of { xn }n i , unless the initial term needs to be emphasized.

^

^

^

^

The following result gives some basic properties of convergent sequences.

Proposition 2.2 . Let ( X , d ) be a metric space and { xn } C X be a sequence.

( i)

If

( ii)

( iii)

{ xn } is convergent to x

If If

G

X and to x' Gl , then x = x' .

{ xn } is convergent , then it is Cauchy.

{ xn } is Cauchy, then it is bounded .

Proof , (i) Suppose { xn } has two different limits x , x' G X . Then for d ( x , x' ) £= > 0 , there exists an N 1 such that 2 d { xn , x )

Consequently, for any n

^

+ d ( xn , x' ) < £ ,

Vn

^ N,

0 < d ( x , x' )

^ N.

d ( x , xn ) 4- d ( xn , x' ) < e =

^

a contradiction.

( ii) Let {xn } converge to x . Then for any e such that

|

d ( xn , x )
0, there exists an N

^1

N.

Vn

^) Nd, (we ,have x ) d( xn

+ xm , x ) < e .

This shows that { xn } is Cauchy.

( iii) Let { xn } be Cauchy. Then for e = 1 such that d ( xn , x N ) < e = 1

> 0, there exists an iV

Wn

^ N.

^1

Mathematical Analysis

12

— A Concise Introduction

Now letting N -l

d (, X k , X N )

M =1+ k= l

one has

M,

d(xn, XN )

1.

Vn

This completes the proof . The following result is a little surprising, but the proof is obvious, which is left to the readers.

.

Proposition 2.3 Let ( X, do ) be a discrete metric space and {xn } C X be a sequence. Then { x n } is convergent to x E X if and only if there exists an N 1 such that Xfi

—x

Vn

> N.

We point out that by changing the metric , the convergence of sequences may be changed. For example, the sequence { }n > l converges to 0 under the standard metric of R. However , this sequence is not convergent under the discrete metric. Further , it is even possible that when the metric is changed , the limit of a convergent sequence might be changed. Here is an example.

^

.

Example 2.4 Let 1

f ( x) =

{ X, 0

Let X

x = 0, *

(0, 1 ) X

= 1.

= [0, 1] and p( x , y ) = \ x — y \ be the standard metric. Define Vx , y e X . d { x , y ) = p( f ( x ) , f ( y ) ) = \ f ( x ) - f ( y ) \

Clearly, d( x , y )

= \ f { x ) - f ( y ) \ = \ f ( y ) - f ( x ) \ = d{ x, y )

0,

Vx , y G [0, 1]

and since / ( •) is one-to-one, we have

d( x , y ) = | f ( x )

-

f ( y )\ = 0

x = y.

Finally,

d( x , y )

= | / (x) - f ( y ) \ < |/ (x ) - f ( z ) \ + \ f ( z ) - f ( y ) \ W x , y , z e [0 , 1]. = d ( x , z ) + d( z , y )

13

Metric Spaces and Limits for Sequences

Hence, d ( - , •) is a metric on X . Now , for sequence

xn =

we have

( n ) = -n -> o

p -, o \

/

and

d

(Vn . l ) =

- - /U)

0. =n n^ Thus the sequence is convergent under p and d, but the limits are different! "

/

Relevant to the above example, we introduce the following definition. Definition 2.5. Let d , d be two metrics on X .

( i) d is said to be stronger than d if for any sequence {xn} C X , and some x £ X , lim d ( xn , x ) = 0

—

n > oo

=>

lim d ( xn , x )

—

n » oo

= 0.

In this case, d is said to be weaker than d.

( ii ) d and d are said to be equivalent if for any sequence {xn} C X , and some x £ X , lim d ( xn , x )

—

n > oo

=0

lim d ( xn , x )

—

n > oo

= 0.

It is clear that if two metrics d and d are strongly equivalent (see (1.9 ) ) , then they are equivalent . Hence, in Rm , convergence of sequences under any pp ( p £ [1, oo] ) are all equivalent. We point out that equivalence of metrics do not have to be strongly equivalent . ( Why? ) Proposition 2.6 . Let { xn } be a sequence in Mm with xn = (x , x , • • • , x ) , and x ( x1, x 2 , • • • , xm ) £ Rm . Then the following are equivalent :

^^

—

( i ) {xn } converges to x with respect to pp , ( ii) {xn} converges to x with respect to pp , ( iii ) For each 1

k

for for

some p £ [1, oo]; every p £ [1, oo];

m , the sequence {x£} is convergent to

xk .

The proof is left to the readers.

Definition 2.7. Let ( X, d) be a metric space, and {xn}n in X.

—

^

i

be a sequence

( i) Let cr : N > N be a strictly increasing map. Then {x 7 ( n ) }n i is called a subsequence of {xn }n i . Sometimes, a subsequence is denoted by { xnk }k 1

^-

(

^

^

14

— A Concise Introduction

Mathematical Analysis

( ii) A point x G X is called a limit point of the sequence {xn}n there exists a subsequence {£< (n ) }n i such that lim d ( xa /v n\ , x ) = 0.

^

j

n

—

^

i

if

oo

Proposition 2.8 . Let ( X , d ) be a metric space , and { xn } C X be a sequence. If {xn} is convergent to x , then any subsequence {xa ( n ) } of { xn } is convergent to x as well. Conversely, if every subsequence {Xo- ( n )} of { xn } has a convergent sub-subsequence with all the limits being x , then the sequence { xn } is convergent to x .

Proof . Suppose { xn } is convergent to x . Then for any e exists an N 1 such that

^

Vn N. d ( xn , x ) < e , Now , let { X ( j i ) } be any subsequence of { xn } ( with cr : N strictly increasing map ) . Noting cr ( n ) n , we have ( £ cr ( n ) ? 3C ) Vn N . Thus , { xa ( n ) } is convergent to x .

^

(j

^

—>

> 0, there

•

N being a

^

*

Conversely, we prove by contradiction. Suppose { xn } is not convergent . Then there exists an e > 0 and two subsequences {xai (n) } and {£ < 2 (n )} such that n 1. ( n ) , X ^- 2 ( n ) ) Consequently, the limits of the convergent subsequences of {x Tl ( n ) } and { £ oo

n

oo

n

oo

i0

*

*

Metric Spaces and Limits for Sequences

15

The following gives a comparison theorem of sequences and their limits.

Proposition 2.10 . Let { xn } , { yn } G c and let

-

Xn

for

some

No

^ 1. Then

N0 ,

Vn

^ Vn — xn

lim

n > oo

lim yn .

n

—

( 2.1 )

oo

Proof . Let

x

= nlimoo xn ,

—*

Suppose x > y instead , then for that

y = lim yn .

— > 0, there exists an JV n > oc

=

£

I xn - x \ + \ yn - y \ < This leads to ( noting

xn — yn

0)

^

n

^ N.

0 < 2£ < x - 2/ = x - x n + x n - 2/n |x - xn | + |?/n y \ < e ,

—

a contradiction.

^ 1, such

+ 2/n - 2/ N,

The above result has a very useful corollary whose proof is obvious.

Corollary 2.11. ( Squeeze Theorem ) Suppose { xn } ,{yn}, { zn } £ f satisfy

°

yn ^ ^ 1. Further , suppose

Xn

for some Afo

Vn

NQ

x. lim — xn — n-+oc zn =

lim

n > oo

Then

lim yn = x .

—

n > oo

Next , we consider sequence with infinite limit.

° . If for any M

Definition 2.12. Let { xn } G £ such that

xn > Af ,

Vn

> 0, there exists an N

^ AT,

1

Mathematical Analysis

16

then we say that xn -» 00, or

xn

— A Concise Introduction

approaches to oo as n goes to infinity, denoted by

oo. — xn =

lim

n >• oo

Likewise, if for any M > 0 , there exists an N

Xn < then we say that xn -> -oo, or

xn

-

M,

approaches to

lim

—

ra > oo

Vn

^Ar1, such that

— oo as n goes to infinity, denoted by

xn = — oo.

The following comparison of sequences is obvious. We leave the proof to the readers.

Proposition 2.13. Let { xn } , { yn } G £ satisfying %n Vn N0 , Vn

°

for

some No

^1

. Then

xn — y oo

Vn

yn ->• -oo

=>

00

xn — y — oo.

Monotone sequences are of special importance and they will play useful roles below. Proposition 2.14. Let { xn } G

f

°.

( i ) Suppose there exists an M G R such that Vn 1. Then { xn } G c , and lim xn = sup xn M,

n-> oo n l is called the supremum ( or the least upper bound ) of the

^

where supn x xn sequence { xn }n i .

^

^

( ii ) Suppose there exists an M eR such that Vn 1. •En 1 Then { xn } G c, and lim xn = inf xn M,

^

—

n > oo

^

^

n l

where infn i xn is called the infimum ( or the greatest lower bound ) sequence { xn }n i .6

^

of

the

^

6 The existence of supremum and infimum relies on the theory of the real number system , which will be presented in the Appendix. F

Metric Spaces and Limits for Sequences

17

Proof ( i) First of all, since { xn } is bounded above, by a result from the Appendix, ,

b = sup xn 6 R exists. Now , for any e

> 0 , there exists an N XN > b — e .

1 such that

By the monotonicity of {xn}, we have b

—

6,

xn

e < XN

Vn

This implies

Vn \ xn - b| < £ , Hence, we have the convergence of xn to 6.

^ N.

N.

(ii) The proof is similar and is left to the readers. We know that c C c C °° , and c £°° . A natural question is what can we say a little more about sequences in \cl We now investigate that .

^

^

Let {xn }n > i £ °°. Then for any k following is well-defined:

^

^ 1, { xn } ^ ^°°, and the n k

. = nsupxn k

Vk

It is not hard to show that { yk } k

^

^

i

is bounded and non-increasing:

Vk 1. ^ Thus, by Proposition 2.14 , the following limit exists: Vk

lim

—

k too

yk

Vk +1

supxn = lim xn . = kinfl yk = kinfl supxn = klim yoo n-+ oo n k n k

^

^ ^

— ^

This is called the limit superior ( limsup, for short ) of {xn}n Likewise, we may define zk

= ninfk xn ,

k

^ which is a non decreasing bounded sequence. -

^i

1

Hence, the following limit

exists, by Proposition 2.14 again:

inf xn = lim xn . xn = klim toon k n —> oo — ^ ^ ^ ^ This is called the limit inferior ( liminf , for short ) of {xn}n i lim zk = sup zk = sup inf

—

k > oo

k

l

k

ln k

^

— A Concise Introduction

Mathematical Analysis

18

The above shows that for any { xn } £ £°° , lim

xn and lim xn always — — exist , thanks to the theory of the real number system , which guarantees n + oo

n

oo

the existence of supremum and infimum for bounded sequences (of real numbers). Further , if { xn }n i is unbounded , lim xn and lim xn can still

^

be defined if they are allowed to be

—

n ± oo

± oo.

n

—

oo

The following gives some relations among limsup, liminf , and limit .

Proposition 2.15 . For any sequence { xn } lim

—

n too

Further , lim

—

n > oo

xn

lim

—

n too

xn ,

lim (

—

n > oo

lim

lim

—

oo

— xn = — lim— xn.

—

0 - xn =

lim

lim

n

Proof . By definition , we have

= ninfk xn

sup xn

^

lim n

—

oo

( 2.3)

—

—

n +oo

Hence,

( 2.2)

oo

xn = nlimoo xn = nlimoo xn .

In particular ,

Zk

)

xn = nlimoo xn .

In this case , n

following hold :

n

xn exists if and only if n -> oo

e £°° , the

^

—

oo

= yk

n k

xn = klimoo Zk

—

Vfc

—

( 2.4)

1.

= nlim — oo xn.

lim

fc

\ xn \ = 0.

oo

r

This proves the first in ( 2.2 ) . The second in ( 2.2 ) can be proved similarly.

Next , suppose lim xn = x . Then for any e > 0, there exists an N n oo such that

—

X

— £ < Xn < X + £

Vn

Hence 5 5

^ iV

VA;

.

iV.

Consequently,

x

lim — ^ lim — xn ^ — xn £

n

oo

n

oo

x + £•

1

Metric Spaces and Limits for Sequences

19

Since s > 0 is arbitrary, we obtain

. x lim xn = nlim —> oo xn — = n—tcc xn

lim

—

n > oo

Conversely, if lim

—

n > oo

x, xn = nlim — oo xn =

then for any e > 0, there exists an N

| j/ fc - x \ < e ,

\ zk

1 such that ^ - x\ < , V /c £

with

x

— £ < Zk = iinfk Xf>

xn < supk ^ = t/fc < x + £ , n e

^

This leads to

^ iV,

^

xn — x \ < £

Vn

5

Hence 5 lim

n —> oo

^ k ^ N.

^ N.

xn = x .

We leave the proof of the rest of the conclusions to the readers.

.

Proposition 2.16 Let { xn } , { yn } G £°° .

( i) The following hold : lim ( xn + yn )

n » oo

n too

+ nlim Vn —t oo

n

n > oo

j/n . + nlim » oo

lim — — lim lim (xn + yn ) — —>

( ii ) Let >0

oo

^ Vn

Then lim

—

n » oo

yn , xn < nlim —> oo

—

Vn

1.

lim

xn

—

n > oo

lim yn .

n-* oo

The proof is left to the readers. Let us now recall the definition of limit point (see Definition 2.7 (ii) ) .

.

Proposition 2.17 Let { xn } G °°. Then both lim n

limit points of { xn } . Further , if

lim xn are ^x is a limit point — °°{ xn} and, then —

lim x n

—

n > oo

> of

. < x < nlim — oo xn )•

xn

n > oo

n i

^

Mathematical Analysis

20

— A Concise Introduction

The proof is left to the readers.

Exercises 1. Find an example showing that a bounded sequence does not have to be Cauchy.

2. Let { xn }n N such that

^ i be

Cauchy. Then either for any N

or there exists an n

^

Vn

Xn ^ N such that

Xn

>

^ 1, there exists an

N,

Vn iV. xn ^ ^ Is it possible that for any N 1 there always exist m , ^ Vn iV? Xn X n , Xn ^ ^ Xn

N such that

2

i

3. If { xn } , { yn } G c, then {max ( xn , yn ) } G c.

4. Let Co be the set of all real sequences that converge to 0. Then Co is a linear space, i.e., for any {xn}, { yn } G Co and any a , /3 G R , { axn -F Pyn } G c0 . Is the conclusion true for c ( the set of all Cauchy sequences ) ? Why?

5. Let {xn}, { yn } G i°° satisfy Xn

< Un

?

n

1.

lim

xn

Then

lim

n — yoo

yn xn < nlim — yoo

n —y oo

lim yn . —

n > oo

For each of the above, find examples that the equalities hold . 6. Let { xn } , { yn } 6 £°° . Then

yn , xn + nlim + j/„ ) < nlim —> oo — yoo lim ( xn + yn ) lim xn + lim yn . yoo n — y oo n » lim ( xn y

n

— oo

— oo

n—

Further , for each of the above, find an example that the strict inequalities hold .

7. Let { xn } G c and { yn } G -£°°. Then lim ( xn

n — yoo

lim yn . + 2/n ) = nlim — yoo xn + n — yoo

21

Metric Spaces and Limits for Sequences

^ i , { yn }n^ i , { zn }n^ i

8. Let { xn }n

£°° such that

lim

—

n » oo

xn = nlim zn = L » oo

xn = nlimoo zn = L.

lim

—

1.

Vn

Un ^ Further , let the following hold: Xn

—

—

n > oo

>

Then

lim yn = L.

= L,

lim

—

n > 00

—

n > oo

From the above, prove the Squeeze Theorem ( Corollary 2.11) . 9. Let 0 < a < 6, x\ = a > 0,

yi

= b > 0, and Vn

%n

Vn+1 ^n + l — y/ XnVn 2 Then lim xn and lim yn exist and they are equal. n > oo n » oo

2.

*

—

—

xn = a. Then nlim n —> oo —> oo 11. Let XQ = 1, xn + i = 1 + j n 10. Let lim

^

further , find the limit .

xn a. = n 0. Show that lim

#1 H- # 2

n

^

“

h

•• •

f~

“

—

n > oo

exists and

12. Suppose { xn } £ £°°\ c. Show that there are two convergent subsequences of { xn } having different limits. 13. Let { xn } e c such that for some subsequence { xnk } k lim xnk = x . Then lim xn = x . n > oo k > oo

—

—

^ i , it holds

14. Let ( X , do ) be a discrete metric space. Let xn G I be a sequence converging to x under do Then there exists an N 1 such that

-

xn = x 3

N.

n

Sets in Metric Spaces

In this section , we look at some sets in metric spaces , and their properties. Let us first introduce the following definition. Definition 3.1. Let ( X , d ) be a metric space.

( i ) Let

xo e

X and r > 0. Define

B ( x0 , r ) = B( X ,d ) ( xo , r ) = { x which is called the open ball centered at

e

XQ

X

\ d { x , x0 ) < r } ,

with radius r .

Mathematical Analysis

22

(ii) A point xo aS

£

— A Concise Introduction

E C X is called an interior point of E if there exists

> 0 such that

B ( xo , 6 ) C E . ( 3.1) The set of all interior points of E is denoted by E° , which is called the interior of E . G X is called a boundary point of E if for any S > 0 , B ( xo , 5) fl E 0 B ( xo , S ) fl Ec 0 . The set of all boundary points of E is denoted by d E , which is called the boundary of E .

( iii) A point Xo

^

^

> 0, E n B ( x0 , 6 ) 0. The set of all adherent points of E is denoted by E , which is called the closure of E . ( iv ) A point

xo G X

is called an adherent point of E if for any 5

^

(v ) A point of E\{x 0}.

xo G X is called a limit

( vi) A point such that

xo G

point of E if it is an adherent point

E is called an isolated point of E if there exists a 6 > 0

(vii) A set E is said to be open if E = E° i.e., for any xo £ E , there exists a 0 such that ( 3.1) holds , which is also equivalent to E n dE = 0.

( viii) A set E is said to be closed if E E which is equivalent to d E C E.

—

( 3.2 )

(3.3)

( ix ) A subset E of F is said to be dense in F if E D F . (x) A set E is said to be disconnected if there are two non-empty sets C? i , C?2 Q X such that G\ n c?2 = ( 3.4 ) E Gi u G 25 Gi n (?2 , E say . is connected In the case E = X, that If E is not disconnected we we call that A is a disconnected metric space and connected metric space , respectively, if E = X is disconnected and connected , respectively.

—

Metric Spaces and Limits for Sequences

23

For the connectedness above, we look at the following simple situation: The set S = ( 1, 0 ) U (0, 1) is disconnected . In fact , by letting G\ = ( 1, 0 ) and G 2 = ( 0 , 1) , we see that the last two relations in ( 3.4 ) hold . On the other hand , let us consider S = ( 1, 1) which should be not disconnected . By letting G\ = ( 1, 0] and G 2 = ( 0 , 1) , although S = G\ U G 2 and Gi n G 2 = 0, but Gi fl G2 = {0} 7^ 0, i.e., the last relation in (3.4 ) fails. The point that we want to make here is that G\ fl G 2 = 0 is not enough for disconnectedness.

—

—

—

—

Example 3.2. ( i) Any finite set A is closed , and all of its points are isolated points.

( ii) The sets 0 , N , Z, and R are closed . All points in N and Z are isolated , and all points in R are limit points. ( iii) The set Q is not closed and its closure Q = R. Therefore, Q is dense in R . All points in Q are limit points of Q. There are no isolated points in Q. Proposition 3.3. Let ( X , d ) be a metric space.

(i) Let

Si

C

52 C X .

Then

Si

C

$2,

SI

C

S2° .

( 3.5 )

( ii) Let Si , S 2 C X be two sets. Then { Si

n s2 ) c Si n s2,

(S1 US2 )

= Si u s2

( 3.6 )

and

( S i n S2 )°

= S? nS£ ,

(Sius2 n s f u s2°.

( 3.7)

Proof . The proof of ( i) is clear. We now prove ( ii) . First of all,

Si n S2 c Si , S2 0 Si u S2 .

( 3.8)

Thus, by ( i) , we have

(Si n S 2 ) c Si

n S2

S1 US2

C

(S1 US2 ) .

Now , we prove the equality in (3.6). For any x G (Si U S 2 ) , there exists a sequence { xn } C Si U S2 converging to x . Clearly, there must be a subsequence of { xn } either in Si or in S2 . ( Why? ) Hence, x is either in Si or in S 2 . Therefore, (Si U S2 )

= Si u s2 .

Mathematical Analysis

24

—

A Concise Introduction

Next , from ( 3.8) , by ( i) again , we have

( Si n S2 ) °

C

St n S2°

S US C ( SiUS2 ) ° .

^ ^

We now prove the equality in ( 3.7). If there is an x E S£ fl S2 which is an open set , then there exists a S > 0 such that £ (*, £ ) c s? ns2° c Si ns2 . This leads to x G ( Si fl S2 ) ° . Hence,

(Sins2 ) ° = Si° ns2° . This completes the proof of ( ii). Note that if we take

Si = ( 0, 1 ) , S2 = ( 1 , 2 ) then

(Si nS2 ) = 0 Also, if we take

^ { } = Si n S 1

Si = ( 0 , 1 ] ,

2.

^ = [1, 2) 2

then

SI u S°2 = (0 , 1 ) U ( 1 , 2)

= ( S 1 U S2 ) ° . Hence, the two inclusions in ( 3.6 ) and 3.7) may be strict , in general. Proposition 3.4. Let ( X , d ) be a metric space and E Cl. Then ( 3.9) U dE . E = E U x implies x G E . ( ii)

If

If

•••

••

•••

•••

The proof is straightforward and is left to the readers.

Note that the intersection of infinitely many open sets might not be open , and the union of infinitely many closed sets might not be closed . Here are counterexamples:

°° 71

1

1

=1 OO

1 -

'

( 0, 1 ) . nJ = =1 It is clear that the open ball B ( xo , r ) is an open set , and its closure 71

B ( x0 , r ) = { x

X

| d( x , x o )

is closed , called the closed ball centered at

XQ

r}

with radius r .

Example 3.6 . ( i ) Consider R 2 with Euclidean metric p2 - Let X x e R}. Then { X , p2 ) is a metric space. Let

= { { x , 0) |

E = { { x , 0 ) | -1 < x < 1} C X . Then E is open in X , but it is not open in R 2 . ( Why? )

( ii) Let R be the metric space with the standard metric p. Let X = ( 1, 1). Then ( X , p ) is a metric space. Note that X is not closed in ( R , p ) . But , X is always closed in ( X , p ) . ( Why? )

—

We now introduce the following definition. Definition 3.7. Let ( X , d ) be a metric space and Y CX , Let E C Y . We say that E is relatively open ( resp. relatively closed) in Y if it is open ( resp. closed ) in the metric space ( Y, d ) .

Mathematical Analysis

26

— A Concise Introduction

The following gives a representation of relative open sets and relative closed sets. Proposition 3.8. Let ( X , d ) be a metric space and E C Y Cl

(i) Set E is relatively open (resp. relatively closed ) in Y if and only if there exists an open (resp. closed ) set V C X such that E V C\ Y .

—

( ii) For any E C Y , if E and E stand Y , respectively, then

for

the closures of E in X and in

E = E fl Y .

(3.11)

Proof , ( i) Suppose E is relatively open in Y . Then for any x G E , there exists a 5( x ) > 0 such that B y { x , 5( x ) ) C E.

Let

v = U B*(*, 0 such that

B x ( x , 5 ) C V. Thus

BY ( X , 5 ) = B x ( x , 5 ) n Y

CV

n Y = E.

Consequently, x is an interior point of E in (Y , d ). Thus, E is open in (Y , d ) .

Metric Spaces and Limits for Sequences

27

Now , let E be relatively closed in Y . Then Y \ E is relatively open in Y . Thus, by what we have proved , there exists an open set V in X such that

Y\ E =V

n Y.

Hence, V c = X \ V is closed in X , and

= Y \ (V n Y ) = Y n (V n Y ) c = Y n ( Vc u Y c ) = y n vc . This proves the relatively closed case. E

( ii) For any set E C Y , E is the smallest closed set in X containing E . Now E D Y is a closed set in Y containing E . Clearly, it is the smallest one. Hence, ( 3.11) holds. The following result gives another interesting characterization of the disconnectedness / connectedness , which will be useful later.

Proposition 3.9 . Let ( X, d) be a metric space and E C X . Then E is disconnected if and only if there are two disjoint open sets V\, V2 C X such that

E r \1 / 0,

E r V :2 7^ 0,

E C V1 U V2 .

( 3.12 )

Note that in the above, we do not assume if E is open or closed. Proof . Sufficiency. Suppose there are two disjoint open sets such that ( 3.12 ) holds. Let

Gi = E n V i ,

(?2

Vi , V2 C X

—ED

Consequently,

E

= E fl ( Vi U V2 ) = ( E n Vi ) U ( E D V2 ) = Gl u G2.

Further , since V\ fl V2 = 0,

Gl n G 2 = ( E n Vi ) n ( E n v2 ) = E n ( Vi n v2 ) = 0. We claim that Gi fl G 2 = 0. In fact , for any x E G 2 = E fl V2 C V2 , since V2 is open , there exists a 6 > 0 such that B ( x , 6 ) C V2

.

On the other hand , if this x G Gi as well , then there exists a sequence xn Gi = E n Vi such that xn > x . Hence, for n large and for the above S > 0,

—

•

x n e G1 n B ( x , S ) c E n V\ n v2 c V\ n v2 = 0

Mathematical Analysis

2 8

—

A Concise Introduction

a contradiction. This shows that G\ D G 2 Therefore, E is disconnected .

= 0.

Likewise, Gi fl G2

= 0.

Necessity. Suppose E is disconnected . Then there are non-empty sets Gi , G 2 C X such that .£/

For any x\

G

— Gi u G ,

Gi n G 2 G 2 . Thus, 2

G1, xi

d ( xi , G 2 )

^

— 0,

Gi n G 2 = 0.

d ( xi , x 2 ) > 0 , = d (xi , G2 ) = 2inf GG 2 X

Vxi

G

Gi .

VX 2

G

G2 .

Likewise

^(^ , GI ) = d( x , Gi ) =


0 , Gi

xi

Now , we define

^

6

xi

Both Fi and F2 are open and

E n Fl D Gi ,

£ n F2 D G 2 ,

2

Fl U F2 2 Gi U G 2 = £.

We now claim that Fi and F2 are disjoint . In fact , if there exists an x G Fi DF2 , then there are x\ G Gi and X 2 G G 2 such that

^

x G J5 xi ,

Id (a: , G )) fl B ( , i

X2

2

^

Without loss of generality, we assume d ( xi , G 2 )

d ( x2 , Gi )

).

d ( x 2 , Gi ). Then

d ( x 2 , Gi ) d ( xi , x 2 ) d ( xi , x ) + d ( x 2 , x ) 1 1 2 - d ( xi , G 2 ) + -d ( x 2 , Gi ) - d ( x 2 , Gi ) , which is a contradiction. Therefore,

Fi and F2 are disjoint .

Proposition 3.10. Let ( X , d ) be a metric space and E C X be a nonempty set . Then E is disconnected as a set in X if and only if it is disconnected as a metric space , with the induced metric d .

Proof . As before, for any G C E , we let G and G be the closures of G in X and in E , respectively. Necessity. Suppose E is disconnected as a set in X . Then by definition , there are non-empty sets Gi and G2 such that

E = Gi U G2 J

GI PI G 2

— 0,

Gi n G 2

—0

*

Metric Spaces and Limits for Sequences

29

The above implies that

= G\ n G% — G\ n G 2 n E = ( Gi n F ) n G 2 = Gi n G2 where G\ = G\ n E is the closure of G\ in E . Likewise, 0 = Gi n G2 — Gi n ( G 2 n E ) = Gi n G 2 , with G 2 = G 2 fl E being the closure of G2 in E . Hence, by definition , E a 0

?

XN

XS

is disconnected metric space.

Sufficiency. Suppose E is a disconnected metric space. Then there are non-empty sets Gi , G 2 C E such that

GI n G2 = 0.

Gi nG 2 = 0 ,

E = Gi U G 2

Then, by Proposition 3.8 ( ii) , we have

G\

— Gi n F,

G2

—G

2

n E.

Gi U G2 = E , Gi n G2 — Gi n E n G 2 — Gi n G2 — 0

Consequently, noting

and

—

Gi n G2 Gi n E n G 2 = G1 n G2 — 0 Hence, by definition , E is disconnected in X.

«

The following gives some interesting characterizations of disconnectedness of metric spaces.

.

Proposition 3.11 Let ( X , d ) be a metric space. Then the following are equivalent :

(i) X is disconnected .

= V\ U V2. (iii) There are two disjoint closed sets Fi , F2 Cl such that X = F1 UF2 . (ii ) There are two disjoint open sets

(iv ) There is a proper subset Proof , ( ii) => (iii) . Since X disjoint . Then

of

VT , V 2 £ X

such that X

X which is both open and closed in X .

= V\ U V2

F1 = X \V1 = V2 ,

with both V\ and V 2 open and

F2 = X \V 2

which are both closed and disjoint , and

Fi U F2 = V2 U V\ = X .

= Vi ,

Mathematical Analysis

30

—

A Concise Introduction

( iii) => ( iv ) . When X = F1 UF2 with both F\ and F2 closed and disjoint. Then Ff = F2 is open and also closed. This means that X has a proper subset which is both closed and open.

( iv ) => (ii) . If G C X is a proper subset of X , which is both open and closed , then G and Gc are two disjoint open sets and

X

= G U Gc.

( i ) => ( iii) . By definition , there are disjoint non-empty subsets G\ , G 2 such

x — G\ u c?2

)

G1 n G 2

— 0,

G\ n G2

—0

*

0. We claim that G\ = G\. In fact , if there exists an G2 , which contradicts G\ fl G2 = 0. Hence, G\ is closed. Similarly, G 2 is also closed . Thus, (iii) holds.

Then G\

x

G

n G2 =

Gi \ Gi

C

( iii) => (i) . If X disjoint , then G\ n G 2

= G\ U G 2

—G

1

with G\ and G 2 both being closed and

n G2 — 0

—G nG

G\ n G 2

\

Hence, X is disconnected .

2

—0

*

Exercises 1. Let ( X, d ) be a metric space and E C X be a finite set. Then E is closed.

2. Let ( X, do ) be a discrete metric space. Then any subset E of X is both open and closed. 3. Let ( X, d ) be a metric space and E C X. Then E° is the largest open set contained in E .

4. Let ( X, d ) be a metric space and E C X. Then F is the smallest closed set containing E . 5. Let ( X, d ) be a metric space and E C X. Then F is closed if and only if for any convergent sequence { xn }n i C F, the limit x of { xn }n i is in F.

^

^

6. Let ( Xfc , dfc ) be metric spaces k = 1, 2, • • • , m. Let X with the product metric

d(( x

, xm ) , ( y i ,

---

= Xi x • • • x Xm

m

, t/m ) )

^

=5

fe = i

djfe ( xfc , 2/ fe ).

Metric Spaces and Limits for Sequences

31

Let Ek C Xk , 1 k m. If each Ek is open ( resp. closed ) , then E • • • x Em is open ( resp. closed ) in ( X , d ) . E\ x

=

7. Show that for any open set G C R , there exists a finite or an infinite sequence of disjoint open intervals (afc , £> /e ) , k 1 such that

G=

UKA ).

Extend the above result to the case of Rm . 8. Let ( X, d ) be a metric space. Suppose G\ yG 2 £ -X . Then "

X

inf

xi GGi , a;2 eG? 2

4

d ( xi , X2 )

=

_inf

#I £

_ d ( xi , X 2 ) .

GI , X 2 CT 2

Properties of Metric Spaces

We now consider more properties of metric spaces.

.

Definition 4.1 ( i ) A set X is said to be countable if there exists a bijection f : N » X , i.e., / ( N ) = X and f ( £ ) = f ( n ) if and only in £ = n.

—

( ii) A set X is said to be at most countable if it is either finite or countable. ( iii) A set X is said to be uncountable if it is not at most countable. Practically, X is countable if and only if X can be written as follows:

X

.

Proposition 4.2 ( i)

If

= {xi , x2 , 3, - - - }. aj

( 4.1)

X is at most countable, then so is any subset

of

X.

( ii)

If

X and Y are at most countable , so are X UY and X x Y .

( iii ) Let X oo

so is

(J Xfc .

^ , k = 1, 2,

•••

be a sequence of at most countable sets. Then

k=l

Proof , ( i) If X is finite, the conclusion is trivial. Let X be of form ( 4.1) . Let Y be a subset of X . In the case that Y is finite, it is of course at most countable. In the case that Y is infinite, we can find a subsequence { x* ( n ) }n> l such that

Y

— {^

( n ) }n l *

(j

^

— A Concise Introduction

Mathematical Analysis

32

Thus, Y is countable.

(ii) We need only prove the case that

X

= { x i , X 2 , x% ,

Y

•

= (yi , 3/2 , 2/3, -

•

Clearly,

XUY

= { x 1 , y 1 , x 2 , y2 , x 3 , y 3 ,

- - - }.

Hence, X U Y is countable. Also, one has

x X Y = { ( x i , y i ) , ( x i , y2 ) , ( x2 , y i ) , ( x i , y3 ) , ( x‘2 , y2 ) , ( x1 , y3 ) , - - - }. This implies that 1x 7 is countable.

(iii) Let

Xi =

i

= 1, 2 , 3, • • • .

Then oo

[ j X i = { x\ , x\, x\ , x\, xl , x\ , - - - }.

i =1

oo

Therefore,

( J Xi ^

2

is countable.

=1

For any x G R, let [x] be the largest integer that is less than or equal to x . We define the sequence ao , ai , • • • as follows:

cto = [ x\

G

Z,

-

=»

e {0, 1,

- - - , 9}

—

do

— 1—0

-

, 9} =>

{0, l , - - , 9}

oi = [10( x - a0 ) ]

a2 =

e [0, 1) ;

x - a0 ai\l

102 (x

x

an = [

n—1

( — 53 ) ]

l 0" x

fc = 0

^

x

&1

“

ao U2

102

G

e

^—

[° 10

° i02 ) >

n 1

6 {0 , l ,

"

»

fc = 0

»

°fc

10s

G

1 °’ 10" ) ’

Hence, for any x G R , one has

*=

Ea10*

fc

do G

Z, dk

G {0 , 1, 2, • • •

, 8, 9}, k

^ 1.

( 4.2 )

f

33

Metric Spaces and Limits for Sequences

Clearly, for any given real number , the above representation is unique, making use of the convention that any rational number is represented by the corresponding repeating decimal number . Hence, we refer to ( 4.2 ) as the canonical decimal representation of real numbers. Proposition 4.3. For any a < b , the set [a , b] is uncountable.

Proof . Without loss of generality, we consider [0, 1]. Suppose [0, 1] is countable. Thus, we let

[0 , 1] = {ai , a 2 ,

- - - }.

We use the canonical decimal representation for each

an :

= O.611&12 &13 = 0.621622623 • • as = O.631632633 • ,

a\ a2

* *

*

•

•

5 5

where bij G {0 , 1, 2 , • • • , 8, 9}. Now , we let C

= O.C1C2 C3 • • •

with

e {0, 1, 2, - . . , 8, 9} \ {6 n } c2 G {0, 1, 2 , • • • , 8, 9} \ {622} C3 G {0 , 1 , 2 , • • • , 8, 9} \ {633}

d

Then c (£ {a1, o2 , o3 ,

- - - } = [0, 1]

a contradiction. Hence, [0 , 1] is uncountable.

Definition 4.4. (i) A metric space ( X , d ) is said to be complete if every Cauchy sequence in ( X , d ) is convergent in ( A, d ) . A normed linear space X is called a Banach space if it is complete under the metric induced by the norm. An inner product space is called a Hilbert space if it is complete under the metric induced by the inner product .

(ii) A metric space ( X , d ) is said to be compact if every sequence in ( X , d ) has at least one convergent subsequence.7 A subset Y C X is said to 7Such a compactness definition only works for metric spaces , and it does not work for general topological spaces. This notion is also referred to as the sequential compactness.

34

Mathematical Analysis

— A Concise Introduction

be compact if subspace ( Y, d ) is compact . Or , equivalently, for any sequence in Y , there exists a convergent subsequence whose limit is in Y.

( iii ) A set Y in a metric space ( X, d ) is bounded if there exists a n x G l and an r > 0 such that Y C B( x , r ) .

A set Y is totally bounded if for any e > 0, there exists an integer N > 0 such that N

YC

1J B( xn , e )

71

for some

, ,

X\ X 2 • • •

=1

, xn E X.

Example 4.5 . ( i) Let ( X, do ) be a discrete metric space. Then it is complete.

( ii) Any finite dimensional normed linear space is complete. (iii) Let oo

{

P = (** ) *> 11

< fc = 1

OO

j, l < p < o o .

We are able to show that £ p is complete. The detailed proof is left to the readers. Proposition 4.6. Let ( X, d ) be a metric space , and ( Y, d ) be a subspace of ( X , d ) .

( i)

( ii)

plete.

If If

( Y, d ) is complete , then Y is closed in X . ( X , d ) is complete, and Y is closed in X , then ( Y, d ) is also com-

The proof is straightforward . The following result is called the completeness of real numbers.

Theorem 4.7. A sequence { xn } E £° is convergent Cauchy, i.e., c c.

if

and only if it is

—

Proof . We already know that if { xn } is convergent , then it is Cauchy (see Proposition 2.2, ( ii )).

Metric Spaces and Limits for Sequences

35

Conversely, let { xn }n i be Cauchy. Then it is bounded and for any e > 0 , there exists an N 1 such that \ x k - X j \ < e , VkJ N .

^

^

Hence XN

xn

£

~

XN

+s

Wn

Consequently, XN

—

£

lim

—

n > oo

Thus 0

lim

—

n » oo

xn xn -

lim

xn

lim

xn

n -* oo

n

—

foo

^ N.

XN

+ £•

2e.

Since £ > 0 is arbitrary, we obtain that lim 71

—» OO

, xn = nlim —¥ oc xn

which leads to the convergence of {xn }, by Proposition 2.15.

We should point out that to be rigorous, one should go through the theory of the real number system (see Appendix ) . As a matter of fact , Proposition 2.15 relies on the existence of the supremum and infimum , which is a result of the real number system. Actually, there is a one-to-one correspondence between the set of real numbers and the set of the so-called equivalent Cauchy sequences. See the Appendix for details.

For compactness, we have the following simple result . Proposition 4.8 . Let ( X , d ) be a metric space.

( i) Every

finite

( ii ) If Yi , > 2 > also compact .

* * *

subset of X , including the empty set , is compact . j

Yn

are compact sets

of

X , then Y\ U I 2 U

( iii) If Yi , I 2 , • • • ,Yn are compact sets of X , then Y\ x Y2 x compact in Xn = X x X x • • • x X .

—

•••

U Yn is

•••

x Yn is

Proof . The proofs of (i) ( ii ) are obvious. Let us prove ( iii) . We only prove the case n = 2. For any sequence yn = (y , y%) E Y\ x Y2 , by the compactness of Yi , we have a convergent subsequence {2/ 1 ( n ) }n 1 of {Vn }n i and some y 1 G Y\ such that

^

^

n —y

00 .

^

^

36

Mathematical Analysis

— A Concise Introduction

Next , by the compactness of > 2 ? there exists a subsequence { y1 and a y 2 e 2 such that

*

d ( yl 2 ( r i ( n ) ) ^ y 2 ) 0 Consequently, ( yl , y 2 ) G Y\ x Y2 and d ( ( (ffi (n )) « ( « n (n)) ) » ( tfStf 2 ) ) This proves the compactness of Y\ x F2 . (

wia

. Sa

77,

^

l

—^ oo .

^0

n -» oo.

The following result gives deeper property of compact sets.

Proposition 4.9 . Let ( X, d) be a compact metric space. Then X is complete and bounded ; and any Y C X is compact if and only if it is closed .

Proof . Suppose ( X , d ) is compact . If ( X, d ) is not bounded , then pick any x\ G X , and we can find

x 2 G X \ B( xi , 1) . Next , we can find

xs e X \ B ( x i , 1) U B( X 2 , 1) , since S(xi , 1) U 5(x 2 , 1) is a bounded set . We continue this procedure and obtain a sequence xn with

—

n 1

xn & X \ (J B ( xk , 1) . k =1

Then we see that

d ( xi , xk ) 1 Vi j. sequence a convergent , subsequence, a contradicHence this does not have tion , proving the boundedness of X . Next , let { xn }n i be a Cauchy sequence in X . By the compactness of X , there exists a^subsequence { xnk } k l converging to some limit point

^

^

x G X . Then we can show that the whole sequence is convergent to the same limit x . Hence, X is complete. The conclusion for Y is clear.

Note that the converse of the above is not true, i.e., bounded and closed set is not necessarily compact. For example, if ( X, do ) is a discrete metric space and Y C X is an infinite set. Then Y is bounded and closed , but not compact . However , we have the following positive result. Theorem 4.10 . ( Bolzano- Weierstrass ) Let {#n}n > i be a bounded sequence in Rm with the £ p -metric pp . Then it has a convergent subsequence.

Metric Spaces and Limits for Sequences

37

Proof . We only prove the case m = 1. Since { xn }n i is bounded , L = lim xn exists. Then by Proposition 2.17, this L is a limit point of n—¥ oo {^n}n > i , i.e., for some subsequence xUk > L.

^

—

•

Corollary 4.11 . ( Heine—Borel ) Let Rm be the metric space with the £ p -metric pp for some 1 p oo. Let E C Rm . Then E is compact if and only if E is bounded and closed . Proof . => It follows from Proposition 4.9.

0 such that B ( y , 5 ) C V . Now, we let

5( y )

^

— sup{ 0 | B( y , S ) C V , for some A E A} > 0. \

Define

S0 = inf{ 0 such that d{ ynk , y )


1 such that

N

N

N

n= l

n= 1

n= l

c

\J v n = \ j K = [ f ] K n ) = K Nt

i

which implies

—K

'

0

jv

n Kft = 0

n Ki C

a contradiction. Hence, ( 4.5) holds.

Now , suppose ( 4.6) holds, then for any x, x E f ] 71

1 ifn ^ n - oo.

—

diam ( Kn ) > 0,

d ( x, x )

, we have

>

Hence, ( 4.7) holds. Next result is relevant to the total boundedness.

.

Theorem 4.15 L e t ( X , d ) b e a metric space and Y C X . Then Y is compact if and only if Y is complete and totally bounded .

Proof . Necessity. Suppose Y is compact . Then Y has to be complete. Next , for any e > 0, we have

Y C

|J

B{ x,e ).

xeY

Then by finite open cover theorem , we see that Y is totally bounded.

Sufficiency. Let { yn }n i Q Y be a sequence. For k = 1, Y is covered by finitely many open balls of radius 1. Among them , there is a ball, denoted by B ( x\ , 1) , containing a subsequence of { yn }n i We denote it by

^

£ ( xi , l ) n y,

^-

1. ^ , 1) fl Y is compact , it is covered by finitely many balls of Vn , I

Next , since B { xi radius Then there exists an of {2/ n , i }n i such that

^

2/71 , 2

X2

E

^

Vn

B ( x i , 1) and a subsequence { yn ,2 }n

e H ( x2 , ) n y,

n

1.

^i

42

Mathematical Analysis

— A Concise Introduction

Continue this procedure, we can find { xk }k {Vn ,k }n>1 such that 1

xk G B ( x k - i ? fc-!) , 2

Now , we let zk that

= V k ,k -

yn , k G B { x k ,

1

^

2

Note that

1

^

) D Y.

k

-i

i- i

k 0

k =0

'

j

^ ^= ^

This means that { x k }k

) n Y.

n

1.

Then { z k } k > l is a subsequence of { y n } n 6 S (a:fe ,

d ( x i , Xi+ j )

Q X and subsequences

^i

^^

^ i such

1.

( 4.8)

1 2^ + fe

1

< 2i _1 '

^ i is a Cauchy sequence. By ( 4.8) we see that lim d ( x k , zk )

—

k KXD

= 0.

Hence, { zk }k i is also Cauchy. Then by the completeness of T , one has the convergence of { zk }k i This proves the compactness of Y .

^

^-

The following example shows that in an infinite dimensional space, bounded closed set is not necessarily compact , in general.

Example 4.16. Recall £ p ( p E [l , oo ) ) . Look at the closed unit ball cenr tered at 0: oo

j

B ( 0 , 1) = ( xfc )

G

P

\ xk \ P

| k= l

l

}-

It is bounded and closed . We claim that it is not compact . In fact , let us look at the following sequence: Xn

— {^

nk

)

Vn

1

where

5n k Then for any n

^

m , one has

Hence, the sequence { xn }n proves our claim.

-

1

k = n,

0

k

^

—2

P

n.

.

^ i does not have a convergent

subsequence. This

Metric Spaces and Limits for Sequences

43

Definition 4.17. Let ( X, d ) be a metric space. Then it is said to be separable if there exists a countable set Xo such that X0 = X.

In other words, X admits a countable dense subset. Proposition 4.18. Let ( X, d ) be a compact metric space , Then X is separable. Proof . Since X is compact , it is totally bounded . Consequently, for any 1, there exists a finite set { x£ | 1 k Kn } such that

n

k= l

Now , let

Xo = {*£

1

k

Kn , n

}

1

Then XQ is countable and it is dense in X . Hence, X is separable.

It is clear that if ( X, d ) is separable, then any subset Y C X is also

separable.

Example 4.19 . ( i ) If ( Xfc , d& ) ( 1 k n ) is separable, then X\ x X i< x • • • x Xn is separable under the corresponding product metric.

( ii) Any subset Y of Mn is separable under the usual product metric of -metric (1 p oo ) .

^

( iii) Let ( X , do ) be a metric space with the discrete metric, and X be an uncountable set . Then ( X , d0 ) is not separable. Exercises

1. Show that the set Q of all rational numbers is countable and the set R \ Q of all irrational numbers is uncountable.

2. Prove the Bolzano-Weierstrass Theorem ( Any bounded sequence in Rm contains a convergent subsequence) by the Nested Compact Sets Theorem.

3. Prove the Heine-Borel Theorem by the Nested Compact Sets Theorem. 4. Let 5 =

l

n

n e N} U {0}. Then S is closed .

5. Show that any discrete metric space ( X, do ) must be complete.

Mathematical Analysis

44

—

A Concise Introduction

6. Let ( X , d ) be a complete metric space and

A-n+i C. Ani Vn 1 diam ( An ) = sup d ( x , y ) x,yeAn

An C X

—* 0,

satisfy

—

n > oo.

Let an G An . Then the sequence {an } converges. Find the difference between this result and the Nested Compact Sets Theorem. 7. Show that Q fi [0 , 1] is totally bounded. 8. Show that £ p is separable for p G [1, oo ) and £°° is not separable. 9. Let ( X, do ) be a discrete metric space. Let E C X be a non-empty subset . Then E is compact if and only if E is a finite set.

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

(i) Suppose Ki ,

••

, Xn

is compact .

C X are compact sets. Then 2

( ii ) Suppose {X , A ^ | K\ is compact.

G

=1

A} is a family of compact subsets in X . Then

P

AGA

11. Let K\ and K 2 be two disjoint compact sets in a metric space ( X, d ) . Then

a=

XX

- d(Xl X2 ) > 0.

K I ,X 2 K 2

>

Moreover there are x\ G K\ and x i G K 2 such that
Y Let x0 be a limit point of E . We say that f ( x ) is convergent to L e Y as x —» #o if for any e > 0, there exists a 5 > 0 such that

.

•

d Y { f { x ) , L ) < s,

Vx G E , 0 < d x ( x , xo ) < 6.

We denote the above by lim X

± XQ

/ (x) = EBxlimtxo f ( x ) = L .

—

( ii ) Let / : X —> Y . We say that / ( •) is continuous at every e > 0 , there exists a 5 > 0 such that

,

d y { f { x ) f { x0 ) )

XQ G

X if for

V d x ( x , x0 ) < S.

2 ) and ( y, dy ) = (R , p) . Any / : Rm >• R , with m > 1, is called a multivariable function. The following gives some equivalent conditions for a function being continuous at a given point X Q .

—

Theorem 1.2 . Let ( X, dx ) and ( Y , d y ) b e metric spaces. Let f : X and X Q G X . Then the following are equivalent : 45

-» Y

Mathematical Analysis

46

— A Concise Introduction

-

(i) / ( •) is continuous at xo (ii) Whenever {xn }n i C X is convergent to xo under dx , the sequence ( {/ £n )}n i is convergent to f ( x0 ) under d y . ^ (iii) For every open set V C Y containing f ( xo ) , there exists an open set V C X containing XQ such that

^

f ( U ) C V. The proof of the above result is straightforward. Such a result leads to the following which is about the continuity of a function f : X -> Y o n X .

.

Theorem 1.3 L e t ( X , d x ) a n d ( Y , d y ) be metric spaces. Let f : X -> Y . Then the following are equivalent :

( i) / ( •) is continuous on X . (ii) Whenever { xn } C X is convergent to some xo under dx , the sequence { f { xn ) } is convergent to f ( xo ) under dy ( iii) For every open set VC Y , f ~ x ( V ) is open in X , where

-

f -\V )

= {x G X | f ( x ) V }. (iv ) For every closed set F in Y , f 1 { F ) is closed in X . Proof , (i) => (ii). Let xn G X with xn > XQ . By the continuity of / ( •)

for any e > 0 , there exists a S > 0 such that d y { f { x ) , f ( x0 ) )

Also , there exists an N

d x { x , x0 ) < ( iii) . Let V C Y be an open set . If / 1( Vr ) is not open, then there exists an xo e f ~ 1 (V ) and a sequence xn e f ~ 1 (V ) c converging to xo Hence f ( xn ) —> f { xo ) . Since V is open, there exists an e > 0 such that

_

-

,

B y ( f ( xo ) e ) C V.

Therefore, there exists aniV f (Xn)

l such that ^ B y ( f ( x ), e 0

)

Vn

> N.

Functions on Metric Spaces

47

This implies that

n

1 1 *« e r ( Bj' ( / (*o ) , e) ) cr (

a contradiction .

( iii) => ( iv ) . For any closed set F C Y , F c is open. Thus,

r l ( Fc ) = r\F )c , is open , leading to that /-1 (F) is closed .

( iv ) => ( hi ) . The proof is the same as ( iii ) => ( iv ) . ( iii) => ( i) . Let xo X . For any e > 0, B y ( f ( xo ) , s) is open. Hence ~1 f ( B y ( f ( x0 ) , e ) ) is open , and x0 e f

1

( BY ( f { x0 ) ,e ) ) .

Thus, there exists a 5 > 0 such that

B x ( x0 , S ) C f -1 ( BY ( f ( x0 ) , e) ) . This means

f ( B x ( x0 , S ) ) C

BY ( f ( x0 ) , e ) ,

.e.

i

d y { f { x ) , f ( x0 ))

< e,

V d x ( x , x0 ) < S.

This completes the proof .

Proposition 1.4. L e t ( X , d x ) , ( Y , d y ) , a n d ( Z , d z ) b e metric spaces. Let f : X » y , g : V -> Z. If / ( •) is continuous at Xo e X and g ( ) i s Z is continuous at x$ . Consequently, continuous at f ( xo ) , then go f : X if / ( • ) is continuous on X and g ( ) i s continuous on Y , then ( g o / ) ( • ) is continuous on X .

-

—

-

The proof is straightforward .

Now , we look at an important special case (Y , d y ) = ( M, p). Note that the order structure. Due to such in R , there is an additional structure a structure, we can say a little more on any bounded , not necessarily continuous function / : X » R. Suppose X Q e X and / ( •) is bounded in a ball B ( xo , r ) , for some r > 0. For any S G (0, r] , we have the existence of

—

/ (so ; S ) = X

inf

) { XO } B ( X O ,6 \

f ( x),

f + ( x0 ; S ) =

sup

X

. B ( X O ,6 )\{ X Q }

/ (*)

•

Mathematical Analysis

48

—

A Concise Introduction

Clearly, 8 f ( xo ; R be a bounded function. (i ) We say that / ( •) is lower semi-continuous at xo if (i.i ) / ( x0 ) lim / ( x ) .

—

a: > cco

( ii) We say that / ( •) is wpper semi- continuous at xo if f ( x0 ) lim / ( x ) . (1.2 ) X ^XQ ( iii) If / ( •) is lower ( resp. upper ) semi-continuous at every point in X , we say that / ( •) is lower ( resp. upper ) semi-continuous on X . To get some feeling about the lower and upper semi-continuous functions, let us present the following example.

.

Example 1.6 Let 1

< p( x ) =

x=0 x O;

V» (*) =
0, there exists a 5 > 0 such that Vx £ B ( xo , 5) . f { x ) > f ( xo )

( iii) For every a £ R , the set (/ > a ) = { x is

open.

£

X

| / (x) > a }

Functions on Metric Spaces

49

Proof , ( i) => ( ii ). By the lower semi-continuity of / ( •) at f ( x0 ) inf lim f ( x ) lim /( )• 540 XGB( XO ,5) \{ XO } * X * Xo Then for any e > 0, there exists a 6 > 0 such that

—

—

f { x0 )
a ) , we have f ( x0 )

Then by ( ii) , for

£

, there exists a S > 0 such that

=

f ( x ) > / ( x0 )

> a.

-£

= f ( x0 ) -

f ( x0 ) - a 2

/ ( x0 ) + a > a 2 Vx G B ( x o , S ) .

This implies that

F( £o , a ) . Thus, ( / > a ) is open ; ( iii) holds. (iii) => ( i) . For any xo G X and any e > 0, x0 e ( / > / ( x 0 ) - e ) . Then there exists a 0 such that / ( x ) > / ( x0 ) - e ,

Vx

G

B( XQ , 5) .

Hence lim X

^XQ

/ (x )

/ ( x0 ) - e ,

-

for any e > 0 , which leads to the lower semi-continuity of / ( •) at xo Likewise, we have the following result for the upper semi-continuous functions. Proposition 1.8. Let ( X, d ) b e a metric space and f : X bounded function. Then the following are equivalent : ( i) / ( •) is upper semi-continuous. any £ > 0, there exists a 5 f ( x ) < f { x0 ) + £ , Vx G L? ( xo , 5 ) . ( iii) For every a G R , t h e s e t

(ii) For every xo

G

X,

for

( f < a) = { x e X is

open .

| / (x) < a}

—>

•

R be a

> 0 such that

Mathematical Analysis

50

— A Concise Introduction

The following is pretty obvious.

—

Corollary 1.9. Let ( X , d ) be a metric space and f : X > R be a bounded function. Then / ( •) is continuous if and only if it is both lower and upper

semi-continuous.

Next , we look at two interesting functions.

Example 1.10. ( i ) The following is called the Dirichlet function:

/ ( X) =

1

x

0

X

eQ e R \ Q.

Since for any x £ R ,

_ (x ) lim yx /

x

—

hm f ( x )

=1

x

—

yx

= 0,

we see that lim_ f ( x ) fails to exist for any x G l, which implies that the X YX Dirichlet function is nowhere continuous. It is upper semi-continuous at every rational point , and lower semi-continuous at every irrational point .

—

( ii) The following is called the Riemann function: I

/ (*) = {

if x

o^

—|£ Q, with q > 0 and p and q are co-prime,

if x £ R \ Q.

Since for any x G l, f ( x ) = 0, lim yx

( why? ) — we see that this function is continuous at every x

irrational point and is

discontinuous at every rational point .

We now introduce the following definition which is concerned with functions from a metric space to itself. Definition 1.11. Let ( X , d ) be a metric space, and / : X x e X is called a fixed point of / ( •) if the following holds: f {x)

—> X . A point

= X.

The following result is called the Contraction Mapping Theorem.

Theorem 1.12. Let ( X , d ) be a complete metric space and f : X satisfy the following: d ( f ( x ) , f ( y)) where a

ad ( x , y )

e ( 0 , 1 ) is a constant . Then / ( ) •

Vx , y

e X,

admits a unique

—> X (1.3)

fixed

point .

51

Functions on Metric Spaces

Proof . Pick any

xo E X , define the sequence =

£ 71

*

f ( %n

—)

1.

n

l

(1.4)

We have the following estimate: d { xn , %n

— ) ^ Oid { xn— , Xn — ) ^ ^ 2

1

l

OL

— d ( xi , x0 )

n 1

Vn

1.

Consequently,

t

d ( x n i X n+ £ )

an

t

^

^^ ^

OLU

^^

fc= i

i

— ) ^ = ^n+ — d(o: , x ) 1 — cxf ^ — d( x , )

d ( xnJtk , ^n + fc

afc d (xi , xo ) = a n

fc = 0

^

u t

0

/c l

i

0

fc i

1- a

—

— a d (xi , x ) —> 0

^^

l

00

1 £0

.

This means that {xn} is Cauchy. By the completeness of ( X , d ) , one has xn y x for some x G X . Then / ( x ) = x. Now , if there exists another x £ X such that / ( x ) = x, then

—

d ( x, x )

ad ( x , x ) .

= d ( / (x) , / (x ) )

Since a < 1, we must have x = x , proving the uniqueness. In the above, (1.4) is called a Picard iteration. Let us consider two functions f : X -> Y and g : X -» Z . We can define their direct sum f ® g : X -» Y x Z by the following:

( f ® 9 ) ( x ) = ( f ( x ) , g ( x ))

X.

Vx

The following can be proved easily.

Proposition 1.13. L e t ( X , d x ) , ( Y , d y ) , and ( Z , dz ) be metric spaces. Let f : X > Y and g : X Z.

—

-

( i) / ( •) and g ( ) are both continuous at xo £ X (resp. on X ) if and only if / 0 Y must be

( ii) Suppose ( X, d x ) = ( Mm , p z ) and (F, dy ) are discrete. Then f : X -A Y is continuous if and only if / is a constant function.

—

, / ( x )) , for all x e Mm . 3. Let / : Mm » R e , with / ( # ) = ( /i ( x ) , ^ Then / ( •) is continuous if and only if each /* ( •) is continuous, 1 i L * * *

4. Let ( X, d ) be a metric space. Then the metric d : X x X continuous.

5. Let ( X, dx ) and ( F, d y ) be two metric spaces and

—> R is

/ : X —> Y

such

that

dY { f ( x) J ( y ))

L d x ( x, y )

Vx, y e X ,

for some L > 0. Then / ( •) is continuous. 6. Let G C Mm be a domain. Then G is pathwise connected, i.e., for any G G, there exists a continuous function 7 : [0 , 1] > G such that 7( 0) = x and 7(1) = y .

—

7. Find a counterexample showing that a connected set in Rm might not be pathwise connected . 8. Prove Proposition 1.13.

—

9. Let f\ : R > R ( A E A ) be a family of continuous functions. Let

F ( x ) = sup f x ( x )

x G M.

AGA

Show that F( - ) is lower semi-continuous. 2

Properties of Continuous Functions

In this section , we will present several important properties of continuous functions.

Functions on Metric Spaces

2.1

53

Compactness preserving and uniform continuity

The following result is concerned with the compactness preserving property of continuous functions. Theorem 2.1. Let ( X , d x ) and ( Y , d y ) be metric spaces. Let f : X be continuous.

(i) Let K

C

—> Y

X be compact . Then

I

f ( K ) = {/ (*) x e K }

is compact in Y . (ii ) Let y = R , and ( X , d x ) be compact . Then / ( •) is bounded and it attains its maximum and minimum. Proof , ( i) Let { y n }n i he any sequence from f ( K ) . Then there is a sequence { x n }n i C K such that

^

^

f { x n ) = Vn

n

1.

Since K is compact , there exists a subsequence { x n k } k x e K . Then by the continuity of / ( •) , we have Vnk = f ( x n k ) -* f ( x ) which shows the compactness of f { K ) .

^ i such that x U k —>

f ( K ),

(ii) See the next theorem. The next theorem generalizes ( ii) of the above theorem .

.

—

Theorem 2.2 L e t ( X , d ) be a compact metric space and f : X ^ R be lower (resp. upper ) semi-continuous. Then / ( •) attains its minimum ( resp. maximum ) on X . Proof . Let

xn G

X be a minimizing sequence , i.e., lim f ( x n ) = inf f ( x ) .

n

—t o o

x£ X

Since X is compact , there exists a subsequence x U k such that lim d ( x n k , x ) = 0,

—

k » oo

for some x E X . Then by the lower semi-continuity of / ( •) , one has inf f { x )

x£ X

f {x)

Thus, x is a minimum of / ( •).

lim f ( x n ) = inf f ( x ) .

—

n > 00

xGX

Mathematical Analysis

54

— A Concise Introduction

The case of upper semi-continuous function can be proved similarly. We now present an interesting example.

.

.

Example 2.3 Let X = £ l Let B = B ( 0, 1) be the closed unit ball in X centered at 0. Define h : B > R as follows:

—

oo

•

— 5=> ( |* | —

h{ x )

l) + 5

fe

2

Vz = (:xk )

G

B

( 2.1 )

5

fc l

where a+

— a V 0 = max{a , 0}. Note that for any x = ( xk ) G B, *fci ^ Ei = oo

k 1

Thus, there are at most one non- zero terms in the series ( 2.1) defining /i ( - ). Now , we show that h( ) is continuous on B . To see that , we let x = ( xk ) G B be given, and let xn = ( x£ ) G B such that

-

oo

E= \ xn Xk \ -> 0. We may assume that there is a X 1 such that ^ 1 Pl ( xn , x ) =

~

k 1

I*f c iI
0 .

fc l

-

This proves the continuity of h ( ) at such a point x. Next , for xn we have h ( xn )

= n —> oo ,

as n

oo.

= ( Snk ) k ^ i

5

Functions on Metric Spaces

55

-

Therefore, /i ( ) is continuous and unbounded on the bounded and closed set B . We now define 9{ x )

1

x

= 1 h( x ) +

e B.

Then g ( - ) is continuous and

inf g ( x ) = 0,

X

g( x )

B

> 0, x

E

B.

This means that continuous function g ( - ) does not attain its minimum over the bounded and closed set B. Recall from Chapter 1, Example 4.16 that in i 1 , the closed unit ball 5(0, 1) is not compact . Hence , the above example tells us that if the domain space X is not compact , a continuous function defined on a bounded and closed set might not be bounded and might not achieve its minimum and maximum .

Next , let us look at another example which will lead us to another issue. Example 2.4. ( i) Prove the continuity of f ( x ) = x 2 on [0 , 1].

Let

XQ

.

E [0 , 1] One has

\ x 2 — XQ \ = \ x — X o \\ x + X o \ 2 \ x — X o \ . Hence , for any e > 0 , by taking 5 = § > 0, one has If ( x ) - f ( xo ) | = \ x 2 - X Q \ 2 \ x - x0| < 25 = s , V|x — xo | < 6. We note that S =|is independent of X Q . ( ii ) Prove the continuity of g ( x ) = - on ( 0 , 1].

Letxo E (0, 1]. For \ x — xo \
0 , by taking 5 = min { — 2 XQ X \ g ( x ) - g ( x0 ) \ = \ X X o \ -

-

Now , for \ x

we have x >

’

— }, 2

XQ

2

x0

we have

^ —^ \ x — xo| < e, 2

— xo \ < — ’

V| x

—

XQ

| < S.

This proves the continuity of g ( ) on (0, 1] . In this proof , 5 not only depends on > 0, but also depends on xo We claim that for this function , one cannot find a S > 0 that is independent of XQ . In fact , if there existed a

-

Mathematical Analysis

56

—

A Concise Introduction

5 = 5 ( e ) independent of xo , then , say, for e = 1, we would have a 5 > 0 such that 1

1

\ 9( x ) - g { - ) I

|nx

1 l/n

x

— 1| < I

X

Vx

But this is impossible , since we may take x = -~~T with n so that 2 1 n+1 0< < 8. n 1 n n( n 1) Then for such an x lb

—

1>

|nx

— 1|

X

— —n

< 8.

^ 1 large enough

—

n+1 2

n— 1 2 n— 1

X

Vn

1

which is a contradiction. The above examples show that sometimes, even for one variable functions, continuity could be quite different. The following notion reveals such a difference.

Definition 2.5. Let ( X, dx ) and ( Y, dy ) be metric spaces. A map / : X —> Y is said to be uniformly continuous on X if for any e > 0, there exists a 5 = 5 ( e ) > 0 , only depends on £, such that dy ( / (*) > A*') )

Vx , x'

0. For each > 0 such that

XQ

E

X , by the continuity of / ( •) , there

exists a 5 ( X Q )

dy ( / (*) , / (*o ) )

|