The Foundations of Analysis: A Straightforward Introduction: Book 2: Topological Ideas

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THE FOUNDATIONS OF ANALYSIS: A STRAIGHTFORWARD INTRODUCTION BOOK 2 TOPOLOGICAL IDEAS

THE FOUNDATIONS OF ANALYSIS: A STRAIGHTFORWARD INTRODUCTION BOOK 2 TOPOLOGICAL IDEAS

K. G. BINMORE Professor of Mathematics London School of Economics and Political Science

CAMBRIDGE UNIVERSITY PRESS Cambridge London New York New Rochelle Melbourne Sydney

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Siio Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521233507 ©Cambridge University Press 1981 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1981 Re-issued in this digitally printed version 2008

A catalogue record for this publication is available from the British Library ISBN 978-0-521-23350-7 hardback ISBN 978-0-521-29930-5 paperback

CONTENTS

Book 1:

Logic, Sets and Numbers Introduction

1

Proofs

2

Logic (I)

3

Logic (11)

4

Set operations

5

Relations

6

Functions

7

Real numbers (I)

8

Principle of induction

9

Real numbers (11)

lOt

Construction of the number systems

llt

Number theory

12

Cardinality

Book 2:

Topological Ideas Introduction

13 13.1 13.3 13.5

13.9 13.11 13.14

V

Distance The space IR" Length and angle in !R n Some inequalities Modulus Distance Euclidean geometry and 1R n

XI

4 5 7 7 8

vi

Contents

13.17t 13.18 13.19t 13.20

Normed vector spaces Metric space Non-Euclidean geometry Distance between a point and a set

14 15 16 17

14 14.1 14.2 14.3 14.7 14.15

Open and closed sets (I) Introduction Boundary of a set Open balls Open and closed sets Open and closed sets in IR"

21 21 21 22 25 29

15 15.1 15.4 15.8t 15.11

Open and closed sets (11) Interior and closure Closure properties Interior properties Contiguous sets

31 31 32 33 34

16 16.1 16.2 16.7 16.13 16.17t

Continuity In trod ucti on Continuous functions The continuity of algebraic operations Rational functions Complex-valued functions

39 39 39 41 44 46

17 17.1 17.2 17.6 17.8 17.15 17.18 17.21t 17.25t

Connected sets Introduction Connected sets Connected sets in IR 1 Continuity and connected sets Curves Pathwise connected sets Components Structure of open sets in IR"

47 47 47 49 50 52 53 56 57

18 18.1 18.4t 18.9t

Cluster points Cluster points Properties of cluster points The Cantor set

60 60 62 63

19 19.1

Compact sets (I) In trod ucti on

66 66

tThis material is more advanced than the remaining material and can be omitted at a first reading.

Contents

vii

19.2 19.5 19.12

Chinese boxes Compact sets and cluster points Compactness and continuity

67 69 73

20t 20.lt 20.2t 20.4t 20.7t 20.15t 20.16t 20.20t

Compact sets (11) Introduction Open coverings Compact sets Compactness in IR" Completeness Compactness in general metric spaces A spherical cube

77 77 77 78 78 83 84 86

21 21.1 21.2 21.3 21.4 21.6 21.8 21.9t 21.15t 21.18t

Topology Topological equivalence Maps Homeomorphisms between intervals Circles and spheres Continuous functions and open sets Topologies Relative topologies Introduction to topological spaces Product topologies

89 89 90 91 92 94 95 95 100 102

22 22.1 22.2 22.3 22.4 22.9 22.12 22.15 22.16 22.19 22.23 22.25 22.34t

Limits and continuity (I) Introduction Open sets and the word 'near' Limits Limits and continuity Limits and distance Right and left hand limits Some notation Monotone functions Inverse functions Roots Combining limits Complex functions

106 106 109 109 110 113 116 118 119 121 123 125 128

23t 23.1t 23.3t 23.5t 23.11t 23.12t 23.20t

Limits and continuity (11) Double limits Double limits (continued) Repeated limits Uniform convergence Distance between functions Uniform continuity

130 130 132 132 136 138 145

viii

Contents

24 24.1 24.2 24.3 24.4 24.5 24.8 24.11 t 24.12t

Points at infinity Introduction One-point compactification of the reals The Riemann sphere and the Gaussian plane Two-point compactification of the reals Convergence and divergence Combination theorems Complex functions Product spaces

149 149 150 153 154 157 163 165 165

25 25.1 25.2 25.7 25.12 25.18 25.23

Sequences Introduction Convergence of sequences Convergence of functions and sequences Sequences and closure Subsequences Sequences and compactness

169 169 170 172 175 176 179

26 26.1 26.2 26.5t 26.11 t

Oscillation Divergence Limit points Oscillating functions Lim sup and lim inf

181 181 182 184 186

27 27.1 27.2 27.8 27.13t 27.16t 27.18t

Completeness Cauchy sequences Completeness Some complete spaces Incomplete spaces Completion of metric spaces Completeness and the continuum axiom

190 190 190 193 195 197 199

28 28.1 28.7 28.8 28.11 t 28.15t 28.19t 28.26t

Series Convergence of series Absolute convergence Power series Uniform convergence of series Series in function spaces Continuous operators Applications to power series

201 201 204 204 208 209 212 215

29t 29.lt 29.2t

Infinite sums Commutative and associative laws Infinite sums

218 218 220

Contents

ix

29.4t 29.9t 29.17t 29.23t

Infinite sums and series Complete spaces and the associative law Absolute sums Repeated series

221 223 226 230

JOt 30.1t 30.2t 30.4t 30.8t 30.11 t 30.12t 30.14t

Separation in IR" Introduction Separation Separating hyperplanes Norms and topologies in IR" Curves and continua Simple curves Simply connected regions

234 234 235 235 239 241 242 243

Notation

245

Index

246

The diagram on p. x illustrates the logical structure ofthe books. Broken lines enclosing a chapter heading indicate more advanced material which can be omitted at a first reading. The second book depends only to a limited extent on the first. The broken arrows indicate the extent of this dependence. It will be apparent that those with some previous knowledge of elementary abstract algebra will be in a position to tackle the second book without necessarily having read the first.

Book I Logic, Sets and Numbers

r-----------,

110. Construction 1 1of number systemsl

19. Real

I numbers (11)

L-----1-----.l

L---T---

1

I

r----- ------,

I

l

I

1

I

..---L----

1

I I

___ _)17. Connected 1 sets L----

18. Cluster points

I I

r----1--~~-----r----------~--~

I 20.

Compact :sets (11)

L----

l

----

11. Number theory

I

:

L----------_.J

----,

: 30. Separation I : in IR"

l

L--------....J

Book2 Topological Ideas

__ ...J

L----r--- J 26. Oscillation

• 28. Series L __ - - - - - - -

H';.-~:~~;el

1sums 1 L------...J

INTRODUCTION

This book is intended to bridge the gap between introductory texts in mathematical analysis and more advanced texts dealing with real and complex analysis, functional analysis and general topology. The discontinuity in the level of sophistication adopted in the introductory books as compared with the more advanced works can often represent a serious handicap to students of the subject especially if their grasp of the elementary material is not as firm as perhaps it might be. In this volume, considerable pains have been taken to introduce new ideas slowly and systematically and to relate these ideas carefully to earlier work in the knowledge that this earlier work will not always have been fully assimilated. The object is therefore not only to cover new ground in readiness for more advanced work but also to illuminate and to unify the work which will have been covered already. Topological ideas readily admit a succinct and elegant abstract exposition. But I have found it wiser to adopt a more prosaic and leisurely approach firmly wedded to applications in the space IR". The idea of a relative topology, for example, is one which always seems to cause distress if introduced prematurely. The first nine chapters of this book are concerned with open and closed sets, continuity, compactness and connectedness in metric spaces (with some fleeting references to topological spaces) but virtually all examples are drawn from IR". These ideas are developed independently of the notion of a limit so that this can then be subsequently introduced at a fairly high level of generality. My experience is that all students appreciate the rest from 'epsilonese' made possible by this arrangement and that many students who do not fully understand the significance of a limiting process as first explained find the presentation of the same concept in a fairly abstract setting very illuminating provided that some effort is taken to relate the abstract definition to the more concrete examples they have met before. The notion of a limit is, of course, the single most important concept in mathematical analysis. The remainder of the volume is therefore largely devoted to the application of this idea in various important special cases. Much of the content of this book will be accessible to undergraduate students during the second half of their first year of study. This material has xi

xii

Introduction

been indicated by the use of a larger typeface than that used for the more advanced material (which has been further distinguished by the use of the symbol t). There can be few institutions, however, with sufficient teaching time available to allow all the material theoretically accessible to first year students actually to be taught in their first year. Most students will therefore encounter the bulk of the work presented in this volume in their second or later years of study. Those reading the book independently of a taught course would be wise to leave the more advanced sections (smaller typeface and marked with a t) for a second reading. This applies also to those who read the book during the long vacation separating their first and second years at an institute of higher education. Note, incidentally, that the exercises are intended as an integral part of the text. In general there is little point in seeking to read a mathematics book unless one simultaneously attempts a substantial number of the exercises given. This is the second of two books with the common umbrella title Foundations of Analysis: A Straightforward Introduction. The first of these two books, subtitled Logic, Sets and Numbers covers the set theoretic and algebraic foundations of the subject. But those with some knowledge of elementary abstract algebra will find that Topological Ideas can be read without the need for a preliminary reading of Logic, Sets and Numbers (although I hope that most readers will think it worthwhile to acquire both). A suitable preparation for both books is the author's introductory text, Mathematical Analysis: A Straightforward Approach. There is a small overlap in content between this introductory book and Topological Ideas in order that the latter work may be read without reference to the former. Finally, I would like to express my gratitude to Mimi Bell for typing the manuscript with such indefatigable patience. My thanks also go to the students of L.S.E. on whom I have experimented with various types of exposition over the years. I have always found them to be a lively and appreciative audience and this book owes a good deal to their contributions. June 1980

K. G. BINMORE

13 13.1

DISTANCE

The space 1R"

Those readers who know a little linear algebra will find the first half of this chapter very elementary and may therefore prefer to skip forward to §13.18. The objects in the set IR" are the n-tuples

in which x 1 , x 2 , ... , x" are real numbers. We usually use a single symbol x for the n-tuple and write

The real numbers x 1 , x 2 , ... , x" are called the co-ordinates or the components of x. It is often convenient to refer to an object x in IR" as a vector. When doing so, ordinary real numbers are called scalars. If x = (x 1 , x 2 , ... , x") and y=(y 1 , y 2 , ••• , Yn) are vectors and ex is a scalar, we define 'vector addition' and 'scalar multiplication' by x+y=(xt +yl' x2+Y2····· xn+Yn) CXX = (cxx 1 , CXX 2 , ••• , CXXn). These definitions have a simple geometric interpretation which we shall illustrate in the case n=2. An object XEIR 2 may be thought of as a point in the plane referred to rectangular Cartesian axes. Alternatively, we can think of x as an arrow with its blunt end at the origin and its sharp end at the point (x 1 , x 2 ).

X2

--------.,

x

I I I

I I

I XI

x as a point

I

x as an arrow

2

Distance

Vector addition and scalar multiplication can then be illustrated as in the diagrams below. For obvious reasons, the rule for adding two vectors is called the parallelogram law.

x+y I

I

I

I

I I I

I I

/

I

Y1

The parallelogram law is the reason that the navigators of small boats draw little parallelograms all over their charts. Suppose a boat is at 0 and the navigator wishes to reach point P. Assuming that the boat can proceed at 10 knots in any direction and that the tide is moving at 5 knots in a south-easterly direction, what course should be set?

,,~

N I

I

'

',

I

',

''

'~

t(x

+ y)

,."'/P '

I

/

"'

/x+y /

II

I

I

I /

I

~---------,~~~~----------~E I I

5 ~·

~

I

I

I I

I

I

X

I

I

I

I

I

I

I

~I tx

The vector x represents that path of the boat if it drifted on the tide for an hour (distances measured in nautical miles). The vector y represents the path of the boat if there were no tide and it sailed the course indicated for

3

Distance

an hour. The vector x + y represents the path of the boat (over the sea bed) if both influences act together. The scalar t is the time it will take to reach P.

13.2

Example Let x = (1, 2, 3) and y = (2, 0, 5). Then

x+y=(1, 2, 3)+(2, 0, 5)=(3, 2, 8) 2x=2(1, 2, 3)=(2, 4, 6).

It is very easy to check~· is a commutative group under vector addition. (See §6.6.) This simply means that the usual rules for addition and subtraction are true. The zero vector is, of course, 0=(0, 0,.

0

0).

.,

The diagram below illustrates the vector y-x=(y 1 -x 1 , y 2 -x 2 , ... , Y.-x.) in the case n = 2.

~

----------

I I I

I

I

I

I

I

I

I

I

It is natural to ask about the multiplication of vectors. Is it possible to define the product of two vectors x and y as another vector z in a satisfactory way? There is no problem when n = 1 since we can then identify ~ 1 with IR. Nor is there a problem when n = 2 since we can then identify ~ 2 with C (§ 10.20). If n ~ 3, however, there is no entirely satisfactory way of defining multiplication in ~ ". Instead we define a number of different types of 'product' none of which has all the properties which we would like a product to have. Scalar multiplication, for example, tells us how to multiply a scalar and a vector. It does not help in multiplying two vectors. The 'inner product', which we shall meet in § 13.3, tells how two vectors can be 'multiplied' to produce a scalar. In ~ 3 , one can introduce the 'outer product' or 'vector product' of two vectors x and y. This is a vector denoted by x 1\ y or x x y. Unfortunately, x 1\ y = -y 1\ x.

4

Distance

Multiplication is therefore something which does not work very well with vectors. Division is almost always meaningless.

Length and angle in IR"

13.3

The Euclidean norm of a vector x in IR" is defined by llxll={x/+x/+ ... +x/} 112 . We think of llxll as the length of the vector x. This interpretation is justified in IR 2 by Pythagoras' theorem (13.15).

The inner product of two vectors x and y in IR" is defined by (x,y)=X1Y!+x2y2+ ··· +XnYn· It is easy to check the following properties: (i) (x, x)=llxW (ii) (x, y)=(y, x) (iii) (ocx + py, z) =oc(x, z) + p(y, z).

The geometric significance of the inner product can be discussed using the cosine rule (i.e. c 2 = a2 + b2 - 2ab cos y) in the diagram below. y y-x \ - - - - - - - - - - - - - - -

\ \ a

\

\

\

\

\ 0

X

Rewriting the cosine rule in terms of the vectors introduced in the right-

5

Distance

hand diagram, we obtain that

llx -yll 2 = llxll 2 + IIYII 2 -211xii·IIYII cosy. But,

llx -yll 2 = <x -y, X-y) = <x, X-y)-
<x, y) = llxii·IIYII cos y. Of course, this argument does not prove anything. It simply indicates why it is helpful to think of

<x, y) llxii·IIYII as the cosine of the angle between x and y.

13.4 Example Find the lengths of and the cosine of the angle between the vectors x = (1, 2, 3) and y = (2, 0, 5) in IR 3 . We have that

llxll = {1 2 + 22 +3 2 P12 = J14, IIYII = {2 2 +02 + 52 P12 = J29, <X, y) 1 • 2 + 2 • 0 + 3 • 5 17 llxii·IIYII j14j29 J14x29. 13.5

Some inequalities

In the previous section y was the angle between x and y. The fact that !cos yl ~ 1 translates into the following theorem.

13.6

Theorem (Cauchy-Schwarz inequality) IfxEIR" and yEIR", then

l<x, y)l ~ llxii·IIYII· Proof Let IXEIR. Then O~llx-tXyll 2 =<x-tXy,

X-tXy) =llxii -21X<x. y)+1X 2 IIYII 2 • 2

6

Distance

It follows that the quadratic equation llxii 2 -21X(x, y)+1X 2 IIYII 2 has at most one real root (§10.10). Hence 'b 2 -4ac;;;;O'- i.e.

4(x, y) 2 -411xii 2 IIYII 2 ;;;;o. It is a familiar fact in Euclidean geometry that one side of a triangle is shorter than the sum of the lengths of the other two sides. x+y

0

This geometric idea translates into the following theorem.

13.7

Theorem (Triangle inequality) If xE~n apd yE~n. then llx+yll~llxll+ IIYII-

Proof

llx+yll 2 =(x+y, x+y) =llxll 2 +2(x, y)+IIYII 2 2

~llxll +211xii.IIYII+IIYII =

13.8

2

(theorem 13.6)

(llxll + IIYII) 2 •

Corollary If XE!Rn and yEW, then

llx- Yll ~ llxii-IIYIIProof It follows from the triangle inequality that

llxll = ll(x -y)+yll;;;; llx -yll +IIYII-

7

Distance

13.9

Modulus

The modulus lxl of a real number x coincides with the Euclidean norm of x thought of as a vector in IR1 1 . We have that

llxll={x 2 } 112 =lxl={

x(x~O)_

· -x (x
One has to remember that yi/ 2 represents the non-negative number whose square is y (§9.13). The modulus lzl of a complex number z=x+iy is identified with the Euclidean norm of (x, y) thought of as a vector in !R1 2 • We have that

lx+iyl = {x 2 + Y2 } 112 = ll(x, y)ll. 13.10 Then

Theorem Suppose that u and v are real or complex numbers.

luvl = lul·lvl. Proof We need only consider the complex case. If u =a+ ib and v=c+id, then

luvl 2 -lul 2 lvl 2 = (ac- bd) 2 +(be+ ad) 2 - (a 2 + b2 )(c 2 + d 2 ) = 0. 13.11

Distance The distance d(x, y) between two vectors x and y in IR" is defined

by

d(x, y)=llx-yllIn interpreting this idea in IR", it is better to think of x and y as the points at the end of the arrows rather than the arrows themselves.

x-y

0

8

Distance

13.12

Examples

(i) The distance between the vectors x = (1, 2, 3) and y = (2, 0, 5) in IR1 3 is lly-xJI ={(2-1) 2 + (0-2) 2 + (5 -W} 112 =3.

(ii) The distance between 3 and 7 (regarded as vectors in IR1 1 ) is 13-71=1-41=4. 4

0

13.13

3

7

Exercise

(1) Let x = (0, 1, 0) and y = (1, 1, 0) be vectors in IR1 3 . Calculate the quantities

(i) x+y (iv) llxll

(ii) x-y (v) llx-yll

(iii) 2x (vi) (x, y).

What is the length of the vector x? What are the distance and the angle between x and y? (2) Calculate the moduli of the following real and complex numbers: (i) -3 (ii) 0 (iii) 4 (iv) 3 + 4i (v) 4- 3i (vi) i. (3) If a and b are any real numbers, prove that Jal < b if and only if -bO, prove that a=O. (4) Let x and y be elements of IRl". Prove that d(x, y) ~I llxii-IIYII J.

[Hint: Use question 3.] (5) Let x and y be elements of IRl". Prove that (x, Y)=HIIx+yll 2 -llx-yll 2 }. (6) Let xEIRlm and yEIRl". Then (x, y) may be regarded as an element of IRlm+n.

13.14

Prove that

Euclidean geometry and IRl"

In § 10.1, we explained that it is the models in terms of which a system of axioms can be interpreted which make that system interesting. But this is a two-way process. It is equally true that a model is interesting because of the systems of axioms which it may satisfy.

Distance

9

The space IR1 2 is particularly interesting because it serves as a model for the axioms of Euclidean geometry in the plane. The axioms given by Euclid himself are incomplete by modern standards in that certain of his theorems cannot be deduced from his axioms alone but depend also on implicit geometrical assumptions which are systematical)y taken for granted but never explicitly stated. (Some of these are mentioned briefly later on.) When we refer to the axioms of Euclidean geometry, we mean the axiom system given by David Hilbert in his famous book The foundations of geometry published at the beginning of this century. The geometric terms which appear in Hilbert's axioms are the words point, line, lie on, between and congruent. To show that IR1 2 is a model for Euclidean plane geometry one has to give a precise definition of each of these words in terms of IR1 2 and then prove each of Hilbert's axioms for Euclidean plane geometry as a theorem in IR1 2 • This will demonstrate, in particular, that the axioms for Euclidean geometry are consistent. An attempt at describing this programme in detail is beyond the scope of this book. (Interested readers will find the book Elementary geometry from an advanced standpoint by E. E. Moise (Addison-Wesley, 1963) an excellent reference.) Instead, we shall merely explain how some of the very familiar ideas of elementary geometry are expressed in terms of the space IR1". A point is simply an element xEIR1". The line 1 through the distinct points a and b is the set

l={<Xa+f3b: <XEIR1, /3EIR1 and <X+/3=1}. This is easier to illustrate with a picture if we rewrite it in the form

l={a+f3(b-a): {3EIR1} and think of 1 as the line through the point a in the direction of the vector b-a.

I a+ (j(b- a) I I I 1

I I

I

I I

!

I I I I

I

_ _.J3(b- a)

____..{'-- - -

:.-----

v-a

0

If, in the above, f3 is restricted to be non-negative we obtain the definition of the half-line from a through b. If a= 0, such a half-line is called a ray. If both <X and f3 are restricted to be non-negative we obtain the definition of a

10

Distance

line segment. Thus the line segment L joining a and b is given by L

= {oca + f3b: oc ~ 0, f3 ~ 0 and oc + f3 = 1}.

ray through b

b

a

b

a

from a through b

line segment joining a and b

0

Some authors do not insist that the endpoints of a line segment belong to the line segment. Where it matters whether or not the endpoints of a line segment L belong to L we shall therefore sometimes stress the fact that they do by calling L a closed line segment. We have already discussed how length and angle are introduced into IR" via the norm and the inner product of the space. Two vectors x and y are said to be orthogonal if and only if

(x, y)=O. Like most important ideas in mathematics, the word orthogonal has numerous synonyms of which some are perpendicular, normal and 'at right angles'.

13.15 (Pythagoras' theorem) Let x and y be elements of!R". Then x and y are orthogonal if and only if

!lx + Yll 2 = llxll 2 + IIYII 2 -

l

IIYII

j X

11

Distance Proof We have that

and hence

llx +YII 2 = <x +y, x +y) = llxll 2 +2<x, y) + IIYII 2 llx+yll 2 =llxii 2 +IIYII 2 if and only if <x, y)=O.

A circle in IR 2 with centre I; and radius r > 0 is a set of the form

S={x: llx-l;ll=r}.

circle in

IR 2

In IR 3 , the set S represents a sphere and in IR • we call S a h ypersphere. The inside of a hypersphere - i.e. the set

B={x: llx-l;ll
d

c

b

a I

12

Distance

Both balls and boxes are examples of convex sets. A set Sin !Rn is convex if and only if, whenever xES and yES, it is true that exx+fJyES provided that ex~ 0, f3 ~ 0 and ex+ f3 = 1. In geometric terms, this means that, whenever two points are in the set, so is the line segment which joins them .

•• convex sets

non-convex sets

A plane in IR 3 through the point ~ with normal u :f. 0 is a set of the form

P={x:

(x-~,u)=O}.

u

piece of hyperplane in IR 3

13

Distance

In IR", we call the set P a hyperplane. A hyperplane in IR" with normal u # 0 is therefore a set of the form

H={x: (x,u)=c}, where c is a constant. The hyperplane passes through any point ; which satisfies (;, u) =c. In IR 3 a hyperplane is an ordinary plane. In IR 2 a hyperplane is a line. In IR 1 a hyperplane is just a point.

hyperplane is a line

It is 'geometrically obvious' (at least in IR 1, IR 2 and IR 3) that a hyperplane splits IR" into two separate pieces (called half-spaces) and that any line segment joining two points from different half-spaces must cut the hyperplane. This fact is one of the assumptions that Euclid failed to give proper expression to in his axiom system. A proof of the result for IR" will have to wait until chapter 17 (exercise 17.28 (6)).

13.16

Exercise

(1) Let x and y be elements of IR". Prove that

llx +YII 2 + llx -yll 2 =211xll 2 +211YII 2 Interpret this result geometrically for n = 2 in terms of the diagonals of a certain parallelogram. (2) Let x and y be elements of IR". Prove that

llxii=IIYII <=>(x-y, x+y)=O. Interpret this result geometrically in terms of a triangle inscribed in a circle. (3) Prove that there is a unique line through any two points in IR". Show that, if two lines meet, then their intersection consists of a single point. Prove that, in IR 2 , a hyperplane is the same thing as a line. (4) The parallel postulate for Euclidean plane geometry asserts that given any line I and any point P not on I, there exists a unique line through P which does not meet I. Prove this for IR 2 • (5) Let us say that a closed box S = 11 x 12 x ... x 1" in IR" is proper if none

14

Distance

of the intervals I k consists of just a single point. Prove that every open ball B contains a proper closed box S and that every proper closed box S contains an open ball. (6) Prove that open balls, closed boxes, lines, half-lines and hyperplanes are all convex sets in IR". Prove that hyperspheres are not convex. t(7) Prove that the intersection of any collection of convex sets is convex. What about unions? t(8) A cone in~· is the union of a collection of rays. Prove that the intersection of any collection of convex cones is a convex cone. t(9) Taking for granted the truth of the assertion in the last paragraph of § 13.15 about hyperplanes separating ~· into distinct 'half-spaces', explain why a line which enters a triangle in ~ 2 through one of its sides must leave the triangle through one of the other sides (Pasch's theorem). t(lO) A function f: ~·__,~·is linear if and only if, for each x and y in~· and all real numbers IX and {3, /(lXX + {Jy)= !Xj(x)+ {Jf(y). Prove that, for each

Sin~·,

S convex=f(S) convex. t(ll) Prove that a function/: ~·-->~ 1 is linear if and only if there exists an ee~" such thatf(x)= (x, e) for each xe~". t(12) Anticipating the definition of§ 13.17, prove that a normed vector space X (with real scalars) admits an inner product satisfying (i), (ii) and (iii) of§ 13.3 if and only if 2

llx + Yll 2 + llx -yll = 211xll 2 + 211YII

2

for each xe X and each ye X. [Hint: Use exercises 13.16(1) and 13.13(5).]

13.17t

Normed vector spaces

The space ~" is an example of a normed vector space. Fora vectors pace, one firstneedsacommutativegroup X (§6.6)toserveas the vectors. The group operation is then called vector addition. Next one needs a field to serve as the scalars. A meaning then has to be assigned to scalar multiplication in such a way that the following rules are satisfied: (i) (ii) (iii) (iv)

1X(x+y)=1XX+1XY (IX+ {J)x =lXX + {Jx (1X{3)x = 1X({Jx) Ox=O; 1x=x.

For a normed vector space, we usually insist that the field of scalars be ~ or C. A norm function from X to ~ is required which satisfies the following properties: (i) llxii~O

(ii) llxii=Ox=O (iii) IIlXXII = I1XIIIxll (iv) llx + Yll ~ llxll + IIYII

(triangle inequality)

15

Distance

for all XE X, yE X and all scalars. (Here llxll denotes the image of x under the norm function.) The chief reason for giving the requirements for a normed vector space at this stage is to provide a summary of the properties of IR" which will be important in the next few chapters. Other normed vector spaces will be mentioned hardly at all. It should be noted, however, that the properties of infinite-dimensional normed vector spaces are often counter-intuitive. For example, a hyperplane in an infinitedimensional normed vector space does not, in general, split the space into two separate half-spaces.

13.18

Metric space A metric space is a set X and a function d: X x X-+IR which

satisfies

(i) (ii) (iii) (iv)

d(x, y) ~ 0 d(x,y)=O=-x=y d(x, y) = d(y, x) d(x, z) ~ d(x, y) + d(y, z).

(triangle inequality)

The function d is said to be the metric for the metric space and we interpret d(x, y) as the distance between x and y. The metric space in which we shall be most interested is the space IP1 ". As we have seen in§ 13.11, the metric d2 : IP1" x IP1 .--+IR in IP1" is defined by d(x, y)= llx -yll·

Strictly speaking, one should refer to d as the Euclidean metric. It is trivial to check that the Euclidean metric satisfies the requirements for a metric given above. In particular, the triangle inequality for a metric follows from theorem 13.7 since llx-zll = ll(x-y)+ (y-z)ll ~ llx -yll + IIY -zll-

Any subset f of IP1" is also a metric space provided that we continue to use the Euclidean metric in f. We then say that f is a metric subs pace of IP1 ". For example, the interval (0, 1] is a metric subspace of IP1 1 provided that the metric d: (0, 1] x (0, 1]-+IR is defined by d(x, y) =

lx- Yl·

d(x,y) or---~----~--------------~ X y

More exotic examples of metric spaces can be obtained by starting with any normed vector space X and defining d: X x X ->IR by d(x, y)=llx-yll·

16

Distance

But there are also many interesting metric spaces which have no vector space structure at all. Since we have been discussing the fact that IR 2 (with the Euclidean metric) is a model for Euclidean geometry (§ 13.14), we shall give PoincanYs model for a non-Euclidean geometry as an example of such a metric space.

13.19t

Non-Euclidean geometry

Perhaps the most well-known of Euclid's postulates is the parallel postulate (quoted in exercise 13.16(4)). This was always felt to be less satisfactory than Euclid's other assumptions in that it is less 'intuitively obvious' than the others. Very considerable efforts were therefore made to deduce it as a theorem from the other axioms. All these efforts were unsuccessful. Finally, it was realised that the task is impossible and Gauss, Lobachevski and Bolyai independently began to study a geometry in which the parallel postulate is false but all the other assumptions of Euclidean geometry are true. Gauss did not publish his work and Lobachevski published before Bolyai. Hence the non-Euclidean geometry they studied will be called Lobachevskian geometry. (Sometimes it is called 'hyperbolic geometry'.) This Lobachevski, incidentally, is the Nikolai lvanovich Lobachevski of the immortal Tom Lehrer song but the scandalous suggestions made in this song are totally unfounded! In Lobachevskian geometry there are many parallel lines through a given point parallel to a given line. This may seem intuitively implausible as a hypothesis about the 'real world' because we have been trained from early childhood to think of space as Euclidean. In fact, Einsteinian physics assures us that, in the vicinity of a gravitating body, space is very definitely not Euclidean. The purpose of this long pre-amble is to explain the interest of the following metric space which was introduced by the great French mathematician Poincare. This space provides a model for Lobachevskian geometry. Its existence therefore demonstrates that the axioms of Lobachevskian geometry are consistent. What is more, since the parallel postulate is true in IR 2 but false in the Poincare model, it must be independent of the other axioms of Euclidean geometry. In particular, it cannot be deduced from them. The set X in Poincare's metric space is the set .'C= {(x, y): x 2 + y 2 < 1} in IR 2 • But the metric used in X is not the Euclidean metric.

/

I

/

_,...- ---

/ I

B ...........

',

'\

\

\

I

I

I

c'I

I

I

I

I

'\,,

A

' ' '-...._

Distance

17

Let P and Q be points of X. If P and Q lie on a diameter of the circle C which bounds X, let L be this diameter. If P and Q do not lie on a diameter, let L be the unique circle through P and Q which is orthogonal to C. If A and B are as indicated in the diagram, we then define the Poincare distance between P and Q by

Ilog(;:~;~) I·

d(P, Q)=

One can think of the points in the diagram below as the footprints of a little man who tries to walk from the centre of X to its boundary. Each step is of equal size relative to the Poincare metric. He will therefore never reach the bounqary and therefore, as far as he is concerned, X extends indefinitely in all directions.

//_,------

',

e

I

I

I I

........

\ •

\

\ I

I I 1 I

I

•

I

X \

\ '

', ..... ......._

/

___

I I

/ _.... ..... /

/1

Poincare defined a 'straight line' in X to be one of our curves L. This is perfectly reasonable since, relative to the Poincare metric, the shortest route from P to Q is along L. He defined the angle between two 'straight lines' L and M to be the ordinary Euclidean angle between Land M. With these definitions X is a model of Lobachevskian geometry. The diagram below shows two 'straight lines' M and N through the point P 'parallel' to the 'straight line' L.

I

I

I

I

I I

I

\

\ \

13.20

', .....

Distance between a point and a set

If I; is a point in a metric space X and S is a non-empty set in X,

18

Distance

we define the distance between

~

d(~,

and S by

S) = inf d(~, x). xeS

Obviously,

d(~,

S) is always non-negative.

13.21 Example Consider the point 3 and the set S = [0, 1] in IR. We have that

d(3, S)=d(3, 1)=2.

d(3, S)

• 0

13.22 Theorem Let empty set in X. Then

3

~

be a point in a metric space X and let S be a nond(~,

S)=O

if and only if, for each e > 0, there exists an

XE S

such that

d(~,

x) <e.

19

Distance Proof Everything in the set

D={d(x,

~):xES}

is non-negative. Thus 0 is a lower bound for D. The statement d(~, S) = 0 is therefore equivalent to the assertion that no e > 0 is a lower bound for D i.e. not 3e>0 1:/xES (d(x,

~)~e).

But this is equivalent to

1:/e>O 3xES (d(x,

~)<e)

as required (§3.10).

It is important to take note of the fact that

d(~,

S) = 0 does not imply that

~ES.

13.23

Examples

(i) Consider the point 1 and the set S = (0, 1) in IR. For each e > 0, we can find an xES such that x> 1-e. X

1-e

Since 1- e < x < 1, d(1, x) <e. It follows from theorem 13.22 that

d(1, S)=O. 0

1

s (ii) Consider the point 0 and the set S = { 1/n: nE N} in IR. Given any e > 0, we can find an nE N such that n > 1/e (because N is unbounded above). But then

d(O, 1/n)= 1/n<e and hence

d(O, S)=O. I

0

I

5 4

l

I

3

2

..... \t !/.

pointsors-

~·

20

Distance

13.24

Exercise

( 1) Find d( (i)

e. S) for the following points eand sets S in IR. e= 1; s = (O, 2)

(ii) e=O; S=[1, 2] (iii) e=t; S=N (iv) e=O; S=(O, 1) (v)

e= 1; s = (2, 3)

(vi) e=2; (vii)

e=J2;

S=(O, 1)u(3,4) S=O. [Hint: Use theorem 9.20.]

(2) Find d(~. S) for the following points ~ and sets S in IR 2 • (i) ~=(1, 1); S={(x,y):O<x d(~. S)=O (ii) S c T => d(~. S) ~ d(~. T) (iii) d(~. SuT) = min {d(~ S), (5)

d(~.

T)}.

What can be said of d(~. Sn T) in terms of d(~. S) and d(~. T)? 11 be points in a metric space .X and letS be a non-empty set in .i. Prove that

Let~ and

d(~. S)~d(~.

TJ)+d(TJ, S).

(6) Let S 1 , S 2 , .•• , Sn be non-empty sets of real numbers. Write s = sl X s2 X ..• X sn and suppose that X= (xl, X2, ... ' Xn)eiRn. Prove that d(x, S) = 0 if and only if

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

14

14.1

OPEN AND CLOSED SETS (I)

Introduction

In the next two chapters we prove a number of theorems about the properties of sets in metric spaces. Most of these theorems are true in any metric space ·whatsoever. Others are true only for the metric space IR ". However, with just an occasional exception, all examples will be of sets in IR". A systematic attempt to give examples from a wider range of spaces at this stage would submerge the essential ideas in a sea of technicalities. We begin by discussing open and closed sets. These generalise the idea of open and closed intervals which we met in §7.13. The intervals (a, b), (a, oo), (- oo, b) are open and the intervals [a, b], [a, oo ), (- oo, b] are closed. (The sets (/) and IR count both as open and as closed intervals.) The difference between open and closed intervals lies in the treatment of their endpoints. In the general case, we shall have to talk about the boundary points of a set rather than the end points of an interval. Our first priority will therefore be to give a precise mathematical definition of a boundary point of a set Sin a metric space X.

14.2 ~

Boundary of a set

Let Sbea set in a metric space X. A boundary point of the setS is a point of X such that

and

(i)

d(~,

S) = 0

(ii)

d(~,

es) = o.

Note that a boundary point of S need not be a point of S. The set of all boundary points of S is denoted by oS. Observe that S and eS have the same boundary - i.e. oS= o( eS).

21

Open and closed sets (I)

22

not boundary points

Some special consideration of the case S = (/) is necessary. In this case S) = 0 is meaningless and hence cannot be true. Thus the empty set has no boundary points- i.e. i3(/) = (/). Similarly, i3 X=(/). d(~,

14.3

Open balls

The open ball B with is defined by

centre~

and radius r>O in a metric space X

B = {X : d(~, X) < r} . 3

In IR , an open ball is the inside of a sphere. In IR 2 , an open ball is the inside of a circle. In IR 1 , an open ball is an open interval. z y

~ ~{jffj

'-------+-X X

y

14.4 Theorem LetS be a set in a metric space X and let ~be a point in X. Then ~ is a boundary point of S if and only if each open ball B with centre ~ contains a point of s and a point of es.

Open and closed sets (I)

23

point of S

Proof The fact that each open ball B with centre ~ and radius r>O contains a point of Sand a point of es is equivalent to the assertion that, for each r>O, there exists xES and yEeS such that d(~, x)
eS) = 0 -

i.e. ~ is a boundary point of S and of

14.5 Examples (i) The boundary of the set F in IR 2 defined by F={(x, y): O~y~l}

is the set oF={(x, y): y=O or y= 1}.

Note that every boundary point ofF belongs to F- i.e. oFcF. y

eF

oF

(ii) The boundary of the set G in IR 2 defined by

G = {(x, y): x 2 + y 2 < 1} is the set oG = {(x, y): x 2 + y 2 = 1}. Note that no boundary point of G·belongs to G - i.e. oG c e G.

eS.

24

Open and closed sets (/) y

ec (iii) The boundary of the set H in IR 2 defined by H = {(x, y): x 2 + y 2 < 1 and y;;;O}

is the set

8H={(x,y):x 2 +y 2 =1 and y;;;O}u{(x,O): -1~x~1}. Note that some boundary points of H belong to H and some do not. y

(iv) The boundary of the set S = {1/n: nE N} in IR 1 is

8S=Su{O}. points of S

0 d/\~ • ••••••! !• !• •I 4

3

2

Each point of S is clearly a boundary point of S. For example,

d(t, S)=d(t, t)=O d(t, es)~d(t, (t, 1))=0 and thus tEaS. Similarly, 0 is a boundary point of S because d(O, S) = 0 d(O,

(example 13.23 (ii))

eS) = d(O, 0) = 0.

Finally, no other point of IR 1 can be a boundary point of S. For example,~ is not a boundary point of S because d(~, S)=d(~, 1)=±#0.

Open and closed sets (I)

25

Exercise

14.6

( 1) Sketch the following sets in IR1 2 and determine their boundary points (i) (ii) (iii) (iv)

P={(x,y):O~x~1 and0
Q={(x, y): x< y} R={(x,y):y~x 2 }

S={(1, 1), (2, 2), (3, 3)}.

(2) Sketch the following sets in IR1 1 and determine their boundary points. (i) (0, 1) (ii) [0, 1] (iii) (0, 1] (iv) [0, 1) (v) (0, oo) (vi) [0, oo) (vii) N (viii) Q. t(3) LetS be a .finite set in IR". Prove that aS=S. (Note that this need not be true in a general metric space. For example, if X= S.) t(4) Show that the open ball Bin IR" with centre I; and radius r>O has boundary

aB= {x: d(x, l;)=r}. (Note that this need not be true in a general metric space. For example, if X is the metric subs pace of IR 1 consisting simply of the two points ~ and ~ + r.) t(5) Find the boundary points of the set S in IR 2 defined by S={(1/m, 1/n): mEN and nEN}. t(6) Let A and B be sets in a metric space :r. Prove that a(AuB)caAuaB. [Hint: Use theorem 14.4.] Deduce that a(AnB)caAnaB.

14.7

Open and closed sets

Let S be a set in a metric space X. The set S is open if and only if it contains none of its boundary points - i.e.

asc es. The set S is closed if and only if it contains all of its boundary points - i.e.

oScS. It may be helpful to think of S as a fenced plot of land in a city. The fence around the plot can then be thought of as the boundary of S. As all city dwellers know, the neighbours are not to be believed when, on moving into a new house, they explain that all of the fence is built on your land (and hence you are responsible for its upkeep). In fact, some of the fence is built on your land and some on your neighbour's land. One can think of the part of the fence on your land as the part of the boundary of S which belongs to S. The part of the fence built on the neighbour's land can be regarded as the part of the boundary of s which belongs to An open set can then be thought of as a plot for which all the fence belongs to the neighbours. A closed set is a plot for which all of the fence belongs to the owner.

es.

26

Open and closed sets (I)

When drawing pictures, we use a dotted line to indicate the boundary of an open set and a firm line to indicate the boundary of a closed set.

,. ..... --- ......... /

\

// 1

I

I I

open

1

I

I \

I

' .... ___ _

I // /

As with most city plots of land, sets are usually neither open nor closed.

14.8 (i) (ii) (iii) (iv)

14.9 closed.

es

Example Consider the sets of examples 14.5. The set F is closed. The set G is open. The set H is neither open nor closed. The set S is neither open nor closed.

Theorem A set S in a metric space X is open if and only if

es

is

Proof The theorem follows immediately from the fact that S and have the same boundary.

es

14.10 Theorem A set F in a metric space X is closed if and only if, for each XE X, d(x, F)=O

=>

xEF.

Proof (i) Suppose that d(x, F)=O => xEF. If I; is a boundary point ofF, then d(l;, F)= 0 and d(l;, eF)= 0. Since d(l;, F)= 0, it follows that I;E F. Thus F contains its boundary points and hence is closed. (ii) Suppose that F is closed. If x~F, then XE eF and hence d(x, eF)= 0. If it is also true that d(x, F)= 0, then x is a boundary point of F and so, since

Open and closed sets (I)

27

F is closed, XEF. This is a contradiction. Hence xrtF = d(x, F)'IO - i.e. d(x, F)=O => XEF.

14.11 Corollary A non-empty closed set F of real members which Is bounded above has a maximum.

Proof We have to show that supFEF. But d(supF, F)=O by exercise 13.24(3). Hence the result by theorem 14.10.

14.12 point

Theorem A set G in a metric space X is open if and only if each is the centre of an open ball B entirely contained in G.

~E G

/

/ I

I I I I / /

/

I I

-----

a:

....

'

''

~~~/ /

I \

''

.....

_____ G

/ / /

Proof (i) Suppose that ~EG is the centre of an open ball B entirely contained in G. Then ~ cannot be a boundary point of G because B contains no point of eG (theorem 14.4). Thus G contains none of its boundary points and so is open. (ii) Suppose that G is open. Let ~E G. All open balls B with centre ~ then contain a point of G. Since ~ is not a boundary point of G, it follows from theorem 14.4 that at least one open ball B with centre ~ contains no point of eG- i.e. BeG.

14.13 (i) (ii) (iii) sets is

Theorem In a metric space X: The sets (/) and X are open. The union S of any collection W of open sets is open. The intersection T of any finite collection {G1 , G2 , ... , Gk} of open open.

Proof (i) The set (/) is open because o(/) = (/) c e(/) = X. The set X is open because oX=(/)c eX=(/). (ii) Let ~ES. We shall use theorem 14.12 and prove that S is open by finding an open ball B with centre ~ such that B c S. Since ~ES, there exists a GEW such that ~EG. Since G is open we can

28

Open and closed sets (/)

find an open ball B with centre result follows.

~

such that Be G. But G e S and so the

(iii) Let ~ET. We shall use theorem 14.12 and prove that T is open by finding an open ball B with centre ~ such that Be T.

Since

~ET,

we have that

~EGl' ~EG 2 , •.• , ~EGk.

... , Gk are all open, we can find open balls B 1 , B 2 ,

Because the sets G1 , G2 , Bk each with centre~

••• ,

such that BieGi

(j=1, 2, ... , k).

One of the open balls B 1 , B 2 , •.• , Bk has minimum radius. Let this open ball be B. Then Be Bie Gi (j = 1, 2, ... , k) and hence k

BeT=~ Gi. Thus B is an open ball with centre

~

such that Be T.

14.14 Theorem In a metric space X: (i) The sets (/) and X are closed (as well as open).

29

Open and closed sets (I)

(ii) The intersection S of any collection W of closed sets is closed. (iii) The union T of any finite collection {F 1 , F 2 , , .. , F k} of closed sets is closed. Proof The theorem follows from theorems 14.9 and 14.13. (i) The sets (/) and X are closed because e(/) = X and e X=(/) are open. (ii) The set S is closed because

es=e(n F)=u Fe'il'

eF

Fe'il'

is open by theorem 14.13. (iii) The set T is closed because

eT=e(~~ Fj)=~

eFj

is open oy theorem 14.13.

14.15

Open and closed sets in

~·

Of the various types of sets in ~" introduced in § 13.14, only open balls are open sets in the sense of§ 14.7. Of the other types of sets considered in §13.14: single points lines (and half-lines) closed line segments hyperspheres closed boxes hyperplanes

and

are all closed sets in the sense of§ 14.7. We prove only that hyperplanes are closed. The proofs for the other types of set in ~" are left as exercises.

14.16

Theorem Any hyperplane in

~"

is closed.

Proof Consider the hyperplane H through the point ~ with normal u#O. This is given by H = {x: (x -~, u) =0}. Suppose that d(y, H)=O. We propose to use theorem 14.10 and therefore seek to prove that yE H. By theorem 13.22, given any s > 0, there exists an XEH such that d(y, x)=IIY-XII<s. But since XEH, (y-~,

u)=(y-x,

u)+(x-~,

u)=(y-x, u).

Hence, using the Cauchy-Schwarz inequality (13.6), l(y-~, u)l~lly-xll·llnll<sllnll·

Open and closed sets (I)

30

Since this is true for every s>O, it follows that Thus H is closed.

14.17

(y-~,

u)=O and so yEH.

Exercise

(1) Determine which of the sets of exercise 14.6(1) and (2) are open and which are closed. (2) Prove that the only sets in~· which are both open and closed are 0 and ~·.

(3) Let S be the set of exercise 14.6(5). Find a point ~f!S such that d(~, S) = 0. Deduce that S is not closed. (4) Prove that open balls in a metric space X as defined in § 14.3 are open sets as defined in § 14. 7. (5) Prove that a set consisting of a single point in a metric space X is closed. Deduce that any finite set in a metric space X is closed. (6) Let Sl, S2, s. be closed sets in ~ 1 • Prove that S=Sl X s2 X X s. is a closed set in ~·. [Hint: Use exercise 13.24(6).] (7) Prove that the types of set in ~" said in § 14.15 to be closed actually are closed. (8) A flat (or affine set) in ~· may be defined to be the intersection of a collection of hyperplanes. Prove that a flat is always closed. (9) The sets S and T in ~· are defined by 0

0

0,

0

S={x:

(x,u)~c};

0

0

T={x: (x,u)
(where u # 0 and c are constant). We call S a closed half-space of~" and Tan open half-space. Prove that S is closed and T is open in the sense of §14.7. t(lO) Give examples to show that a convex set in IR" may be (i) open (ii) closed (iii) neither open nor closed (iv) both open and closed. t(ll) Prove that any open set Gin a metric space X is the union of the collection of all open balls it contains. t(l2) Find a sequence (/k) of open intervals and a sequence (Jk) of closed intervals in IR such that, if 00

00

l=ll Ik; J=U Jk, k=l

then I is not open and J is not closed.

k=l

15

15.1

OPEN AND CLOSED SETS (11)

Interior and closure

Let E be a set in a metric space X. The interior defined by

E of the set

E is

E=E \oE. The closure

E is defined by E=EuoE.

oE

The interior

E is obtained from E by removing all the boundary points of

E is obtained from E putting in all the boundary points of E which happen not to be elements of

E which happen to be elements of E. The closure

E.

15.2

Example Consider the set Sin IR 2 defined by S={(x,y):O~y<1}.

This set is neither open nor closed. Its interior

S={(x, y): 0< y< 1}.

31

S is the

set

Open and closed sets (I/)

32

Its closure S is the set

S={(x, y): 15.3

O~y~l}.

Exercise

(1) Prove that, for any set E in a metric space X, (i) E is open <=>E=E (ii) E is closed <=> E =E. (2) Determine the interior and closure of each of the sets of exercises 14.6(1) and (2). (3) Prove that, for any set E in a metric space X, (i) e(Ji)=eE (ii) e(E)=(e£) 0 • [Hint: e(E)= e(Ene(8E)) and e(E)= e(Eu8E).] (4) Let H be a hyperplane in IR". Prove that (i) 8H=H (ii) fl=H (iii) H=0. [Hint: We have seen that H is closed (theorem 14.16). Hence 8H c H. If f,E H and H has normal u, show that d(f,, f.+ tu)= iti·iiull- Deduce that HcaH.] (5) LetS= {x: <x, u)
(ii) S= T (v) f= T

(iii) S=S (vi) T= S.

[Hint: For (iv), (v) and (vi), recall question (3).] t(6) Let B1 ={x: d(~, x)O) be balls in IR". Prove that

B1 =B 1 (ii) B 1 =B 2 (iii) [Hint: Use exercise 14.6(4).]

(i)

B2 =B 1

(iv)

B2 =B 2 .

15.4

Closure properties

15.5

Theorem For any set E in a metric space X,

xEE <=>d(x, E)=O Proof (i) If XE E = EuaE, then XE E or XE a E. In the former case d(x, E)=d(x, x)=O. In the latter case, we have by definition that d(x, E)=O and d(x, eE)=O. Thus XEE =>d(x, E)=O. (ii) If d(x, E)=O, then either xEEcE or else xEeE. In the latter case d(x, eE)=d(x, x)=O and so XE8EcE. Thus d(x, E)=O => XEE.

Open and closed sets (I I)

15.6

33

Corollary Let S and T be sets in a metric space X. Then

ScT=Scf. Proof Let xES. By theorem 15.5, d(x, S)=O. But SeT implies d(x, S)'?;d(x, T). Hence d(x, T)=O. Thus xE T.

15.7t Theorem Let E be a set in a metric space X. Then closed set containing E.

E is the smallest

Proof To show that E is closed, we use theorem 14.10. We therefore assume that d(~, E)=O and seek to deduce that ~EE. By theorem 13.22, given any e>O, we can find an xEE such that d(~, x)<e. But xEE =d(x, E)=O by theorem 15.5. Hence d(~, E)~d(~,

x)+d(x,

E)=d(~,

x)

(exercise 13.24(5)).

Thus d(~,

E)
Since this is true for any c: > 0, we conclude that d(~, E)= 0 and hence ~E E by theorem 15.5. Now suppose that F is any closed set containing E. Since EcF, we have from corollary 15.6 that E cF. But F is closed and therefore F = F (exercise 15.3(1 ii)). It follows that E c F for each closed set F containing E. Thus E is the smallest closed set containing E.

15.8t

Interior properties From exercise 15.3(3i) we know that

Thus any result about closures leads to a corresponding result about interiors. For example,

SeT= es=>eT= es=>eT= e(eS)ce(eT) and so S c T

= Sc t

15.9t Theorem Let E be a set in a metric space X. Then set contained in E.

E is the largest open

Proof Note first that E=e(eE) and hence is open because it is the complement of a closed set. If G c E, then Gc E. If G is also open it follows that

Open and closed sets (Il)

34

G c E (because then G =G). This argument shows that contained in E.

15.1 Ot

E is

the largest open set

Exercise

(1) A flat (or affine set) F is the intersection of a collection of hyperplanes in IR". Prove that (i) F=F (ii) F=0. Show that the same results hold if F is a half-line or a closed line segment. (2) Let E be a set in a metric space X. Prove that xeE if and only if there exists an open ball B with centre l; such that B c E. (3) Let E be a set in a metric space X. Prove that aE is closed. [Hint: aE =En e(E).] (4) Let E be a set in a metric space X. Prove that (i) E =

1'1

(ii)

F

EcF

E= U

G.

GeE

F closed

G open

(5) Let A and B be sets in a metric space X. Prove that

(i)

AvB=AvB; (AnB) ·=AnB

(ii) A c B ~ A c B; A c B ~A c B (iii) AnBcAnB; (AvB) ·cAvB. Give examples to show that c cannot be replaced in general by (6) Let E be a set in a metric space X. Prove that

in (iii).

=

Show also that: (i) a(£) c aE

(ii) a( E) c aE.

Give examples to show that c cannot be replaced in general by case. (7) Let C be a convex set in IR" with a non-empty interior. Prove that (i)

C=C

(ii)

=

in either

C=C.

Give examples of (non-convex) sets C for which these equations are false. Give an example of a. closed convex set C for which C=0. (8) Let C be a convex set in IR". Prove that Cand Care also convex. (9) Let E be a set in a metric space X. Let W denote the collection of all sets which can be obtained from E by applying the operations e, - and • to E any finite number of times in any order. Show that the maximum number of distinct elements that Wean contain is 14.

15.11

Contiguous sets

We shall say that two sets A and B in a metric space X are contiguous (i.e. 'next to each other') if a point of the closure of one of the sets

Open and closed sets (IJ)

35

belongs to the other. Thus A and B are contiguous if and only if

Ar.B#(/)

Ar.B#(/).

or

We say that the sets A and B are separated if and only if they are not contiguous.

' '

I

I

\

''\

/~ I

' ...... __ ....

I

1

I

I

I

\

/

,/

I

''

A andB separated

A and B contiguous

To say that A and Bare contiguous, means that either the sets overlap or else a boundary point of one set belongs to the other. Using the metaphor of §14.7 which interprets sets as plots of land in a city, the owners of A and B are neighbours. Theorem Two sets A and B are contiguous if and only if there 15.12 exists a point in one of the sets which is at zero distance from the other. Proof This follows from theorem 15.5. We have that

aEAnB ~aEA and d(a, B)=O. Similarly,

bEAnB ~bEB and d(b, A)=O. 15.13

Examples

(i) The intervals A=(O, 1) and B=[l, 2] in IR 1 are contiguous. We have that d(l, A)=O. Alternatively one can note that

Ar.B= [0, l]n[l, 2] = {1} # (/). The diagram illustrates the owner of set A walking from his house aE A to visit his neighbour whose house is at bEB. Observe that this visit does not involve crossing any point outside AuB.

.

~---

*-----------. {

1

a

A

b

B

36

Open and closed sets (IJ)

(ii) The intervals A= (0, 1) and B = (1, 2) in IR 1 are separated. The diagram illustrates the owner of A jumping over the point 1 in order to get to B.

a

0

b

A

2

B

In formal terms AnB=[O, 1]n(1, 2)=0 AnB=(O, 1)n[1, 2] =0.

15.14 Theorem Two open sets G and H m a metric space X are contiguous if and only if they overlap.

-------

----;.....c_ 1/

.....

/

/

/

I

/

I

(

G

\ \

'

',

\

\

'

\

I

\

H

I

I

I

'..........

.... _

''

\

I

'

',,

..... ,

/1

I

I

/

_ ..... /

-;:-.-c.."'----

------

Proof Two sets which overlap are obviously contiguous. We therefore need only prove that, if G and H are contiguous open sets, then

GnH#-0. Suppose that GnH=0. Then GceH. Hence GceH=eH (corollary 15.6) because eH is closed. Thus GnH =0. Similarly, Gnfl =0. Thus G and H are not contiguous.

15.15 Theorem Let A, B, G and H be sets in a metric space X. If A c G and BcH, then A and B contiguous

= G and H contiguous.

37

Open and closed sets (IJ)

,.~

--

/ /

/

G

.....

/'

--- ' '

\

\

'

I I I

I I I I I

I

\

'

AnB#(/)

Proof Suppose that AnB # (/). Then GnH => AnB # Gnfi#0.

=

0. Similarly,

15.16t Theorem Two sets A and B in a metric space X are separated if and only if there exist disjoint (i.e. non-overlapping) open sets G and H such that A c G and BcH.

Proof (i) Suppose that G and H are disjoint open sets for which A c G and B c H. Then A and B are separated by theorems 15.14 and ~ 5.15. (ii) Suppose that A and B are separated. Define the sets G and H by G={x:d(x, A)
The sets G and H are obviously disjoint. We prove that G and H are open using theorem 14.12. Let ~E G and write r=t{d(~, B)-d(~, A)} >0.

Then the open ball with centre ~ and radius r > 0 lies entirely inside G. To prove this, we suppose that d(~, x)O. Observe that d(~, B)~d(~,

-i.e. Also

d(x, d(x,

x)+d(x, B) x). A)~d(x, ~)+d(~, A). B);;;d(~, B)-d(~,

Hence d(x, B)-d(x,

A);;;d(~, B)-2d(~, x)-d(~, >d(~, B)-2r-d(~,

A)

A)

=0.

Thus XE G and so G is open. Similarly, H is open. It remains to show that A c G and B c H. Since A and Bare separated, AnB = Hence, given any aEA, d(a, B)>O (theorem 15.12). Thus, for each aEA, d(a, A)=O
and hence AcG. Similarly, Be H.

0.

38

Open and closed sets (II)

15.17

Exercise

(1) Decide which of the following pairs of sets in IR 2 are contiguous and which are separated. (i) A={(x,y):x>O};B={(x,y):x 0} ; B = {(x, y): y = 0 and x ~ 0} (iii) A= {(x, y): x 2 + y 2 < 1} ; B = {(x, y): x 2 + y 2 > 0} (iv) A= {(x, y): x>O and y~O}; B= {(x, y): x~O and y
16

16.1

CONTINUITY

Introduction

A child takes a lump of modelling clay and by moulding it and pressing it, without ripping or tearing, shapes it into the figure of a man. We say that the original shape is transformed 'continuously' into the final shape. What is it that distinguishes a continuous transformation from any other? Most people would agree that the essential feature of a continuous transformation is that pieces of clay which were 'next to each other' at the beginning should still be 'next to each other' at the end. In this chapter, we shall express this idea in a precise mathematical definition and explore some of its immediate consequences.

16.2

Continuous functions

Let .'l and 'IJ be metric spaces and let S c .'r. A function f: S -+'lj is continuous on the set S if and only if, for each pair A and B of subsets of S, A and B contiguous=> f(A) and f(B) contiguous.

39

40

Continuity

If f: S --+?j, then f(f - 1 (T)) = T for each Tc f(S) (exercise 6.4(5)). It follows that an equivalent formulation of the definition is that f is continuous on S if and only if, for each pair C and D of subsets off(S), C and D separated=

f- 1(C) and f- 1(D) separated.

Our first theorem records the unremarkable fact that, if one continuous transformation is followed by another, then the result will be a continuous transformation.

Theorem Let X, ?J and f be metric spaces and let Se X, Tc?J. 16.3 Suppose that g: S -+?,I and f: T--+ f are continuous on S and T respectively and that g(S) c T. Then the composite function fag: S--+ f is continuous on

S.

Proof The diagram renders the proof obvious.

16.4 Theorem Let X and ?J be metric spaces and let S c X. Then f: S--+?J is continuous on the setS if and only if, for each xES and each E c S, d(x, E)= 0

= d(f(x), f(E)) = 0.

Continuity

41

Proof (i) Suppose that f is continuous on S. If d(x, E)= 0, then {x} and E are contiguous by theorem 15.12. Hence {f(x)} and f(E) are contiguous. Thus d(f(x), f(£))=0 by theorem 15.12 again. (ii) Suppose that d(x, E)=O=>d(f(x),f(£))=0. Let A and B be contiguous subsets of S. Then a point of one set is at zero distance from the other b~ theorem 15.12. Suppose aEA and d(a, B)=O. Then d(f(a), f(B))=O and hence f(A) and f(B) are contiguous by theorem 15.12 yet again.

If f: S -+IR", the formula

f(x) = (f1 (x), ... J.(x)) defines n real-valued functions f 1 : S -+IR ,/2 : S -+IR, ... ,f,: S -+IR. We call these functions the component functions off and write

16.5 Theorem Let .'! be a metric space and let S c.'!. A func~ion f: S -+IR" is continuous on the set S if and only if each of its component functions is continuous on S.

Proof This follows immediately from theorem 16.4 and exercise 13.24(6).

16.6

Corollary The projection function Pk: IR"-+IR defined by

is continuous on IR ".

Proof The identity function /: IR"-+IR" defined by /(x)=x is obviously continuous. But l=(PI' P 2 , •• • , P.) and hence Pk is continuous by theorem 16.5.

16.7

The continuity of algebraic operations

It is easy to see that constant functions and identity functions are continuous. We shall obtain some slightly less trivial examples of continuous functions from these by using the operations of addition, sub-

42

Continuity

traction, multiplication and division. But first we must prove that these operations are themselves continuous. We begin with a simple lemma.

16.8 Lemma Let X and ?J be metric spaces and let Se X. Thenf: S-+?J is continuous on S provided that for each xES there exists a y > 0 and a c > 0 such that d(x, y) < y

= d(f(x),f(y)) ;£ cd(x, y)

for each yES.

Proof Suppose that d(x, E)=O. Given any e>O, we shall prove that there exists a yEE such that d(f(x),f(y))<e. Thus d(f(x),f(E))=O by theorem 13.22. Choose e0 = min {eC- 1 , y}. Then e0 > 0 and hence there exists yE E such that d(x, y)<e 0 by theorem 13.22. But e0 ;i:y. It follows that d(f(x), f(y)) ;£ cd(x, y) < ce 0 ;£ e as required.

16.9 Theorem Let a and b be real numbers and let L: IR" x IR"-+IR, M: IR x IR -+IR and D: IR x (IR \ {0})-+IR be defined by (i) L(x 1 , x 2 ) = ax 1 + bx 2 (ii) M(x 1 , x 2 )=x 1 x 2 (iii) D(xp x 2 ) = X 1 /x 2 • Then L, M and D are continuous on their domains of definition.

Proof (i) We have that llx 1 -Ydl;£11(xl, X2 )-(y 1 , y 2 )1_1 and IIX 2 -Y 2 II;£11(xp X2 )-(y 1 , Y2 )11 because

Hence IIL(X 1 , Xz)-L(yp Yz)ll=lla(xi-Y 1 )+b(xz-Yz)ll ;£lal·llx 1 -YIII+Ibl·llxz-Yzll ;i:{lal+lbl}-ll(xp X2 )-(Y 1 , Yz)ll2

Thus Lis continuous on IR " by lemma 16.8. (ii) We use lemma 16.8 with y = 1. Assume that ll(xp

X 2 ) - ( Y1,

Yz)ll < 1.

43

Continuity Then

and

In particular,

IYzl =I Yz -Xz +xzl~l Yz -xzl +lxzl < 1 + lxzl· Put C= 1 + lxil + lx 2 1. Then IM(xp Xz)- M(yp Yz)l = lxixz- YiYzl = lxixz -xi Yz +xi Yz- Yi Yzl ~ lxil·lxz- Yzl +I Yzl·lxi- Yil ~cll(xi, Xz)- (yi, Yz)ll. (iii) We use lemma 16.8. If (xi> x 2 )EIR x!R \ {0}, then x 2 ;60 and we take

y=!lx 2 1. Assume that Then

and

In particular,

IYzl =I Yz- Xz + Xzl ~ lxzl-1 Yz- Xzl >tlxzl· Put c=2lx 2 l- (lxil+lx 2 1). Then 2

I::- ~: I xiyz-xixz+xixz-YiXzl XzYz

I

<2lx 2 l- 2 {lxil·l Yz -xzl + lxzi"IYi -xi I}

and the result follows from lemma 16.8.

16.10 Corollary Let X be a metric space and letS eX. Ifji: S-+IR" and S -+IR" are continuous on the set S, then so is the function F:S -+IR"

f 2:

44

Continuity

defined by

F(x) = af1 (x) + bf2 (x). Proof This follows immediately from theorems 16.3 and 16.9 since F=LO (f1 ,f2 ).

16.11

f 2 : S-+lP!

Corollary Let .'l be a metric space and let S c .'l. If f 1 : S -+lP! and are continuous on the set S, then so is the function F: S-+lP! defined

by

F(x) = f 1 (x)f2 (x). ProofVve have that F=Mo (/1 , /2 ). Corollary Let .'l be a metric space and let Se .'l. Hf1 : S-+IP! and {0} are continuous on the set S, then so is the function F: S-+IP! defined by

16.12

f 2 : S-+IP! \

F(x)=f1 (x)/f 2 (x). Proof We have that F=D o (/1 , /2 ).

16.13

Rational functions

Suppose that P(xp x 2 , •• • , xn) is an expression obtained from the variables x 1 , x 2 , • •• , xn and a finite number of real constants using a finite number of multiplications and additions. Then we say that P: IP!n-+IP! is a polynomial. If Q: IP!n-+IP! is a polynomial and

Z={x: Q(x)=O} then the function R :IP!n \Z-+IP! defined by

R(x)=

~i:;

is called a rational function.

16.14

Example

ThefunctionP:IP! 2 -+IP! defined by P(x, y)=x 2 +3xy+x 2 y 3 + 1

is a polynomial. The function R: lP! 2

\

{

(0, 0)} -+lP! defined by

Continuity

45

is a rational function.

16.15 Theorem All rational functions are continuous on their domains of definition. Proof Constant functions and the identity function are continuous. So are the projection functions (corollary 16.6). The fact that rational functions are continuous therefore follows from corollaries 16.10, 16.11 and 16.12.

16.16

Exercise

(1) For each of the following functions/: IR -+IR, find non-empty contiguous sets A and B in IR, for which f(A) and f(B) are not contiguous. (i) f(x)={x

x-1

(ii) f(x)=g

(x~

1) (x> 1)

(xEO) (x~Q).

Deduce that in neither case is f continuous. (2) Let eEIR". Prove that the functionsf:IR"-+IR 1 and g:IR"-+IR 1 defined by (i) f(x) = llxll (ii) g(x) = <x, e) (see § 13.3) are continuous. (3) A function f: !R"-+!Rm is said to be linear if and only if f(r:xx+f3y)= r:x/(x) + f3f(y) for all real r:x and f3 and all x and y in IR ". Prove that a linear function / 1 : IR" -+IR 1 satisfies / 1 (x) = <x, e) for some eE!R". Hence show that any linear function/: IR"-+!Rm is continuous. (4) Let X be a metric space and let S be a non-empty set in X. Prove that the function f: X -+!R defined by f(x)=d(x, S) is continuous on X. [Hint: Begin by showing that ld(x, S)-d(y, S)l ~d(x, y) using exercise 13.24(5).] (5) Let X and ?J be metric spaces and letS c Tc X. If f: T -+'lj and g: T -+'lj satisfy f(x) = g(x) for each x E S, prove that g continuous on T implies f continuous on S. t(6) Prove theorem 16.9 with IR replaced everywhere by C.

46

Continuity

16.17t

Complex-valued functions

In the above discussion we have confined our attention to real polynomials and real rational functions. However, it is worth noting that all of the results of this chapter remain valid if IR1 is replaced throughout by C and the word 'real' replaced by 'complex'. No other change in the proofs is necessary.

17

17.1

CONNECTED SETS

Introduction

The characteristic feature of an interval in IR 1 is that it is 'all in one lump'. A little man standing on any point of an interval I would be able to walk to any other point of I without having to cross any points of the complement of I.

-----z,----~

• y

•

X

I

What sets in IR" have the same property? One can think, for example, of a set S in IR 2 as a country entirely surrounded by water. When does S consist of a single island and when does it consist of a whole archipelago of islands? Before trying to answer this question, we must of course first express it in precise mathematical terms.

17.2

Connected sets

A set S in a metric space X is connected if and only ifS cannot be split into two non-empty separated subsets. This means that if A and B are any two non-empty subsets of S satisfying AuB=S, then A and B are contiguous.

connected set

disconnected set

If one thinks of a connected set S in IR 2 as a country entirely surrounded by water, the definition says that, however one divides S into two provinces

47

48

Connected sets

A and B, these two provinces will be 'joined up'. A little man will be able to go from A to B without getting his feet wet.

17.3 Theorem A metric space X is connected if and only if the only sets in X which are both open and closed are (/) and X. Proof Observe that (i) AuB= X and AnB=(/)

=>

AuB= X and AnB=(/)

=>

B=eA AuB=X. AuB= X and AnB=(/) B=eA AnB=(/).

=>

(ii) AuB= X and AnB=(/)

=> => =>

Taking these two results together we obtain that AuB= X and AnB=(/) ~ B=B and B= eA ~

B is closed and A= eB.

It follows that X can be split into two disjoint, separated sets A and B if and only if A= eB and B is simultaneously open and closed. The theorem is an immediate consequence.

17.4

Corollary The metric space !Rn is connected. Proof See exercise 14.17(2).

17.5 Theorem Let W be a collection of connected sets in a metric space X all of which contain a common point ~- Then

T=U

S

is connected. Proof Let A and B be non-empty subsets of Tsuch that AuB= T. Either ~EA or ~EB. Suppose that ~EA. Then all of the sets SnA with SEW are non-empty. Suppose that SnB=(/) for all SEW Then

0=U SnB=BnU S=BnT=B SEW

SEW

because BeT. This is a contradiction because B#(/). Hence there exists an

49

Connected sets S 1 EW such that S1 nBi=0. Also S1 nAi=0 and

(S 1 nA)u(S 1 nB)=S 1 n(AuB)=S 1 n T=S 1 • Because S1 is connected, it follows that S1 nA and S1 nB are contiguous. Thus A and B are contiguous by theorem 15.15.

17.6

Connected sets in IR 1 An interval I in IR 1 is a set with the property that, if aE I and bE I,

then

a
a

b

I

It seems 'geometrically obvious' that a connected set in IR 1 is the same thing as an interval. However, the proof is rather more intricate than one would expect.

17.7

Theorem A set E in IR 1 is connected if and only if it is an interval.

Proof (i) Suppose E is connected. If aE E, bEE and a< c < b, we have to prove that cEE. If crj:E, then the sets A=En(-oo, c) and B=En(c, oo) are non-empty and AuB=E. But A and Bare separated by theorem 15.14. This is a contradiction and it follows that E is an interval. points in A

/1~

• •• • • G

• ••

= (--oo, c)

points in B c

A\

)I( • • •

• ••

H= (c, oo)

50

Connected sets

(ii) Suppose that E is an interval. Split E into two non-empty subsets A and B with AuB= E. We need to show that A and Bare contiguous. Let aE A and bE B. We shall assume that a;?, b. (If not, reverse the roles of A and B.) Let S=An[a, b] and T=Bn[a, b]. IfS and Tare contiguous, then so are A and B by theorem 15.15. Since E is an interval, [a, b] c E and thus SuT= [a, b]. In particular, c =sup S satisfies a;?, c;?, b and hence cE S or cE T. points of S a

•

/~\~ c !Ill 11111111•

Ill

b

•

E

By exercise 13.24(3), d(c, S) = 0. Hence, if cE T, then S and T are contiguous by theorem 15.12. If cljT, then cES. Moreover, because c is an upper bound of S, (c, b] is a non-empty subset of T. Thus d(c, T) = 0 and hence we arrive again at the conclusion that S and Tare contiguous.

17.8

Continuity and connected sets

17.9 Theorem Let X and ?J be metric spaces and let S c X. If f: S-+?J is continuous on the set S, then

S connected

=f(S) connected.

Proof Let C and D be non-empty subsets of f(S) such that CuD =f(S). We need to show that C and D are contiguous. If C and Dare separated, then so aref- 1 (C) andf- 1 (D) (§16.2). But this contradicts the hypothesis that S is connected because f- 1(C)uf- 1(D)= S and neither of the sets f- 1 (C) or f- 1 (D) is empty. f(S)

51

Connected sets

17.10 Theorem Suppose that I is an interval in IR and that f: 1-+IR is continuous on I. Then f(l) is an interval.

Proof This is an immediate consequence of the previous two theorems.

17.11 Corollary (Intermediate value theorem) Suppose that f:[a, b] -+IR is continuous on [a, b] and ..1. lies between f(a) and f(b). Then there exists a ~E[a, b] such that

f(b)

f(a)

----

b

a

17.12

Example Show that the equation 17x 7 -19x 5 -1 =0

has a solution

~

which satisfies -1 < ~ < 0. y

X

52

Connected sets Proof The function/: IR--+IR defined by f(x)= 17x 7 -19x 5 -1 is a

polynomial and hence is continuous on IR. Thus it is continuous on [ -1, 0] c!R. Now f( -1)= 1 andf(O)= -1. Since -1 <0< 1, it follows from the intermediate value theorem that f m= 0 for some ~ between - 1 and 0.

17.13 Example Letf:[a, b]--+[a, b] be continuous on [a, b]. Then, for some ~ E [a, b], f m=~. Thus a continuous function from [a, b] to [a, b] 'fixes' some point of [a, b].

Proof The image of [a, b] under f is a subset of [a, b]. Thus The function g: [a, b]--+IR defined by g(x)=f(x)-x is continuous on [a, b] by corollary 16.10. But g(a)~O and g(b)~O. By the

f(a)~a andf(b)~b.

intermediate value theorem, it follows that there exists a

~E[a,

b] such that

g(~)=O

17.14 Note The previous example is the one-dimensional version of Brouwer's fixed point theorem. This asserts that, if B is any closed ball in W andf:B--+B is continuous on B, then there exists a ~EB such that/(~)=~. Thus, water proof result

17.15

if a spherical tank full of water is shaken around, one tiny drop of will end up where it started. This is by no means 'obvious' and the in the general case is quite difficult. It is, however, a very important with many applications.

Curves

What is a curve? A physicist might say that a curve is the path

53

Connected sets

traced out by a particle as it changes its position with time. This idea leads us to the following definition. Let .i be any metric space. We shall say that a path in .i is a function f: [0, 1]-+ .i which is continuous on I= [0, 1]. One can think of x = f (t) as the position in .i of a particle at time t. The diagram illustrates the case .i=IRz. [(I)

We shall call the set f(I) a curve. From the intuitive point of view, this is not a very satisfactory definition since it classifies as curves all sorts of sets which do not look at all like curves 'ought to look'. Many mathematicians therefore prefer not to use the word 'curve' except with some qualifying adjective. If f (0) =a and f (1) = b, we shall say that the curve f (I) joins a and b. 17.16

Example The closed line segment in IR" given by

L={ta+(1-t)b:

O~t~

1}

is a curve which joins the points a and b. We have that L=f(I)wheref: I -+IR" is defined by f(t)=ta+ (1-t)b.

17.17

Theorem A curve in a metric space .i is connected. Proof This is an immediate consequence of theorems 17.7 and

17.9.

17.18

Pathwise connected sets

A pathwise connected set is a set with the property that each pair of points in the set can be joined by a curve which lies entirely in the set.

54

Connected sets

curve joining a and b

17.19 Theorem A pathwise connected set E in a metric space X is connected. Proof Let ;EE and let W be the collection of all curves which contain ; and are subsets of E. The hypothesis of the theorem ensures that

SE Ut

Hence E is connected by theorems 17.5 and 17.17.

I

I

I

I

I

I

I

I I

s3 / /

I

/

It may be helpful to think of E in the previous theorem as an island and of the paths emanating from ; as railway lines linking the capital ; to the rest of the island. The theorem supplies us with a useful means of generating examples of connected sets. But, before considering these examples, there is a natural question which needs to be answered. Are all connected sets pathwise connected? The diagram below illustrates an example due to Brouwer of a set E in IR! 2 which shows the answer to be negative. The set E consists of the union of a circle and a spiral which winds around the circle infinitely often. The

55

Connected sets

circle and the spiral are contiguous and so E is connected. But no curve lying within the set E joints the points a and b.

17.20

Example Consider the set S in lll 2 defined by

S= {(x, y): x~O}. For each point uES, it is true that the closed line segment joining 0 and u lies in S. Hence S is connected.

0

17.2tt

Components

17.22t Theorem Let C and Dbecontiguoussetsin a metric space .'L. IfC and Dare connected, then so is CuD. Proof Suppose that A and B are non-empty subsets of CuD such that AuB=CuD. We must show that A and Bare contiguous. If AnC# and BnC#, then AnC and BnC are contiguous because C is connected. Hence A and Bare contiguous by theorem 15.15. If BnC =, then either AnD# (in which case A and B are contiguous because D is connected) or else

56

Connected sets

AnD= 0 (in which case A= C and B = D and so A and B are contiguous by hypothesis). Similarly if AnC = 0.

Let E be a set in a metric space X. If l;e E, then the component of E containing I; is the largest connected subset S of E which contains I;- i.e. if U is the collection of all connected subsets of E containing 1;, then

S=U

T.

TeU

If one thinks of E as a country entirely surrounded by water, then it archipelago of which the constituent islands are the components of E.

IS

an

components of£

17.23t Theorem Any set E in a metric space X is the union of its components. Any two distinct components of E are separated. Proof The first sentence follows immediately from the definition because each l;e S determines a component. Of course, in general, many points will determine the same component. Suppose that C and D are distinct components. Let ye C. Then C is the largest connected subset of E containing y. But, if C and D are contiguous, then CvD is a connected subset of E larger than C. Hence C and D are separated.

17.24

Examples

(i) The components of the set E in lll 2 defined by E={(x, y): x~O or x> 1}

are

E 1 ={(x, y):

x~O}

and

E 2 ={(x,y):x>l}.

Connected sets

57

m, ....

(ii) The components of the set S={1jn: neN} in ~ 1 are the sets {1}, {t}, Each point of S therefore constitutes a separate component of S. We say that such a set is 'totally disconnected'. (iii) The components of the set T={O}uS are the sets {0}, {1}, {t}, Notice that any two distinct components are separated but that {0} is not separated from the union of the other components.

m, ....

17.25t

Structure of open sets in W

Theorem 14.13 asserts that the union of any collection of open sets is open. It follows that the set G=(O, 1)u(l,2)u(4, oo) is an open set in lll

0

1

•

I

2

4

We shall prove that all open sets in lll 1 are of this general type. We begin with the following theorem.

17.26t

Theorem Any component of an open set

Gin~·

is itself an open set.

Proof LetS be a component of G. Let ~ES. Since S c G, ~EG. Because G is open there exists an open ball B with centre~ such that B c G (theorem 14.12). Observe that it must also be true that BcS. Otherwise BuS would be a connected subset of G larger than the largest connected subset containing ~ (i.e. S). Thus S is open by theorem 14.12.

58

Connected sets

17.27t Theorem Any open set G in IR 1 is the union of a countable collection of disjoint open intervals. Proof We know from theorem 17.23 that fEW.

where W is the collection of components of G. These components are connected by definition and hence are intervals by theorem 17. 7. By theorem 17.26, the components are open. The components are therefore open intervals. Moreover, distinct components are separated. In particular, distinct components are disjoint. (See theorem 15.14.) The only thing left to prove is that W is a countable collection. This is very easy. From theorem 9.20, we know that each open interval contains a rational number. We may therefore construct a function f: w~o. which has the property that f(I)EI for each I E W.

f

~~

f

f(l)

The function f is injective because the intervals I in W are disjoint. Since countable, it follows from theorem 12.7 that W is countable.

17.28

Exercise

(1) Which of the following sets in IR 2 are connected?

(i) {(x,y):O<x~landO~yO or y~O}

Q

is

59

Connected sets

{(x, y): x~ 1 or x1} {(x,y):xy=1} (x) {(x, y): xy=O}.

(iii) (iv) (v) (vi) (vii) (viii) (ix)

(2) Prove that any convex set in IR" is connected. (See§ 13.14.) Deduce that open balls, closed boxes, lines, half-lines and hyperplanes are all connected sets in IR". [Hint: Use exercise 13.16(6).] Show also that halfspaces (exercise 14.17(9)) are connected. (3) Let I be an interval in IR 1 and suppose thatf: I -+IR 1 is an injection which is continuous on I. Prove thatfis either strictly decreasing on I or else strictly increasing on I. [Hint: Intermediate value theorem.] (4) Let S be the set in IR 2 defined by S={(x,y): 1~x 2 +y 2 ~2}. Give an example of a continuous function/: S-+S which has no fixed point. (5) The sets A and Bin IR 2 are defined by A= {(x, y): y 2 ~ 1}; B = {(x, y): x 2 ~ 1}.

Explain why a function f: IR 2 -+IR 2 cannot be continuous on the set A if f(A)=B. t(6) Let H be a hyperplane in lhl". Prove that lhl" \H has two components and that

these are the open half-spaces determined by H. (See exercise 14.17(9).) Explain why any line segment which joins two points from different half-spaces contains a point of H. (See §13.14.)

18

18.1

CLUSTER POINTS

Cluster points

A cluster point of a set E in a metric space

~

is a point

~E ~

such

that ~EE \ {~}.

A cluster point of a set E may or may not be an element of E. Equally, an element of E may or may not be a cluster point of E. Those elements of E which are not cluster points of E are called isolated points of E .

.({ - • S,.r

....----?isolated points

•

~

of£

cluster points of£

Cluster points are often referred to as 'points of accumulation' or as 'limit points' of the set. The latter usage is unfortunate and we prefer not to employ the word 'limit' except when discussing convergence. (See §26-.2.) The tables below list a number of equivalent assertions about cluster points and isolated points. The proofs of these equivalences provide useful practice in the techniques of chapters 14 and 15 and so are the subject of one of the problems in the next set of exercises.

60

61

Cluster points TABLE I

(i)

TABLE

~

is a cluster point of E. (ii) ~EE but is not an isolated point of E. (iii) d(~, E \{~})=0. (iv) Each open ball with centre ~ contains a point of E other

11

~ is an isolated point of E. (ii) ~E E but is not a cluster point of E.

(i)

(iii) ~EE and d(~, E \ g})>O. (iv) There exists an open ball containing no point of E except ~.

than~·

(v) Each open ball with centre ~ contains an infinite subset of E. To deduce lv from Iiv, consider a sequence of smaller and smaller open balls with centre~· Note that Iv implies that every point of a .finite set is an isolated point.

18.2

Examples

(i) The set S = {1/n: nE N} in~ 1 has a single cluster point, namely 0. Note that O~S. Every point of S is an isolated point of S. (ii) The set N in ~ 1 has no cluster points. All its points are isolated. (iii) The cluster points of the set Tin ~ 2 defined by

T={(x, y):x 2 + y 2 < 1}u{(2, 2)} are the points of {(x, y): x 2 + y 2 ~ 1}. The point (2, 2) is an isolated point of T. (2, 2)

T

18.3

Exercise

(1) Find the cluster points and the isolated points of the following sets in ~I

62

Cluster points (i) (0, 1) (ii) [0, 1) (iii) (- oo, 0] (iv) N (vi) IR (viii) {1, 2, 3} (ix) [0, 1]u{2} (vii) 0

(v) Q

(x) [0, 1]u(1, 2]

(2) Repeat the above question for the following sets in 111 2 • (i) (ii) (iii) (iv) (v) (vi) (vii)

A={(x,y): lxi+IYI~1} B=[(x, y): x~O} C={(x,y): x 2 +y 2 <1} D=N 2

E={(1/m, 1/n): mEN and nE N} F={(x, y): x 2 +(y-2) 2 ~ 1}u{(x, y): (x-2) 2 + y 2 ~ 1} G = {(x, y): x 2 < 1} u{ (x, y): y 2 > 1}.

(3) Prove the equivalences given in tables I and II of §18.1.

18.4t

Properties of cluster points

18.5t

Theorem Any isolated

point~

of a set E in IR" is a boundary point of E.

Proof By § 18.1 (IIiv), there exists an open ball B with centre ~ all of whose points are in e E except for ~E E. By theorem 14.4, ~is a boundary point of E. (Note that this theorem need not be true in a general metric space. For example, if X=E.)

18.6t Theorem Let E be a set in a metric space X. A boundary point 11 of E which is not a cluster point of E is an isolated point of E.

Proof If 11 is not a cluster point of E, then Iiv does not hold and hence there exists an open ball B with centre 11 containing no point of E except (possibly) 11· Since 11 is a boundary point, any open ball B with centre 11 contains a point of E. Hence fiE E. Thus 11 is an isolated point of E by § 18.1 (IIii).

18.7t

Theorem A set E in a metric space X is closed if and only if it contains

all its cluster points.

Proof (i) Suppose E is closed and

~

is a cluster point of E. Then

~EE \{~} cE=E.

(ii) Suppose that E contains all its cluster points. Since it contains all its isolated points (by definition), it follows from theorem 18.6 that E contains all its boundary points and hence is closed.

63

Cluster points

18.8t

Exercise

(1) Let E be a set in a metric space X and let r be a positive real number. If d(x,

y)~r

for each distinct pair of points x and y in E, prove that E has no cluster points. (2) Show that any set E in a metric space X with no cluster points is closed. [Hint: §2.10.] Show also that every point of E is isolated. Give an example of a closed set with a cluster point and a set with only isolated points which is not closed. Show that a closed set F in a metric space X whose elements are all isolated points has no cluster points. (3) Use the previous results to show that 1\J is a closed set in IR 1 and that any finite set in a metric space X is closed.

18.9t

The Cantor set In mathematics, an example which shows that unexpected things can happen is said to be pathological. The Cantor set is a magnificent instance of a pathological example in that it shows that lots of unexpected things can happen simultaneously. From the point of view of this chapter, the Cantor set is interesting because it is a perfect set. This means that it is a closed set in IR 1 with no isolated points. The fact that perfect sets exist is not particularly surprising. Any closed interval in IR 1 is a perfect set. However, the Cantor set is also totally disconnected (see example 17.24(ii)). This means that each of its components consists of just a single point. The Cantor set is constructed from the closed interval [0, 1] as follows. Delete from [0, 1] its open middle third (t, f). This leaves the set F 1 =[0, tJu[~, 1] which is closed, being the finite union of closed sets (theorem 14.14). Delete the open middle thirds of each of the closed intervals [0, tJ and [t, 1] which make up F 1 . This leaves the set F 2 =[0, ~]u[~, tJu[~, ~]u[~, 1]

which is closed and satisfies F 2 c F 1• Now delete the open middle thirds of each of these intervals and so on. We thus obtain a sequence (F.> of closed sets which satisfy FI=>F2=>F3=> ....

The Cantor set is defined by

Observe that F is closed by theorem 14.14. Obviously F contains all the end points of the closed intervals which make up the sets FP F 2 , F 3 , .••• Thus F contains all the points 0, 1,

t. f, t. ~. ~. ~. f7, ....

64

Cluster points ~

0

M'ff'£%%~~ I

0

9

2

9

!

~

3

3

~ffffff~ I

2 I

2

7

8

I

0 27 27 9

927273

~~

Wf.l

7

9 ~~

2 19 20

~

7

3 27 27 9

~~~I 9 27 27

~t:m

Er{~

~

At first sight it may seem that F contains only this countable set of rational numbers. But F is in fact an uncountable set. It contains not only the points listed above but all cluster points of the set of these points. Some of the properties of the Cantor set are listed below. The proofs are left for the next set of exerciies. (i) F is closed (ii) F is ·uncountable (iii) F is totally disconnected (iv) F is nowhere dense - i.e. F contains no open ball (v) F is perfect (vi) F has 'zero length'. The last property can be interpreted as meaning that, for each e > 0, intervals I 1, I 2 , ... exist such that F c I 1 ui 2 u ... and

where I(/k) denotes the length of the interval I k"

18.1 Ot

Exercise

(1) Let F be the Cantor set as constructed in §18.9. Prove that xEF if and only if

x=

I

n~ I

a. 3"

where each of the coefficients a. is 0 or 2. Use exercise 12.22(2) to deduce that F is uncountable. [Hint: Observe that the left-hand endpoints of the closed intervals which make up F N are the numbers of the form N

I~

n~ I

3"

where each a. is 0 or 2. Moreover, each of these closed intervals is of length 3- N and

L"'a...!'.::;; L" ' 2-=3-N.

n~N+I

3" -.~N+I 3"

J

(2) Prove assertions (i)- (vi) of§ 18.9. [Hint F = F c F.(n = 1, 2, ...).]

Cluster points

65

(3) By theorem 14.9 the complement of the Cantor set is a collection of disjoint open intervals. Obtain explicit formulae for the endpoints of these intervals and check directly that there is only a countable number of these. (See theorem 17.27.)

19

19.1

COMPACT SETS (I)

Introduction

Of the different types of interval in IR1 1, perhaps the most important in analysis are those of the form [a, b] = {x: a~x ~b].

Such intervals are said to be compact. We shall always assume in discussing a compact interval [a, b] that a~b -i.e. that compact intervals are nonempty. Why are compact intervals so important? This is not an easy question to answer without anticipating the theorems which we shall prove in this and later chapters. However, some readers will already know of the vital theorem which asserts that any continuous function f: [a, b] -+IR1 attains a maximum and a minimum value on [a, b]. This theorem is fundamental for a rigorous development of the differential and integral calculus. In this chapter we shall generalise the idea of a compact interval by introducing the notion of a compact set. In seeking such a generalisation, the natural way to begin is by asking: what distinguishes a compact interval from other intervals? The obvious answer is that an interval I is compact if and only if it is closed and bounded. Should we therefore define a set S in. a metric space X to be a closed and bounded set? The meaning of 'closed' was discussed in chapter 14 and it is easy to provide an appropriate definition of 'bounded'. We say that a set S in a metric space X is bounded if and only if there exists an open ball B of which S is a subset.

S bounded

66

Compact sets (I)

67

As we shall see, it would be adequate to define a compact set S to be a closed and bounded set provided that we stick very firmly to the space IR ". But this definition would not be of any use in more general metric spaces. The second, and more important reason from our point of view, is that it is NOT the fact that compact intervals are closed and bounded which makes them so useful. What makes them important is another more subtle property which we discuss in the next section. This will require some preliminary remarks on oriental culture.

19.2

Chinese boxes

The next theorem is named after the works of art produced by Chinese and Indian artists consisting of innumerable filligree boxes, one inside another, all carved from a single block of ivory. Recall from § 13.14 that a closed box in IR" is a set S of the form I 1 xI 2 x ... xI., where each of the sets Ik is a compact interval in IR 1 . The diagram illustrates a nested sequence of closed boxes in IR 2 •

A sequence (Sk) of sets is said to be nested if and only if

for each kEN.

19.3 Theorem (Chinese box theorem) Let (Sk) be a nested sequence of closed boxes in IR ". Then

Proof We begin with the proof for IR 1 . Suppose therefore that (lk) is a nested sequence of compact intervals Ik = [ak, bk]. Put A= {ak: kEN} and B= {bk: kEN}. Since (Ik) is nested, each element of B is an upper bound for A.

68

Compact sets (I)

Let ~=sup A. Then ak~~;;i:bk for each kEN. Hence ~Elk for all kEN. This proves the theorem in the case n = 1. Now let (Sk) be any nested sequence of closed boxes of the form sk =I k(1) X I k(2) X ... X I k(n) in IRl n. The first part of the proof demonstrates the existence of a real number ~ 1 E I k
... '

~n)EJk(1)

for each kEN. Hence there exists a

X

~ESk

Jk(2)

Jk(n)

X .•. X

for each kEN.

'"r

I L:~

2

I

f--Sk =I

......

,

L L

~.

J

19.4 Example We give another proof of the fact that IRl is uncountable using the Chinese box theorem. Suppose that [0, 1] is countable and that f: N -+[0, 1] is a surjection. Split [0, 1] into three congruent compact subintervals [0, H G. f], [~, 1]. At least one of these subintervals does not containf(1). Call this subinterval I 1 . Split the compact interval I 1 into three congruent compact subintervals in the same way. At least one of these subintervals does not contain /(2). Call this subinterval I 2 •

69

Compact sets (I) ~

f(J)

•

f(2)

~~g~l

In this way we construct inductively a nested sequence intervals such that

of compact

for each kEN. It follows that 00

f(N)c:U elk= k= I

00

en Ik. k= I

This contradicts the assumption that f: N ~ [0, 1] is a surjection because the intersection of the sequence contains at least one point of [0, 1] by the Chinese box theorem.

19.5

Compact sets and cluster points

The fact that compact intervals satisfy the Chinese box theorem is a far more significant fact than that they are closed and bounded sets. Given this fact, it is therefore sensible to seek to use the Chinese box theorem as the basis for the definition of a compact set in a general metric space .X. What we want is a definition of a compact set which ensures that any nested sequence
1\Kn n =I

There is no shortage of different (but equivalent) ways of defining a compact set in a metric space .X. We shall give two definitions.

70

Compact sets (/)

The first of these definitions is given below. It is not the more elegant of the two definitions: nor does it generalise in a satisfactory way to spaces other than metric spaces. But it is a definition whose significance is easy to understand and which is fairly easy to apply (given the material which we have covered so far in this book). The second definition of a compact set in a metric space X is given in the next chapter. This should be regarded as the 'proper definition' of a compact set. The two definitions are equivalent in a metric space and most of the next chapter is concerned with proving this. A set S in a metric space X is compact if and only if each infinite subset E of S has a cluster point ~ES.

points of E

Note that this condition is satisfied 'vacuously' in the case when S is finite. If expressed in formal terms, the condition is that 'l:f E c S (E infinite=>£ has a cluster point). But, as we know from §2.10, P=Q is true when P is false. Thus all finite sets are compact. Our first priority is to show that a compact interval in !Rl 1 is compact in the sense defined above. For this purpose we require the following important theorem. Theorem (Bolzano--Weierstrass theorem) 19.6 Every bounded infinite set E in IRl" has a cluster point. Proof Since E is bounded, it lies in some closed box S (exercise 13.16(5)). The box Scan be covered by a finite number of sub-boxes each of whose dimensions are half that of the original box S. Since E is infinite, at least one of those sub-boxes contains an infinite subset E 1 of E. Let S 1 be the sub-box containing E 1.

Compact sets (I)

71

We can now repeat the process with S 1 replacing S and E 1 replacing E. Using an appropriate inductive argument we obtain a nested sequence (Sk) of closed boxes each of which contains an infinite subset of E. By the Chinese box theorem, there exists a ~ which belongs to each of these boxes. Let B denote any open ball with centre~· Since the dimensions of Sk are 2 -k times those of S, a clos.ed box S K will be a subset of B provided that K is sufficiently large. Thus B contains an infinite subset of E and so ~ is a cluster point of E (§18.1(1v)).

sk

•

•

•

•

• •

•

• • •• •

• • • •

•

•

•

B ~

• • •

•

•

•

•

19.7 Note It is of great importance to remember that the BolzanoWeierstrass theorem (and hence the Heine-Borel theorem which follows) are FALSE in a general metric space X. This point is taken up again in chapter 20. 19.8 Theorem (Heine-Borel theorem) A set K in IR" is compact if and only if it is closed and bounded. Proof The proof that a compact set in IR" is closed and bounded is quite easy and, since we prove the same result for a general metric space later on (theorems 20.12 and 20.13), we shall therefore consider only the deeper part of the proof- i.e. that a closed, bounded set in IR" is compact. Suppose that K is a closed, bounded set in IR" and let E be an infinite subset of K. From the Bolzano-Weierstrass theorem it follows that E has a cluster point~· By theorem 18.7, the fact that K is closed implies that ~EK. Thus K is compact.

19.9 Theorem (Cantor intersection theorem) Let (Fn) be a nested sequence of non-empty closed subsets of a compact set K in a metric

72

Compact sets (I)

space X. Then

Proof If one of the sets Fn is finite, the result is trivial. Otherwise we can construct an infinite subset E of K consisting of one point from each of the sets Fn·

Since E is an infinite subset of a compact set K it has a cluster point ~- Since all but a finite number of points of E belong to each Fn, ~must be a cluster point of each of the sets Fn. But each Fn is closed. Hence, by theorem 18.7, ~EFn for each nE N. Thus

19.10 Example The sequence (Fn) of sets constructed in §18.9 is a nested sequence of non-empty closed subsets of the compact set [0, 1]. By the Cantor intersection theorem, its intersection F is non-empty. In fact, as is explained in § 18.9, F is uncountable.

19.11

Exercise

(1) Which of the sets given in exercise 18.3(1) are compact? (2) Which of the sets given in exercise 18.3(2) are compact?

73

Compact sets (/)

(3) Give examples of a nested sequence of non-empty closed sets in IR 1 and a nested sequence of non-empty bounded sets in IR 1 which have empty intersections.

19.12

Compactness and continuity

19.13 Theorem Let .'l and ?J be metric spaces and let S c .'l. If f: S -?J is continuous on the set S, then

S compact=> f(S) compact. Proof Suppose that S is compact. Let E be an infinite subset of f(S). We need to show that E has a cluster point in f(S). Define g :f(S)-s so thatf(g(y)) =y (see example 6.9). Then D = g(E) is an infinite subset of S. Since S is compact, it follows that D has a cluster point ~ES- i.e. d(~, D \{~})=0. Now let 11 = f@. Then 11Ef(S). Also since f is continuous on S, d(~, D \{~})=0 =>d(11,f(D \{~}))=0

by theorem 16.4. But f(D \

{~}) =

E \ {11} and so 11 is a cluster point of E.

The next theorem is of fundamental importance in optimisation theory. It asserts that any real-valued continuous function achieves a maximum and minimum value on a compact set K. It is worth noting that this theorem need not be true if K is not compact. Consider, for example, the function f: IR -~R defined by f(x)=x. This does not achieve a maximum nor does it achieve a minimum on the open set (0, 1).

74

Compact sets (I)

19.14 Theorem Suppose that K is a non-empty compact set in a metric space X and that f: K -.jR is continuous on the set K. Then f achieves a maximum and a minimum value on the set K - i.e. there exists ~E K and 'lE K such that, for any xE K, !@ ~f(x) ~(1]).

K

Proof By theorem 19.13,/(K) is compact. A compact set in IR 1 is closed and bounded by the Heine-Borel theorem (19.8). Because f(K) is a closed, non-empty set of real numbers which is bounded above it has a maximum (corollary 14.11). For siPlilar reasons,f(K) has a minimum.

19.15 Corollary Suppose that K is a non-empty compact set in a metric space X and that yE X. Then there exists a ~E K such that d(y, K) = d(y,

~).

Compact sets (/)

75 y

Proof The function/: X--+IR defined by f(x) =d(y, x) is continuous on X by exercise 16.16(4). Hence it achieves a minimum on K by theorem 19.14.

19.16 Corollary Suppose that F is a non-empty closed set in IR" and that yEIR". Then there exists a C:,EF such that d(y, F)=d(y, C:,).

Proof Let li be the closed ball with centre y and radius d(y, F)+ 1. Then linF is a non-empty, closed and bounded set in IR". It follows from the Heine-Borel theorem (19.8) that linF is compact. The result therefore follows from corollary 19.15. (Note that the result does not hold in a general metric space.)

19.17

Exercise

(1) The sets A and Bin IR 2 are defined by

A= {(x, y): lxl +IYI;;;; 1}; 2

B={(x, y):O<x;;;; 1}. 2

Explain why a function f: IR ->IR cannot be continuous on A if f(A)=B. (2) Let I be an interval in IR 1 and let f: IR 1 --->IR 1 be continuous on I. Give counter-examples to each of the following (false) propositions: (i) I closed => f(l) closed (ii) I closed => f(I) bounded (iii) I open => f(I) open (iv) I bounded => f(I) bounded (v) I open and bounded => f(l) has no maximum. (3) Which of the propositions above are necessarily true whenf: IR 1 --+IR 1 is continuous on a compact interval J containing I. t(4) The distance d(S, T) between two non-empty sets Sand Tin a metric space X

is defined by d(S, T)=inf {d(x, y):xES and yET}.

76

Compact sets (/)

Show that, if S and Tare compact, then there exist seS and te T such that d(S, T)=d(s,t). Show that the same result holds when X =IJ;l"ifSiscompact but Tis only closed. Give examples of closed sets Sand Tin IJ;l 2 for which the result is false. t(5) Let Sc:IJ;l and suppose that/: S-+IJ;l is continuous on the setS. The graph off is the set Tin IJ;l 2 defined by T={(x, f(x):xeS}. Prove that (i) S connected => T connected (ii) S compact => T compact. Give examples of non-continuous functions f: S-+IJ;l for which (i) and (ii) are true. t( 6) Let f: IJ;l -+IJ;l be a bounded function whose graph is a closed set in IJ;l 2 • Prove that f is continuous on IJ;l,

20t

20.1 t

COMPACT SETS (11)

Introduction

In this chapter we give the 'proper definition' of a compact set (see§ 19.5). Most of the chapter is then concerned with proving that this definition is equivalent to that of the previous chapter for a metric space .'L. This is quite a lengthy piece of work and many readers will prefer to skip this chapter for the moment. The book has been written with this possibility in mind.

20.2t

Open coverings A collection U of sets is said to cover a set E if and only if

EcU S. Seil

,.,.------- ......... I

/

/

SI

'

I

I \

' t

\

I

' -----< ' --- / /

20.3

Example Let o: [0, 1]-->(0, oo). Then the collection U={(x-o(x), x+o(x)): xe[O, 1]}

covers the set E = [0, 1].

77

78

Compact sets (H)

20.4t

Compact sets A set K in a metric space X is said to be compact if any collection U of open sets which covers K has a finite subcollection Y c U which covers K.

20.5

Example Suppose the function b :[0, 1] -+(0, oo) of example 20.3 is given

by

x+1

b(x)=- . x+ 2

Then a finite subcollection Y of U which covers [0, 1] is fj- {( _.!. 2• .!.) 2, (.!. 3• .2.)} 3

- i.e. one needs only the intervals (x- b(x ), x + b(x )) with x = 0 and x = 1.

20.6t

Exercise

(1) A function f: IR1-+IR1 has the property that, for each xE [0, 1], there exists a b > 0 such that f is either increasing or else decreasing on (x- b, x +b). Prove that f is either increasing or decreasing on the whole interval [0, 1]. (Use the definition of §20.4 but assume the Heine-Borel theorem.) (2) Prove that any closed subset of a compact set in a metric space X is compact. (This may be deduced fairly easily from both of our definitions of compactness.) (3) A collection V of sets has the finite intersection property if and only if the fact that each finite subcollection of V has a non-empty intersection implies that V has a non-empty intersection. Prove that any collection of closed subsets of a compact set K in a metric space X has the finite intersection property. Deduce the Cantor intersection theorem (19.9).

20.7t

Compactness in 1R"

In this section we shall show that both our definitions for compactness are equivalent in the space !RI". This will involve proving a number of interesting theorems en route. The logical scheme we shall use is illustrated on p. 79. In this section we shall prove only those implications marked with a single line. The other implications are either obvious or have been proved already. (The fact that the Bolzano-Weierstrass property implies the Chinese box property, for example, is the substance of the proof of the Cantor intersection theorem (19.9).) The implications indicated with firm lines will be proved for any metric space X. The implications indicated with broken lines will be established only for !RI". As we know, the Bolzano-Weierstrass theorem is false in a general metric space. This is

Compact sets (Il)

79

Kis compact

... I

1 Lindelof's 1 theorem

I Kis closed

K has the Chinese box property

Kis countably compact

K has the Kis BolzanoWeierstrass bounded property .___ _ _ __. Bolzano- ' - - - - - - - ' Wei ers trass theorem

-f--

also true of Lindelof's theorem. However, in §20.16 we shall see that the implication labelled with Lindelof's theorem can be proved in any metric space using a somewhat more complicated method. Three of the terms in the diagram are undefined as yet. We therefore begin with definitions of these terms. A set K in a metric space X has the Bo/zano--Weierstrass property if and only if each infinite subset E of K has a cluster point 1;. A set K in a metric space X has the Chinese box property if and only if any nested sequence (Fk) of closed, non-empty subsets of K has a non-empty intersection- i.e.

The proof that we gave of the Cantor intersection theorem (19.9) shows that any set K with the Bolzano-Weierstrass property has the Chinese box property. A set K in a metric space X is said to be countab/y compact if and only if each countable collection U of open sets which covers K has a finite subcovering j which covers K. This definition is, of course, exactly the same as that for a compact set (§20.4) except for the insertion of the word 'countable'. It is therefore not particularly surprising that a compact set and a countably compact set are precisely the same thing in a metric space.

80

Compact sets (I I)

It is obvious that a compact set K in a metric space X is countably compact. For the converse implication, we need it to be true that every collection U of open sets which covers K has a countable subcollection which covers K. A set K for which this is true will be said to have the Lindelof property. Let E be a set in a metric space X. We say that D is dense in E if and only if E c i5 - i.e. for each xe E,

d(x, D)=O.

•

Example The set IQ is dense in IR 1 • This follows from theorem 9.20. Hence Q" is dense in IR" by exercise 13.24(6).

20.8

20.9t

Theorem Let E be a set in a metric space X. Then E has the Lindelof property provided that there exists a countable set D which is dense in E. Proof Let U be any collection of open sets which covers E. We must show that a countable subcollection covers E. Let $ denote the collection of all open balls with centres in D and rational radii which are subsets of at least one of the sets GeU. Then $ is countable. (Consider D x IQ and apply theorem 12.10.)

/ BE$

(

/

,.--- ....... ' GE U

~\ ~ I d ~ ~~/ J '~ //

/I / - rational -r---radius

///_,/ {I '

---- -

' ..... _

'

\

/

//

Bo

/

B

81

Compact sets (Il) If GEU, then each ~EG lies in an open ball BE$. Hence

EcU GcU B. GeU

Befll

The first sentence of this paragraph is proved in the following way. Since G is open, there exists an open ball B 0 with centre~ and radius r>O such that B0 cG (theorem 14.12). Since D is dense in E, there exists a dED such that d(~, d)
d)+d(d, y)
and so yEB 0 cG. Now let f: 51-+U be chosen so that Bcf(B). Then f($) is a countable subcollection of U and EcU BcU .f(B)= U Beffi

Beffi

G.

Gef(fll)

.,.,---... ....... I

Bo~----/

I

/ ""

I I I

.......

I

\

I I

\

)

I I

\

/

' ' ....... B

---

I

'\

[(B)

\

\

I

I

I

I

--1-"' ....-?-.... , II /I~ \ I I~"'~ )-"'r ( B I "1--.....

I I

\ \

/

I

-.L.-.~.

//

\

/

~

,_..,.... / /

\

I

\

....--\ J

/I

I

' ......___ ,..,../ ........x.._

'........

/

---

other sets GEU which contain B

We have shown that f($) is a countable subcollection of U which covers E.

20.10t

Corollary (Lindelof's theorem) Any set E in

~·

has the Lindelof

property.

~·

Proof The set IQ" is countable by exercise 12.13(3). Since IQ" is dense in (example 20.8), IQ" is dense in E. The result therefore follows from theorem 20.9.

20.llt

Corollary Any countably compact set K in

~·

is compact.

82

Compact sets (II) Proof This follows immediately from Lindelof's theorem.

20.12t

Theorem Any countably compact se~ K in a metric space X is bounded. Proof Let

~

be any point of X and let U be the nested collection of open

balls of the form

{x: d(~, x)
20.13t

Theorem Any countably compact set K in a metric space X is closed.

Proof Let ~ be any point not in K and let U be the nested collection of all open sets of the form {x:d(~,x)>l/n}

where nE N. Then U is countable and covers K. Since K is countably compact, a finite subcollection fi of U covers K. Let B = X \ S be the ball of minimum radius such that SE·fi. Then B contains no point of K. Hence ~ is not a cluster point of K. Thus K is closed.

20.14t Theorem Any closed set K with the Chinese box property in a metric space X is countably compact.

Proof Let U be a countable collection of open sets with the property that no finite subcollection covers K. We shall deduce that U does not cover K. Thus K is countably compact. Let the sets in U be the terms of the sequence Gk>· Since {G1 , G2 , ••• , Gk} does not cover K,

<

k

k

Fk=KneU Gi=U (KneGk) j= I

is non-empty.

j=l

83

Compact sets (Il)

But
00

0-td1

Fk=Kr~eU Gi

k=l

j=l

and so U does not cover K.

This theorem completes the proof of the implications of the diagram of §20.7.

20.15t

Completeness In§ 19.5 we took note of the fact that the Bolzano-Weierstrass theorem is

not true in a general metric space X. The same is therefore true of the Heine-Borel theorem. Indeed, it can be shown that if the 'sphere' S = {x: llxll = 1} in a normed vector space X is compact, then X is necessarily finite-dimensional. Since S is always

closed and bounded, this shows that the Heine-Borel theorem is always false in an infinite-dimensional normed vector space. Why does the Bolzano-Weierstrass theorem fail in a general metric space? There are two reasons. The first is that it may be impossible to prove a suitable analogue of the Chinese box theorem. When no such analogue exists there is no point in seeking to obtain some substitute for the Bolzano-Weierstrass theorem. The deficiency is too severe. However, there is a second reason why the BolzanoWeierstrass theorem may fail in a general metric space X. Observe that, in the proof of the Bolzano-Weierstrass theorem given in§ 19.5, it is necessary that, given any 1>0, it be possible to cover the bounded set E with a finite number of closed boxes whose sides are all of length at most I. One can then deduce from the fact that E is infinite, the conclusion that at least one of the boxes contains an infinite subset of E. If one needed an infinite collection of such boxes to cover E, the proof would fail. This observation leads us to the following definition. We say that a set S in a metric space X is totally bounded if and only if, given any r > 0, there exists a: finite collection 5, of open balls of radius r>O which covers S.

/

.....

--

....

/

/

I

I

I I I

\

' .... __ _

I

/

It is obvious that any totally bounded set S in a metric space X is bounded. (If 51

84

Compact sets (II)

contains N elements, S is a subset of any open ball B with radius 2N provided that the centre of B lies in S.) In IR" the converse is also true because, by the Heine-Borel theorem, ifS is bounded, then S is compact. But S is covered by the collection of all open balls of radius r. Hence Sand therefore S is covered by a finite number of open balls of radius r. In a general metric space X, however, a bounded set certainly need not be totally bounded (see §20.20). The second reason we gave for the failure of the Bolzano-Weierstrass theorem in a general metric space X clearly disappears if we replace the word 'bounded' in its statement by the words 'totally bounded'. These considerations lead us to the notion of a complete metric space. The usual definition is given in chapter 27, but this definition is equivalent to the assertion that, in a complete metric space X, every totally bounded set has the BolzanoWeierstrass property. Thus a complete metric space is one in which a suitably amended version of the Bolzano-Weierstrass theorem is true. A large number of important metric spaces are complete and so this is an important idea. An example is given in § 20.20.

20.16t

Compactness in general metric spaces The diagram below summarises the results of this chapter.

K has the Lindelof property

Kis compact

..

~

and ~

Kis closed

7 There exists a countable set which is dense inK

Kis countably compact

A

" and

K has the Chinese box property

...

K has the BolzanoWeierstrass property

......

______

.;------

Kis totally bounded

" "

~---

Kis bounded

We shall prove in this section only those implications indicated with a single firm line. Other implications are either obvious or else have been proved already. The meaning of the different styles of arrows representing implications in the diagram is explained in the following table.

Compact sets (Il)

--

85

True in any metric space

Not yet proved

=====.l

True in a complete metric space

Proved already

---~

True in !Rn

-

~

.

Theorem A set S in a metric space X which has the Weierstrass property is totally bounded.

20.17t

Bolzano~

Proof IfS is not totally bounded, then for some r > 0 no finite collection of open balls of radius r covers S. If x 1 , x 2 , ••• , x._ 1 are points of S one can therefore always find a point x.es such that d(x., x.J~r (k= 1, 2, ... , n-1).

/' /

I

I I

\

''

I

I I I

\

' ' .... __ _

The setS therefore contains an infinite subset with no cluster point (exercise 18.8(1).)

20.18t Theorem IfS is a totally bounded set in a metric space X, there exists a countable set D which is dense in S.

Proof Let c. denote the set of centres of a finite collection of open balls of radius 1/n which covers S. Then given any xeS, d(x, C.)< 1/n. Now let

Then D is countable because it is the countable union of countable sets (theorem 12.12) and D is dense in S because, for each xeS, d(x,

D)~ d(x,

C.)< 1/n.

86

Compact sets (Il)

Since this is true for all ne N, d(x, D)= 0.

20.19t

Theorem Any countably compact set K in a metric space X has the Bolzano-Weierstrass property.

Proof Let E be any infinite subset of K. If E has no cluster point, this is also true of any countable subset F of E. By exercise 18.8(2), F is closed and all of its points are isolated. points ofF

For each xeF we can find an open ball which contains no point of F except x (§18.l(IIiv)). The collection of all such open balls together with the open set eF is countable and covers K. But no finite subcollection covers K.

This theorem completes the proof of the validity of the implications indicated in the diagram of §20.16.

20.20t

A spherical cube

In this section, we give one of our rare examples of a metric space other than IR". We begin by observing that one can think of a sequence <xk) of real numbers as the 'infinite-dimensional' vector

in which the co-ordinates of the vector x are just the terms of the sequence <xk>· Vector addition and scalar multiplication may then be defined just as we did in § 13.1 for vectors in IR". We shall consider the normed vector space 1co consisting of all bounded sequences of real numbers. The norm on 1co is defined by llxll =sup lxkl· kel\l

The distance between points x and y of 1co is therefore given by d(x, y)=llx-yll=sup lxk- Ykl· kel\l

87

Compact sets (Il)

The first thing to notice is that the proof that we ga vefor the Chinese box theorem (19.3) works equally well for I 00 • The same is therefore true of the Bolzano- Weierstrass theorem provided that 'bounded' is replaced by 'totally bounded'. Thus 100 is a complete metric space. Consider the setS= {x: llxll;:;; 1} in 100 • This can also be written in the form S = {x : lxk I;:;; 1 for each kE 1\J}.

The first expression would describe a ball in IR 3 and the second would describe a cube in IR 3 . Hence the title of this section. The set S is clearly a closed and bounded set in 100 • But it satisfies none of the other properties listed in the diagram of §20.16. This is most easily shown by proving that the subset V of S consisting of all its 'corners' does not satisfy the Lindelof property. The set V is the set of all sequences each of whose terms is either 1 or -1. It is therefore an uncountable set (exercise 12.22(2)). The fact that V does not have the Lindelof property simply follows from the fact that the collection U of all open balls with centres in V and of radius 1 is uncountable and covers V but has no proper subcover at all. Note in particular that S is an example of a closed, bounded set which is not compact. For an example of a non-trivial compact set in a space other than IR" see exercise 20.21(6).

20.2lt

Exercise

(1) Prove that 100 as described in §20.20 is a metric space. (2) The normed vector space 12 consists of all sequences x = <xk) of real numbers for which

converges. An inner product is defined on 12 by Cf.)

<x, Y) =

I

XkYk·

k=l

Prove a version of the Cauchy-Schwarz inequality (13.6) and deduce that 12 is a metric space if distance is defined in the usual way for a normed vector space (§ 13.18). (3) A closed box in the set of all sequences of real numbers is a set S of the form S=J 1 xJ 2 xJ 3 x ... where
88

Compact sets (II)

(5) Prove that the set S = { x: llxll ~ 1} is closed and bounded in F but not totally bounded. Deduce that S is not compact in F. [Hint: Look at the set of all points of the form (0, 0, ... , 0, 1, 0, 0, ...).] (6) Prove that the 'closed box' S = { x: lxkl ~ 1/k for each keN} is closed and totally bounded in 12 • [Hint: Given r>O, choose n so that

[ L00 x/]''2
and then use the fact that closed boxes are totally bounded in S is compact in F.

~·.]

Deduce that

21

21.1

TOPOLOGY

Topological equivalence

Suppose that X and 'IJ. are metric spaces and thatf: X-+ 'IJ. is a bijection with the property that bothf: X-+'IJ. andf- 1: 'IJ.-+X are continuous. Such a function is called a homeomorphism. (This term is easily confused with the term homomorphism which means something quite different.) If a homeomorphismf: X-+'IJ. exists, we say that X and 'IJ. are topologically equivalent. The subject of topology is concerned with studying the properties that topologically equivalent spaces have in common. Consider, for example, a teacup made from modelling clay. This can be continuously deformed into a doughnut and the process can then be reversed.

teacup

doughnut

The doughnut and the teacup (regarded as metric subspaces of ~ 3 ) are therefore topologically equivalent. The topological property that they have

ball

89

doughnut

90

Topology

in common is that they both have precisely one hole. Note that a ball is not topologically equivalent to a doughnut. To deform a ball into a doughnut would require making a hole at some stage and this cannot be done without pulling neighbouring pieces of modelling clay apart. But such a deformation is not continuous.

21.2

Maps

The diagram below is a topological map of the mainland of Europe. Notice, for example, that Italy looks nothing like the familiar Wellington boot. This is because quantities like distance and angle are not invariant under homeomorphism. However, contiguity is preserved. The diagram therefore indicates which nations have a common boundary and which do not. A tourist planning a trip by road might find the diagram useful in deciding what visas he will require but he would be unwise to use it in estimating how long the journey will take.

I

Denmark Poland West Germany

Sweden

Norway Czechoslovakia USSR

h Austria France

D

~ Spain

n

Italy

n

D

I

Greece

I

Schematic railway maps and wiring diagrams are more familiar examples of the same idea. The point is that the topological properties of a space are those which specify how the space is 'fitted together' or 'connected up'. From the topological point of view, all other properties of the space are irrelevant. Perhaps the most famous example of a topological property is that which arises in the 'four colour problem'. Suppose that one is asked to colour the map above in such a way that no nations with a common boundary are coloured the same. How many colours will be needed? More generally, what is the minimum number of colours which will suffice for all such

91

Topology

maps? (One must, of course, exclude cases where four or more nations meet at a point like the states of Arizona, Colorado, New Mexico and Utah.) The answer for maps drawn on the surface of a doughnut is seven. This is relatively easy to prove and has been known for a long time. Curiously, the same problem for maps drawn on a plane is very much more difficult. Only recently has it been shown that the answer is four. The proof (by Haken and Appel) has an extra element of interest in that it is the first computer assisted proof of a result of any substance.

21.3

Homeomorphisms between intervals

We now study the topological properties of intervals (regarded as metric subspaces of 1Pl 1 ). If an interval has at least two points it falls into one of only three topologically distinct types. (See exercise 21.5(3).) The intervals ( - 1, 1), ( - 1, 1] and [- 1, 1] are representatives of each oft he three types. In this section we shall explain why ( -1, 1) is not topologically equivalent to ( -1, 1] but is topologically equivalent to ( -1, 1] but is topologically equivalent to !Pl. Suppose thatf: (-1, 1]--+(-1, 1) is a bijection. Let/(1)=~. Then f(( -1, 1)) = ( -1, ~)u(~, 1). Hence f cannot be continuous since it maps a connected set onto a disconnected set. It follows that ( -1, 1] and ( -1, 1) are not topologically equivalent. Consider the function/: ( -1, 1)--+!Pl defined by X

f(x)=-1-z·

-x

This is a homeomorphism between ( -1, 1) and !Pl. (For a proof of this result which avoids the need for any calculation, see theorem 22.22.) It follows that ( -1, 1) and IPl are topologically equivalent.

-I

11 I I

I I

I I I I I I

92

Topology

21.4

Circles and spheres

It is easy to see that a circle C and a line segment L (regarded as metric subs paces of IR 2 ) cannot be topologically equivalent. If f: C-+ L is a continuous bijection, then L is compact because the same is true of C (theorem 19.13). Hence L contains its endpoints A and B. But f- 1 : L-+C then maps the connected set L \{A, B} onto the disconnected set C \ {A', B'} and hence is not continuous. B

However, if just one point is removed from C we obtain a set which is topologically equivalent to a line. The homeomorphism which is usually used to show this is called the stereographic projection. This is illustrated in the diagram below. The point deleted from C (labelled N in the diagram) is referred to as the 'north pole' of the projection. N

c

X

Similarly, a 'punctured sphere' is IR 3 (i.e. a sphere with one point removed) is topologically equivalent to a plane. Again, the stereographic projection provides a suitable homeomorphism.

93

Topology

Let S be the sphere in IR 3 with centre (0, 0, t) and radius t. Take as the 'north pole' the point N=(O, 0, 1) and let P be the plane {(x, y, z): z=O}. Then the stereo graphic projection f: P -+S \ [ N} is given by xz+y2 ) x y f(x, y, 0)= ( 1+x2+y2' 1+x2+y2' 1+x2+y2

and f- 1 : S \ {N}-+P is given by

f

_ 1

(X

Y

(X, Y, Z)= 1-Z' l-Z'

~

Of

Both functions are continuous by theorems 16.5 and 16.15.

21.5

Exercise

(1) Prove that topological equivalence as defined in §21.1 is an equivalence relation as defined in § 5.5. (2) Lf't S be any set in IRn which is topologically equivalent to a closed ball B. Assuming Brouwer's fixed point theorem as quoted in note 17.14, show that for any function/: S-.S which is continuous on S there exists a ~E S such that/(~)=~· (3) Let J denote the collection of intervals in IR 1 which have at least two points. Let U denote the collection of all open intervals in 5 , let V denote the co1lection of all compact intervals in J and let W denote the remaining intervals in 3 . Show that U , V and W are the equivalence classes into which 3 is split by the relation of topological equivalence. (4) Let L be a closed line segment in IR 2 and let S be the closed box [0, 1] x [0, 1]. Prove that the removal of any three points from L produces a disconnected set but that the same is not true of S. Deduce that L and S are not topologically equivalent. t(5) Check the formula given in §21.4 for the stereographic projection between the plane P and the punctured sphereS\ {N}. t(6) A function F: C -.s \ {N} is defined by

Ylz 5mz lzl 2 ) 2 2 ( F(z)= 1+lzl ' 1 +lzl ' 1 +W

·

Prove that F is a homeomorphism between the complex plane C and the punctured sphere S \ [N} (where S \ [N} is regarded as a metric subs pace of IR 3 ). Show that

94

Topology

21.6

Continuous functions and open sets

21.7 Theorem Let X and ?J be metric spaces and letf: X--+?J. Then the following statements are all equivalent. (i) f is continuous on X. (ii) For any XE X and any S c X, d(x, S) =0 = d(f(x),f(S))=O. (iii) For any S c X, f(S)cf(S). (iv) For any F c ?J, F closed = f- 1 (F) closed.

(v) For any G c ?J, G open =

f- 1 (G) open.

Proof We shall show that (i)=(ii)=(iii)=(iv)=(v)=(i). (i)=(ii). This is just theorem 16.4. (ii) = (iii). Assume (ii) and suppose that yEj(S). Then y=f(x) where xES. Since d(x, S) = 0 it follows that d(f(x),f(S)) = 0. Thus y = f(x)Ej(S). (iii)=(iv). Assume (iii) and suppose that F c ?J is closed. Since f(/- 1 (F))cF, /(/-

1

(F))cf(f

1

(F))cF=F.

Hencef- 1 (F) cf- 1 (F). Thereforef- 1 (F) is closed. (iv)=(v). Assume (iv) and suppose that G c ?J is open. Then and hence

eG is closed

is closed. Thusf- 1 (G) is open. (v)=(i). Assume (v) and suppose that C and D are separated sets in ?J. Then disjoint, open sets G and H exist with Cc G and D c H by theorem 15.16. The setsf- 1 (G) andf- 1 (H) are then disjoint, open sets in X by (v). But thenf- 1 (C)cf- 1 (G) andf- 1 (D)cf- 1 (H) are separated sets in X. Thus C and D separated implies f- 1 (C) and f- 1 (D) separated. Hence f is continuous (§16.2).

The most significant of the equivalences in the above theorem is that which asserts that a function f: X -+?,I is continuous if and only if, for each

95

Topology

G c ?J, G open implies f- 1 (G) open. The next few sections are devoted to exploring the consequences of this result.

21.8

Topologies

As we know from the previous section, two metric spaces X and ?J are topologically equivalent if and only if there exists a bijection f: X ~?J such that, for each G c ?J,

G open- f- 1 (G) open.

(1)

Thus, if we wish to determine whether or not X and ?J are topologically equivalent, the only question we need to ask about X and ?J is: what are their open sets? Any bijection f: X ~?J can then be tested to see whether it satisfies (1). Other information about X and ?J may be helpful or interesting but it is not strictly necessary. Thus we know all that there is to know about the topological structure of a space if we have a list of all its open sets. For this reason the collection of all open sets in a space is called its topology.

21.9t

Relative topologies

Notice that theorem 21.7 refers to a function/: X-+?J where X and ?J are metric spaces. Various equivalent conditions are then given for f to be continuous on the whole space X. In contrast, the theorems of chapter 16 always involved a subset Z of X and were concerned with the continuity of a function f: Z -+?j on the set Z. If Z is a subset of X, then it is, of course, true that Z is itself a metric space provided that we use the same definition for distance in Z as is used in X. We say that Z is a metric subspace of X.

z

In chapter 16, we carefully chose a definition for a function f: -+?j to be continuous on the set Z which makes it quite irrelevant whether we regard our

96

Topology

underlying metric space to be .'L or whether we simply throw out all the points of .'L which are not in f and take our underlying metric space to be f. This is because two subsets A and B off are contiguous in ·the metric space .'L if and only if they are contiguous in the metric space f. It follows that theorem 21.7 applies equally well in respect of a function f f -+'/j which is continuous on a subset f of the metric space .'L. One simply takes note of the fact that f is a metric subspace of .'L and replaces .'L at each occurrence by f. This leaves the meaning of item (i) of theorem 21.7 unaltered and the same is true of item (ii). However, very considerable care is necessary with items (iii), (iv) and (v) of theorem 21.7 when the underlying metric space is switched from .'L to f. Such a switch affects the meaning of the words 'open', 'closed', 'closure' etc. This is because these ideas all depend on the notion of a boundary point. Recall that ;e .'L is a boundary point of S if and only if d(;, S)=O and d(;, eS)=O. But, if ;~f, it ceases to be eligible as a boundary point when one switches the underlying space from .'L to f. Even if ;ef it may still cease to be a boundary point when we switch from .'L to f because the nature of es depends on whether we use .'L or f as the universal set. If f replaces .'L as the universal set, es will in general become a smaller set and hence d(;, es) may cease to be zero. Since a subset S of f has fewer boundary points relative to the metric space f than it does relative to the larger metric space .'L, it is easier for S to satisfy the criteria for being open or closed relative to f than it is to satisfy the criteria for being open or closed relative to .'L. Thus the subsets of f which are not open or closed relative to .'L may very well become open or closed when the underlying metric space .'L is replaced by the metric subs pace f. Returning to items (iii), (iv) and (v) of theorem 21.7, this observation simply reflects the fact that it is easier for a function f to be continuous on a subset f of .'L than it is for f to be continuous on the whole of .'L. If f is a metric subspace of the metric space .'L, we define the relative topology of f to be the collection of subsets off which are open relative to f. As explained above, the relative topology off is NOT just the collection of all subsets off which are open relative to .'L. It is a larger collection than this.

21.10 Example Suppose that .'L = IR 2 • Let f 1 = {(x, y): x > 1} and let f 2 =f 1 v{(1,0)}. Take

Note that S 1 cf 1 cf 2 • The set S 1 is neither open nor closed relative to the metric space .'L=IR 2 • The boundary points on its straight edge belong to .'L\S 1 while those on the curved edge belong to S 1 • The set S 1 is closed relative to the metric space f 1• The boundary points of S 1 relative to the metric space f 1 are those on the curved edge of S 1 and these all belong to S 1 • Notice that the points on the straight edge of S 1 do not belong to f 1 and hence are not eligible as boundary points relative to the metric space f 1• The set S 1 is not closed relative to the metric space f 2 • We have that (1, 0) is a boundary point of S 1 relative to the metric space f 2 which does not belong to S 1•

97

Topology

(I, 0)

The set S 2 = S 1 u{ (1, 0)} is also interesting. This is not closed relative to the metric space X. It is not eligible as a closed set relative to the metric space f 1 since it is not a subset of f 1 • It is closed relative to the metric space f 2 • Observe that although (1, O)e f 2 and is a boundary point of S2 relative to the metric space X, it is not a boundary point of S 2 relative to the metric space f 2 • We have that d((1, 0), X\ S 2)=0 but d((1, 0), f 2 \ S 2 )= 1. Of course f 1 is both open and closed relative to the metric space f 1 (theorem 14.14). Similarly, f 2 is both open and closed relative to the metric space f 2 • Observe finally that f 1 is an open set relative to the metric space f 2 • Its only boundary point relative to the metric space f 2 is (1, 0).

The next theorem makes it quite easy to find the relative topology of a metric subspace f of a metric space X provided that one knows the topology of X to begin with.

21.11t

Theorem Let f be a metric subspace of a metric space X. Then His an open set relative to the metric space f if and only if

for some set G which is open relative to the metric space X.

98

Topology

Proof Recall that an open ball B with centre ~E f and radius r > 0 in the metric space f is defined by B={z: d(z, ~) 0 in the metric space X. Thus B' = {x: d(x, ~) < r}, where the variable x ranges over the universal set X. Obviously B = B' n X. (i) Suppose that H is an open set relative to the metric space f. By exercise 14.17(11), H is the union of all open balls B which it contains. Here B denotes, of course, an open ball relative to the metric space f. Let W denote the collection of all such open balls B contained in H. Then Bew

Let

Bew

Then G is open relative to the metric space X because it is the union of sets which are open relative to the metric space X (theorem 14.14). Also H = Gnf.

G

H

(ii) Suppose that H = Gnf where G is open relative to the metric space X. Let W' denote the collection of all open balls B' (relative to the metric space X) which are contained in G. Then

G=U

B'.

B'ew.'

But then

and hence is the union of sets which are open relative to the metric space f.

21.12t

Corollary Let f be a metric subspace of a metric space X. Then H is a

Topology

99

closed set relative to the metric space Z if and only if H=FnZ

for some set F which is closed relative to the metric space X.

Proof We have that Z "-.His open relative to the metric space Z if and only if Z\ H = GnZ where G is open relative to the metric space X. But Z\ H = GnZ is equivalent to H = eGn Z and the result follows. 21.13 Example The set [0, 1) is neither open nor closed relative to the metric space 1Pl 1 • However, [0, 1) is open relative to the metric subspace [0, 2] because [0, 1)= ( -1, 1)n[O, 2] and ( -1, 1) is an open set relative to the metric space 1Pl 1 • The set [0, 1) is closed relative to the metric subspace ( -1, 1) because [0, 1) = [0, 2] n (- 1, 1) and [0, 2] is a closed set relative to the metric space 1Pl 1 • metric subspace of

I

-I

21.14t

0 open

IR 1

metric subspace of

] 2

I

-I

IR 1

0/ I

] 2

closed

Exercise

(1) Let A, B, C and D be the sets in 1Pl 2 given by A={(x,y): 1~x 2 +y 2 ~2}, B={(x, y): y2 <2}, C= {(x, y): y 2 ~ 2}, D={ (x, y): y2 > 2},

Explain why the function f: 1Pl 2 --+1Pl 2 cannot be continuous on 1Pl 2 if any of the following properties hold:

r

1 (i) (B)=C (iii) f(A)=C

r

1 (ii) (A)=D (iv) f(B)=D.

(2) Let X be a compact metric space and let f:X--+'Ij be a bijection which continuous on X. Prove that f is a homeomorphism. [Hint: Use theorem 21.7iv and exercise 20.6(2).] (3) Let S be a set in 1Pl 2 defined by

IS

S={(x, y): 1 <x~2}. Decide whether or not S is open or closed relative to the .three metric spaces considered in example 21.10. (4) Characterise the subsets of [0, 1] which are open relative to [0, 1] regarded as a metric subspace of 1Pl 1 . [Hint: Use theorem 17.27.]

100

Topology

(5) Prove that every subset of N is both open and closed relative to N regarded as a metric subs pace of~ 1 • (The relative topology of N is said to be discrete.) (6) Let A and B be subsets of a metric subspace f of a metric space X. Show that A and B are contiguous relative to the metric space f if and only if they are contiguous relative to the metric space X. (7) Let f be a metric subspace of a metric space X. Show that a subset C of f is connected relative to f if and only if it is connected relative to X. Give an example of a connected set D in ~ 2 such that Dr. f is not connected relative to f ={(x, y): x 2 +y 2 ~ 1}. (8) Let f be a metric subspace of a metric space X. Prove that a subset K of f is compact relative to f if and only if it is compact relative to X. Give an example of a compact set L in ~ 1 such that Lr.f is not compact relative to f = (0, 1). (9) Let f be a metric subspace of X. Prove that f is a connected subset of X if and only if the only subsets of f which are both open and closed relative to f are 0 and f.

2t.t5t

Introduction to topological spaces

We sometimes wish to discuss the topological properties of a space on which a metric is not given. From §21.18, it is clear that this will be no problem provided that we know what the open sets of the space are. This leads us to define a topological space to be a non-empty set X and a collection g of sets in X. The collection g will serve as the topology of X. In order that it be sensible to regard the sets in gas the 'open sets' of X we require that g satisfy the conclusions of theorem 14.13- i.e.

(i) 0Efi, XEfi. (ii) Any union of sets in g is in g. (iii) Any finite intersection of sets in

g

is in

g.

Of course, a topological space is a more abstract notion than a metric space just as a metric space is a more abstract notion than ~·. But it would be naive to suppose that this makes a topological space mathematically more difficult to deal with. On the contrary, the more abstract a space is, the less structure it has and therefore the fewer theorems that can be true of it. The usual pattern in passing from a concrete space to a more abstract space is that many results which were theorems in the concrete space become definitions in the abstract space. They therefore do not need to be. proved. For example, the triangle inequality for distance in ~· which we proved as a theorem (theorem 13.7) with the help of the Cauchy-Schwarz inequality becomes part of the definition of distance in a metric space. These remarks do not mean, of course, that it is not useful to pass from a concrete space to a more abstract space. In the first place, one obtains a theory that is more widely applicable. In the case of topological spaces, this fact is particularly important. In the second place, one often gains considerable insight into the structure of the original concrete space since the process of abstraction forces one to focus on those properties of the original space which really matter for the proof of a particular theorem.

101

Topology

Returning to the idea of topological space, the immediate question is: how many of the properties of a metric space are valid in a topological space? We begin by defining a closed set in a topological space to be the complement of an open set. Thus theorem 14.9 becomes a definition for a topological space. Theorem 14.13 is part of the definition of an open set in a topological space. Theorem 14.14 follows from theorems 14.9 and 14.13 and is therefore true in any topological space. We define the closure E of a set E in a topological space to be the intersection of all closed sets which contain E. Similarly, the interior E of E is defined to be the union of all open sets contained in E. Thus theorems 15.7 and 15.9 become definitions in a . topological space. The following properties of the closure and interior of sets remain valid and are worth remembering: (i) AuB=AuB; (AnBt=AnB

(ii) AcB=>AcB; AcB=>AcB (iii) AnBcAnB; (AuBtcAuB. (See exercise 15.10(5).) We can now define the boundary oE of a set E in a topological space by

If X and 'IJ are topological spaces, weusetheorem21.7(v)toprovideadefinitionforthe continuityofafunctionf: X-+'IJ. Wesaythatf: X-+'IJiscontinuous(on X)ifandonlyif, for each G c 'IJ, G open

=>

f- 1(G) open.

The proof of theorem 21.7 shows that this condition implies the more intuitively satisfying criterion given in§ 16.2. If X and 'IJ are topological spaces, a homeomorphism f: X-+'IJ is a continuous bijection with a continuous inversef- 1 : 'IJ-+ X. If a homeomorphismf: X-+'IJ exists, we say that the topological spaces X and 'IJ are topologically equivalent (or homeomorphic). From the topological point of view, two homeomorphic spaces are essentially the same (see the discussion of §9.21). Let f be a subset of a topological space X with topology :J. Then V={ZnG: GE:J}

is a collection of subsets off which satisfies the conditions for a topology on f. We call V the topology on f relative to X. Thus theorem 21.11 for a metric space becomes a definition for a topological space. The set f with the topology V is called a topological subspace of X. The introduction of the idea of a relative topology means that we do not need a separate definition for continuity on a subset f of a topological space X. We use the definition given above but with f (regarded as a topological subspace of X) replacing X. Similarly, we need only provide definitions of a connected topological space and a compact topological space. A connected topological space X is one in which the only sets which are both open and closed are (/) and X. Theorem 17.3 for a metric space therefore becomes a

102

Topology

definition for a topological space. It is important that theorem 17.9 remains true in a topological space. With our new definitions, the proof is even easier.

21.16t Theorem Let .'L and 'IJ be topological spaces and let f:.'L-+'IJ be a continuous surjection. Then .'L connected => 'IJ connected.

Proof Suppose that E is a set in 'IJ which is both open and closed. Since f:.'L-+'IJ is continuous, f- 1(£) is both open and closed in .'L. But .'L is connected. Thus, f- 1(£) = (/) or f- 1(£) = .'L. Because f is a surjection, it follows that E = (/) or E = .'L. Thus 'IJ is connected. A compact topological space .'L is one with the property that any collection U of open sets which covers .'L has a finite subcollection which covers .'L. This definition is identical with that of §20.4. It is important that theorem 19.13 remains true in a topological space and, again, with our new definitions, the proof is even easier.

Theorem Let .'L and 'IJ be topological spaces and let f:.'L-+'IJ be a 21.17t continuous surjection. Then .'L compact=>'IJ compact.

Proof Let V be a collection of open sets which covers 'IJ. Then U={f- 1(H):HEV} is a collection of open sets which covers .'L. Since .'L is compact, a finite subcollection

E covers .'L. Let 5={f(G): GEE}. Then 5 is a finite subcollection of V which covers 'IJ. Hence 'IJ is compact.

21.18t

Product topologies In the space IR 2 the Euclidean metric d: IR 2-+IR is defined by d(x, y.)= {(xi- Yl)2 + (x2- Y2)2}1/2.

We use this metric because it corresponds to the notion of the distance between two points as understood in Euclidean geometry. However, the Euclidean metric is not the only possible metric which can be used in IR 2. The function m: IR 2-+IR defined by m(x, y)=max{lxl- Y1l, lx2- Y2l} is an example of an alternative metric. So is the function I: IR 2-+ IR defined by

103

Topology

To verify that these functions satisfy the requirements for a metric given in §13.1 is very easy. Of course, an open ball with respect to one of the metrics l and m looks very different from a Euclidean open ball as the diagrams below illustrate.

-----~5~ ~-1

I

I I ~ 1 -r

E={x:

d(~,

x)
M={x:

~ 1 +r

m(~,

L= {x:

x)
/(~,

x)
However, the geometrical shape of these open balls is irrelevant from the topological point of view. We know that any open set in a metric space X is the union of all open balls which it contains (exercise 14.17(11)). But it is evident that, given any one of the open balls illustrated above, we can find an open ball of either of the other two types with the same centre which fits inside the original open ball. Thus a set G is open with respect to one of the metrics if and only if it is open with respect to them all - i.e. the three metrics all generate the same open sets in IR 2 . Now suppose that X 1 and X 2 are topological spaces. What is the appropriate topology for X 1 x X 2? If X 1 and X 2 are metric spaces with metrics d 1 : X 1 -->IR and d 2 : X 2 -->IR, we can proceed as with the space IR 2 =IR x IR. In particular, the functions d: X 1 x X 2 -->IR and m: X 1 x X 2 -->IR defined by (1)

and (2)

are metrics on X 1 x X 2 which generate the same topology on X 1 x X 2 . Why should this topology be more useful than the various other topologies which one might impose on X 1 x X 2? The significant fact about the metrics d and m is that they generate a topology on X,xX 2 which makes the projection functions P 1 :X 1 xX 2 -->X 1 and P2 : X 1 x X 2 --> X 2 continuous. For example, if we use the metric d in X 1 x X 2 , then d 1 (P 1 (x), P 1 (y))=d,(x 1 ,

y,)~d(x,

and hence P 1 is continuous on X 1 x X 2 by lemma 16.8.

y)

104

Topology

~-----------L----------------x, P1 (x) x 1

=

This observation makes it natural to define the product topology on X 1 x X 2 to be the weakest topology (i.e. the topology with the fewest open sets) with respect to which the projection functions P 1 : X 1 x X 2-+ X 1 and P 2 : X 1 x X 2-+ X 2 are continuous. It follows that, if G 1 is an open set in X~> then P 1 - 1 ( G 1) must be an open set in the product topology of XI X x2. Similarly, if G2 is an open set in x2, then p2 -I(G2) must be an open set in the product topology of X 1 x X 2 •

~---------------------XI

Since the intersection of a finite collection of open sets must be open, it follows that Gl x G2 = p 1 -I(Gl)nP2 -I(G2)

must be an open set in the product topology of X 1 x X 2 • Finally, the union of any collection of open sets must be open. Thus, if W is any collection of sets of the form G 1 x G2 (where G 1 is open in X 1 and G2 is open in X 2), then (3) G, xG,eU'

must be an open set in the product topology of X 1 x Xi· The collection f1 of all sets S of the form (3) satisfies the requirements for a topology on XI X x2 given in §21.15. It follows that g is the product topology on X1 x X2.

105

Topology S=

u

Gl

X

G2

G 1 xG 2 EW1

:r I Now return to the case in which X, and X 2 are metric spaces with metrics d1 and d2 respectively. If we use the metric m: X 1 x X 2 --+IR defined by (2), then the open balls Bin X 1 x X 2 are of the form B=B 1 x B 2 where B 1 is an open ball in X 1 and B 2 is an open ball in X 2 • It is therefore apparent from the preceding discussion, that the open sets in X 1 x X 2 generated by the metric m are precisely those which lie in the product topology g. The same is therefore true of the metric d: X 1 x X2 --+IR defined by (1) and of numerous other metrics. From the topological point of view it does not matter which of these metrics we choose to use in X 1 x X 2 since they all generate the same topology. When discussing topological matters, we therefore work with the metric in X 1 x X 2 which happens to be most convenient for the problem in hand. In IR 2 (or IR") this is usually the Euclidean metric. When X 1 and X 2 are more general metric spaces, however, the metric m is often much less cumbersome.

22

22.1

LIMITS AND CONTINUITY (I)

Introduction

Letf:X~'Ij and suppose that ~EX and 11E'IJ. In this chapter, we shall study the meaning of the statement

f(x)~11

as

x~~-

Sometimes this is written in the equivalent form lim f(x)=11· We say that 'f(x) tends to 11 as x tends to ~· or that 'f(x) converges to the limit 11 as x approaches ~·. The diagrams below illustrate the idea we are trying to capture. The first diagram is of a function!: IR 2 ~1R 2 for whichf(x)~'l as x~e. It shows x approaching~ along a path. As x describes this path,f(x) approaches 11· We shall of course want the same to be true however x approaches ~-

;'{ X

/--. __,;~ x)

'

/

The next diagram shows the graph of a functionf: as x~~-

/06

-

f(x)

...\

,.._.....J

f

•f(~)

f(x')

---...

,

.,)

IR~IR

for

whichf(x)~IJ

107

Limits and continuity (I) y

f@

-------------. I I

I --------------T--

f(x)

t

1

7

f(x')

~--------~~--~--~--------~ X

x·-~----x

Note that in both diagrams it is false that 11 = f(l;). It is not the case that one can always find the value of limf(x)

x-e

by replacing x in the formula for f(x) by 1;. This point is of some importance in calculus. The derivative of a function f: IR -.IR at the point ~ is defined by

f'(~) = lim f(x)- f(~) x-~

x-~

(provided that the limit exists). Observe, however, that if we replace x on the right-hand side by ~ we obtain a meaningless expression because the right-hand side is not defined when x = ~- In computing the limit, we are interested in what happens as x approaches ~ not in what happens when x equals ~- Our definition of a limit will take this into account by explicitly excluding consideration of what happens when x = 1;. Sometimes, of course, it will be true that

We then say thatfis continuous at the point 1;. The diagram below illustrates a function f: IR -.IR which is continuous at ~Y

Y

~-------------:"------------------x ~

=f(x)

108

Limits and continuity (I)

Observe that, in drawing the graph off, one does not have to lift the pencil from the paper when passing through the point where x = ~. This provides one explanation for the use of the word 'continuous'. A more adequate explanation is provided by theorem 22.5. We shall also be interested in providing a definition of what it means to say that

through the set S.

,-,

,-...., \

)

'-'>-"---

f( x') ' ,,I ' ..... _,

... -..!....!.-/

"'"'

,,

\....I

When considering a limit through the set S, we are interested only in what happens as x approaches ; through the set S. What happens as x approaches ; from outside the set S is irrelevant and our definition will specifically exclude consideration of such points. The definition we shall give works for any set S and any point ;. Note, however, that it does not make very much intuitive sense to talk about 'x approaching ; through the set S' if ; is not a cluster point of S. There is then no way for 'x to approach ;· without going outside the set S. The diagram below illustrates this point.

We should therefore not be too surprised if the definition yields results which are not in accordance with our intuitions about limits in those cases when ; is not a cluster point of S.

Limits and continuity (I) 22.2

109

Open sets and the word 'near'

We are seeking a precise mathematical definition of the statement f(x)-+1)

as

x-+;.

The intuitive notion which we wish to capture is thatf(x) gets 'nearer and nearer' to 11 as x gets 'nearer and nearer' to ;. This is a rather woolly statement which could mean various things. We choose to interpret it as meaning that 'If x is sufficiently near to ;, then f(x) is near to 1).' This formulation allows us to concentrate on the crucial issue which is: what do we mean in mathematical_terms by the word 'near'. Recall that an open set Gin a metric space .X is a set which contains none of its boundary points. Thus, if ;e G, it cannot lie on the 'edge' of G. Hence G must contain all points of .X which are 'sufficiently near' to ;.

It is therefore natural to frame the definition of a limit in terms of the idea

of an open set. Such a definition has the additional advantage that it still makes sense in a topological space- i.e. a space in which one knows what the open sets are although one may not be provided with a metric.

22.3

Limits

Let .X and 'IJ be metric (or topological) spaces and let f: .X-+'IJ where S is a set in .X. Suppose that ;e .X and 1)E'IJ. Then we say that f(x)-+1)

as x-+;

through the set S if and only if For any open set G containing 1), there exists an open set H containing ; such that xeHnS ::.f(x)EG provided x # ;.

110

Limits and continuity (I)

f

One thinks of the choice of Gas specifying how 'near' we wantf(x) to be to 11· Since we are interested only in the limit through the set S, we restrict attention to xES. We also exclude the case x = ~· With these provisos, the definition says that, if we take x 'sufficiently near' to~- i.e. xEH- thenf(x) will be as 'near' to 11 as we specified - i.e.f(x)E G. We now define the statement f(x)--+11

as

X-+~.

This should mean that f(x) approaches 11 however x approaches appropriate definition is therefore that:

~·

The

For any open set G containing 11. there exists an open set H containing ~ such that xEH =f(x)EG provided x =I~. In order that this definition makes sense, it is necessary thatfbe defined on some open set S containing ~ (except possibly at ~itself). Such an open set S contains all points of X which are 'sufficiently near' to ~· Indeed, our definition of the statement )(x)--+11 as X-+~' is equivalent to the assertion thatf(x)--+11 as X-+~ through the setS for some open setS containing~·

22.4

Limits and continuity

lfj(x)--+f@ as x--+~, we say thatfis continuous at the point~· The next theorem relates this terminology to our previous work on continuity.

22.5

Theorem Let X and 'lj be metric (or topological) spaces and let

Limits and continuity (I)

f: :X-+'Ij where S is a set in if, for each ~ES,

111

:r. Then/ is continuous on the setS if and only f(x)-+f(~)

as

X-+~

through the set S. Proof We give the simplest proof. This depends on the results of chapter 21. An alternative proof using only the ideas of chapter 16 is suggested in exercise 22.24(4). (i) Suppose that f is continuous on the setS. Let~ be any point of Sand let G be an open set containing/(~). Thenf- 1 (G) is an open set relative to S by theorem 21.7. From theorem 21.11, it follows that there exists an open setH in :X such thatf- 1 (G)=HnS. We then have that xEHnS=f(x)EG. This shows that f(x)-+f@ as X-+~ through the set S. (ii) Suppose that f is not continuous on the set S. Then there exists an open set Gin ?J such thatf- 1 (G) is not open relative to S (theorem 21.7). Let ~E/- 1 (G) be a boundary point ofj- 1 (G) relative to S. Then any open set H in :X which contains ~ has the property that H n S contains a point ZE

e /- 1 (G).

Since zEHnS butf(z)~ G, it follows that it is false thatf(x)-+J(e) as x-+e through the set S. 22.6 Corollary Let P: !Rn-+IR and Q: !Rn-+IR be polynomials and suppose that Q@-# 0. Then P(x)

P(~)

---+ - - as Q(x) Q(~)

x-+~.

112

Limits and continuity (I) Proof This follows immediately from theorems 16.15 and 22.5.

Note that the same result holds with

22.7 tion/:

~

replaced throughout by C.

Examples (i) Let S = ~ 2\ {(0, 0)} and consider the rational funcs~~ defined by =x 3 y+2xy 2 +3 !( X, y) 2 2 · X

+y

By corollary 22.6, f(x,

(ii)

LetS=~\ {0}

y)~3

as

(x,

y)~(l.

1).

and consider the rational function F:

s~~

defined by

(1 + x)2 -1.

F(x)

X

We know from corollary 22.6 that F(x)~ 1 as does not help immediately in evaluating

x~

-1 but corollary 22.6

lim F(x) x~o

because F(O) is not defined. Note, however, that for x =1= 0, F(x)=1+2x+x2-1

2+x.

X

If G: ~ ~~ is the polynomial defined by G(x) = 2 + x, it follows that F(x) = G(x)unlessx =0. Since we explicitly ignore what happens whenxequals 0 when evaluating a limit as x approaches 0, it follows that lim F(x) = lim G(x) = lim (2 + x) = 2. x~o

x~o

x~o

22.8 Exercise (1) Evaluate the following limits (i)

lim

(x,y)~(l,l)

y2-x2 -2 2 Y +x

(ii)

lim (x,y) ... (l,-1)

Y2-x2

y+x

(2) Let K={(x, y): O;;;;;x;;;;;1 and O;;;;;y;;;;;1} and letf: ~ 2 ~~ be defined by f(x, y)={1 (x, y)eK 0 (x, y)ltK.

Limits and continuity (I)

(3)

(4)

(5) (6)

113

Prove thatf(x, y)-+1 as (x, y)-+(1, 1) through the set K andf(x, y)-+0 as (x, y)-+(1, 1) through the set e K. Suppose that ~ is not a cluster point of S. Explain why it is true that f(x)-+tt as x -+~ through the set S for alltt. If~ is a cluster point of S, prove that there exists at most one 11 such thatf(x)-+tt as x-+~ through the setS provided that, for each pair of distinct points tt 1 and tt 2 in 9,1., there exist disjoint open sets G1 and G2 such that tt 1 E G1 anc! tt 2 E G2 • Show that such disjoint open sets always exist when 9,1. is a metric space. Suppose that S 1 uS 2 = S. Prove that f(x)-+tt as x -+~ through S if and only if/(x)-+tt as X-+~ through S 1 and through S2 . Deduce that, if f is the function of question 2, then there is no ttEIR for which f(x, y)-+tt as (x, y)-+(1, 1). Suppose that T is a subset of S. If f(x)-+tt as X-+~ through the setS, prove that f(x)-+tt as x -+~through the set T. Suppose that f: IR -+IR and g: IR -+IR are continuous on the set IR and f(x)=g(x) for each xEO. Prove that f =g. [Hint: Let ~ be irrational. Then ~ is a cluster point of Q (why?). Also f(x)-+/(~) as x-+~ through Q and g(x)-+g(~) as x-+~ through Q.]

22.9

Limits and distance

In a metric space X a set G is open if and only if each XE G is the centre of an open ball B which is entirely contained in G. This fact means that, if X and 9,1. are metric spaces, then we can rewrite the definition of a limit given in §22.3 in terms of open balls rather than open sets. Let X and 9,1. be metric spaces and letf: S-+9,1. where S is a set in X. Then f(x)-+tt as X-+~ through the setS if and only if, for each open ball E with centre tt. there exists an open ball A with centre ~ such that xEAnS =f(x)EE provided that x #

~-

f

114

Limits and continuity (I)

If we take the radius of the open ball E to be e and that of Ll to be b, the definition assumes the following familiar form: For any e>O, there exists a b>O such that for each xES, 0
= d(11,f(x)) <e.

It is sometimes helpful to think of f(x) as an approximation to 11· The

quantity d(11, f(x)) is then the error in approximating to 11 by f(x). The definition then asserts that this error can be made as small as we choose by taking x sufficiently close to ~- (Here, of course, it is understood that values of x outside S \ g} are to be ignored.) 22.10 Example Consider the functionf: IR 2-+IR 2 defined by (u, v) = f(x, y) where

u=x+y} v=x-y. The fact that f(x, y)-+(2, 0) as (x, y)-+(1, 1) follows from theorem 22.5 becausefis continuous on IR 2. In this example we give an alternative proof using the criterion for convergence given in §22.9.

Proof Given any e > 0, we shall demonstrate the existence of a () > 0 such that 0 < ll(x, y)- (1, 1)11 < b

= llf(x, y)- (2, 0)11 <e.

We begin by observing that ll(x, y)-(1, 1)11 ={(x-1) 2 +(y-1) 2}112 and llf(x, y)-(2, O)ll=ll(x+y, x-y)-(2, 0)11 = {(x + y _ 2)2 +(x- y)2}1;2 ={[(x-1)+(y-1)] 2 + [(x-1)-(y-1)] 2}112 = {2(x -1) 2 + 2(y -1)2}112

=J2 ll(x, y)-(1, 1)11. Given e > 0 we therefore have to choose () > 0 such that O
=J211(x, y)-(1, 1)11 <e.

The choice of() =e/J2 (or any smaller positive number) therefore completes the proof.

II5

Limits and continuity (/) 22.11

l

Example Consider the functionf: IR 2 --+IR 1 defined by y2-x2 -

y 2 +x 2

f(x, y)=

(x, y)=f.(O, 0) (x, y)=(O, 0).

0

Let L. be the line L.= {(x, y): y=cxx}. For points (x, y)EL., we have that CX2X2- X2

f(x, y) = f(x, cxx) =

2

CX2 - 2

-1

(x =f. 0).

2

2

as

(x, y)--+(0, 0)

cxx +x

ex +1

Hence ()(2

-1

f(x, y)--+ cx 2 +

1

along the line y = cxx - i.e. through the set L •. It follows thatf(x, y) cannot tend to a limit as (x, y)--+(0, 0) through IR 2 • If it did, it would have to tend to the same limit through each of the sets L. (see exercise 22.8(4)). The diagram indicates how the function tends to different limits along different straight lines through (0, 0).

z

X

The part of the surface z = f(x, y) drawn looks somewhat like a piece of spiral staircase.

116

Limits and continuity (I)

22.12

Right and left hand limits

For a function f: IR --+IR we have that f(x)--+q as x--+e if and only if for any e > 0, there exists a ~ > 0 such that O

1/(x)-ql<e.

This definition can be illustrated graphically as below.

___f ______ _

y = f(x)

8

In this diagram R is a rectangle with centre (e, q) of height 2e and width 2~. The definition asserts that, however small we choose to make the height of R, its width can be chosen so as to ensure that the part of the graph off above (e-~. e+~) lies within R (except possibly for the point (e.J(e))). Now suppose that I =(a, b) is an open interval in IR and that f: (a, b)--+IR. We write f(x)--+'1

as

x--+a+

and say that f(x) tends to '1 as x tends to a from the right if and only if f(x)--+'1 as x--+a through the set (a, b). Similarly, we write f(x)--+q

as

x--+b-

if and only if f(x)--+q as x--+b through the set (a, b). f

a

a+b f(x)-+'1 as x-+a+

b

a

b-b f(x)-+'1 as x-+b-

b

117

Limits and continuity (I)

It is useful to note that f(x)--+1'/ as x-+a+ if and only if, for any e>O, there exists a (j > 0 such that a<x
=>

lf(x)-'71 <e.

Also f(x)--+1'/ as x-+b- if and only if, for any e > 0, there exists a that b-b<x
=>

(j >

0 such

lf(x)-IJI<e.

22.13 Theorem Let I be an open interval containing e and suppose that f: I\ {e}-+IR. Then f(x)-+IJ

as

x-+e

if and only if f(x)-+IJ as x-+e- and f(x)--+1'/ as x-+e +. Proof This follows immediately from exercise 22.8(4).

22.14

Example Let f: IR -+IR be defined by

(x~ 1)

f(x)={1-x 2x

(x> 1).

Thenf(x)--+0 as x--+1- andf(x)--+2 as x--+1+. It follows from theorem 22.13 that lim f(x) x-1

does not exist. In particular, f is not continuous at the point 1.

y

2

!=f(x) I I I I I I I I

I I I

X

0

Note that it is not necessary to use an argument involving e and (j to show thatf(x)--+2 as x--+1 +.The function g: IR-+IR defined by g(x)=2x is a polynomial and hence is continuous everywhere. Also f(x) = g(x) when x > 1.

118

Limits and continuity (J)

In calculating a limit as x-d + we ignore what happens when x;:::; 1. Hence lim f(x)= lim g(x)= lim 2x=2. X-tl+

x-l+

Similarly for f(x)-+0 as x-+1-.

22.15

Some notation When the limits exist, the notation f(~-)=limf(x); /(~+)=

lim /(x)

can often be useful. However, it is important not to confuse f(~-) or f(~ +) with/(~). These three quantities are equal only when/is continuous at the point~.

If/(~-)= f( ~) we say that f is continuous on the left at r If/(~+)= f( ~), we say that f is continuous on the right at ~. The diagrams below illustrate some of the possibilities.

i

f(~)=f(~+)

___y

f(U)

I I

f(~-)

I

I

I

lf continuous on I

the right at

I neither f(~---) nor if(~+) is equal to If(~)

~

I

I

I I I

:f((-)andf(~+)

I do not exist

119

Limits and continuity (I)

22.16

Monotone functions Suppose that S is a non-empty subset of IR. We say that f: S->IR

increases on S if and only if

x
=f(x)~f(y)

for each x and y inS. Similarly, f: S->IR decreases on S if and only if x< y

=> f(x)~

f(y)

for each x and y in S. We say that f is strictly increasing on S if and only if x < y =f(x)
for each x and y in S. Finally, f is strictly decreasing on S if and only if x< y

=> f(x)

> f(y).

strictly decreasing function

incr/easing function

f

~ \

L_

s

s

A function which is either increasing or else decreasing on S is said to be monotone on S. (A function can be both increasing and decreasing on S. But

then it must be constant.) A function which is either strictly increasing or strictly decreasing is said to be strictly monotone.

22.17 Theorem Suppose that I= (a, b) is an open interval in IR and that f: I ->IR is increasing on I. (i) Ifjis bounded above with supremum L on I, thenf(x)->L as x->b-. (ii) If f is bounded below with infimum l on I, then f(x)-->1 as x->a +. Proof We prove only (i). From §22.12, we know that, given any s > 0, we have to demonstrate the existence of a o> 0 such that

b-o <x
120

Limits and continuity (1) L

____ T __________ _

I

I I I

I I

I

I

I

I

J

I

___________ j

I I I I

a

b

f(x) < L + e is automatically satisfied because L is an upper bound for f on (a, b). Since L-e is not an upper bound (because L is the smallest upper bound), there exists a yE(a, b) such thatf(y)>L-e. But/increases on (a, b). Therefore, for any x satisfying y < x < b, L-e
The choice

(J =b-y

then completes the proof.

I I

I I

I

I I I

I

I

I

I

y

b

--+-1 b

a

~

22.18 Corollary Suppose that 1 =(a, b) is an open interval in lP! and that f: 1--+IR! is increasing on J. If ~E (a, b), then/(~-) and/(~+) both exist and f(x)~f(~- )~(~)~(~ + )~(y)

provided that

a<x<~
Limits and continuity (/)

121

I~

/(~+)

----------------V I

f(~) /(~-)

J

_______________ _.I I

------~ I

a

y

X

b

Proof The function f is bounded above on the interval (a, ~) by By theorem 22.17, the smallest upper bound off on (a, ~) is f(~- ). It follows that, for any XE(a, ~),

/(~).

f(x)~(~- )~(~).

A similar argument for the interval

22.19

(~,

b) yields the other inequalities.

Inverse functions

Recall that a function f: S--+ T admits an inverse function T--+ S if and only if f is bijective. A function is bijective if and only if it is both surjective and injective. Surjective means that f(S)= T and injective means that f(x 1 )= f(x 2 ) <=>X 1 =x 2 .

f - 1:

In this section, we are interested in the case when S and Tare intervals in IR and f is continuous on S. It then seems 'intuitively obvious' that f is bijective if and only if T =f(S) and f is strictly increasing or strictly decreasing on S. Furthermore, it seems equally clear that the inverse function f - 1 : T--+ S will be continuous on T. y

X

+

s

~--+-------~x--+-~x

s

~--4-----~y----~--Y

T

122

Limits and continuity (1)

But, like so many 'intuitively obvious' theorems, the proof requires some deep results.

22.20 Proposition Let 1 be an interval in IR and suppose that f: 1--+IR is continuous on 1. Thenfis injective if and only iffis either strictly increasing· or strictly decreasing on 1. Proof The fact that a strictly increasing or decreasing function is injective is trivial. The proof of the other half of the theorem is the content of exercise 17.28(3). It requires a simple application of the intermediate value theorem (17.11). The diagram illustrates a function f: I --+IR which is continuous on I but neither increasing nor decreasing on /. If a< b < c and f(a)
f(~)=f(c)

----

a

~

b

c

22.21 Theorem Let I be an interval in IR and let J be a set of real numbers. Suppose that f: I--+ J is strictly increasing or strictly decreasing on I and that J = f(I). Then f is continuous on I if and only if J is an interval. Proof (i) Suppose that f is continuous on /. Then J = j(l) is an interval by theorem 17.1 0. (ii) Suppose that f is not continuous on /. Then, by theorem 22.5, for some ~El, it is false thatf(x)--+f(~) as X-+~ through the set/. From theorem 22.17 it follows that at least one of the quantities f(~-) or f(~ +)exists but is not equal to f(~). But corollary 22.18 then implies than J is not an interval.

Limits and continuity (I)

123

f(~+),.ef(O

I

22.22 Theorem Let 1 and J be intervals in IR1 and suppose that f: 1--.J is surjective, continuous and strictly increasing or decreasing on /. Then f: 1--.J is a homeomorphism- i.e. f is bijective and f- 1 : J --./is continuous on J. Proof The fact that f is bijective follows from the two preceding results. Since f- 1 : J-> 1 is surjective and strictly increasing or decreasing, then it is continuous on J by theorem 22.21.

22.23

Roots

In §9.13 we proved that, if y~O, then there exists a unique such that y=x". We call x the nth root of y and write

x~O

x=y1fn.

A more elegant proof of this result uses the theorems of the preceding section. Let nE N and define f: [0, oo )-> [0, oo) by f(x)=x".

The function f is strictly increasing and continuous on [0, oo ). To use theorem 22.22 we need to prove that it is surjective. We know from theorem 17.10 that J = f([O, oo )) is an interval. But f(O) = 0 and J is clearly unbounded above. It follows that J = [0, oo) and so f is surjective. By theorem 22.22,/admits an inverse functionf- 1 : [0, oo)->[0, oo). We have that y=x" <=>X=f- 1(y) provided x ~ 0 and y ~ 0. Thus f- I.e.

1

:

[0, oo )-> [0, oo) is the nth root function

124

Limits and continuity (1)

Note that theorem 22.22 supplies the extra information that this function is continuous on [0, oo ).

22.24 Exercise (1) Letf: IR 2 ~1R 2 be defined by f(x, y)=(u, v) where u={1+x

(y~x)

y

(y <x)

v={1+y x

(y~x) (y <x).

What can be said about the limits off as (x, y)~(O, 0) through (i) the set S={(x, y):xO} and (ii) the set T={(x, y):x>O and y
I

xy2

x2+y4

((x, y)#(O, 0))

0

((x, y) = (0, 0)).

Prove that g(x, y)~O as (x, y)~(O, 0) along each of the lines y=ax. But show that g(x, y)f-0 as (x, y)~(O, 0) along the parabola x=ay 2 • (3) A function f: IR 2\ {(0, 0)} ~IR is defined by xy f(x, y)=-2--2 · X +y

If 0 < 6
I~~<26{1+J(l-46 )}2

y)~O as (x, through the set

If g(x,

y)~ (0,

S={(x,

1

or

~~~<26{1+J(l-46 2 )}- 1 .

0), prove that f(x,

y):~~~
y)

or

y)~O

~~~
as (x,

y)~(O,

0)

y)}.

(4) Let X and 'lj be metric spaces and let f: S~y where S is a subset of X. If ;EX and ttE 'IJ, prove that f(x)~tt as x ~; through the set S if and only if, for each subset E of S not containing ;, d(;, E)=O => d(q,J(E))=O.

Use this result to deduce theorem 22.5 from theorem 16.4. (5) Let 1 be an interval in IR and let f: 1 ~IR be increasing on 1. Prove that the set of points in 1 at which f is not continuous is countable. [Hint: Recall the proof of theorem 17.27.] (6) Let f: IR~IR be defined by f(x)= 1 +x+x 3 . Show that f has an inverse functionf- 1 : IR~IR and that this is continuous on IR.

Limits and continuity (I)

22.25

125

Combining limits

22.26 Theorem Suppose that X, ?J and f are metric (or topological) spaces and that f: T-+f and g: S-+Twhere Se X and Tc?J. Let ;EX and let fiE T. Ifjis continuous on the set T and g(x)-+11 as x-+; through the setS, then f(g(x))-+f(TI)

as

x ...... ;

through the set S.

Proof Let G be any open set in f which contains f(fl). Since f is continuous on the set T, it follows from theorem 22.5 that there exists an open set H in ?J containing 11 such that

yEHnT=f(y)EG.

( 1)

(Note that we do not need to exclude the possibility that y = 11 because f(fi)E G by assumption.) Also, since H is an open set containing fl, there exists an open set J containing ; such that xEJnS =g(x)EHnT (2) provided x#;. (Note that g(S)cTand so xES=g(x)ET.) Combining (1) and (2), we obtain that XEJ nS = f(g(x))E G provided x # ;. It follows that f(g(x))-+f(TI)

as

x ...... ;

through the set S.

22.27 Corollary Suppose that X is a metric (or topological) space and that g: S-+ [0, oo ). If g(x)-+IJ as x ...... ; through the set S, then {g(x)}t/n__.l]t/n as through the set S for any nE N.

x-+;

126

Limits and continuity (1)

Proof This follows immediately from § 22.23 and theorem 22.26.

22.28 Theorem Suppose that X is a metric (or topological) space and that f: S ->IR" where S is a set in X. Let ~EX and write

Thenf(x)-->11 as

X->~

through the setS if and only if, for each k = 1, 2, ... , n, fk(x)->l]k

as

x ...... ~

through the set S.

Proof This is most conveniently deduced from exercise 22.24(4) and exercise 13.24(6).

22.29

Example Consider the function f: IR 2 ->IR 2 defined by f(x, y) = (g(x, y), h(x, y))

where g: IR 2 ->IR 1 and h: IR 2 ->IR 1 are the functions of exercise 22.24(2) and (3). We have that g(x, y)-->0 as (x, y)-->0 along the line x= y. Similarly, h(x, y)-->0 as (x, y)-->0 along the line x = y. From theorem 22.28 it follows that f(x, y)-->(0, 0)

along the line x = y. On the other hand, g(x, y) and hence

(x, y)-->(0, 0)

+ 0 as (x, y)-->(0, 0) along the parabola x = y

f(x, y)f(O, 0)

along x=

as

as

2

(x, y)-->(0, 0)

i.

22.30 Theorem Suppose that X is a metric (or topological) space and that / 1 : S->IR" and f 2 : S->IR" where S is a set in X. Let ~EX and suppose that

and

/ 1 (x)->l) 1

as

x ...... ~

/ 2 (x)->l) 2

as

x ...... ~

through the set S. Then for any real numbers a and b, af1 (x)+bf2 (x)->al) 1 +bl) 2 through the set S.

as

X->~

Limits and continuity (J)

127

Proof Define F: S ->IR" byf(x) = af1 (x) + bf2 (x). From theorem 22.28, we have that (f1(x),f2 (x))->(11J, 11 2 )

as

X->~

through the setS. But, with the notation of theorem 16.9, F=L o (/1 , /2 ).

Since Lis continuous on IR" x IR" (theorem 16.9), it follows from theorem 22.26 that

through the set S.

22.31 Theorem Suppose that X is a metric (or topological) space and that f 1: S ->IR and f 2 : S ->IR where S is a set in X. Let ~EX and suppose that

and

f 1 (x)->11J

as

x -->~

fz(x)->IJ 2

as

X->~

through the set S. Then f1(x)fz(x)->IJ 117z

as

'.-->~

through the set S.

Proof Define F: S ->IR by F(x) =f 1 (x)f2 (x). From theorem 22.28, we have that U1(x), fz(x))->(IJ 1, 17 2 )

as

X->~

through the set S. But, with the notation of theorem 16.9, F =M o (f1 , /2 ). Since M is continuous on IR x IR (theorem 16.9), it follows from theorem 22.26 that F(x)-->17 117 2 as X->~ through the setS.

22.32

Theorem With the same hypotheses as in the previous theorem, f1(x) '11 ---->fz(x) IJz

as

x->~

through the set S provided that 17 2 # 0.

Proof The proof is the same as that of theorem 22.31 except that, in the notation of theorem 16.9, D replaces M. (Note that, iff(x)-->17 2 as X->~ through the setS and 17 2 #0, then there exists an open set J containing~ such thatf(x)#O for any xEJnS.)

Limits and continuity (J)

128

22.33 Exercise (1) Suppose that g: IR 2 -+IR 1 and h: IR 2 -+IR 1 have the property that g(x, y)-+l as (x, y)-+(0, 0) and h(x, y)-+m as (x, y)-+(0, 0). Find (i)

lim

/ 1 (x, y)

(ii)

(x, y)-(0, 0)

lim

.f2 (x,

y)

(x, y)-(0, 0)

whenf1 : IR 2 -+IR 1 and/2 : IR 2 -+IR 1 are defined by

!t (x, f2(x,

YW 12 y) = [1 + {g(x, YW + {h(x, YWr y) = {g(x, y)h(x,

2

•

Also find

j(x, y)

lim (x, y)-(0, 0)

where f: IR 2 -+IR 2 is given by f=(/1, / 2 ). (2) Let f: IR -+IR and g: IR -+IR be defined by g(x)=3.

f(y)={ 1 (y=3) 2 (y# 3)

Prove thatf(y)-+2 as y-+3 and g(x)-+3 as x-+4 but show that it is false that f(g(x))-+2 as x-+4. (3) Suppose that S c. X and that g: S -+?,I has the property that g(x)-+1} as X-+~ through the setS whilef: ?j-+f has the property thatf(y)-+~ as Y-+1). Prove that f(g(x))-+~

as

X-+~

through the set S provided that either of the conditions (i) f is continuous at the point 11 or (ii) 11 ~ g(S) is satisfied.

22.34t

Complex functions

Let S be a set of complex numbers (§ 10.20) and consider a function f: S-+C. If wand' are complex numbers, we are interested in the statement f(z)-+w

as

z-+'

(1)

through the set S. As we know from § 13.9, the complex number z = x + iy can be identified with the point (x, y) in IR 2 in which case llzll={x 2 +y 2 2 =11(x, y)ll. There is therefore no difficulty in interpreting (1) which simply means that

P'

Ve>03c5>0VzeS, 0<1z-'l
=>

lf(z)-wl<e.

Limits and continuity (I)

129

It is useful to note that theorems 22.30, 22.31 and 22.32 apply equally well replaced throughout by C.

if~

is

22.35 Example Consider the function f: [; -+C defined by f(z) = z3 • We shall prove that z3 -+-i as

z-+i.

Observe that i3 =i 2 i= -i. Hence lz 3 - ( -i)l = lz 3 -i 3 1= lz-il·lz 2 +zi +i 2 l (see exercise 10.24(5)). We have that lil = 1 and, if lz- il < 1, then lzl ~ lil + 1 = 2. Hence, if lz- il < 1, lz 3 - ( -i)l O is given and choose b=min{e/7, 1}. Then, if lz -il
23 t

23.tt

LIMITS AND CONTINUITY (11)

Double limits

We begin with some notation. Let X, 'lj and £ be metric spaces and let and ~E £. Suppose that A is a set in X and B is a set in 'lj and write S =A x B. We shall consider a function

~EX, lJE 'lj

f: S\ [(~,

lJ)}--->£.

'if

B

~----~----~-------+--~x

~

A

In mathematical applications one often comes across statements of the type 'f(x, y)--->~ as X->~ through A and Y--->lJ through B'. A limit specified in this manner

will be called a double limit. It is important to be. aware of the fact that statements of this sort about double limits can be highly ambiguous and that it is therefore always necessary to examine the context carefully in order to determine precisely what the author means. The most straightforward case is that in which x and y are intended to be 'independent variables'. In this case the double limit statement simply means that f(x, y)--->~

as

(x,

y)--->(~,

t])

through the set S =A x B. However, in order to use the definition of a limit given in §22.3 we need to know what sets are open in X x 'lj. We therefore augment our definition of a double limit in this case by observing that the open set~ of X x 'lj are to be those in the product topology of X x 'lj (§21.18).

130

131

Limits and continuity (1 I)

As explained in §21.18, when :r and 'lj are metric spaces with metrics d 1 and d 2 respectively, there are various metrics which we can introduce into :r x 'lj which generate the product topology of :r x 'lj and the choice of which of these metrics to use in :r x 'lj is largely a matter of convenience. When :r = 'lj = IR and so :r x 'lj = IR 2 , we usually use the Euclidean metric but, in many cases, it is easier to use the metric m: :r x 'lj --+IR defined by m((x 1 , yd, (x 2 ,

Y 2))=max{d 1 (x 1 ,

x 2 ), d 2 (y 1 ,

Y2)}.

The use of this m-::tric in the formulation of the definition of a limit given in §22.9 yields the following criterion for the existence of a double limit: For any e > 0, there exists a b > 0 such that for each (x, y)E S =A x B, O<m((~, 1]),

(x,

y))
y))<e.

This can in turn be written in the following less clumsy form: For any e > 0, there exists a b > 0 such that for any XE A and any yE B, d 1 (~,

provided that (x, y) #

x)
(~, 1]).

23.2 Example Suppose thatf: IR 2 --+IR 1 and thatf(x, y)--+( as (x, y)--+(~, IJ). If we use the Euclidean metric in IR 2 , the definition of this statement can be expressed in the form: For any e > 0, there exists a b > 0 such that (1) 2

Alternatively, we can use the metric m in IR and obtain the definition in the form: For any e > 0, there exists a 11 > 0 such that lx- ~I <11 and IY-IJI <11 = lf(x, y)- (I <e

(2)

provided (x, y)#(~, IJ). The first form of the definition says that lf(x, y)- (I< e provided that (x, y)# (~, IJ) is in a sufficiently small disc with centre(~, IJ). The second says that lf(x, y)-(l<e provided that (x, y)#(~, IJ) is in a sufficiently small box with centre (~, IJ). Both definitions are plainly equivalent because, if b=£5 1 works in (1), then 11=bd-/2 works in (2). On the other hand, if 11 works in (2), then 6=11 works in (1).

132

Limits and continuity (IJ)

23.3t

Double limits (continued) We have not yet finished with the statement 'f(x,

y)-+~

as

x-+~

through

A and y -+lJ through B'. In the discussion above we considered its meaning when x

and y were to be understood as 'independent variables'. However, the statement is often used in circumstances when x and y are not 'independent variables'- i.e. some relation between x and y must be satisfied. Sometimes one is told explicitly what this relation is. On other occasions one has to guess the relation from the context. The existence of such a relation means that there is a set R in X x ?J with the property that (x, y) not in R are to be disregarded. Our double limit statement then means that f(x,

y)-+~

as

(x,

y)-+(~.

TJ)

through the set RnS (where S=A xB and it is understood that the product topology is used in X x ?J).

23.4 Example Consider the function f: IR 2 -+IR of exercise 22.8(2). If one is asked to consider the double limit (defined by the statementf(x, y)-+( as x-+1 and y-+ 1 where x + y > 1, then one should interpret this as meaning that f(x, y)-+(

as

(x, y)-+(1, 1)

through the set J = { (x, y): X+ y> 1}. Since J c that ( =0.

23.5t

eK,

it follows from exercise 22.8(2)

Repeated limits The notation limf(x, y) ·-~ Y-'1

is available for a double limit as discussed in §23.1. We shall however prefer the less clumsy notation lim

f(x, y)

(x, y)-(1;, q)

although the latter notation is somewhat less precise in that it takes for granted the use of the product topology in X x ?J. It is important not to confuse either of these pieces of notation with repeated limits lim (lim f(x, y)); X-+~

y--+1)

lim (lim f(x, y)). y--+1)

x-~

The reasons why one needs to be careful in dealing with repeated limits are best explained with the help of some examples.

133

Limits and continuity (ll) 23.6

Example Observe that

and hence . (. x2-y2) hm hm - 2 - -2 =1. x-o

y-0 X

+y

On the other hand, a similar argument shows that x2-y2) lim ( lim - 2 - -2 = -1. y-0 x-o X + y

The order in which the limits appear cannot therefore be reversed in general without changing the result.

23.7

Example Observe that, for each xerr;l, . xy 0 II m - 2 - - 2 = . y-0 X +y

It follows that

lim(lim 2xy 2 )=0. x-o y-OX +y Similarly,

However, the double limit lim

-

(x, y)-(0, 0)

X

xy 2

-2

+Y

does not exist (for the same reason that the similar limit in example 22.11 does not exist; see also example 23.15).

23.8

Example Consider the function h: h(x, y)={x

IJ;l 2 ->IJ;l

(y;;:O)

-x (y
defined by

134

Limits and continuity (11)

and so it is easily shown that lim

h(x, y)=O.

(x, yl-(0, 0)

On the other hand, we have that h(x, y)->x as y-.0+ and h(x, y)-> -x as y-.0-. It follows that

lim h(x, y) y-o

(1)

does not exist unless x = 0 and hence that the repeated limit lim (lim h(x, x-+0

y-+0

y))

does not exist. The results of examples 23.6 and 23.7 are not particularly surprising. We have already seen a number of examples in which f(x, y) tends to different limits as (x, y)-> (0, 0) along different paths and, as the diagrams below indicate, one can think of taking a repeated limit as another special way of allowing (x, y) to approach (0, 0). lim (lim f(x.

y))

llJ

x .....o ,v ..... o

t

t

I

I

___ , __ j_ 1

+

(0, 0)

:

(x. y)

I

L-0~--:(x

y)

,--

+ I

!

t

-------1

!

~~

lim (lim f(x,

y-+0 X-+0

y))

Example 23.8 requires a little more thought. The problem is that the existence of the double limit as (x, y)-> (0, 0) does not guarantee that h(x, y) tends to a limit as (x, y)-> (X, 0) along the line x =X (unless X= 0). As the diagrams below indicate, there is no reason why matters should be otherwise.

(x,y)

(x, y)

(0, 0)

1

I

~: 11

1

"I

(X, 0)

135

Limits and continuity (Il)

However, if we remove this pitfall by restricting our attention to those functions for which the limit (1) does exist, then the existence of the double limit implies the existence of the repeated limit and the two are equal. This result is the content of the next theorem.

23.9t Theorem Let X, 'IJ and f be metric spaces and let ~EX, 1JE'IJ and ~Ef. Suppose that A is a set in X with cluster point ~ and that B is a set in 'IJ. Let f: S""(~, tt)-->f where S =A x B. Suppose that f(x,

y)--+~

as

(x,

y)--+(~,

as

x -->~

tt)

(2)

through the set S and that, for each yE B, f(x, y)--+l(y)

(3)

through the set A where 1: B-->f. Then /(y)--+~

as

(4)

Y-->Tt

through the set B.

---,------------,

I

(~, y)

I

(X, y): I

I

B

T

:

. ;

I I

I

---+------+--------1 (~. 11l :

II

I

I

1

:c A

Proof Let s > 0 be given. By (2) there exists a b > 0 such that for each xEA and each yEB d 1 (~,

provided (x, y) #

(~,

x)
=>- d(~,

f(x, y))<s/2

tt). By (3), for each yE B there exists a dy > 0 such that, for each

XEA,

0
x)
=>-

d(l(y), f(x, y))<s/2.

136

Limits and continuity (I J)

For each yeB, choose xyEA so that 0
d 1 (~,

xy)
d 1 (~,

xy)<6y. Then, if

l(y)) ;';';; d(~,f(xy, y)) + d(f(xy, y), l(y))

d(~,

<e/2+e/2=e

and hence (4) holds.

It is sometimes important to know whether or not it is true that

Iim (Iimf(x, y))=lim (limf(x,y)). •-~

Y-~

Y-~

(5)

•-~

If the equation holds, the two limiting operations are said to 'commute'. Applied mathematicians often write in a manner which would lead one to suppose that (5) is always true. But this is very definitely not the case even when both sides of the equation exist. Example 23.6 is a case in point. However, equation (5) is true sufficiently often for it to be appropriate to show some measure of surprise should it turn out to be false in a particular case. In particular, we have the following result. The proof is trivial.

23.10t

Theorem Let X, 'lj and f be metric spaces and let Sbe an open set in X x 'lj t)). Ifj: S--+f is continuous on S, then

containing(~,

Iim (lim f(x, y))=lim (lim f(x, y)). •-~

Y-"1

Y-~

•-~

To proceed any further with this topic we need to introduce the subject of uniform convergence. In particular, we shall show that, if both sides of (5) exist, then the two sides are equal provided that one of the inner limits is a uniform limit.

23.11 t

Uniform convergence

Suppose that X and f are metric spaces and that f:A x B--+ f where A is a set in X. Let ~EX and suppose that I:B--+f has the property that, for each yeB, f(x, y)--+l(y) as

X-+~

through the set A. Expressed in full detail, this assertion means that 'VyEB 'Ve>O 3b>0 'VxEA, x) d(l(y), f(x, y))<e.

O
(1)

We know from chapter 3 that the order in which the quantifiers 'V and 3 occur irt a statement is significant to its meaning. In particular, if P(u, v) is a predicate, it is in

137

Limits and continuity (Il) general false that

Vu 3v P(u, v)

= 3v Vu P(u, v).

It follows that, if we move the term 'VyeB' from the beginning of our list of quantifiers to the end, we obtain a statement which means something different from the original statement - i.e. the statement

Ve>O 3<5>0 VxeA VyeB, O
x)
= d(/(y), f(x, y))<e

does not mean the same as (1).

pointwise convergence

A

A

(2)

I 38

Limits and continuity (I I)

The difference in the meaning of the two statements is that, in statement (1), the value of b which is asserted to exist may depend both on the value of e and on the value of y. In statement (2), the value of b depends only on the value of e - i.e. the same value of b works uniformly for all values of yE B. If (2) holds, we say that f(x, y)-+l(y)

as

x-+~

through the set A uniformly for yE B. It is sometimes necessary to emphasise that this statement is stronger than (1). When (1) holds, we therefore say f(x, y)->l(y)

as

x-+~

through the set A pointwise for yEB. Note that It IS obvious that uniform convergence implies pointwise convergence. But •he converse is very definitely false.

23.12t

Distance between functions With the notation of the previous section, suppose that f(x, y)-+l(y)

as

X-+~

(1)

through the set A uniformly for yE B. Suppose that e > 0 is given. Then ej2 > 0. Hence there exists a b > 0 such that for each xEA and yEB O
x) d(l(y), f(x, y))<e/2.

Since the final inequality holds for all yEB we may conclude that, for any e>O there exists a b > 0 such that, for each XE A, O
x)sup d(/(y), f(x, y))<e YEB

i.e. sup d(l(y),f(x, y))-+0

as

x-+~

(2)

YEB

through the set A. We have shown that (1):o-(2J and it is even easier to show that (2):o-(1). This observ< ion suggests the following definition. Suppose that F: B-+f and G: B-+ f. Then we define the uniform distance between F and G by u(F, G)= sup d(F(y), G(y)) yEB

where this quantity exists.

(3)

Limits and continuity (ll)

139

B

23.13 Example Consider the function F: [0, 1]->IR defined by F(y)= 8y 2 + 1 and the function G: [0, 1]-.IR defined by G(y)=6y. The uniform distance between F and G is given by u(F, G)= sup jF(y)-G(y)l o;;;y~t

= max 18y 2 -6y+ll. o;:;y;:;t 2

The maximum of H(y)=l8y -6y+ll is either attained at y=O or y=1 or else at a point IJE (0, 1). In the latter case, differentiation yields that 161]-6=0 IJ=i.

Since H(O)= 1, H(1)= 3 and H(il=t, it follows that u(F, G)=3.

t 3

t

140

Limits and continuity (I J)

In terms of the uniform distance notation, statement (2) assumes the less clumsy form, u(l, f(x,• ))__..0

as

x_..~

(4)

through the set A, where f(x,•): B_..f. is the function whose value at ye B is f(x, y). u(l, f(x, •))

If F and G are bounded on B then u(F, G) defined by (3) always exists and the set of all bounded functions on B is a metric space with u as metric. (See § 13.18.) When dealing with bounded functions we may therefore rewrite (4) in the simple form: (5)

through the set A. It is often helpful to 'forget' that f(x,e) and l are functions and to think of them instead as 'points' in the metric space of bounded functions on B.

space of bounded functions onB

141

Limits and continuity (Il)

It is important, however, not to forget that the distance between the 'points' f(x, •) and l is defined by u(l, f(x, •)) =sup d(l(y), f(x, y)).

(6)

yeB

The above discussion is intended to indicate why uniform convergence is in some respects a more straightforward conr-ept than pointwise convergence although it seems at first sight as though the opposite were the case. However, the discussion also has some practical benefits in that it is often easiest when seeking to establish uniform convergence to begin by calculating or estimating (6).

23.14

Example Let A =IR""'- {0} and define g: A x IR -+IR by x3+x2y+l g(x, y)=

X

2+

y

2

We begin by observing that, for each yEIR, as

g(x, y)-+y

x-+0

-i.e. g(x, y)-+y as x-+0 pointwise for yEIR. It follows that, if g(x, y)-+l(y) as x-+0 uniformly for yEIR, then it must be the case that I( y) = y. Consider u(l, g(x, •)) =sup 11( y)- g(x, y)l. yeUl

Taking l(y)= y, we obtain that, for each xEA, ll(y)-g(x, y)l=

I

y-

x3+x2y+l x 2 +i

lxl 3 x2+y2"

Hence, for each xEA, u(l, g(x,•))=sup yeUl X

lxl 3 +Y

- - =lxl. 2 2

Since lxl-+0 as x-+0, it follows that g(x, y)-+y as x-+0 uniformly for yEIR. The diagram over illustrates the set of (x, y) for which it is true that I y- g(x, y)l <e. Observe that the choice b = e ensures that o < lxl < b =>I y- g(x, y)l < e for all yEIR.

142

Limits and continuity (JJ) y

23.15

Example Let A =llli '\ [0} and define

h:

A x IRi ->llli by

xy h(x, y)=-2--2. X

+y

We begin by observing that, for each yEIRi, h(x, y)-->0

as

x-->0

- i.e. h(x, y)-->0 as x-->0 pointwise for yEIRi. It follows that, if h(x, y)-->1( y) as x-->0 uniformly for yEIRi, then I( y)=O. Observe that, for each xEA,

u(O, h(x,• ))=sup yEll

10-~~ X

+Y

-1 -2.

(The geometric-arithmetic mean inequality gives lxyl ~ !(x 2 + y 2 ) and equality is attained when x = y.) Since u(O, h(x, •l)+O as x-->0, it follows that it is not true that h(x, y)-->0 as x-->0 uniformly for yEIRi. The diagram below illustrates the set of (x, y) for which it is true that lh(x, y)l < e when 0 < e
Limits and continuity (II)

143 y

It is clear that, if 0 < 8 < -!, then there is no value of .5 > 0 such that for all XE A and all yEIKl Olh(x, y)l<8. If y # 0, the largest value of .5 > 0 for which it is true that 0
.5 =28{1 +J(l-48 2 )} -I y.

Theorem Let X, ?j and f be metric spaces and let ~EX, T)E?j and ~Ef. Suppose that A is a set in X and that B is a set in ?j. Letf: S\(~, TJ)-->f where S =A x B and let /: B--> f. Suppose that

23.16t

f(x, y)-->/(y)

X->~

as

(7)

through the set A uniformly for yE B and that

l(y) -->~

as

y -->TJ

y)-->~

as

(x,

(8)

through the set B. Then f(x,

y)-->(~,

TJ)

through the set S.

Proof Let XEA and each yEB,

8>

0 be given. By (7) there exists a .5 1 > 0 such that, for each

0
x) d(l(y), f(x, y))<8/2.

By (8), there exists a .5 2 > 0 such that for each yE B 0

d(~,

l(y))<8/2.

144

Limits and continuity (II)

Choose b=min{b 1 , b2 }. Then, if O
1(y))+d(1(y),f(x, y))

<e/2+e/2=e.

The next theorem justifies the sentence at the end of § 23.5 about the circumstances under which one can reverse the order of the two limiting operations in a repeated limit.

Theorem Let X, 'lj and £ be metric spaces and let ~EX, 1]E'/j and ~E £. Suppose that A is a set in X for which ~is a cluster point and that B is a set in 'lj. Let f: S \(~, 1])-+£ where S=A x B. Let 1: B-+£ and let m: A-+£. Suppose that (i) f(x, y)-+1(y) as X-+~ through the set A uniformly for yEB, (ii) 1(y)-+~ as y-+1] through the set B, (iii) f(x, y)-+m(x) as y-+1] through the set B pointwise for XEA. 23.17t

Then

m(x)-+~

as

X-+~

through the set A.

Proof The theorem is simply a combination of theorems 23.9 and 23.16. 23.18

Example Let g be defined as in example 23.14. We have that

(i) g(x, y)-+ y as x-+0 uniformly for yEIR. (ii) y-+0 as y-+0. (iii) g(x, y)-+x as y-+0 for XEIR \ {0}. From theorem 23.17 we may conclude that lim lim g(x, y)=Iim lim g(x, y). x-o y-o This is easily verified directly.

23.19t

Exercise

(1) Consider the function[: IR 2 \{(0, 0)}-+IR defined by x2-y2 f(x, y)=-2--2 · X +y

Prove the following: (i) f(x, y)-+1 as (x, y)-+(0, 0) along the line y=O.. (ii) f(x, y)-+ -1 as (x, y)-+(0, 0) along the line x=O. (iii) f(x, y)-+0 as (x, y)-+ (0, 0) along the line x = y.

Limits and continuity (II)

145

Deduce thatf(x, y) does not tend to a limit when x--+0 and y--+0 independently. (2) Let f be as in question 1. If 0 < s < 1, prove that lf(x, y)l < s if and only if

1- s <

2

(~) < 1 + s .

l+s

1-s

x

Prove that f(x, y)--+0 provided that. x--+0 and y--+0 in such a manner that 1 -lxl < 1+ I Yl

(l'_) 2 < 1 +I Yl . x

1 -lxl

(3) For each of the functions f:(IR\ {0}) x IR --+IR given below, decide whether or not it is true that lim (lim f(x, y))=lim (limf(x, y)). x-o y-o y-0 x-o

. x2-y2 (I)f(x, y)=l+x2+y2 x3+ y4

...

(m) f(x, y)=-2- 4 X

+y

.. x2y4 (n)f(x, y)= x2+y4 . x3+ y6 (1v) f(x, y)=-2- 4. X +y

(4) In each case considered in question 3, find a function 1: IR --+IR such that f(x, y)--+1( y)

as

x--+0

pointwise for yEIR. (5) In each case considered in question 3, decide whether or not it is true that

f(x, y)--+1( y)

as

x--+0

uniformly for (a) yEIR and (b) yE [ -1, 1]. (6) A functionf: IR 2--+IR 1 has the property that

(x, ~x 2 and Yt ~Y2)=>f(x,, Yt)~(x2, Y2). Prove that lim ( lim f(x, y))= lim ( lim f(x, y)). x-~+

23.20t

Y-'1+

Y-'1+

x-~+

Uniform continuity

Let 1: and ?J be metric spaces and suppose thatf:S--+?J where S is a set in 1:. To say thatfis continuous on S is equivalent to the assertion that, for each ~ES, f(x)--+f(~)

as

x--+~

through the set S. This means, in turn, that V~ES d(~,

lls>O 3<5>01ixES,

x)

d(f(~),

f(x))<e.

146

Limits and continuity (11)

In this statement the value of b > 0 which is asserted to exist may depend both on the value of £>0 and on the value of I;ES. If, given £>0, a value of b>O can be found which works for all I;ES simultaneously, then we say that f is uniformly continuous on the set S. Thus f is uniformly continuous on the set S if and only if

't£>0 :lb>O \lxES \II;ES, d(l;, x)d(f(l;),f(x))<E.

23.21

Example The function

f:

[0, oo)->[0, oo) defined by

f(x)=Jx is uniformly continuous on [0, oo ).

Proof It is evident from the diagram that 1Jx-J~1 is largest for x~O, and lx- ~I~ b when ~ = 0 and x =b. (This is easily checked analytically by proving that Jb-Ja~(b-a) 112 when O~a~b.)

~ ~0

0

It follows that, given any any x~O and any y~O,

23.22

- ., E>

X

____

0, there exists a b > 0 (namely b = ~; 2 ) such that, for

Example The function f: (0, oo)->(0, oo) defined by 1

f(x)=-

x

is not uniformly continuous on (0, oo ).

Limits and continuity (I I)

147

y

Proof Given 8 > 0, the largest value of b > 0 for which lx-~l
1 1 =---<s ~

X

satisfies

1

1

----=8

~-b

~

b-

Hence Since this expression tends to zero as

. 8;1'2

.,

1 +8~ ·

~--->0+

it follows that no b>O can work for all

~>0.

23.23t Theorem Let X and 'lj be metric spaces and suppose that f: K--->'lj, where K is a compact set in X. If f is continuous on K, then f is uniformly continuous on K. Proof Since f is continuous on K,f(x)--->f(y) as x--->y through K for each yEK. Let 8>0 be given. Then h>O and so. for each yEK, there exists a b(y)>O such that, for each xE K, d(y, x) < b(y) = d(f(y), f(x)) <±8.

(1)

Let U denote the collection of all open balls with centre yE K and radius !o(y). Then U covers K. From the definition of a compact set (§20.4) it follows that a finite subcollection .'i of U covers K. Let b > 0 be the radius of the open ball of minimum radius in the subcollection :J. Now suppose that x and ; are any points of K which satisfy d(;, x)
(2)

Limits and continuity (JJ)

148

Also, d(y, x)~d(y, ~)+d(~, x)
(3)

Combining (2) and (3), we obtain that d(f(~),f(x))<e.

We have therefore shown that, for any e > 0, there exists a b > 0 such that, for each XEK and each ~EK, d(~,

x)
=>

d(f@, f(x))<e

and so the proof of the theorem is complete.

23.24t

Exercise

(1) Determine which of the following functions f: (0, tinuous on the interval (0, 1). (i) f(x)=x 2 (ii) f(x)=(1-x)- 2 (iii) f(x)=lx-tl (iv) f(x)=x- 113 •

1)-->~

are uniformly con-

.X and ?J be metric spaces and suppose thatf:S->?J where S is a set in .X. Let g 1 : [0, 1]->S and g 2 : [0, 1]->S and suppose that d(g 1 (t), g 2 (t))->O as t->0+. If! is uniformly continuous on S, prove that

(2) Let

Interpret this result geometrically. Show that the result need not be true if f is only continuous on S. (3) Let {a 0 , a 1 , ••• , a"} and {b 1 , b2 , •.. , bn} be two sets of real numbers and suppose that 0=a 0 ~ which satisfies ak_ 1 ~xg(x)=bk

(k=1,2, ... ,n)

is said to be a step function. Prove that any continuous function f:[O, 1] --~ can be uniformly approximated on [0, 1] by a step function - i.e. given any e > 0, there exists a step function g: [0, 1] ...... ~ such that u(f, g)= sup lf(x)-g(x)l<e. O~x~l

24

24.1

POINTS AT INFINITY

Introduction

The notion of 'infinity' is one which has received much attention from philosophers, theologians and assorted mystics of one persuasion or another throughout the ages. By and large, however, their writings have little useful to offer except the observation that 'infinity' can be used to 'prove' almost anything provided that one is willing to argue sufficiently badly. We have already taken note of mathematical howlers of the type oo+oo=oo :. 2xoo=oo 2= 1. Here the fallacy is readily traceable to the fact that the symbol oo is treated as though it were a real number and, whatever oo may represent in a particular context, it certainly never represents a real number. Philosophical howlers tend to be slightly more subtle and to come attractively giftwrapped in verbal packaging. But they share the same fatal flaw as the elementary mathematical howlers in that they assume that 'infinity' can be treated as though it were some more familiar object without offering any justification for this assumption. As mathematicians, we know that there is no point in seeking to uncover the secrets of the universe by deductive reasoning alone. The best we can hope to do is to set up a consistent axiom system from which ore can deduce the existence and the properties of objects which it is sensible to interpret as 'infinite'. These objects may or may not have a referent in the 'real world' but this is none of our concern. Such questions can be safely left to the philosophers. As soon as one tackles the problem from this point of view, it becomes apparent that the word 'infinity' does not apply to a single concept. It applies to a whole bundle of rather diverse concepts and before one can say anything useful on the subject one first has to specify what sort of 'infinity' one has in mind. Chapter 12 contains an account of one approach to the problem of

149

150

Points at infinity

infinity. This theory is the creation of the great German mathematician Cantor. As we have seen, he offered a precise definition of an infinite set and then proceeded to the observation that some infinite sets are 'larger' than others. This led to the introduction of the transfinite cardinals ~ 0 , ~ 1 , ~ 2 , ... for the classification of infinite sets. The set {a, b, c} has cardinality 3 (because it has three elements). The smallest type of infinite set (e.g. Not 0) has cardinality ~ 0 (aleph-zero). The next largest type of infinite set has cardinality ~ 1 and so on. Note, incidentally, the richness of the structures discovered by Cantor. Contrary to popular opinion, the removal of romantic misconceptions from intuitive ideas seldom leads to something less exciting (unless the original idea is essentially empty). One may think of Cantor's infinities as 'set-theoretic infinities'. They are useful for describing the number of elements in a set. But there are other essentially distinct types of infinity. As an example we mention the 'geometric infinities' used in translating statements from Euclidean geometry into statements of projective geometry. Here the Euclidean plane IR 2 is augmented by inventing an extra 'line' called the 'line at infinity' on which all parallel lines meet. A pair of parallel lines can then be treated just like any other pair of lines with a consequent gain in simplicity.

In this chapter we shall be concerned with what might be called 'topological infinities'. These are the infinities which appear in statements like

f(x)-

+ oo

as

x+- oo.

The theory of such infinities is rather humdrum compared with the theories mentioned above. However, it is a theory which requires careful attention since the opportunities for confusion are considerable. Our treatment will be an elementary one but some readers may find it helpful to skip through or to review the topological ideas introduced in chapter 21.

24.2

One-point compactification of the reals

The real number system IR has many satisfactory features but there is one respect in which it is not so satisfactory. As we have seen, compact sets have some very useful properties but IR is not compact because it is not bounded (theorem 20.12). What can be done about this?

151

Points at infinity

The natural mathematical response to such a problem is to ask whether it is possible to fit IR inside some larger system X which is compact. It turns out that it is quite easy to find an appropriate X. Indeed, one can find an X which has only one more point than IR. This is called the one-point compactification of IR. For most purposes, however, it is very much more useful to employ a two-point compactification of IR. (See §24.4.) We begin by considering the one-point compactification. This consists of a compact space IR # which contains IR as a subspace but has only one point more than IR. For reasons which will soon be apparent we denote this extra point by oo. Recall from §21.4 that IR is topologically equivalent to a circle C in IR 2 with one point N deleted. The stereographic projection f: IR --+C \N indicated below provides an appropriate homeomorphism. N

z

(0, 0)

It is now perfectly natural to invent a new object to be mapped onto N. What we use for this new object is entirely irrelevant provided only that we do not use a real number. Whatever our choice for this new object, the symbol we use to denote it is oo and we write IR # =IR u{ oo }. The function f: IR --+C\ {N} is extended to be a function g: IR # --+C by defining

g(x)={f(x) N

X

(xEIR) (x= oo)

y

152

Points at infinity

We want IR # to be compact. But unless IR # is a metric (or a topological) space this assertion will be meaningless. We make the obvious choice of a metric c: IR # --+IR for the space IR # by defining

c(x, y) = d(g(x), g( y)).

(1)

Thus the distance between x and y regarded as elements of IR # is the Euclidean distance between the points g(x) and g( y) in IR 2 • If x and y are real numbers,

If z is a real number, 1 c(z, oo)=c(oo, z)= ( +z 2 ) 112 • 1

In view of (1), the function g: IR #--+C is a homeomorphism. Since C is compact it therefore follows that IR # is compact (theorem 19.13). It may be helpful to think of IR # as being obtained from IR by squeezing the real line into an open line segment, bending this into a circle and then plugging the resulting gap with oo.

schematic diagram of [R#

Notice that the distance c(x, y) between x and y regarded as elements of IR # is not the same as the Euclidean distance between x and y - i.e. c(x, y) =1-lx- yl. It follows that IR (with the usual Euclidean metric) is not a metric subspace of IR #. In what sense, then, is it reasonable to say that IR 'fits inside' IR #? The answer is that IR is a topological subspace of IR #. In fact, a subset of IR is open relative to the space IR # if and only if it is open in the usual sense. This is a simple consequence of the fact that f: IR--+C'\{N} is a homeomorphism and hence both f andf- 1 preserve open sets (theorem 21.7). The fact that IR fits inside IR # as a topological structure but not as a

153

Points at infinity

metric structure means that it is a mistake to attach much significance to the metric c: IR # --+IR. This is useful only as an auxiliary device. What matters is the topology of IR # - i.e. the nature of its open sets. Since our convergence definitions have been framed in terms of open sets, this is, in any case, precisely what we need to know for applications. The easiest way to describe the open sets of IR # involves an abuse of the term 'open ball'. A set G in a metric space .X is open if and only if, for each ~E G, there exists an open ball B with centre ~ such that B c G (theorem 14.12). This criterion remains valid for an open set Gin IR # if we think of an 'open ball' in IR # as an interval of the form (~- (J, ~ + fJ) when its centre is a real number ~ and as a set of the form {oo}u{x:xEIR and lxi>X} when its centre is oo. (The abuse lies in using the Euclidean metric when the centre is a real number and the IR # metric when the centre is oo. When indulging in this abuse we shall write 'open ball' in inverted commas.)

24.3

The Riemann sphere and the Gaussian plane

The discussion given above of the one-point compactification of IR works equally well when employed with IR". The open sets G in the space IR"# =IR"u{ oo} are those with the property that, for each ~EG, there exists an 'open ball' B with centre ~ such that B c G. Here an 'open ball' is to be understood as a Euclidean ball when its centre ~EIR" -i.e. B={x: llx-~llX}u{oo} when its centre is oo. The case n = 2 is particularly important since IR 2 can be identified with

154

Points at infinity

the system C of complex numbers (§ 10.20). The one-point compactification of !R1 2 is called the Riemann sphere or the Gaussian plane depending on whether one prefers to think of !1;11 2 as squeezed and wrapped into a sphere with oo plugging the gap or as a plane with oo floating ethereally in some nameless limbo. The link between these viewpoints is, as befor-e, the stereographic projection. (See §21.4.)

'open balls' on Gaussian plane

We mentioned in §24.2 that the one-point compactification of IR1 is not the most useful possible compactification of IR1. But this ceases to be true for IR1n when n ~ 2. In particular, one always uses the one-point compactification of !R1 2 when working in complex analysis and many authors on this subject write as though oo were a member of C -i.e. they use C where we would write C #.However, they are seldom entirely consistent in this practice.

24.4

Two-point compactification of the reals

Our discussion of the one-point compactification was based on the fact that the stereographic projection provides a topological equivalence between IR1 and C \ {N}. A somewhat more natural topological equivalence, however, is that which holds between IR1 and the open interval ( -1, 1). (See §21.3.) The diagram below illustrates a homeomorphism which can be used to demonstrate this topological equivalence.

X

Points at infinity

155

It is natural here to augment~ by inventing two further objects which we shall denote by - oo and + oo (rather than one extra object as in §24.2). We shall write

[-oo,

+oo]=~u{-co,

+eo}.

We then define an open set in [-eo, +eo] to be a set G with the property that, for each eEG, there exists an 'open ball' B with centre such that B c G. Here an 'open ball' is an interval of the form (e- <5, + <5) when its centre is a real number but is a set of the form

e

e

e

(X, +oo)u{ +eo}

when its centre is

+ oo

and a set of the form

(-oo, Y)u{-co} when its centre is -eo.

-00·

~

~

w~~

'open

balls'~n

/

•+oo

[ -00, + oo]

This definition for an open set in [-eo, + oo] makes the function g : [- oo, +eo]--> [- 1, 1] illustrated below a homeomorphism. Thus [- oo, +eo] is topologically equivalent to [ -1, 1]. In particular, [-eo, +eo] is compact. -I

g(x)

Moreover, ~ is a topological subs pace of [-eo, +eo]. In fact, a subset of~ is open relative to [- oo, + oo] if and only if it is open in the usual sense. When discussing the one-point compactification of ~, we made no attempt to extend anything but the topological structure of ~ to ~ #. In particular, there is no way in which the order structure of~ can sensibly be extended to ~ #. But this is not true of the two-point compactification of~

156

Points at infinity

and the order structure of IRl extends in a very natural way to [- oo, + oo]. An ordering ~ on [- oo, + oo] is obtained by requiring that, for all XE [ - 00,

+ 00], -00 ~X~+ 00

and that, for all real numbers x and y, x ~ y if and only if this is true in· the usual sense. This usage has considerable notational advantages. For example, in [- oo, + oo] all non-empty sets have suprema and intima. If S is a nonempty set of real numbers which is bounded above in IRl (i.e. which has a real number as an upper bound), then sup S has its usual meaning. If S is unbounded above in IRl, then its smallest upper bound in [- oo, + oo] is + oo and hence supS=+ oo. Thus sup S = + oo is a shorthand way of saying that S is unbounded above in IRl. Note also that sup 0=- oo. The reason is that all elements of [- oo, + oo] are upper bounds for 0 (because xE 0 is always false and therefore xE0=>x ~Y is always true). With + oo and - oo available, one can therefore use the notation supS without first needing to check that S is non-empty and bounded above. Similar remarks, of course, apply in the case of inf S. There is also some advantage in extending the algebra of IRl to [- oo, + oo]. One cannot, of course, extend all of the algebraic structure of IRl to [- oo, + oo] since then [- oo, + oo] would be a continuum ordered field and we know from §9.21 that IRl is the only such object. However, one can extend some of the algebraic structure of IRl to [- oo, + oo]. We make the following definitions.

X+ ( + 00) = ( + 00) +X = + 00 } x-(- oo)= -(- oo)+x= + oo

(provided xi= - oo)

(ii) x+(- oo)=(- oo)+x=- oo } x-( + oo)= -( + oo)+x=- oo

(provided xi= + oo)

(i)

(iii) x( + oo)=( + oo)x= + oo} x(- oo)=(- oo)x=- oo (iv) x/+oo=O} x/- oo =0

(prov1.d ed x> O)

. (provided x i=

± oo ).

Perhaps the most significant feature of these definitions are those items which are omitted. Observe that (+oo)+(-oo), (+oo)-(+oo), O(+oo),

157

Points at infinity

0(- oo ), + oo I+ oo and other expressions are not defined. There is no sensible way to attach a meaning to these collections of symbols and we do not attempt to do so. This fact means that there is little point in attempting algebraic manipulations involving + oo and - oo. Definitions (i}-(iv) should therefore be regarded as handy conventions rather than as a basis for a serious mathematical theory. Note also that division by zero remains unacceptable. As an example of the use to which these conventions may be put, consider two non-empty sets S and T of real numbers and two non-empty sets P and Q of positive real numbers. We then have that (i) sup(-x)=-(infx) xeS

(ii)

sup (x. y)eS x T

(iii)

sup ~x.

y)eP x Q

xeS

(x +

y) =(sup x) +(sup y) xeS

yeT

(xy) =(sup x) (sup y) xeP

yeQ

without any need for assumptions about the boundedness of the sets concerned provided that the suprema and intima are allowed the values + oo or - oo (See exercise 9.10(9).) Note, however, that some vigilance is necessary if one or more of the sets is empty. For example, ifS=(/) and T is unbounded above, the right-hand side of (ii) becomes meaningless. The system [- oo, + oo] with the structure described in this section is called the extended real number system. It should be pointed out that some authors use IR # (which we have used for the one-point compactification) to stand for [- oo, + oo]. Since other authors use IR # for yet other purposes, some element of confusion is inevitable in any case.

24.5

Convergence and divergence

Let S be a set of real numbers. Usually it would be convenient to regard S as a set in the space IR but, in what follows, we shall wish to regard S as a set in the space [- oo, + oo]. Next consider a function f: S-+[- oo, + oo] for which f(S)c!R. Again, one would normally regard f as taking values in IR but here it is convenient to regard f as taking values in [- oo, + oo].

158

Points at infinity e+oo I

I

I I I

~

f

I

•

I

~s I I I I

e-oo

If I;E [-eo, + co] and IJE [-eo, +eo], it makes sense to ask whether or not f(x)->1J as x--->1; through the set S. The definition is the same as always. (See §22.3.) For any open set G in [- co, +eo] containing IJ, there must exist an open set H in [- co, + eo] containing ~ such that

xEHnS

=>

f(x)EG

provided that x ;6 ~. As we found in §22.4, it is more helpful for technical purposes to rephrase this definition in terms of 'open balls' in [-eo, + co]. We obtain that f(x)->1J as x-> ~ through S if and only if: For each 'open ball' E in [- co, +eo] with centre IJ, there exists an 'open ball' Ll in [-eo, +eo] with centre ~ such that XELlnS

=>

f(x)EE

provided x ;6 ~. If ~ and 1J are real numbers, this definition is exactly the same as the familiar definition of §22.4. But if~ or 1J are + co or -eo then something new is obtained. If we insert the relevant definition for an 'open ball' given in §24.4 and bear in mind thatftakes only real values, we obtain the criteria listed in the table below. In this table l and a are to be understood as real numbers. There is little point in committing these criteria to memory since it is easy to work out in any particular case what the form of the definition must be. There is some point, however, in acquiring a feeling for the intuitive meaning of the different statements. Consider, for example, the statement j(x)->l as X->+ eo'. One can think of f(x) as an approximation to l. Then lf(x) -11 is the error involved in using

159

Points at infinity

f(x) as an approximation for I. The statement 'f(x)-->1 as x-> + oo' then tells

us that this error can be made as small as we choose (i.e. less than e) by making x sufficiently large (i.e. greater than X). Similarly, 'f(x)->- oo as x->- oo' means that we can make f(x) as negative as we choose (i.e. less than Y) by making x sufficiently negative (i.e. less than X). f(x)->l as x->a

'v'e>O 3b>O 'v'xES, O
through S f(x)->l as X->+ oo

through S

'v'e>O 3X 'v'xES, x>X ~ lf(x)-ll<e 'v'e>O 3Y'v'xES, x
f(x)->

+ oo as X->a through S

'v'X 3b>0 'v'xES, OX

f(x)->- oo as x->a

'v'Y 3b>0 'v'xES, O
through S f(x)->l as X->- oo

through S f(x)->

+ oo

'v'X 3Y'v'xES, x>Y ~ f(x)>X

as X->+ oo

through S f(x)->

+ ooas X->- oo

'ifX :lY'ifxES, xX

through S

'v'Y 3X 'v'xES,

f(x)->- oo as X->+ oo

x>X ~f(x)
through S

'v'Y 3X

f(x)->- oo as X->- oo

V XES,

x<X ~f(x)
through S

f

f(x)-+l as x-++

24.6 Example Consider the f(x) = 1/x. We have that (i) 1/x-->0 as x-> + oo (ii) 1/x-->+oo as x-->-0+

oo

function

f: IR"' { 0} ->IR

defined

by

160

Points at infinity

(iii) 1/x--+-oo as x--+0(iv) 1/x--+0 as x--+- oo. To prove (i) we have to show that, for each e>O, there exists an X such that

x>X

=l~l<e.

Clearly the choice X= 1/e suffices. To prove (iii), we have to show that, for each Y, there exists a D> 0 such that 1

-D<x
Any value of D> 0 is adequate for this purpose when Y ~ 0. When Y < 0, the choice D= - 1/Y suffices.

X

Note that in spite of (ii) and (iii) we say thatf(x) diverges as x approaches 0 from the right and that f(x) diverges as x approaches from the left. The reason is that there exists no real number 1 for whichf(x)--+1 as x--+0+ and no real number m for which f(x)--+m as x--+0-. If we restrict our background space to be the real number system IR, it is therefore untrue to say that f(x) converges as x--+0+ or x--+0-. We next consider a setS in IR""' and a function/: S--+!Rm"'. (Recall that IR""' is the one-point compactification of IR".) We are interested in the case when Sc!R" andf(S)c!Rm. If ~EIR""' and fiEIRm"', we have thatf(x)--+11 as X-+~ through S if and only if: For each 'open ball' E in !Rm"' with centre fl, there exists an 'open ball' A in IR""' with centre ~ such that xEAnS =f(x)EE provided x #

~-

161

Points at infinity

If we insert the relevant definition for an 'open ball' given in §24.2 and bear in mind thatftakes values only in !Rim, we obtain the criteria listed in the table below. In this table, I and a are to be understood as vectors in !Rim and !RI" respectively. Ve>O 3()>0 VxES O llf(x)-lll<e Ve>O 3X VxES llxii>X => llf(x)-lll<e VX 3b>0 VxES O llf(x)II>X 1:/ X 3¥1:/xES llxll > Y => llf(x)ll >X

f(x)-+1 as x-+a through S f(x)-+1 as x-+ oo through S f(x)-+ oo as x -+a through S f(x)-+oo as x-+oo through S

Example Consider the function/: IR1 2'\_{(0, 0)}-+IR1 2 defined by

24.7

-y ) X f(x, y)= ( x2+y2' x2+y2 · We have that (1)• (ii)

(

X - 2- - -y - -) --+(0, 0) X +y 2 X2 +y 2

(~ - y )--+oo X +y X2 +y 2

as

as

(x, y)-+ oo

(x, y)--+(0, 0).

Proof To prove (i), we have to show that, for any e > 0, there exists an X such that

ll(x, y)ll >X

=>

llf(x, y)- 011 < e

-i.e.

(x2+ y2)>X2

=>

(~)2 +( 2- y 2)2 <e2 X +y X +y

-i.e.

(x2+y2)>X2

=>

---<e2. x2+y2

1

The choice X= 1/e clearly suffices. To prove (ii) we have to show that, for any X, there exists a (j > 0 such that 0 < ll(x, y)- (0, 0)11 < (j

If X~ 0 any value of that

(j

=>

llf(x, y)ll >X.

> 0 is adequate. Otherwise we require a

(j

> 0 such

162

Points at infinity

and the choice b = 1/X clearly suffices. The diagram illustrates (i). The function value f(x, y) lies in the disc with centre (0, 0) and radius e > 0 provided (x, y) lies outside the disc with centre (0, 0) and radius X.

This example, incidentally, is easier if complex notation is used. Since

x+iy

(x-iy) (x+iy) (x-iy)

x-iy x 2 +y 2 '

we can identify the given function with the function f: I[\ {0}--->IC defined by f(z)= 1/z. To prove (i) we then have to show that, for any e>O, there exists an X such that 1 lzl >X=>-< e. lzl

When working with 1Rl 1 there is sometimes room for confusion over whether the one-point or two-point compactification is in use especially since it is not uncommon for authors to write oo where we have been writing + oo. Indeed, in earlier chapters, we have been using the notation (a, oo) = {x: x > 0} rather than (a, + oo) since the latter notation seems painfully pedantic. In those exceptional cases when one wishes to make use of the one-point compactification of IRl it is therefore as well to employ notation of the type

f(x)--->1 or

as

lf(x)l---> oo

lxl--->oo

as

X->~

in order to avoid any possible confusion. Similar notation is also sometimes useful when working in IRln in order to emphasise the use of the onepoint compactification.

Points at infinity 24.8

163

Combination theorems

In the previous section we have explained how some problems concerning the divergence of functions f: !R:n--+!R:m can be replaced by convergence problems by replacing IR: by [- oo, + oo] and !R:n by W"'. A natural next question is to ask to what extent the combination theorems of §22.25 carry over to the new situation. There is no problem at all when [- oo, + oo] or IR: n # replaces the metric space X in these theorems. Thus, for example, theorem 22.30 shows that, if / 1(x)--+1 1 as x--+- oo and / 2(x)--+1 2 as x--+- oo, then aj1(x)+bf2 (x)--+al 1 +bl 2

as

x--+-oo.

However, there are a number of cases not covered by the theorems of §22.25 The most important of these cases is the subject of the next result.

24.9 Proposition Let X be a metric (or a topological) space and let S--+[- oo, + oo] and/2: S--+[- oo, + oo], where S is a set in X. Let ~EX and suppose that

j~:

and

/ 1(x) -+Ill

as

x --+~

/2(x) --+1] 2

as

x --+~

through the set S. Then, if a and b are real numbers (i) (ii) (iii)

aj1(x)+b/2(x)--+aiJ1 +hrr2 / 1(x)f2(x)--+IJ1'12 / 1(x)//2(x)--+IJ d'12

as x--+~ through the set S, in each case for which the right-hand side is a meaningful expression in the sense of §24.4.

It is sometimes useful in addition to note that, if 0 < 17 1 ~ + oo,j2(x) > 0 for xES and 17 2 =0, then

/ 1(x)//2(x)--++oo through S. Similarly, if

-00~17 1

as

x--+~

<0,/2(x)>0 for xES and 1] 2 =0, then

/ 2(x)//2(x)--+- oo

as

x --+~

through S. It is dangerous to assume the truth of results other than those explicitly mentioned above. For example, it is tempting to assume that, ifj1(x)--+ + oo as x--+0+ and / 2(x)--++oo as x--+0+, then / 1(x)//2(x)--+1 as x--+0+. However, as the examples below indicate, this result is false in general. (i) / 1(x)=x 2, / 2(x)=x

164

Points at infinity

(ii) f 1 (x)=2x, / 2 (x)=x (iii) / 1 (x)=x, f 2 (x)=x 2 . 24.10 Exercise (1) Prove the following results:

(i) sup (2x + 3) = + oo

(ii) sup (2x + 3) = - oo

x> 0

(iii)

inf

x2<0

(xy)=-oo (iv)

x>O,y
sup

(y-x)=+oo.

x>O,y>O

(2) Let f: IR \ {1} -->IR be defined by x+1 f(x)=- . x- 1 Prove the following results: (i) f(x)-->1 (ii) f(x)--> 1 (iii) f(x)--> 1 (iv) f(x)--> + oo (v) f(x)-->- oo (vi) lf(x)j-->oo (vii) xf(x)--> + oo (viii) xf(x)-->-00 (ix) lxf(x)l--> oo

as as as as as as as as as

X->+ oo X->- oo Ixi-> oo X-> 1 + X-> 1x-->1 X->+ oo x-->-00 lxl--> oo.

(3) (i) Let f: IR 2 -->IR be defined by f(x, y) = (1 + x 2 + y 2 ) -t. Prove that

f(x, y)-->0 as (x, y)-->00. Consider also the function g: IR 2 -->IR defined by g(x, y)=(l +(x- y) 2 )- 1 . Prove that g(x, y)-->1 as (x, y)-->oo along the line x= y. Does g(x, y) approach a limit as (x, y)-->oo? (ii) The function h: (0, oo)-->IR 2 given by h(t)=(1jt, t) defines a curve in IR 2 • Sketch this curve. Prove that

as t-->0+ as t--> + oo. (4) A function f: IR -->IR has the property that f(x)-->1 as X->+ oo and f(x)-->m as X->- oo where l and m are real numbers. Define g: [ ~ oo, + oo] -->IR by f(x) (xEIR) g(x)= l . (x= + oo) m (x=-oo). (a) h(t)-->00

(b) h(t)--> oo

l

If f is continuous on IR, explain why g is continuous on [- oo, oo]. Deduce the following results:

(i) The set f(!R)u{l, m} is compact. (ii) The function f is bounded on IR. (iii) If there exists an xEIR such that f(x)~l and

f(x)~m,

Points at irifinity

165

then f achieves a maximum on IR. t(iv) The function f is uniformly continuous on IR (§23.20).

(5) The metric space IR # with metric c: IR # --+IR was introduced in §24.2. Explain why a set G in IR # is open if and only if, for each ~ E G, there exists an 'open ball' B with centre ~ such that B c G. Here 'open ball' has the meaning assigned in §24.2. (6) Prove proposition 24.9 Suppose that f 1 (x)--+ + oo as x--+- oo and f 2 (x)--+ + oo as x--+- oo. Give examples to show that neither of the following statements need be true. (i) / 1 (x)-/2 (x)--+0 (ii) / 1 (x)//2 (x)--+1

24.11 t

as as

x--+- oo x--+- oo.

Complex functions

We have studied the one-point compactification of IR 2 and observed that it should always be assumed that this compactification is in use when IR 2 is identified with the system I[ of complex numbers. Since I[ has a rich algebraic structure (i.e. I[ is a field), it is natural to seek to extend some of this structure to I[ # just as we extended some of the algebraic structure of IR to [- oo, + oo ]. There is no point, of course, in seeking to extend the order structure of I[ since I[ has no order structure. (See § 10.20.) In this context, we make the following definitions: (i) X+ 00 = 00 +X= 00 (ii) X • 00 = 00 • X= 00 (iii) x/ oo = 0 (iv) x/0= oo

(x# oo) (x#O) (x#oo) (x#O).

With these definitions propositiOn 24.9 remains valid when I[# replaces [ -oo, +oo]. Note that oo+oo, O·oo, oojoo and 0/0 are not defined. In particular, we do not define oo+oo=oo. For example, z--+oo as z--+oo and -z--+oo as z--+oo but z+( -z)--+0 as z--+oo. Even more than in §24.4, it is necessary to warn against attempting algebraic manipulations on the basis of definitions (i}-(iv). Item (iv), for example, does not represent a genuine 'division by zero' but merely indicates a handy convention.

24.12t

Product spaces

The one-point compactification of IR" is not the .only possible compactification of IR". An obvious and important alternative is to regard IR" as sitting inside the space [- oo, + oo ]". In the case of IR 2 this amounts to thinking of the plane as squeezed into the interior of a square and regarding the edges of the square as the 'points at infinity'.

166

Points at infinity r---------------~(+oo,+oo)

(x. y) (-=,y)

~

•

(x. y)

(0, 0)

(+=, 0)

(+=. y)

IR2 L---+-----------_.(+oo,-oo) (X, -oo)

(X,

-oo)

(+oo, -oo)

Schematic diagram of [ -=, + =] 2

In [- oo, + oo ]", of course, one uses the product topology (see §21.18). One can describe this by defining an 'open ball' B with centre ~ = (~ 1 , ~ 2 , ... , ~.) in [- oo, + oo ]" to be a set of the form B=Bl

X

82

X ••• X

B.,

where each set B; (i = 1, 2, ... , n) is an 'open ball' in [- oo, + oo] with centre~;· A set G is then open in [- oo, + oo ]" if and only if each ~E G is the centre of an 'open ball' B such that B c G. Note that IR" is a topological subs pace of [- oo, + oo ]". In fact a subset of IR" is open relative to [- oo, + oo ]" if and only if it is open in the usual sense. Next consider a set S in [- oo, + oo ]" and a function f: S--+ [ - oo, + oo ]m. We are interested in the case when S c IR" and f(S) c IR m. If ~E[- oo, + oo]" and lJE[- oo, + oo]m, we have thatf(x)--+lJ as X-+~ through S if and only if: For each 'open ball' E in [- oo, + oo ]m, with centre '1 there exists an 'open ball' L\ in [- oo, + oo ]" with centre ~ such that XE L\nS =f(x)E E provided x =1 ~. The table below gives some sample criteria in the case m= 1 and n = 2. In this table I, a 1 and a 2 are to be understood as real numbers. f(x)--+1 as x--+a through S f(x)--+1 as x--+( + oo, a 2 ) through S f(x)--+ + oo as x--+(al' - oo) through S

\fe>O 3<5>0 lfxES, lx 1 -a 1 lO 3X 3<5>0 lfxES,

x 1 >X and lx 2 -a 2 l0 3YifxES, lx 1 -a 1 lX

f(x)--+-oo asx--+(+oo, +oo) \fY 3X lfxES through S x 1 >X and x 2 >X =f(x)
Points at infinity 24.13

167

Example Consider the function f: IR 2 --+IR defined by (xy)2 1 +(xy)2.

f(x, y)

This does not tend to a limit as (x, y)--+oo. Vfe have that f(x, y)--+t as (x, y)--+oo along the curve xy = 1 but f(x, y)--+0 as (x, y)--+ oo along the line x = 0. However, . IliD

(x,

-1

(xy)2

y)~(+w. +w) 1 +(xy)

2-.

(1)

To prove this we have to show that, given any e > 0, there exists an X such that, for any x>X and any y>X,

I

2 (xy) 1 +(xy)2

I

1 <e.

But

and hence the choice X= e- 114 clearly suffices. It is worth noting that, although (1) exists, . IliD (x,

(xy)2

y)~w

1 +(xy)

2

(2)

does not exist. The reason for the difference is that in considering (1) we look at regions like that indicated on the left below while, when considering (2), we look at regions like that on the· right.

Finally, note that many authors would write (1) in the form . IliD

x~w

(xy)2 1 +(xy)

-1 2-.

y~w

One has to be careful not to confuse this notation with (2) (nor to confuse it with the notation for repeated limits). (See §23.5.)

168

Points at infinity

24.14t

Exercise

(1) Prove that

(i)

x-y) lim ( lim - = -1 y-+oox+y

x-+oo

(ii)

. ( hm . x-y) hm - = 1. y-+oo x-+oox+y

Discuss the existence of

(. . ) . m

Inn

(x,y)-(+oo. +oo)

(X- y)

-- . x+y

(2) Let f: IR 2 -+IR 2 and consider the following statements (i)

lim ( lim f(x, y))=l x-+oo

(ii)

y-+oo

lim ( lim f(x, y))=l Y-+oo

(iii)

x-+oo

lim

f(x, y)=l

(x, y)-( + oo, + oo)

(iv) (x,

lim f(x, y)=l. y)-oo

Show that (iv)=>(iii) and (iv)=>(ii). Under what circumstances is it true that (iii)=>(ii) and (iii)=>(i)? Under what circumstances is it true that (ii)=>(i)? [Hint: See theorems 23.9 and 23.17.] (3) Consider the function f: IR 2 -+IR defined by yx+1 f(x, y)= x+ 1 .

Prove that f(x, y)-+ y as x-+ + oo uniform! y for 0 ~ y ~ 1 - i.e. prove that, for any e > 0, there exists an X such that, for any x >X and any y satisfying 0 ~ y ~ I, lf(x, y)- Yi <e.

(The point is that the same X must work for all y satisfying 0 ~ y ~I. Thus X must depend only on e.)

25

25.1

SEQUENCES

Introduction

In this chapter we shall study the general properties of sequences of points in a metric space. As we shall see, this topic allows us to tie together many of the ideas introduced in earlier chapters and, to some extent, this chapter will therefore provide a useful opportunity for some revision of these ideas. Sequences of real numbers are particularly important but we have not felt it necessary to stress the importance of their special properties since most readers are likely to have studied these special properties at length elsewhere. Formally, a sequence of points in a metric space ?J is a function/: N-+?J. We call f(k) the kth term of the sequence. The notation

will be used to denote the sequence whose kth term is Yk· Thus
<

Intuitively, it is useful to think of a sequence Yk> as a list. We therefore sometimes refer to a sequence by listing its first few terms. The first few terms of the sequence (k 2 + 1) for example are 2, 5, 10, 17, 26, 37, ... 169

170

Sequences

25.2

Convergence of sequences

that

Suppose that
T)E 'lj.

Yk-+1)

as

k-+ oo

if and only if lts>O 3K,

k > K = d( 1), y k) < B.

This definition is notning new. If f: N -+'lj satisfies f(k) = Yk> the definition simply asserts that f(k)-+1) as k-+ + oo through the set N (see §24.5).

From each of the results of chapter 22, we can obtain a theorem about sequences as a special case by taking X= [- oo, + oo ], S = N and I;= + oo. Some of these special results are of sufficient importance to make it worthwhile listing them below. Theorem Suppose that <xk) is a sequence of points in ~n and that I; is a point in ~n. We write xk =(xlk, x 2 k, ••. , xnd and I;=(~ 1, ~ 2 , ... , ~n). Then xk -+I; as k-+ oo if and only if 25.3

xjk-+ej

for each j= 1, 2, ... , n. Proof Theorem 22.28.

as

k-+oo

171

Sequences 25.4

Example Consider the sequence

< cos k

sink)

(xk> = ( -k-, -k- >

of points in IR 2 • We have that

Icosk k j::;~_.o -k Isi~ k ~~~--o

and

as

k->oo

as

k->oo.

Hence, by theorem 25.3, xk-->(0, 0) as k->oo.

xz/

/

........

------

......

..... XI

I

I I I

' ' , ...'_____ ; ..... _- ) .... •x4 /

x3

25.5 Theorem Suppose that (xk> and (yk) are sequences of points in !Rn and that xk-->1;

and

Yk->l)

as as

k->oo k-> oo.

Then, for any real numbers a and b,

axk+byk->al;+bl)

as

k->oo.

Proof Theorem 22.30.

25.6

Theorem Suppose that (xk> and (yk) are sequences of real numbers

and that xk->~

and Then;

Yk--> 1J

as as

k->oo k--> oo.

172

Sequences

(i) xk yk--+

e'l as k--+

00

and, provided that '1 # 0,

e

.. ) ---+-as xk k --+ oo. (n Yk '1

Proof Theorems 22.31 and 22.32.

This theorem remains true if IR is replaced throughout by C.

25.7

Convergence of functions and sequences

25.8 Theorem Let X and ?J be metric spaces and suppose that f: S --+?j where S is a_ set in X. Then the statements (1) f(x)--+fl and

as

x --+;

through S;

(2) For each sequence (xk) of points of S"\ {;}, lim xk=;) = (lim f(xk)=fl); ( k-oo k-oo

are equivalent. Proof (i) (1)=(2). Suppose that {1) holds. Then, given any e>O, there exists a <5>0 such that for any xES \ {;}

d(;, x)<J = d(fl, f(x))<e.

(3)

Now suppose that (xk) is a sequence of points of S\{;} such that xk--+; as

k--+ oo. Then there exists a K such that

(4)

173

Sequences

Combining (3) and (4), we obtain that k>K => d(TJ, f(xk))<e

and hence f(xk)--+TJ as k--+oo. (ii) not (1) =>not (2) (see §2.10). Suppose that it is false that f(x)--+TJ as 'v'e>O 3!5>0

X-+~

'v'xES\.{~}, (d(~,

through S. If it is false that

x)d(tt,f(x))<e),

then it is true that 3e>0 'v'b>O 3xES\.{~}, (d(~, x)O and all b>O. Given b= 1/k (k = 1, 2, 3, ... ), we can therefore find xkES\{~} such that d(~,xk)<1/k

and

d(tt,f(xk))~e.

From the first of these inequalities it follows that the second, it follows thatf(xk).frtt as k--+oo.

xk--+~

as k--+oo and, from

25.9 Corollary Let X and 'lj be metric spaces and suppose thatf: S--+'lj where S is a set in X. Then f is continuous on S if and only if, for each sequence (xk) of points of S which converges to a point of S, lim f(xk)= f(lim xk)· k-+oo

k-+oo

25.10 Example In §9.2 we explained how Archimedes trapped the area of a quarter-circle of radius 1 between two sequences (A.) and (B.) of rational numbers. The terms of these sequences satisfy A 1 =t, B 1 = 1 and (i) A.+ 1 = {A. B.} 112

(ii)

B;+\ =2{A;} 1 +B; 1 }.

Using the fact that the geometric and harmonic mean of two real numbers lies between these numbers, it is trivial to prove that t~A.~A.+t~B.+ 1 ~B.~l. From this it follows that (A.) and (B.) converge (see exercise 25.11 (4)) and that their respective limits A and B satisfy t~A ~ B ~ 1. Applying theorem 25.5 to the equation (ii), we obtain that B- 1 =2(A- 1 +B- 1 ) and hence A= B. It therefore makes sense to interpret the common limit of the two sequences as the area of the quarter-circle. To prove that the common limit is n/4, however, is considerably more difficult. The function/: [0, oo )--+IR defined by f(x)=.J xis continuous on [0, oo). We

Sequences

174

have that AnBn __.AB as n __. oo and hence, by corollary 25.9, j(AnBn)---j(AB)

as

n__. 00.

Since An+ 1 = j(AnBn), it follows that A= j(AB) and hence the result A= B is obtained again.

25.11

Exercise

(1) Determine whether or not the sequence (xk) converges in IR 2 in each of the following cases.

(iv) xk =(cos k, sin k). (2) Write down definitions for the statements 'xk__. + oo as k__. oo' and 'xk__.- oo ask__. oo' in the case when (xk) is a sequence of real numbers. Show that k 2 __. + oo as k---oo and that -Jk__.- oo as k---oo. Discuss the sequences (( -1l) and (k( -1l). (3) Let (xk) be a sequence of points in IR". Write down a definition for the statement 'xk__.oo as k__.oo' based on the one-point compactification of IR". For which of the sequences of question 1 is it true that xk __. oo aF k__. oo? (4) A sequence (xk) of real numbers is said to increase if and only if xk~xk+ 1 for each kEN. Prove that an increasing sequence (xk) of real numbers is unbounded above if and only if xk__. + oo as k__. oo. Prove that an increasing sequence of (xk) of real numbers is bounded above if and only if xk__.x as k__. oo and that the limit x is the supremum of the sequence. What are the corresponding results for decreasing sequences? (5) A sequence (xk) is defined by x 1 = 1 and xk+ 1 = f(xk) where f:[O, oo )---IR is defined by f(x)=x+1. x+2 Prove that 0<xk+ 1 ~xk (k=1, 2, ... )and explain why it follows that (xk) converges. If xk__.x as k---oo, justify the conclusion that x=f(x) and hence show that J5-1 X=--2-.

175

Sequences

(6) Prove that the sequence (xk) of rational numbers defined by x 1 = 2 and

xk+! =Kxk+

:J

decreases and is bounded below. Deduce that (xk) converges and prove that the limit is

J2.

25.12

Sequences and closure We begin with the following lemma.

25.13 Lemma Let S be a non-empty set in a metric space X and let I;E X. Then d(l;, S) = 0 if and only if there exists a sequence (xk) of points of S such that xk-->1; as k-> oo.

Proof (i) Suppose that d(l;, S) =0. By theorem 13.22, given any a>O, there exists an xES such that d(l;, x)O, it follows that there exists an xkE S such that d(l;, xk)< 1/k-->0 as

k->oo.

Hence xk-->1; as k->oo. (ii) Suppose that there exists a sequence (xk) of points of S such that xk-->1; as k->oo. Then, given any a>O, there exists a K such that k>K=> d(l;, xk)
25.14 Theorem Let S be a set in a metric space X and let I;E X. Then I;E S if and only if there exists a sequence (xk) of points of S such that xk-->1; as k->oo.

176

Sequences

Proof By theorem 15.5, /;ES ~d(l;, S)=O and so the theorem follows immediately from lemma 25.13.

25.15 Corollary A set Sin a metric space X is closed if and only if each convergent sequence of points of S converges to a point of S. Proof By exercise 15.3(1), S is closed if and only if S = S.

25.16t Note It is of some importance to take note of the fact that theorem 25.14 and corollary 25.15 are false for a general topological space X. Analogues of these results are valid but the notion of a sequence must be replaced by the more general notion of a net.

25.17

Exercise

(1) If <xk) is a sequence of points in IR" such that xk-+1; as k-+oo, explain

why (i) <xk, u)-+<1;, u) (ii) llxkll-+111;11

as

as

k-+oo

k-+oo.

[Hint: Use corollary 25.9.] Deduce from corollary 25.15 that the sets S= {x: <x, u) ~c} and T= {x: llxll ~r} are closed. (2) Let A and B be two non-empty sets in a metric space X. Prove that A and B are contiguous if and only if a sequence of points in one of the sets converges to a point of the other. Use this result and corollary 25.9 to prove that, ifS is a connected set in X and f: S -+?,J is continuous on S, then f(S) is connected. (See theorem 17.9.) (3) Let S be a set in a metric space X and let /;EX. Prove that I; is a cluster point of S if and only if there exists a sequence <xk) of distinct points of S such that xk-+1; as k-+oo. (Distinct means that xi=xk ~j=k.)

25.18

Subsequences

Suppose that g: N-+ N is strictly increasing. Then the sequence fog: N -+?J, is said to be a subsequence of the sequencef: N -+?J,. If we 'think of the subsequence illustrated in the diagram as a list, then its

first few terms are

177

Sequences

Note that this is obtained from the original sequence

by omitting some of the terms. If
· (If f(k)= y and g(l)=k" then f(g(l))= Yk 1.) For example, if by deleting terms as indicated below:

t, 5, v1, 17, ~. 37, ~. 65, ~. 101, ...

25.19 Theorem Let
Proof Letf: N--+'IJ be defined by f(k)=yk and let g: N--+N be defined by g(0 = k1• We have that f(k)--+tt as k--+ + oo through N and that g(0--+ + oo as I-++ oo through N (see exercise 25.22(1)). It follows from exercise 22.33(3ii) that f(g(0)--+tt

as

I-++ oo

through N -i.e. Yk I --+fl as 1--+oo. 25.20

Example We have that 1/k--+0 as k--+ oo. From theorem 25.19 it

178

Sequences

follows without further calculation that 1j2k->0

as

k->oo

because (1/2k) is a subsequence of (1/k).

25.21 Example Consider the sequence (( -1l) of real numbers. If (-1l->l as k->oo, then (-1) 2 k->[ as k->oo and (-1) 2 k+ 1 ->l as k->oo. But ( -1) 2 k= 1 and ( -1) 2 k+ 1 = -1. Hence I= 1 = -1. From this contradiction we deduce that (( -1l) diverges. --

25.22

Exercise

(1) Suppose that (k 1) is a strictly increasing sequence of natural numbers.

Prove by induction that k1 ~I for all lE N and deduce that k1-> + oo

as

1-->oo.

(2) Show that, for any mEN, the sequence of real numbers

is increasing and bounded above. Deduce that the sequence converges to a real number cP(m). By considering an appropriate subsequence, prove that

?J. If 11 is a cluster point of E={yk: kEN}, prove that (Yk> has a subsequence (yk) with the property that

(3) Let (yk) be a sequence of points in a metric space

Yk 1->fl

as

1-->oo.

(4) Let (yk) be a sequence of points in a metric space 9J and let E = {Yk: kEN}. Ifyk->fl ask-> oo, prove that Eu{ 11} is closed. If (yk) has no

convergent subsequences, prove that E is closed. (5) Let (xk) be a sequence of real numbers. If each set En={xk: k>n} has a maximum, prove that (xk) has a decreasing subsequence. If at least one of the sets En has no maximum, prove that (xk) has an increasing subsequence. Deduce that any bounded sequence of real numbers has a convergent subsequence. [Hint: Use exercise 25.11 (5).] (6) Let ((xk,yk)) be a bounded sequence of points in 1Rl 2 • Prove that ((xb yd) has a convergent subsequence. [Hint: Begin with a convergent subsequence (xk I >of (xk) and consider (Yk,).]

Sequences

25.23

179

Sequences and compactness

The Bolzano-Weierstrass theorem (19.6) was central to our discuss;ion of compact sets in chapter 19. We therefore begin by giving a version of the Bolzano-Weierstrass theorem for sequences.

Theorem (Bolzano-Weierstrass theorem) Any bounded sequence 25.24 <xk) of points in IR" has a convergent subsequence. Proof This may be deduced from theorem 19.6 as follows. If E = {xk: kEN} is infinite, then theorem 19.6 asserts that E has a cluster point ~- This cluster point must be the limit of a subsequence of <xk) (exercise 25.22(3)). If E is finite, then <xk) has a subsequence all of whose terms are

equal. This subsequence therefore converges. Alternatively, the theorem may be proved directly as indicated in exercise 25.22(5) and (6).

In chapters 19 and 20, we considered a number of different definitions of compactness and the above version of the Bolzano-Weierstrass theorem suggests yet another. We say that a set Kin a metric space X is sequentially compact if and only if every sequence of points in K has a subsequence which converges to a point of K.

25.25 Theorem A set K in a metric space X is compact if and only if each sequence <xk) of points of K has a subsequence which converges to a point of K- i.e. in a metric space, compactness and sequential compactness are the same.

Proof We use the definition of compactness given in § 19.5- i.e. a set K in a metric space X is compact if and only if each infinite subset E of S has a cluster point ~ES. (i) Let K be compact and let <xk) be a sequence of points of K. If E = {xk: kEN} is finite, then <xk) has a subsequence all of whose terms are equal to one of the terms of <xk>· Hence <xk I > converges to a point of K. If E is infinite, then E has a cluster point ~E K. By exercise 25.22(3), <xk) has a subsequence which converges to ~(ii) Let K be sequentially compact and let E be an infinite subset of K. Since E is infinite, there exists a sequence <xk) of distinct points of E. Because K is sequentially compact, <xk) has a convergent subsequence whose limit ~ lies in K. But, by exercise 25.17(3), ~ is a cluster point of E.

<xk,>

180

Sequences

25.26t Note It is important to bear in mind that sequential compactness is not the same as compactness in a general topological space. It is true that a countably compact set (§20.7) in a topological space is sequentially compact but even the converse of this assertion is false without some subsidiary assumptions.

Example Suppose that .X and ?J are metric spaces and that S c .X. 25.27 Ifj: S-+?J is continuous on the setS, then S compact= f(S) compact. We have already seen two proofs of this very important theorem. (See theorems 19.13 and 21.17.) A third proof may be based on theorem 25.25. Let (yk) be any sequence of points inf(S). We seek to show that (yk) has a subsequence which converges to a point ofj(S). Write Yk = f(xd. Since S is compact, (xk) has a subsequence which converges to a point ~ES. Suppose xk I -+~ as [-+ oo. By corollary 25.9,/(xk I )-+/(~) as [-+ oo. It follows that the subsequence (Yk 1) converges to the pointf(~)Ef(S). 25.28

Exercise

(1) Let (Yk) be a sequence of points in a metric space .X and let E={yk: kEN}. Ifyk-+fl, prove that Eu{ra} is compact. t(2) Let X and ?J be metric spaces and let f: K -+?J be continuous on the compact set Kin X. Use theorem 25.25 to show that/ is uniformly continuous on K (see §23.20).

[Hint: The contradictory of the statement that f is uniformly continuous on K begins with '3e > 0 V b > 0'. Use this assertion with b = 1/k for each kEN.] t(3) Given an example of a bounded sequence in /00 (see §20.20) which has no convergent subsequence.

26

26.1

OSCILLATION

Divergence

Suppose that X and 'IJ are metric spaces and thatf: S-*'IJ where S is a set in X. Let 11E'IJ and let I; be a cluster point of S.

Oz(X)-->- oo as x-->0+

I

\

\

\ \

i\

/

//

181

o3 oscillates 'finitely' as x-->0+

\

\

\

/

o4 oscillates 'infinitely' as x-->0+

182

Oscillation

Ifj(x)--+11 as X-+~ through the setS, we say that the function converges as x approaches ~ through S. If the function does not converge as x approaches ~ through S, then it is said to diverge. Divergent functions can exhibit a variety of different behaviour as the diagrams on p. 181 illustrate. We saw in chapter 24 that the behaviour of the functions g 1 and g 2 can be treated by the same machinery as one uses for convergent functions. Although these functions do not converge as x approaches 0 from the right when regarded. as mappings to the space IR, they do converge when regarded as mappings to the space [- oo, + oo]. The functions g 3 and g4 are more interesting. These are examples of oscillating functions. To discuss these, we require the notion of a limit point.

26.2

Limit points

Suppose that X and ?J are metric spaces and thatf: is a set in X. Let ttE ?J and let ~ be a cluster point of S. Suppose it is true that, for some AE ?J, f(x)--+1.

as

S--+?J where S

X-+~

through the set T where T is a subset of S for which ~ is a cluster point. Then we say that A. is a limit point off as x approaches ~ through S.

Observe that there is a distinction between a 'limit' and a 'limit point'. A limit exists only whenf(x) converges as x approaches~ through S. But limit points may exist even though f(x) diverges as x approaches ~ through S.

183

Oscillation Examples (i) Consider the function f: IR -->IR defined by

26.3

f(x)={

1 (x~O) -1 (x
This function diverges as x approaches 0 but f(x)-->1 as x-->0+ and f(x)--> -1 as x-->0-. Hence the function has two limit points as x approaches 0. These are 1 and - 1. (ii) Consider the function/: IR 2 -->IR 1 of example 22.11. We have seen that 1)(2

-1

f(x, y)--> oc 2 +

1

as

(x, y)-->(0, 0)

along the line y = ocx. Examining each ocEIR in turn, we find that every point of [0, 1) is a limit point of f(x, y) as (x, y) approaches (0, 0). Also

f(x, y)--> 1 as

(x, y)-->(0, 0)

along the line x = 0. Hence 1 is also a limit point. Since -1 ~f(x, y)~ 1 for all (x, y)EIR 2 , the function can have no other limit points. It follows that the set oflimit points off(x, y) as (x, y)-->(0, 0) is equal to [ -1, 1].

Exercise

26.4

(1) Suppose that X and ?J are metric spaces and thatf: S-->?J where S is a

set in X for which ~ is a cluster point. Prove that 'i.E?J is a limit point off as x approaches~ through the set S if and only if there exists a sequence (xk) of points of S such that xk # ~ (k = 1, 2, ... ) and xk -->~as k--> oo for which f(xk)-->1.

as

k--> oo.

[Hint: Use theorem 25.8]. (2) Find all real limit points of the following sequences of real numbers:

(i)

(ii) (( -l)kk) (v) (1 + (-1)k)

(k)

(iv) (( -1l)

(iii) (1/k) (vi) (k+(-1)kk).

These sequences may also be regarded as taking values in [- oo, + oo]. Find all limit points from [- oo, + oo] of the given sequences. (3) Find all real limit points as x -->0 + f: (0, oo)-->IR:

of the following functions

. 1 (1). f( x)=sill-

.. f (x)=x Sill. 1 (n)

X

X

... f

1 .

1

X

X

(m) (x) =- Sill -

. f (x)=1-sill. 1 (1v) x

184

Oscillation 1

1 .

1

X

X

X

(v) f( x)=--- Sill-

. 1 1 . 1 (v1) f(x)= 2 - - Sill-. X X X

These functions may also be regarded as taking values in [- oo, + oo]. Find all limit points from [- oo, + oo] of the given functions as x--+0 +. (4) Find all real limit points as (x, y) approaches (0, 0) of the function f of · exercise 22.24(3). (5) Give examples of functions f: (0, oo )--+IR for which the set L of real limit points as x--+0 + is as indicated below: (i) L=(/) (ii) L=IR (iii) L= [0, 1] (iv) L= [0, oo) (v) L={-1, 1} (vi) L=N (vii) L={O} but f(x)-\-+0 as x--+0+.

Show that in the case of each of your examples except (iii) and (v), + oo or - oo is also a limit point. (6) Suppose that f: (0, oo) --+IR is continuous on (0, oo ). Let L be the set of real limit points as x --+0 +. Prove that L is connected.

26.5t

Oscillating functions

26.6t Theorem Suppose that X and 'IJ are metric spaces and letf: S->'IJ where S is a set in X for which ~ is a cluster point. Then the set L of limit points off as x approaches ~ through S is given by 00

L=(l f(SnAk) k=l where the set Ak is obtained by removing radius 1/k.

~

from the open ball with centre

~

and

Proof (i) Suppose that AE L. Then there exists a sequence (xk) of points of S \ {~} satisfying xk --~as k-> oo such thatf(xk)->A. ask-> oo (exercise 26.4(1)). Given any J, there exists a K such that

Hence l.Ef(AJnS). It follows that 00

Le(} f(AknS). k=l

(ii) Suppose that A.Ef(AknS) for each kEN. Then for each kEN, there exists a sequence (xk, 1) of points of AknS such that f(xk,,)->l. as 1->oo. For each kEN, choose lk so that d(l., f(xk, 1)< 1/k. Then xk,lk --~ as k-> oo and f(xk, 1)->l. as k-> oo.

185

Oscillation Thus A.E L and so 00

( l f(AknS)c:L. k= I

Theorem Let X and ?J be metric spaces and suppose thatf: S-+?J where S is a set in X for which ~ is a cluster point. Then the set L of limit points off as x approaches ~ through S is closed. 26.7t

Proof By the previous theorem, L is the intersection of a collection of closed sets. Thus L is closed by theorem 14.14. 26.8t Theorem In addition to the assumptions of the previous theorem, suppose that f(S) lies in a compact subset K of ?J. Then the set L of limit points off as x approaches ~ through S is non-empty.

Proof The sets SnAk are non-empty for each kEN because ~ is a cluster point of S. It follows thatf(SnAd is non-empty for each kEN. Thus
Theorem With the assumptions of the previous theorem, f(x)-+tt

through S if and only if L= {tt}:

as

X-+~

186

Oscillation

Proof If f(x)--+11 as X-+~ through S, then it is trivial to prove that L={11}· If it is false thatf(x)--+11 as x--+~ through S, then there exists a sequence (xk) of points of S \.{~} such that xk--+~ ask--+ oo butf(xk) +11 ask--+ oo. We deduce the existence of an e0 >0 and a subsequence (xk,) for which d(11,f(xk,))Geo (/EN). But (f(xk1)) has a convergent subsequence by theorem 25.25. Hence L#{11}·

The assumption that f(S) be a subset of a compact set K c:. ?J. which is made in theorems 26.8 and 26.9 is not so restrictive as it may seem at first sight. For example, if '11 = IR, one can always begin by replacing IR by the compact space [- oo, + oo].

26.10 that

Example Consider the sequence (( -1)kk -k) of real numbers. We have

( -1) 2 k2k- 2k--+0

as

k--+ oo

(-1) 2 k+ 1(2k+1)-(2k+1)--+-oo

as

k--+oo.

The sequence therefore has only one rea/limit point namely 0. However, one cannot conclude from theorem 26.9 that the sequence converges because the range of the sequence does not lie in a compact subset of IR. Note that, if we regard the sequence as taking values in the compact space [- oo, + oo], we find that the set L of limit points contains two elements namely 0 and - oo. Thus again theorem 26.9 does not apply.

Suppose that X is a metric space and thatf: S--+IR where S is a set in X for which ~ is a cluster point. Let L denote the set of limit points from [- oo, + oo] as

x

approaches ~ through S. Since [- oo, + oo] is compact, we know from theorem 26.8 that L# ~- If L consists of a single point, then either fconverges as x approaches~ through S or else f diverges.to + oo or diverges to - oo as x approaches~ through S. If L consists of more than one point, we say that f oscillates as x approaches ~ through S. If f oscillates as x approaches ~ through S and Lis a compact subset of IR, it is customary to say that f 'oscillates finitely'. (Note that, since L is closed, it follows that Lis a compact subset of IR if and only if Lis a subset of IR which is bounded in IR .) Otherwise f is said to 'oscillate infinitely'.

26.11 t

Lim sup and lim inf

Let X be a metric space and suppose that f: S--+IR where S is a set in X for which ~ is a cluster point. The set L of limit points from [- oo, + oo] as x approaches ~ through S is nonempty and closed by theorems 26.7 and 26.8. It follows that L has a maximum element A and a minimum element A. (corollary 14.11). We call A the limit superior (or lim sup) off as x approaches~ through S. We call A. the limit inferior (or lim inf). Sometimes A is referred to as the upper limit and A. as the lower limit.

Oscillation

187

The function 9 3 illustrated in §26.1 is an example of a function for which both A and A are finite (i.e. A and A are elements of~). The function 94 of §26.1 is an example of a function for which A is finite but A= + oo.

Theorem Let X be a metric space and suppose that/: S--+~ where S is a set in X for which ~ is a cluster point. Then the function f oscillates as x approaches ~ through S if and only if A=/= A. Moreover, given any YJE [- oo, + oo ],

26.12t

f(x)--+'1

as

x --+~

through S if and only if A= '1 =A.

Proof This is an immediate consequence of theorem 26.9.

26.13t Proposition With the assumptions of the preceding theorem, A is the limit superior off as x approaches ~ through S if and only if (i) 'VI> A 38>0 'VxES, (ii) 'VIO 3xES,

O
x)
O
and f(x)>l.

Item (i) above asserts that nothing larger than A is a limit point. Item (ii) guarantees that A lies in the closure of the set L of limit points. Since Lis closed, this means that AE L. Similar criteria, of course, hold for the limit inferior A. The notation A=lim supf(x);

A=lim inf f(x)

x~;

x~;

is often used. The reason for the choice of this notation will be evident from the statement of the following result.

26.14t Proposition Let X be a metric space and suppose thatf: S--+~ where S is a set in X for which ~ is a cluster point. Let A 0 denote the set obtained by removing ~ from the open ball with centre ~ and radius o> 0. Then (i) A= lim { sup f(x)} 0-+0+

xeSnA 0

(ii) A= lim { inf f(x)}. li~O+

xeSnAb

188

Oscillation

The proof of (i) above depends on the observation that the function F: (0, oo)-+[- oo, + oo] defined by

F (o)= sup f(x) xeSnA 0

increases on (0, oo). This guarantees the existence of the limit as 0--+0+. The identification of the limit with A is then achieved with the help of proposition 26.13.

26.15t

Exercise

(1) Determine the lim sup and lim inf for each of the examples given in exercises

26.4(2), (3) and (4). (2) Suppose that (ak) is any sequence of real numbers. Prove that AE[- oo, + oo]

is the lim sup of the sequence if and only if (i) For any subsequence (ak,> of (ak), ak 1-+'7 as 1--+oo ~ '7;:i;A

and

(ii) There exists a subsequence (ak,> of (ak> such that ak 1-+A

as

1--+oo.

What is the corresponding result for the lim inf of a sequence? (3) Suppose that (ak) is any sequence of real numbers. Prove that AE [- oo, + oo] is the lim sup of the sequence if and only if and

(i) VL>A, {k: ak;;:;L} is finite (ii) VI< A, {k: ak;;:; l} is infinite.

What is the corresponding result for the lim inf of a sequence? (4) Suppose that (ak) and (bk) are any sequences of real numbers. Prove that lim sup (ak + bd;::;; (lim sup ak) + (lim sup bk) k-+oo

k-+oo

k-+oo

whenever the right-hand side makes sense. Give an example for which the lefthand side is - oo and the right-hand side is + oo. Show that the two sides are equal whenever (bk) is a convergent sequence of real numbers. What are the corresponding results for the lim infs of the sequences? (5) Suppose that (ak) and (bk) are any bounded sequences of real numbers. Prove that lim inf (ak+bk);:i;(lim inf k-+oo

k-+oo

ak) + (lim su.p bk);;;;lim sup (ak+bk). k-+oo

k-+oo

Show also that lim sup (-ad= -(lim inf ak)· k-+oo

k-+oo

Deduce that, for any bounded sequence (ck) of real numbers,

Oscillation

189

(6) Suppose that K, prove that there exists an H such that ak < Hrk for all kEN. Deduce that

27

27.1

COMPLETENESS

Cauchy sequences

A Cauchy sequence (xk) of points in a metric space .X is a sequence with the property that, for any e > 0, there exists an open ball E of radius e and a K such that

E

The definition of a convergent sequence asserts that the terms of the sequence (xk) can be forced as close to the limit ~ as we choose by taking k sufficiently large. The definition of a Cauchy sequence asserts that the terms of the sequence (xk) can be forced as close to each other as we choose by taking k sufficiently large. This latter point is more evident if the definition is rewritten in the form:

Ve>O 3K, (k>K and l>K)

27.2

= d(xk, x)<e.

Completeness

A complete metric space .X is a metric space in which every Cauchy sequence converges. Our first task is to link this notion with the ideas introduced in §20.15 by showing that a complete metric space is one in which a suitable analogue of the BolzanoWeierstrass theorem (19.6) is true.

190

191

Completeness

27.3t Theorem A metric space X is complete if and only if every totally bounded set in X has the Bolzano--Weierstrass property.

Proof (i) Suppose that X is complete and that S is a totally bounded set in X. Let E be an infinite subset of Sand let (xk) be a sequence of distinct points of E. The definition of a Cauchy sequence is equivalent to the assertion that the range of the sequence is totally bounded. It follows that (xk) is a Cauchy sequence and hence converges. Its limit is then a cluster point of E (exercise 25.17(3)). (ii) Suppose that every totally bounded set in X has the Bolzano--Weierstrass property. Let (xk) be a Cauchy sequence. Then its range is totally bounded and hence is either finite or else possesses a cluster point. In either case (xk) has a convergent subsequence (exercise 25.17(3)) and therefore converges (exercise 27.7(3)). Not only is an analogue of the Bolzano--Weierstrass theorem true in a complete metric space, but we also have the following analogues of the Heine-Borel theorem and the Chinese box theorem.

27.4t Theorem A set Kin a complete metric space X is compact if and only if it is closed and totally bounded.

Proof See §20.16. 27.5t Theorem Let (Bk) be a nested sequence of non-empty closed balls in a complete metric space X whose radii tend to zero. Then

is non-empty (and in fact consists of a single point).

Proof Note that a closed ball need not be compact in a general complete metric space X. (See §20.20.) The Cantor intersection theorem (19.9) therefore does not apply. Let xkE Bk. Given any e > 0, let B K be the first ball with radius less than e. Then k>K

=>

XkEBK

and hence (xk) is a Cauchy sequence. Since X is complete, (xk) therefore converges. But each of the sets Bk are closed and the limit of (xk) therefore belongs to each of these sets (corollary 25.15).

Theorem 23.17 was concerned with the conditions under which one can reverse the limiting operations in a repeated limit. If X is a complete metric space, this theorem takes a more satisfactory form.

192

Completeness

27.6t Theorem Let X, 'IJ and f be metric spaces and suppose that f is complete. Let ~EX and l)E'IJ. Suppose that A is a set in X for which ~ is a cluster point and that B is a set in 'IJ· Letf: S \ (~, tt)-+ f where S =A x B. Let l: B-+ f and let m: A-+f. Suppose that (i) f(x, y)-+l(y) as x-+~ through the set A uniformly for yEB, and (ii) f(x, y)-+m(x) as y-+l) through the set B pointwise for xEA. Then there exi"sts a (iii) (iv)

l(y)-+~

~E f

such that

as y-+l) through the set B, and as X-+~ through the set A.

m(x)-+~

-,-------------, I I

S

I I

I

I

I

I I

B

I

I

1

I I

l

t

I

~-~~--

t

I

I

__ I, _ _ _ _ _._ ____ ___._ : ___ j I

1J

I

I

(~, 1))

~-+----~~---------+--X

A

Proof Let e>O be given. From (i) we have that there exists a <5>0 such that for each XEA and each yE B, 0 < d(x,

~) <

t5 => d(l(y),f(x, y)) < e/4.

From (ii) we have that, for each XE B there exists a O
tt)<~. =>

It follows that if XE A satisfies 0
~.

such that

d(m(x),f(x, y))<e/4.

~) <

t5 and yE B satisfies 0
d(l(y), m(x))<e/2. We may deduce that, if x 1 EA and x 2 EA satisfy 0
~)
and 0
~)
d(m(x 1), m(x 2)) <e. From this it follows that, if (xk) is any sequence of points of A \g} such that xk-+~ as k-+oo, then (m(xk)) is a Cauchy sequence in X. Item (iv) then follows from

193

Completeness

theorem 25.8 and the fact that X is complete. We can then appeal to theorem 23.17 to obtain item (iii). 27.7 Exercise (1) Prove that any convergent sequence of points in a metric space X is a Cauchy sequence. Prove that any Cauchy sequence in a metric space X is bounded. (2) Prove that any sequence (xk) in a metric space X which satisfies d(xk, xk+d~2-k (keN) is a Cauchy sequence. Prove that any Cauchy sequence (xk) in a metric space X has a subsequence (xk,) which satisfies. 1 d(xk,I xk l +)~2l-

(/eN).

(3) Prove that any Cauchy sequence in a metric space X which has a convergent subsequence is itself convergent. t(4) Let 'IJ be a set in a metric space X. Then 'IJ may be regarded as a metric subspace of X. Prove the following: (i) If X is complete, then 'IJ is complete if and only if 'IJ is closed in X. (ii) If 'IJ is compact in X, then 'IJ is complete. t(5) Let X be a metric space for which it is true that every nested sequence of nonempty closed balls whose radii tend to zero has a non-empty intersection. Prove that X is complete. (See theorem 27.5.) t(6) If X is a metric space, a function!: X-+ X is called a contraction if and only if there exists a real number Cl(< 1 such that d(f(x), f(y)) ~ Cl(d(x, y)

for each xEX and each ye X. Letf: X-+X be a contraction. Prove that: (i) f is continuous on X. (ii) Any sequence (xk> of points of X which satisfies xk + 1 = f(xk) is a Cauchy sequence. (iii) If X is complete, then f has a fixed point - i.e. for some ~E X,f@ = ~. (See example 17.13.)

27.8

Some complete spaces

The results of the previous section show that complete spaces are of some interest. In this section we shall consider a few specific examples of complete spaces. Since the Bolzano-Weierstrass theorem holds in ~n, it follows from theorem 27.3 that ~n is complete. The proof of the next theorem establishes this result in a more direct fashion.

27.9

Theorem The metric space

~n

is complete.

194

Completeness

Proof Let (xk) be a Cauchy sequence in !Rn. By exercise 27.7(1), (xk) is bounded and hence, by theorem 25.24, (xk) has a convergent subsequence. The conclusion therefore follows from exercise 27.7(3).

Let S be a non-empty set and let 1J be a metric space. We shall denote the set of· all bounded functions f: S -+?j by YJ (S, ?J). The set YJ (S, ?J) becomes a metric space if we define the distance between two 'points' f and g of YJ (S, ?J) by

u(f, g)= sup d(f(x), g(x)). XES

We call u the uniform metric on YJ (S, ?J). This is not a new notion. We introduced the same idea in §23.12 in connection with uniform convergence and in §20.20 we studied the special case lJC=YJ(N, IR).

27.10t

Theorem If 1J is complete, then so is the metric space 93 (S, ?J).

Proof Let (h> be a Cauchy sequence in 93 (S, ?J). Then, for each xES, (fk(x)> is a Cauchy sequence in ?J. Since 1J is complete, it follows that, for each xES, (fk(x)> converges- i.e. there exists a function f: S -+?j such that fk(x)-+f(x) as k-+ oo. In the language of §23.11 and §23.12, we have shown that (fk(x)> converges pointwise for XES. What we need to show is that converges in the space 93 (S, ?J) - i.e. (fk(x)> converges uniformly for XE S. Given any a> 0, there exists a K such that for any k > K and any l > K, u(J;, fk) < e/2. It follows that, for each XE S,

d(J;(x), fk(x)) < e/2.

But, for each xES,fk(x)-+ f(x) as k-.oo. Thus, for any l> K d(J;(x), f(x)) ~ e/2.

(See exercise 16.16(4).) We deduce that, for any l>K

u(J;, f)~ e/2 <e. It follows that J;-+f as I-+ oo in the space 5.1 (S,

?J).

27J 1t

Theorem Suppose that 1J is a complete metric space. Then the set C of all continuous functions in YJ (S, ?J) is closed in YJ (S, ?J). Proof Suppose that is a sequence of functions in C such that fk-.f as k-.oo. To prove that C is closed, we need to show that fEC (corollary 25.15). Let a> 0 be given. Then there exists a K such that

195

Completeness

Let ~E Sand let J > K. Sincef1 is continuous on S, there exists a (j > 0 such that for

each xES d(~,

x)d(f1 @,f1 (x))
It follows that, if d(~, x) < b, then

d(f@,f(x));;;; d(f@Ji~)) + d(fi~),fjx)) + d(fjx),f(x))

< t:/3 + t:/3 + t:/3 = t: and so f is continuous on S- i.e. jEC. We shall use the notation e(S, ?J) to denote the space of all continuous functions f: S->?J. Such functions need not in general be bounded. However, ifS is a compact set in a metric space X , then a continuous function f: S-> ?J is bounded and, moreover, u(f, g)= max d(f(x), g(x)). xeS

To obtain this result we apply theorem 19.14 to the continuous function F: S->IR defined by F =do (f, g).

27.12t

Theorem Suppose that ?J is a complete metric space and S is compact.

Then e(S,

?J) is complete. Proof This follows from theorem 27.11 and exercise 27.7 (4i).

27.13t

Incomplete spaces

It is by no means the case that all metric spaces are complete. As a typical example of an incomplete space we shall consider the system Q of rational numbers (regarded as a metric subspace of IR). Some more examples are given in exercise 27.15. The fact that Q is not complete follows from the fact that Q is not a closed subset of!R (see exercise 27.7(4i)). It is instructive, however, to consider some specific examples for which the properties of a complete space fail to hold in Q.

27.14

Examples Consider the sequences

<xk>

and

defined by

x1=2,

y 1 =t and

xk+l =~ ( xk+

:J; y;)l =~(Y;

1

+

2y~-l).

These are both sequences of rational numbers. The sequence <xk> decreases and (See exercise 25.11(6).) Similarly, the sequence converges (in the space IR) to increases and converges (in the space IR) to

J2.

J2.

196

Completeness x3

x2

I I

Ill ell! I

(i) Since (xk) converges in the space IR, it is a Cauchy sequence in IR (exercise 27.7(1)). It follows that (xk) is a Cauchy sequence in the space 0. But (xk) does not converge if it is regarded as a sequence of points in 0. There is no rational number~ such that xk-+~ as k-+oo. In fact, xk-+J2 as k-+oo and j2 is irrational. (ii) Let lk = {r: rEO and xk ~ r ~ h}· Then lk = [xk, Yk]nO and hence is closed in the metric space 0 (corollary 21.12). It follows that (h) is a nested sequence of closed boxes in 0 (§ 19.2) but

k=!

- i.e. the Chinese box theorem fails in the space 0. (iii) Consider the set S={r: rEO and O~r~j2}. This set is closed in the metric space 0. It is also totally bounded. But it is not compact in 0. For example, the continuous function!: 0 -+IR defined by f(x)=x does not achieve a maximum on S.

27.15t

Exercise

(1) Find the uniform distance between the functionsf: IR -+IR and g: IR -+IR defined by 1 f(x)= 1 +x2;

X

g(x)= 1 +x2

where these functions are regarded as points in the metric space 8l (IR, IR). (2) Let X be the set of all continuous functionsf: [a, b]-+IR. Let the metric in X be defined by d(f, g)=

r

lf(x)-g(x)ldx.

Prove that X is not complete. (3) Let S be a set in a metric space X and let ~E X. Let 'IJ be a complete metric space and suppose that L~(S, 'IJ) is the subset of 8l (S, 'IJ) consisting of all bounded

functions f: S-+'IJ such that there exists an ttE'IJ for which f(x)-+tt as X-+~ through S. Prove that L~(S, 'IJ) is closed in 8l (S, 'IJ). [Hint: Use theorem 27.6.] (4) Explain why the open interval (0, 1) (regarded as a metric subspace of IR) is not complete. Give an example of a Cauchy sequence in (0, 1) which does not converge in (0, 1). Let.f: IR xiR-+(0, 1) be defined by f(x, y)=xy. Show thatf(x, y)-+y as x-+1 through (0, 1) uniformly for yE (0, 1) and that f(x, y)-+x as y-+ 1 through (0, 1) pointwise for xE (0, 1). Explain the relevance of this result to the assumption in theorem 27.6 that f is complete. (5) Let P.: [0, 1] -+IR be defined by x2 x3 x" P.(x)=1 +x+-+-+ ... + 1 . 2! 3! n.

197

Completeness

Prove that (P") converges in the space $([0, 1], IR). 8' denote the metric space of all real polynomials where the metric is defined by

(6) Let

u(P, Q)= max IP(x)-Q(x)i. O;>x;>l

Prove that 8' is not complete. [Hint: See the previous question.]

27.16t

Completion of metric spaces

When a space lacks a certain desirable property, a natural mathematical response is to seek to fit the space inside a larger space which does have the desirable property. In particular, if X is an incomplete metric space, can we fit X inside a larger complete metric St)ace? In this section we answer this question in the affirmative by constructing such a complete metric space. We shall, in fact, construct the smallest complete metric space inside which X can be fitted. This is called the completion of X and denoted by X*. The construction is very simple. The defect in X is that it has Cauchy sequences (xk) which do not converge. We therefore need to invent a limit for each such Cauchy sequence. We must be careful, however, not to invent too many objects. In particular, if two Cauchy sequences (xk) and (yk) satisfy (1)

then we shall want both sequences to converge to the same limit. We therefore begin by introducing an equivalence relation - on the set S of Cauchy sequences of points of X by writing

if and only if (1) holds. The set X* is defined to be the set of all equivalence classes of S defined by this equivalence relation. We shall use the notation [ (xk)] to denote the equivalence class containing the Cauchy seqt!ence (xk>· The original space X is fitted inside the new space X* by identifying an element x of X with the element [(x)] of X*. We need to show that X* is a complete metric space. First it is necessary to introduce a metric into X* which is consistent with that of X. We therefore define d: X*-+IR by d([(xk)], [(yk)])= lim d(xk, Yk). k-ro

Note that the limit on the right-hand side exists because (d(xk, Yk)) is a Cauchy sequence of real numbers whenever (xk) and (yk) are Cauchy sequences in X. (See exercise 27.20(1).) But every Cauchy sequence of real numbers converges because IR is complete (theorem 27.9).

198

Completeness

27.17t

Theorem For any metric space X, the metric space X* is complete.

Proof We have to prove that every Cauchy sequence in X* converges to a point of X*. We proceed by showing that, if (X) is a sequence in X* satisfying

(2) then there exists an X EX* such that Xr-+ X as j-HD. Since each Cauchy sequence in X* has a subsequence which satisfies (2) (exercise 27.7(2)), the result will then follow from exercise 27.7(3). Recall that the elements of X* are equivalence classes of Cauchy sequences in X. All subsequences of a given Cauchy sequence in X clearly belong to the same equivalence class. From exercise 27.7(2) it follows that we may write Xi=[(xkU>)],

where (xk U>> is a sequence in X which satisfies d(xk(j)• xk+IU>)<2-
and hence (3)

provided that l ~ k. Now (2) asserts that lim d(xkUl, xkU+Il)<2-U+l). k-oo

We may therefore deduce the existence of a strictly increasing sequence (ki) of natural numbers such that, for any l ~ ki, d(xiUl, xlu+ ll) < 2 -u+ 1>.

(4)

From (3) we also have that d(x U+l) x kj

'

U+ll)<2-.

kj+ 1

(5)

-

We are now in a position to define X EX* by X= [ (xk.u>>J. J

For this to be an acceptable definition, it is necessary that (xk Ul) is a Cauchy 1 sequence in X. This fact follows from exercise 27.7(2) because d(x U> x kj

'

kj+l

U+l>):::::d(x U> x U+l>)+d(x U+l> x -

kj

'

kj

.

kj

'

kj+l

)

< 2-(j+ I)+ 2-(j+ I) =2- j

by (4) and (5). It remains to show that X 1--+X as 1--+oo. From (4) we have that, for d(x (I) X (j))::;; d(X kj

'

kj

-

X (I+ I))+ ... + d(x (j-l) X (j))

(I) kj

'

kj

<2-(l+l>+ ... +2-j <2-1.

kj

'

kj

eachj~l,

Completeness

199

But X 1= [ (xk j <1>)] and thus it follows that d(Xz, X)~2- 1 ->0

27.18t

as

1->oo.

Completeness and the continuum axiom

After the previous discussion it is natural to consider first the completion iiJ! * of the rational number system. Is iiJ! * the same as the real number system IR? Before seeking to answer this question, we should remind ourselves that iiJ! and IR are not just metric spaces. Both are also ordered fields. The question is therefore only meaningful if we can extend the algebraic structure of iiJ! to Q* (as well as the metric structure). In particular, we need to be able to add, multiply and to order the equivalence classes which constitute the elements of Q*. The appropriate definitions are the obvious ones. We define (i) [(xk)]+[(yk)]=[(xk+Yk)] (ii) [(xk)]·[(yk)]=[(xkyk)] (iii) [(xk)]>[(h)] = 3K(k>K =>Xk>h).

With these definitions iiJ! * becomes an ordered field and

To prove this is a somewhat tiresome but essentially trivial task and so we shall omit the details. Recall that iiJ! is the 'smallest' ordered field. Thus iiJ! * is the 'smallest' complete ordered field. We know from theorem 27.9 that IR is complete and thus IR * c IR. (These remarks, of course, take for granted that structures isomorphic to iiJ!, iiJ! * or IR are to be identified with Q, Q* or IR respectively- see §9.21.) The next theorem shows that the system iiJ! * and the system IR are the same - i.e. the real number system is the completion of the rational number system. Recall that IR is an ordered field which satisfies the continuum axiom. This asserts that every non-empty set which is bound above has a smallest upper bound.

27.19t

Theorem The system

Q*

satisfies the continuum axiom.

Proof Let S be a non-empty set in iiJ! * which is bounded above. Let a1 ES and let b 1 be an upper bound of S. We construct an increasing sequence (ak) of points of S and a decreasing sequence (bk) of upper bounds of S such that O~bk+ 1

-ak+ 1 ~2-k+ 1 (b 1 -a 1 ).

The construction is inductive. If ck =t(ak + bk) is an upper bound of S, we take bk + 1 = ck and ak + 1 = ak. If ck is not an upper bound we take bk + 1 = bk and choose ak+ 1 ES so that ak+ 1 ~ck. The sequences (ak) and (bk) are both Cauchy sequences (exercise 27.7(2)) and

200

Completeness

hence converge. They have a common limit smallest upper bound of the set S.

~

which is easily shown to be the

~~~points of S

27.20t

Exercise

(1) Let (xk) and (yk) be Cauchy sequences in a metric space X. Prove that (d(xk, yk)) is a Cauchy sequence of real numbers and hence converges. (2) Check that IQ* is an ordered field with the definitions of addition, multiplication and an ordering given in §27.18. (3) Every ordered field X contains the system N of natural numbers (or else a system isomorphic to N). We say that X is Archimedean if N is unbounded above in X. (See §9.16.) The system~ is Archimedean (theorem 9.17) and hence so are IQ and IQ*. Prove that an Archimedean ordered field X is complete if and only if it satisfies the continuum axiom (i.e. X=~). (Note that by the assertion that the Archimedean ordered field X is complete we mean that X is 'complete' with respect to the 'metric' d: X x X--+ X defined by d(x, y)=lx-yl.)

28

28.1

SERIES

Convergence of series

Suppose that (ak) is a sequence of points in a normed vector space X. (See § 13.17.) The K th partial sum of the sequence is defined by K

sK=

L ak=at +a2 + ... +aK.

k=l

If the limit of the sequence (sK) exists, then we say that the series 00

L ak=at +a2 +a3 + ...

k=l

converges and we write 00

s=

L ak

k=l

if and only if sK-+s as K-+oo. If the sequence of partial sums does not converge, we say that the series diverges.

28.2

Example If lxl < 1, then 00

L xk=(1-x)-t. k=O To prove this result we observe that K

L xk= 1+x +x 2 + ... +xK

k=O

1-xK+t 1 ----+-1-x 1-x

as

K-+ oo.

If lxl ~ 1, then the sequence of partial sums diverges and so the infinite series does not exist.

201

202

Series

28.3 Theorem (Comparison test) Suppose that X is a complete normed vector space and that (1)

is a convergent series of non-negative real numbers. If llakll~bk

(kEN),

then the series

converges.

Proof The partial sums of the sequence (1) are a Cauchy sequence (exercise 27.7(1)). Given any a>O, it follows that there exists anN such that, for any K>l>N, K

ltK-tJI

=

I

k=J+l

bk
where t K denotes the Kth partial sum of the series (1). It follows that, for any K>l>N, K

I

llsK-sJII=II akll k=J+l K

~

I

k=J+I

K

llakll ~

I bk
and so <sk) is a Cauchy sequence. Since X is complete, we may conclude that <sk) converges.

Many analysis textbooks are stuffed full with tests for establishing the convergence of series. Most of these tests, however, are useful only for somewhat obscure series and we shall therefore restrict our attention to two of the more important.

28.4 Proposition (Ratio test) Let
203

Series

converges if bk+ 1 1 (1.) I.1m sup--< k~co

bk

and diverges if

. f -b-> bk + 1 1 (11.. ) I.Imm . k~co

k

Proof If (i) holds, there exists a p < 1 and an H such that bk < H pk (kEN). The convergence of the series then follows from the comparison test. If (ii) holds, then bk-f.O as k-HfJ and so the series diverges .

Proposition (Root test) Let
converges if (i) lim sup bk 11k < 1 k~co

and diverges if (ii) lim sup b/fk > 1. k~co

Proof The same remarks apply as in the proof of proposition

28.4.

28.6

Example Consider the case bk = k- •. Then

lim bk+ 1 = 1 · bk '

k-...oo

lim b/fk = 1. k -...oo

It follows that neither the ratio test nor the root test are helpful m

determining the convergence or divergence of the series Cl)

1

2:-. k"

k= 1

As it happens, the series converges for ex> 1 and diverges for exercise 28.10(3).)

ex~

1. (See

204

Series

28.7

Absolute convergence A series

converges absolutely if and only if

converges. The comparison test shows that in a complete space any absolutely convergent series converges. The ratio test and the root test can therefore be used with bk = lla~c.ll to establish the convergence of certain series in a complete space. But note that both tests (and the comparison test) are only able to establish absolute convergence. However, series exist which converge but do not converge absolutely. (See exercise 28.10(4).)

28.8

Power series

If (a~c.) is a sequence of real (or complex) numbers and ( is a real (or complex) number, then the series

is called a power series (about the point (). For what values of z does this power series converge? The root test is useful here. Suppose that lim sup la~c.l 11k = p. k~oo

Then lim sup la~c.(z-(tl 11k=piz-(i. k~oo

If 0 < p < + oo, it follows that the power series converges when lz- (I< 1/p. If p = 0, the power series always converges and, if p = + oo, it

converges only when z=(. This argument shows that there exists an RE [0, + oo] such that the power series con verges when lz- (I < R and diverges when lz- (I > R. It follows that the set S of values of z for which the power series converges satisfies

{z: lz- (I
Series

205

Thus, if z is a real variable, S is an interval with midpoint(. We say that S is the interval of convergence of the power series. If z is a complex variable, S is a disc with centre (. We call S the disc of convergence of the power series. For obvious reasons, R is called the radius of convergence of the power senes.

-R-...-R--..

interval of convergence

disc of convergence

A given boundary point~ of S may or may not be an element of S- i.e. the power series may or may not converge at ~- The above theory yields no information about this question. Note finally that we have shown that a power series converges absolutely at interior points of S (i.e. for lz- (I< R). If it converges at a boundary point of S, however, there is no reason why the convergence should necessarily be absolute.

Example The power series

28.9

oo

zk

k=1

k

I <-1l- 1 has radius of convergence 1. We have that

. I (-lt-1 11/kk -1.

hm

k-oo

The series obtained by writing z = 1 converges but not absolutely. The series obtained by writing z = -1 diverges.

28.10

Exercise

(1) Suppose that
and that the series

conyerges. Prove that ak-->0 as k-> oo. Give an example of a sequence

206

Series

diverges. (2) Suppose that
exists if and only if the sequence of partial sums is bounded above. (3) Prove that

diverges by demonstrating that

where sK denotes the Kth partial sum. Prove that, if IX> 1, then w

1

1

1

J1 k" = 1 + 2" + 3" + ... converges by demonstrating that t 2 i- 1 ~2"- 1 /(2"- 1 -1)

UEN)

where tK denotes the Kth partial sum. (4) If
aK-aK+1~1

L (-1)k ak

k=K

l~aK.

Show that, if also ak--+0 as k--+oo, then

converges. [Hint: Show that the sequence of partial sums is Cauchy.] Deduce that the series S=

w

(-1)k-1

k=1

k

L

207

Series

converges and that

! ~ s ~ 1.

Prove also that

3 11111111 -s= 1 +---+-+---+-+---+ ... 2 3 2 5 7 4 9 11 6

(5) Suppose that
L: ak

s=

k=1

converges if and only if

is finite and that, in this case, s = t. (6) Suppose that A and B are non-empty sets and that f: A x B--+IR1. Prove that sup (sup f(a, b))= sup aEA

bEB

f(a, b).

(a,b)EAxB

Let be a 'double sequence' of non-negative real numbers and let J be the collection of all finite subsets of N. Prove that

exists if and only if t= sup GEj X j

L

aik

(j,k)EG

is finite and that, in this case, s = t. What can be said of U=

I (I aik)?

k= 1

(See exercise 23.19(6).)

j= 1

208

Series

28.11t

Uniform convergence of series Under what circumstances is it true that 00

lim x~~

00

I

fk(x)=

k=!

I

limfk(x)?

k=!

x~~

This is a problem about the circumstances under which two limiting operations 'commute'. We discussed such problems in chapter 23. In particular, we introduced the notion of uniform convergence to provide an appropriate criterion. Suppose that, for each kEN,fk: S-+'IJ where 'IJis a normed vector space. We say that (1)

converges uniformly for xES if and only if the sequence of partial sums converges uniformly for xES. Similarly, (1) converges pointwise if and only if the sequence of partial sums converges pointwise for xES.

28.12t Theorem Suppose that X is a metric space and that 'IJ is a complete normed vector space. Let S c X and suppose .that, for each kEN, the functions fk:S-+?J satisfy

through the set S and that

converges uniformly for xES. Then 00

I k=l

00

fk(x)-+

I

fJk

as

x-+~

k= I

through the set S.

Proof This is an immediate consequence of theorem 27.6. 28.13t Corollary Suppose that X is a metric space and that 'IJ is a complete normed vector space. Let Se X and suppose that, for each kEN, the function fk: S-+'IJ is continuous on S. Then the functionf: S-+'IJ defined by f(x)=

I

fk(x)

k=!

is continuous on S provided the series converges uniformly for xES.

Series 28.14

209

Example The series

I

f(x)=

xe-kx

k=O

converges pointwise for each xe[O, oo). For x>O,/(x) may be evaluated using the formula for a geometric progression. We obtain that x -{x(l-e-y f( )- 0

1

(x>O) (x=O).

Observe thatf(x)-+1 as x-+0+ but/(0)=0. Thus f is not continuous on [0, oo) and hence the series does not converge uniformly for xE [0, w ).

28.15t

Series in function spaces

Recall from §27.8 that iJ (S, 'IJ) denotes the space of all bounded functions F: S -+'lj with metric u defined by u(F, G)= sup d(F(x), G(x)). xeS

If 'IJ is a normed vector space, then we may regard YJ (S, 'IJ) as a normed vector space by defining vector addition and scalar multiplication in YJ (S, 'IJ) by (F + G)(x) = F(x) + G(x)

(aF)(x) = aF(x). The norm on iJ (S, 'IJ) is, of course, defined by IIFII = supiiF(x)ll xeS

so that u(F, G)=IIF-GII. We proved in theorem 27.10 that, if 'IJ is complete, then so is YJ (S, 'IJ). In particular, if 'IJ is a complete normed vector space, then so is YJ (S, 'IJ). Suppose that 'IJ is a normed vector space and that, for each keN, j~:S-+'Ij is bounded - i.e.
I

J;,(x)

k=l

converges uniformly for xeS is the same as saying that the series

I h.

k=l

converges in the space YJ (S, 'IJ). This is a useful observation because it allows us to extend many of the preceding

210

Series

results of this chapter to the case of uniformly convergent series. In particular, we obtain the following version of the comparison test.

28.16t Theorem (Weierstrass 'M test') Suppose that, for each keN, fk: S-+'IJ, where 'IJ is a complete normed vector space. Suppose also that
converges and that, for each keN and each xe S, (1)

Then the series OC!

L

Jk(x)

k=l

converges uniformly for xeS.

Proof Condition (1) implies that

li.ficli ~ bk

(ke 1\!).

Hence the series OC!

I.h

k= I

converges in the space YJ (S,

'IJ) by

the comparison test.

28.17 Example Consider the functionfk: [0, oo)-+~ defined by Jk(x)=x 2e-kx. This achieves a maximum at x=2k- 1 • Hence, for each xe[O, oo) and each keN,

It follows from the Weierstrass 'M test' that the series OC!

f(x)=

L

x2e-kx

k= I

converges uniformly for xe [0, oo ). From corollary 28.13 we may conclude that f is continuous on [0, oo). In fact, for each xe [0, oo ), we may evaluate f(x) as in example 28.14 to obtain x2(1-e-x)-l f(x)= { 0 Observe that f(x)-+f(O) as x-+0+.

(x>O) (x=O).

211

Series 28.18t

Exercise

(1) Prove that the series

converges pointwise for x;?;O. Show that f is not continuous on [0, w) and deduce that the series does not converge uniformly for x;?; 0. Does the series converge uniformly for (a) O~x~l,

(b) l~x~2?

(2) Prove that the series

f(x)=

I k=O

sin kx 2 k

converges uniformly for xEihl. Explain why f is continuous on lhl. (3) Suppose that (uk) and (vk> are sequences of real numbers and let K

I

U K=

(K>J).

uk

k=J+I

Prove that K-1

K

I

I

ukvk=vKUK+

k=J+I

Uk(vk-vk+d

k=J+I

(Abel's 'partial summation' formula). (4) Suppose that (ak) is a decreasing sequence of positive real numbers which tends

to zero and that ihl satisfies

for all KEN and all XES. Prove that

converges uniformly for xES. [Hint: Use question 3.] (5) Establish the identity

. . . cos tx-cos(n+t)x sm x+sm 2x+ .. . +sm nx= . 2 sm 21 x and deduce that the series f(x)=

I k=

I

sin kx k

converges uniformly for xE[b, 2n-b] provided that b>O. [Hint: Use question 4.] Deduce that f is continuous on (0, n:).

212

Series

(6) Suppose that the real power series 00

L akxk

f(x)=

k=O

has interval of convergence I and let J be any compact subinterval of I. Prove that the series converges uniformly for xei. [Hint: To show that the series converges uniformly on [0, X]ci, take uk=akXk and vk=(xX- 1 )k in question 3.] Deduce that the function F: [ -1, 1)-->IR defined by xk

cn

F(x)=

L k=

I

k

is continuous on [ -1, 1).

28.19t

Continuous operators Suppose that X and ?J are metric spaces and that T: X -->?J is continuous

on X. Then we know from corollary 25.9 that

T(lim k-.oo

X~= lim }

T(Xd

k-+oo

provided that the left-hand side exists. It follows that

provided that the left-hand side converges. This is a useful result, particularly in the case when X is a function space. We then usually call the continuous function T: X-->?J a continuous operator. We shall illustrate the usefulness of the result by examining the case in which X is the space of all continuous functions f: [a, b] -->IR (usually denoted by C[a, b]) and T is the 'integration operator' on C[a, b].

28.20t Theorem Let C[a, b] denote the space of all continuous functions [a, b]-->IR and define the operator T: C[a, b]-->IR by

f:

T(f)=

r

f(t)dt.

Then T is continuous on C[a, b].

Proof We have that IT(f)-T(g)i=

r r f(t)dt-

1

g(t)dt 1

213

Series

~

f

lf(t)-g(t)ldt

~(b-a)

max lf(t)-g(t)l a;£t ;;i:b

=(b-a)llf -gll. It follows that T(f)-. T(g) as j-.g and hence that T is a continuous operator on C[a,b].

28.21t

Corollary For each kEN, letfk: [a, b]-.IR be continuous on [a, b] and

suppose that (1)

converges uniformly for xE[a, b]. Then (2)

Proof To say that (1) converges uniformly for xE[a, b] is the same as saying that CO

I.h

k=l

converges in C[a, b]. The result therefore follows immediately from theorem 28.20.

28.22 Example It is worth noting that, without the uniformity condition, corollary 28.21 does not hold. Consider, for example, the function F k: [0, 1] -.IR iH.ustrated below.

1

2k

j o_k-~-

We have that, for each xE[O, 1], Fk(x)-.0 as k-.oo

~i.e.

(Fk> converges pointwise to

214

Series

the zero function. But

It follows that the series whose sequence of partial sums is
28.23t Theorem Suppose that X and 9J are metric spaces and that T: X-+'IJ is continuous and bijective. Then

provided that the right-hand side exists.

Proof Since T is continuous

Hence

We illustrate this result by taking X= C[a, b] and 9J = Ce[a, b] where the latter notation denotes the set of all F: [a, b]-+!R which can be written in the form F(x)= J:f(t)dt

(a~x~b)

wherefis a function in C[a, b]. The role of the operator T: X-+'IJ will be assumed by the 'integration operator' I: C[ a, b]-+ Ce[ a, b] defined by I(f)= F. The fundamental theorem of calculus asserts that the 'differentiation operator' D: Ce[ a, b]-+ C[ a, b] defined by D(F) = f where

f(x)=F'(x)

(a<x
is inverse to the integration operator- i.e.

It is easily shown that I is continuous on C[a, b] and we therefore obtain the following theorem in which C'[a, b] denotes the set of all f: [a, b]-+!R which have a continuous derivative on [a, b].

Series 28.24t

215

Theorem Suppose that, for each keN, gkeC'[a, b] and that 00

L

9k'(x)

k=l

converges uniformly for xe [a, b] and

converges for some ee [a, b]. Then, for each xe(a, b)

v(J gk(x~= J(vgk(x~. 1

(3)

1

Proof Apply theorem 28.23 with yk=fk, wherefk: [a, b]-R is defined by fk(x)=gix)-gie).

28.25

Example Consider the function Gk: [0, 1] ..... R illustrated below.

0

We have that Gke C'[a, b] and that
(keN).

It follows that the series whose sequence of partial sums is (Gix)) fails to satisfy (3) of theorem 28.24.

28.26t

Appli~ations

to power series

A trivial consequence of the Weierstrass 'M test' (28.16) is that a real (or complex) power series

converges uniform! y for lz-" ;;a r provided that the constant r is chosen so that r < R where R is the radius of convergence of the power series. Exercise 28.18(6) is a more subtle result. This demonstrates that a real power series converges uniformly on any

216

Series

compact subinterval of its interval of convergence. We may deduce the following results.

28.27t

Proposition (Abers theorem) The sum of any real power series is continuous on its interval of convergence.

28.28t

Proposition If c and d are any points in the interval of convergence of a real power series f(x)=

I

ab-~)k,

k=O

then

28.29t

Proposition If y is any interior point of the interval of convergence of a real power series f(x)=

I

ak(x-~)\

k=O

then f is differentiable at y and f'(y)=

I

kak(y-~)k-1.

k=!

The second and third propositions are true also of complex power series (provided we replace 'interval of convergence' by 'disc of convergence'). Abel's theorem, however, is false for complex power series. The diagram below represents the disc of convergence of a complex power series. If the power series converges at the point y, then it is true that f(z)-f(y) as z-y along curves like C 1 but it is in general false thatf(z)-f(y) as z-y along 'tangential' curves like C2 •

,..,..---- ................ '

'

\

~

y

217

Series 28.30

Example If lxl < 1, we have that

By proposition 28.28, if IYI < 1, then

f

log(1+y)= y -dx - = fy(oo I (-1)kxk) dx o 1+x o k=o

oo

=I

k+ I

(-1)k

_Y- .

k+1

k=O

The latter power series has interval of convergence ( -1, 1] and, by Abel's theorem (28.27), its sumf(y) is therefore continuous on (- 1, 1]. Butf(y) = log(1 + y) for yE (- 1, 1). It follows that f (1) =log 2 - i.e. 1

1

1

1

log 2=1-2+3-4+5- ...

28.31 t

Exercise

(1) Prove that, for lxl < 1, 1

00

--2 = I (-1tx2k. 1+x k=O Hence obtain a power series expansion for arctan x which is valid for -1 <x~ 1. Deduce that n 1 1 1 1 -=1--+---+-4 3579···

(2) Obtain the general binomial theorem by differentiating (l+x) (3)

-·

;, a:(a:-1) ... (a:-k+ 1) k k! x

=k~o

(-1<x
The power series 00

f
I

k=O

00

ak
I

bk
k=O

both converge for lz- (I< r and f(z) = g(z) for lz- (I< r. Prove that ak = bk (k =0, 1, 2, ... ). [Hint: Differentiate.]

29t

29.1t

INFINITE SUMS

Commutative and associative laws

The operation of addition in a vector space X satisfies certain laws. (See § 13.17.) In this section we wish to concentrate on the commutative law and the associative law. The commutative law asserts that for any XE X and yE X,

x+y=y+x. The associative law asserts that for any

X,

XE

yE

X and

ZE

X,

(x +y)+z=x + (y +z). The two laws together imply that, in evaluating a finite sum, it does not matter in which order the requisite additions are performed: the result will always be the same. For example, ((1 + 2)+ 3)+4= (3 + 3)+4=6 +4= 10

(4+ 1)+ (3+2)=5 +5 = 10. This means that we can write

1+2+3+4=10 without needing to specify the method of summation which is to be used. It is natural to ask to what extent these laws extend to the infinite case. We begin with the commutative law. For this purpose we require the notion of a permutation. A permutation 4> on a set A is simply a bijective function cf>: A-->A. We shall be concerned with permutations 4> on the set F\1 of natural numbers. Given such a permutation 4>: F\1--> F\1 and a sequence a rearrangement of because it has the same terms as but placed in a different order. An appropriate generalisation of the commutative law to series would assert that, for any permutation 4> on F\1, 00

I k=l

00

ak=

I

aq,(kl

k=l

provided that one side or the other exists. Unfortunately, this result is false m

218

219

Infinite sums

general. Exercise 28.10(4) provides a counterexample. If we write 00

1-t+t-i+t-i+. ·.=I ak,

(1)

k=l

then 00

1 +t-t+t+t-i+ ... = I a<J>(k>• k=l

where 4>: N--+ N is the permutation defined by c/>(3k)=2k;

c/>(3k-1)=4k-1;

c/>(3k-2)=4k-3.

But although both infinite series converge, they are not equal. Next we turn to the associative law. This is concerned with the manner in which the terms in a sum are bracketed. We begin with the following instructive example:

0=0+0+0+ ... =(1-1)+(1-1)+(1-1)+ ... =1-1+1-1+1-1+ ... = 1 + ( -1 + 1) + ( -1 + 1) + ( -1 + 1) + ... =1+0+0+0+ ... =1. Clearly something is wrong here. The problem is not hard to locate. The removal of the brackets at the fourth step is invalid. In fact, the series 1-1 + 1-1 + 1-1 + ... does not converge. The removal of brackets is therefore very definitely not in general permissible in an infinite series. One might hope however that the insertion of brackets in a series might be acceptable. This is certainly the case if we choose only to bracket neighbouring sets of terms as indicated below:

In this case, the sequence of partial sums of the new series is s3 , s5 , s9 , •••• This is a subsequence of <sK) and hence converges to the same limit. But what happens if we choose to bracket sets of non-neighbouring elements? To take an extreme case, we might choose to bracket all the even terms and all the odd terms separately as below:

If we attempt to bracket series (1) in the fashion indicated, we obtain

(1 +t+t+-t+ ... )+( -t-i-i- ...). But neither of these bracketed series converges. (See exercise 28.10(3).) All these negative results are rather depressing. But no cause for despair exists. As mathematicians, we know that, if matters do not work out very well with one definition, the thing to do is to try another definition. This is the subject of the next section.

220

Infinite sums

29.2t

Infinite sums

As we saw in the previous section, series do not satisfy the appropriate generalisation of the commutative law. It is therefore natural to seek an alternative definition for which the commutative law does hold. Suppose that X is a normed vector space and that a: I-> X is a function. If SE :r. there is, of course, no difficulty in defining s= L a(i)

(1)

iel

in the case when the 'index set' I is finite. In the case when I is infinite, we shall define (1) to mean that

For any e > 0, there exists a finite set F cl such that, for each finite set G, Fe Gel ~11 s-

L a(i)ll<e. ieG

A vector s defined in this way will be called an infinite sum. (Note incidentally that the convergence notion introduced here differs somewhat from those we have met earlier. It is an example of the convergence of a net.) Since the definition takes no account of any ordering which may exist on the index set I, it is clear that infinite sums must satisfy the commutative law. The next theorem puts the proposition in formal terms.

29.3t

Theorem Suppose that ,P is a permutation on J, then

:r is a normed vector space and

that a: I-> :r. If

L a( i) = L a( f/J(i)) iel

iel

provided one side or the other of the equation exists.

Proof We shall suppose that it is the left-hand side which exists. (Otherwise replace a by b =a o ,p and ,P by ,p ~ 1 .) Let e > 0 be given. Then there exists a finite set F c I such that, for any finite set G, FcGcl ~ 11

L a(i)- L a(i) ll<e. iel

Hence, for any finite set H,

ieG

Infinite sums

221 =>

L a(i)-

11

iel

=>

~

L a(i)- L

11

a(i) 11 <e

ie.P- (H)

iel

a((i)) 11 <e

ieH

and the result follows.

Next it is natural to consider the associative law. The definition of an infinite sum is not tailored to meet the requirements of this law and so we need to work a little harder to obtain a result. As a preliminary we need to relate the idea of an infinite sum to that of a series.

29.4t

Infinite sums and series

We begin with a result for infinite sums which is analogous to the theorem which asserts that, if an infinite series exists, then its terms tend to zero (exercise 28.1 0(1 )).

29.5t

Theorem Suppose that X is a normed vector space. Let a:

1-+ X and

suppose that

s= L a(i) ie/

exists. Then, given any e >0, there exists a finite set F c I such that, for each jE I, j$F

=> lla(j)ll<e.

Proof Let e > 0 be given. Then F c I such that for any finite set G FcGci

e/2 > 0 and hence there exists a finite set

=>lis- L a(i)ll<e/2. ieG

Let jEI\F. Then

lla(j)ll=ll

L

a(i)-s+s- L a(i)ll

iefu{j}

~11

L

ief

a(i)-sll+lls-l:a(i)ll

iefu{j}

<e/2+e/2=e.

ief

222

Infinite sums

29.6t

Corollary Suppose that X is a normed vector space. Let a: I-> X and

suppose that s= Ia(i) ie/

exists. Then the set H = {i: a (i) # 0} is countable.

Proof Let kEN. By the previous theorem the set Gk= {i: lla(i)ll;;:; 1/k}

is finite. But

and so H is the countable union of finite sets. It follows that H is countable by theorem 12.12.

The definition of an infinite sum given in §29.2 seems to be one of considerable generality in that the index set may be any set whatsoever while the index set in the case of an infinite series is restricted to be N. But the last result shows that this generality is largely illusory. An infinite sum of an uncountable collection of non-zero objects never exists. In future, we shall therefore always assume that the index set I is countable. Indeed, it will often be convenient to assume that I= N. This assumption can be made without loss of generality when I is countable because there then exists a bijection cf>: N-> I. Thus I may be replaced by N provided that a: I-> X is replaced by the sequence b: N-> X where b =a o cf>.

29.7t Theorem Suppose that X is a normed vector space and that
(1)

provided that the left-hand side exists.

Proof Let e>O be given. If the left-hand side of (1) exists, then there exists a finite set F c I such that for any finite set G, F c Gc N

=> J1

I

I M< e.

bk-

keN

keG

Let K 0 =max F. Then, provided that K>K 0 , the set G={1, 2, 3, ... , K} satisfies FcGc N. Hence K

K>K 0 =>11IbkkeN

and the equation (1) follows.

I

k =I

bkll<e

Infinite sums

223

Theorem 29.7 says that, if an infinite sum exists, then so does the corresponding series and the two are equal. Of course, as we know from §29.1 it is in general false that the convergence of a series implies the existence of the corresponding infinite sum. There is, however, an important special case for which the convergence of a series does imply the existence of the corresponding infinite sum. This is the case of a series with non-negative terms.

29.8t

Proposition Suppose that (bk) is a sequence of non-negative real num-

bers. Then

provided that the right-hand side converges.

The proof follows from exercise 28.10(5). An immediate consequence is that series of non-negative real numbers satisfy the commutative law although, as we have seen, this is certainly not the case of series in general.

29.9t

Complete spaces and the associative law

To proceed any further with the study of infinite sums, we need to restrict our attention to the case when X is a complete normed vector space. (Such a space is called a Banach space.) This is no surprise since most of the results of the previous chapter required completeness as well. Recall that a complete metric space is one in which every Cauchy sequence converges (§27.2). Alternatively, it is a metric space in which every totally bounded set has the Bolzano- Weierstrass property. We shall wish to use the fact that a version of the Heine-Borel theorem holds in a complete space. This asserts that every closed and totally bounded set is compact (see theorem 27.4). Some examples of complete normed vector spaces were discussed in §27.8. The most notable example of such a space is, of course, the space IR".

29.10t Theorem Suppose that X is a complete normed vector space and that§ is the collection of all finite subsets of I. Let a: I--. X and suppose that s=Ia(i) iel

exists. Then the set S={I a(i): IEH

HE§}

is totally bounded and hence its closure is compact.

224

Infinite sums

Proof We have to show that, for each e > 0, there exists a finite collection of open balls of radius e which covers S. Let e > 0 be given. Then there exists a finite set F c I such that, for each finite set G, FcGci

=>

III a(i)-sll<e. ieG

Define the set C by C={s- I a(i): EcF}. ieE

Then C is a finite set. The collection of all open balls of radius e whose centres are elements of C is therefore finite. It covers S because, for each HE j, 11

I a(i)-s+ I ~H

29.1lt

a(i) 11 =

~F\H

11

I

a(i)-s 11 <e.

~HvF

Theorem Suppose that .'L is a complete normed vector space and that

a :I-+ .'L. Then the infinite sum Ia(i)

(1)

iel

exists if and only if, for any e > 0, there exists a finite set F c I such that, for any finite set H, H cl \F

=> 11

I a(i)ll <e.

(2)

ieH

Proof It is quite easy to prove that the existence of (1) implies criterion (2). We therefore leave this as an exercise. There then remains the problem of showing that criterion (2) implies the existence of (1). With the notation of the previous theorem, let SE={.I a(i): HE jand EcH}. IEH

Then {SE: EE5} is a collection of closed subsets of the compact setS. Since this collection has the finite intersection property (exercise 20.6(3)) it follows that it has a non-empty intersection. Let sESE for each EE 5. Let e > 0 be given. Then, by (2), there exists a finite set F 0 c I such that, for any finite set H, Hci\F 0

=> 11

I a(i)ll<e/2. ieH

Also, since sESFo• there exists a finite set F satisfying F 0 c F c I such that

Infinite sums

225

Jls- L a(i)ll<e/2. ieF

Now suppose that G::::;F. Then G\Fc::.I\F 0 and so

lis- L a(i)ll;£lls- L a(i)ll+ll L a(i)ll ieG

ieF

ieG'.F

<e/2+e/2=e. Hence (1) exists.

Theorem 29.11 is a variant of the result which says that, in a complete space X, every Cauchy sequence converges. It is not surprising therefore that it has some useful corollaries.

29.12t Corollary Suppose that X is a complete normed vector space and that a: I-+ X. If J c::. I, then the existence of the infinite sum La(i) iel

implies the existence of La(i). ieJ

29.13t Corollary Suppose that X is a complete normed vector space and that a: I-+ X. If J and K are disjoint sets with union /, then L a(i) = L a(i) + L a(i) ieJ

ieJ

leK

provided that one side or other of the equation exists.

29.14t Corollary Suppose that X is a complete normed vector space and that a: I-+ X. Then the existence of the infinite sum La(i) le/

implies that, for any e>O, there exists a finite set Fe::. I such that for any set G, Fc::.Gc::.J =>IlL a(i)- L a(i)ll=ll L ie/

ieG

ie/\G

a(i)ll<e.

226

Infinite sums

The proofs of the corollaries are easy and we give them as exercises (29.22(1), (2) and (3)). Having obtained these results, we are now in a position to prove a version of the associative law in complete spaces.

29.15t

Theorem Suppose that X is a complete normed vector space and that a: I---> X. Let W denote any collection of disjoint sets with union I. Then (3)

provided the left-hand side exists. Proof Let e>O be given. Since the left-hand side of (3) exists, it follows from corollary 29.14 that there exists a finite set F c I such that, for any set G, FcGc/

=>

11 I a(i)- I a(i)ll<e. iel

ieG

Let 5 = { J: J E W and F n J # f/J} and suppose that (j is a finite collection of sets satisfying 5 c (j c W. Then the set

G=U

J

JE{J

satisfies F c G c I and therefore, using corollary 29.13,

IIIa(i)- I {. Ia(i)}II=III a(i)- I a(i)ll<e. iEf

JE{J

iEJ

iE/

iEG

Equation (3) follows.

Theorem 29.15 asserts that, if an infinite sum exists, then one can bracket its terms in any manner whatsoever without altering its value. The insertion of brackets in infinite sums therefore creates no problem. But the problem of removing brackets remains- i.e. the existence of the right-hand side of (3) does not in general guarantee the existence of the left-hand side. The infinite sum

IO

kEN

certainly exists. But, if we write 0 = (1-1) in this sum and then remove the brackets, we obtain

which does not exist. We noted at the end of §29.4 that, for the special case of series of non-negative real numbers, the commutative law does not fail. It is also true that the removal of brackets from infinite sums of non-negative real numbers creates no problems.

227

Infinite sums

Proposition Suppose that (bk) is a sequence of non-negative real numbers and that W is any collection of disjoint sets with union N. Then

29.16t

provided that the right-hand side exists.

The proof follows trivially from exercise 28.10(6). Propositions 29.8 and 29.16, taken together, mean that the problem of adding up a sequence of non-negative real numbers is very much simpler than the general problem. Roughly speaking, if any method of adding up a sequence of non-negative real numbers yields a finite result, then all methods will yield the same result. Is there a wider class of sequences with this pleasant property? We explore this question in the next section.

29.17t

Absolute sums

We met the notion of an absolutely convergent series in §28.7. A similar notion can be defined for infinite sums. If X is a complete normed vector space and a: I--> X, we say that l:a(i) iel

exists absolutely if and only if (1)

Llla(i)ll iel

exists. Since the terms of the latter sum are non-negative real numbers, the sum exists if and only if any method of adding up its terms yields a finite result. (See propositions 29.8 and 29.16). In particular, in the case when I= N, the infinite sum (1) exists if and only if the series

L lla(i)ll i= 1

converges.

Proposition (Comparison test) Suppose that X is a complete normed vector space and that a: I--> X and b: I -->IR. Suppose that the infinite sum

29.18t

l:b(i) iel

exists and that lla(i)ll;::; b(i)

(iE I).

228

Infinite sums

Then the infinite sum Ia(i) ieE

exists. Proof This is the same as the proof of theorem 28.3 except that theorem 29.11 is used instead of the fact that Cauchy sequences converge.

29.19t Corollary Suppose that X is a complete normed vector space and that a: I-> X. Then the existence of Illa(i)ll iel

implies the existence of Ia(i). ie/

Proof Take b(i)= lla(i)ll in proposition 29.18.

29.20 Corollary Suppose that X is a complete normed vector space and that a: I-> X. Let W denote any collection of disjoint sets with union I. Then the existence of

(2) implies the existence of the infinite sum Ia(i).

(3)

iel

Proof By proposition 29.16, the existence of (2) implies that (3) exists

absolutely.

An immediate application of this result is to the problem of removing brackets from infinite sums.

29.21 t Corollary Suppose that X is a complete normed vector space and that a: I-> X. Let W denote any collection of disjoint sets with union I. Then Ia(i)= I iel

Jeut

{Ia(i)} ieJ

provided the right-hand side exists absolutely.

229

I rifinite sums

Proof This follows immediately from corollary 29.20 and theorem 29.15. We shall use this result in §29.23 to study the problem of reversing the order of summation in repeated series.

29.22t

Exercise

X is any normed vector space and that a: I-+ X. If J and K are disjoint sets with union I, prove that

(1) Suppose that

Ia(i)= L a(i)+ ieJ

ie!

L a(i) ieK

provided the right-hand side exists. (2) Show that the equation of question 1 holds when X is a complete normed vector space provided the left-hand side exists (i.e. prove corollaries 29.12 and 29.13). (3) Suppose that X is a complete normed vector space and that a: I-+ X. Show that.

Ia(i) iel

exists if and only if, for any e>O, there exists a finite set G such that for any set Hcl\G

11 Ia(i)ll<e. ieH

(4) Let be a sequence of real numbers. Write ak + =max {ak, 0} and ak- = max { - ak, 0}. Prove that the infinite sum

exists absolutely if and only if any two of the series (i)

L

k=1

ak

(ii)

L

ak +

(iii)

k=1

L

ak

k=1

converge. If (i) converges but neither (ii) nor (iii) converge, then the series (i) is said to converge conditionaliy. Prove that, in this case, given any ~EIR there exists a permutation : N -+N such that 00

~=

L

k=1

a
(The latter result is called Riemann's theorem.) (5) If a: I -+IR, prove that the infinite sum

L a(i) iel

230

Infinite sums

exists if and only if it exists absolutely. [Hint: Use the previous question.] Obtain the same result when a: I --->IR". Let (ak) be the sequence in 100 (see §20.20) whose terms are all zero except for the kth which is equal to 1/k. Prove that

exists but not absolutely. Deduce that the converse of corollary 29.19 is false in general (although, as question 4 shows, the converse is true for .'L=IR"). (6) Prove that a normed vector space .'L is complete if and only if, for every a: I---> .'L, the existence of

I

lla(i)ll

iel

implies the existence of

I

a(i).

iel

29.23t

Repeated series One sometimes has to deal with repeated series of the form (1)

It follows from corollary 29.21 that if

(2) exists, then

I00{00I j=l

k=i

} I00{00 I

ajk =

k=i

}

ajk .

j=i

This is, of course, a special case of a much stronger result. If (2) exists, then any method of adding up the terms of (1) will yield the same answer.

29.24

Example It follows from the binomial theorem (exercise 28.31(2)) that (3)

provided that lzl < 1. It is instructive to see how this result may be deduced from the theory developed in this chapter. We begin with the formula for a geometric progression (1-z)- 1 = 1 +z+z 2 +z 3 + ... which is valid for

lzl < 1. Thus

(4)

231

Infinite sums

and so 00

(1-z)-2= I

00

I

zl

(5)

j= 0 I= j

provided that lzl < 1. But we know that the power series (4) exists absolutely for lzl < 1. Thus (5) exists absolutely and so we can reverse the order of summation and obtain oo

oo

I

(1-z)-2=I I

I

I

zi=I zl

1=0 j=O

1.

j=O

1=0

Thus 00

(1- z)- 2 = I

(I+ l)z1 = 1 + 2z + 3z 2 + ...

1=0

provided that lzl < 1. Some explanation of these calculations may be helpful. Given that 00

00

exists, the same result will be obtained regardless of whether the rows or columns in the array below are summed first.

+ '3

a30 + a31 + a32 + ...

+ '2

a2o + a21 + a22 + ...

+ rl

alO + all + au + ...

+ 'o

s

aoo + aol + ao2 + ...

+

+

+

+

a3o

a31

a32

a33

+

+

+

+

a2o

a21

a22

a23

+

+

+

+

aw

au

au

an

+

+

+

+

aoo

ao1

ao2

ao3

s = c0

+ cl

+ c2

+ c3

+ ...

In (5) we are concerned with the special case in which the terms in the array above the main diagonal are all zero as on page 232.

232

Infinite sums

0

0

0

0

0

~22

0

~I I

~()() +

~33

0

0

0

a33

+ a23

0

0

a22

a23

au

an

ao2

ao3

+

s ~

+ au + an

0

aOI + ao2 + a03

I

+ +

+ +

Then 00

S=

I

00

ri=

00

00

I 1=0

00

ai,

c,=

ai,

j=O l=j

00

I I

00

I I

=

j=O 1=0

j=O

S=

00

I I

I

00

I I

ai,=

1=0 j=O

ai,.

1=0 j=O

But summing an array by first adding the rows or columns is not the only way in which one can proceed. One can also sum, for example, by diagonals as indicated below.

Observe that d,

I

=

aik"

j+k=l

This method is particularly suitable for multiplying power series. We have that

=

I 1=0

I

aibkzi+k

j+k=l

00

=I

1=0

z'

I

j+k=l

aibk

233

I n.finite sums oo

=

L:

I

z'

1=0

L:

ambl-m

m=O

provided that lzl
oo

=

L 1=0

29.25t

oo

I

z1

L

1=

m=O

L

(1+1)z1•

1=0

Exercise

(1) Use the method of example 29.24 to show that, for each natural number n, 00

(n+k-1)! (1- z)-" = k~O k!(n -1)!

zk

provided lzl < 1. (2) Prove that

(3) Prove that 00

1 +1 -(- 1 ...} L: x(x+1) ... (x+k)=e{1- 1- - k=o x 1! (1 +x) 2! 2 +x) provided that xell\l "-.7L..

30t

30.lt

SEPARATION IN 1R n

Introduction

In this chapter we return to the study of the geometry of IR" which we discussed to some extent in chapter 13. Recall that, in §13.14, we observed that the axioms given by Euclid for his system of plane geometry are incomplete by modern standards. In particular, many of Euclid's proofs tacitly assume that, if H is a line in IR 2 , then IR 2 \ H has precisely two components and that, if x and y lie in different components, the line segment joining them cuts H.

This fact is not hard to prove. Given any hyperplane H = {x: <x, ~) = c} in IR ", the half-spaces S 1 ={x: <x, ~)c} are convex, open and disjoint. It follows that S 1 and S 2 are the components of IR" \H. (See theorem 15.14 and § 17.21.) If xE S 1 and yE S 2 , then the line segment L joining x and y must cut H. Otherwise, LcS 1 uS 2 and so S 1 uS 2 is connected (theorem 17.19). An analogous argument shows that any hyperplane (§ 13.14) splits IR" into two distinct components. It is natural to ask: what other sets have this property? This is one of the questions we shall discuss in this chapter. The chief point that needs to be made is that it is vital not to dismiss the answers to such questions as 'geometrically obvious'. Even when the 'geometrically obvious' results are actually true (which is certainly not always the case), they are often exceedingly difficult to prove. The Jordan curve theorem (30.13) is a notable example. It says that any simple 'closed' curve has an 'inside' and an 'outside'. But the proof Df the Jordan curve theorem is much too hard to present in this book. Not only is it long and complicated, it also makes use of some quite subtle ideas from algebraic topology. Once one has seen the proof, one becomes very much less ready to dismiss the result as 'obvious'. It is also worth noting that many results true in IR" do not extend to infinitedimensional spaces. For example, a hyperplane H in an infinite-dimensional space X need not be closed. But it can be shown that, in this case, H = X and X \ H is connected! Only when H is closed does X\ H split nicely into two components.

234

235

Separation in 1R"

30.2t

Separation

The word 'separate' is used with various meanings in different branches of mathematics. In§ 15.11, we said that two sets A and Bare 'separated' if and only if they are not contiguous. This usage is sometimes extended by saying that a topology :J 'separates' points of X if and only if, for each XE X and yE X with xi= y, there exist disjoint open sets S and T for which XE S and yET. (This is not automatically true in a topological space. A topological space for which it is true is called a Hausdorff space. See theorem 15.16 and exercise 22.8(3).) In this chapter we shall use the word 'separate' in a related but not identical sense. We shall say that a set H in a metric (or topological) space X separates X into two components if and only if X\ H has two components S and T. If A and Bare two sets in X, we shall say that H separates A and B if and only if H separates X into two components Sand T for which AcS and BeT

30.3

Example The

line

H = {(x, 0):

x E IR}

separates

the

sets

A={(x, y): (x-1) 2 +y 2 ;;i; 1} and B={(x, y): (x+ 1jl +l;;:;; 1} in IR 2 • H

30.4t

Separating hyperplanes

We begin with the observation that, if Y is a non-empty, closed set in IR" and ~EIR", then there exists a point zEYwhich is nearest to ~-i.e. 11~-zll=d(~, Y). (See corollary 19.16.)

236

Separation in

~n

In the special case when Yis convex (as in the diagram above), it seems reasonable to expect that z will be characterised by the fact that the angle IJ is obtuse for all yE Y. This expectation is confirmed by the following theorem.

30.5t

Theorem (Projection theorem) Let Y be a closed, convex, non-empty set

in~·. Then for any ~E~", there exists a unique zEYwhich satisfies 11~-zll=d(~, S)".

This value of z is characterised by the fact that <~-z, y-z)~O

(1)

for all yE Y.

Proof Recall from our discussion of the 'cosine rule' in § 13.3 that (2)

It follows that, if (1) holds, then 11~-YII~II~-zll for all yEYand hence z is the nearest point of Y to ~Suppose, on the other hand that z is the nearest point of Y to ~- (The existence of such a z is guaranteed by corollary 19.16.) IfyEY, then U=Ot:y+ (1-0t:)z=z+Ot:(y -z)EY provided that 0 ~ Ot: ~ 1 because Y is convex. Replacing y in (2) by u, we obtain that 11~ -ull 2 = ll~-zll 2 +Ot: 2 IIY -zll 2 -20t:(~ -z, Y-z).

If (~-z, y-z)>O, then we can ensure that 11~-ull
11~ -yll ~~~~-zll + IIY -zll 2

2

2

>ll~-zll •

2

Hence y cannot be the nearest point of Y to

~-

Theorem 30.5 is usually invoked in the case when Yis actually a vector subspace of In this case, the theorem tells us that the nearest point zE Y to ~ is characterised by the fact ihat ~- z is orthogonal to Y.

~·.

y

Separation in llln

237

We may then define a function P: IR!n-+y by writing P(;)=z. For obvious reasons, P is said to be the orthogonal projection of IR n onto Y. Suppose that Y is a convex, non-empty set in IR!n. Then f is closed and convex. (See exercise 15.10(8).) Suppose that ;~f and let z be the nearest point in f to ;. We know from theorem 30.5 that, for all yE Y,

(;-z, y-;+;-z);:i;O. Taking u = (;- z)/11;- zjj, it follows that (u, y-;);2;

-11;-z11

for all yEY. Thus

is the equation of hyperplane which has the property that Ylies entirely inside one of the two half-spaces determined by the hyperplane. Note that the hyperplane passes through the point ; and has normal u where Ilull= 1. (See § 13.14.)

(u,x-1;)=0 (u, x-1;) < 0

In the next theorem we extend this result to the case when ; is a boundary point of a convex, non-empty set Y. We show that, in this case, there exists a hyperplane through ; such that Y lies entirely inside one of the two closed half-spaces determined by the hyperplane. Such a hyperplane is called a supporting hyperplane of the set Y.

supporting hyperplanes

30.6t

Theorem Let Y be a non-empty, convex set in IR!n and suppose that ;~Y.

238

Separation in IR"

Then there exists a u # 0 such that

for all yEY.

Proof Since ~rt=Y, each open ball of radius 1/k with centre ~ contains a point ~kEe(Y) (see exercise 30.10(5)). Let uk be chosen so that llukll=l and oo and u as I-> oo. Then, from the continuity of the inner product, as required.

Theorem (Theorem of the separating hyperplane) Suppose that A and Bare non-empty, convex sets in !Pl" for which (i) A #0 and (ii) AnB=0. Then there exists a u # 0 such that 30.7t

for all a EA and all bE B.

Proof We apply the previous theorem to the convex set C={a-b:aEA and bEB}. Conditions (i) and (ii) ensure that 0 1= C. We obtain that there exists a u # 0 for which
for all a-bEC. Hence
a)~
b) for each aEA and each bEB.

If the real number c is chosen so that

sup
a)~c~inf
b),

bEB

then theorem 30.7 tells us that the hyperplane
(u, x)= c

239

Separation in 1R"

The theorem of the separating hyperplane has many important applications. It is used, for example, in proving Von Neumann's famous maximin theorem of two person zero-sum game theory. As an illustration of these ideas, we give a minor application to the problem of comparing the topologies generated by different norms in IR".

30.8t

Norms and topologies in IR"

In § 13.17 we introduced the idea of a normed vector space. The natural norm to use in the space IR" is the Euclidean norm

llxll = {x1 2+x/+ ... +x/} 112 . In §21.18, however, we came across some other norms which are compatible with the vector space structure of IR". We shall use the notation

and

llxll, = lx1l + lxzl + · · ·+ lxnl to denote these alternative norms. In chapter 23 we found that the first of these alternative norms is sometimes more convenient to use than the Euclidean norm. The reason, of course, that this norm could be substituted for the Euclidean norm in this context is that both norms generate the same topology in IR" and, when considering limits, it is only the topologies in the spaces concerned which matter. (See §21.18 and §23.1.) There is an infinite collection of norms which are compatible with the vector space structure of IR" and it is natural to ask whether all these different norms generate the same topology on IR". The next proposition answers this question in the affirmative. Note, however, that the corresponding result is very definitely false in infinite-dimensional spaces.

30.9t

Proposition All norms on IR" generate the same topology on IR".

Proof We shall use the usual notation for the Euclidean norm on IR". We shall suppose that we are also given an alternative norm on IR" and use the notation lllxlll to denote this norm. As indicated in §21.18, it is enough if we can show that the set S = {x: lllxlll < 1}

contains an open Euclidean ball B, = {x: llxll < r} and is contained in another open Euclidean ball BR={x: llxiiO while a:XI{;S for large a:>O. First suppose that S is unbounded (with respect to the Euc/idean metric). Each unbounded convex set in IR" contains a half-line (proof?). Let X#O be a point on this half-line. Then aXES for all a> 0 and this is a contradiction. Hence S is bounded and so there exists an R for which S c BR·

240

Separation in IRl"

Next suppose that 0 is a boundary point of S (with respect to the Euc/idean metric). Since S is convex, it follows from theorem 30.6 that there exists an X# 0 such that <X, s);;;;O for all sES. But, if IXXES and IX>O, then <X, 1XX)=1XIIXII 2 >0. This is a contradiction. We conclude that 0 is an interior point of S (with respect to the Euc/idean metric) and hence there exists an r>O such that B,cS.

(X,

30.10t

x> = 0

Exercise

(1) Prove that any hypersphere (§13.14) separates IR" into two components.

(2) Prove that any ball, box or line segment S (§ 13.14) has the property that IR" \ S is connected. (3) Find hyperplanes which separate the following sets in IR 2 : (i) A={(x,y):x 2 +/;;;;1};B={(x,y):y;;:;1} (ii) A= {(x, y): x>O and xy> 1}; B= {(x, y): x>O and xy< -1}. (4) Find two non-empty, convex sets A and Bin IR 2 with AnB=0 which cannot be separated by a hyperplane. (5) Let S be a convex set in IR". (i) If ~E as, prove that ~Eo( eS). Give an example of a non-convex set s for which this result is false. (ii) If S is unbounded, prove that S contains a half-line. Give an example of a convex set S in [00 for which this result is false. (6) Prove that in a Hausdorff space (§30.2) every compact set is closed. Prove that IR" is a Hausdorff space.

Separation in 1R n

30.11t

241

Curves and continua

In the remainder of this chapter, we focus our attention on IR 2 • We have seen that lines and circles separate IR 2 into two distinct components. What can be said about other curves? First, something needs to be said about curves in general. In § 17.15, we defined a curve in IR 2 to be the image f(l) of a compact interval I under a continuous function f: I --+IR 2 • As we remarked at the time, this definition is not very satisfactory from the intuitive point of view since it fails to classify as a curve various 'one-dimensional' objects such as the example of Brouwer's (§17.18) consisting of a spiral wrapped around a circle.

Brouwer's set E is compact and connected. Such a set is called a continuum (although this usage is not consistent with the familiar description of IR as 'the continuum'). The fact that E fails to qualify as a curve is something to which one can accommodate oneself. But much worse things can happen than this as Peano discovered in 1890. We describe a version of his construction given by Hilbert. Let I= [0, 1] and let S be the square [0, 1] x [0, 1]. For each nE N, Hilbert constructed a continuous function j~: I--+ S. The curves j~ (!), JP) and [ 1 (/) are illustrated in the diagram below.

s

s

s

:I I I

I -- --T---I

I

I

T

tl

I

r L

tl

I

.,

I

The sequence of functions converges uniformly to a continuous functionf:J --+S. As is evident from the construction, the curve f(l) passes through every point of S i.e. f(l) = S. We say that f(l) is a 'space-filling curve'.

242

Separation in 1R"

This construction shows that 'two-dimensional' objects can be curves according to the definition given at the head of this section. This is, of course, highly counterintuitive. (Even worse, the function f: I -.S has the property that it is continuous on I but differentiable at no point of I.) Peano's discovery is by no means an oddity. The Mazurkiewicz-Moore theorem asserts that every locally-connected continuum is a curve. (A set E in IR 2 is locally. connected if for each eE E and each e > 0 there exists a b > 0 such that for each xE E satisfying lle-xll
30.12t

Simple curves

We shall say that a curve C is a simple arc if it is topologically equivalent (§21.1) to the compact interval [0, 1]. This means that there exists a homeomorphism f: [0, 1]-+C. A curve C will be called a Jordan curve(or simple 'closed' curve) if it is topologically equivalent to the unit circle U in IR 2 .

c

Jordan curve

0

The word 'simple' is used to indicate that we are discussing curves which do not 'cross themselves'. Hilbert's space-filling curve is therefore very definitely not simple. Simple arcs and Jordan curves also admit characterisations similar to that provided by the Mazurkiewicz-Moore theorem for general curves. Thus a simple arc is a continuum with the property that the removal of any point with two exceptions produces a disconnected set. (The exceptions are the endpoints.) A Jordan curve is a continuum with the property that the removal of any single point produces a connected set but the removal of any two distinct points produces a disconnected set. It is clear what the separation properties of simple arcs and Jordan curves 'ought' to be. If A is a simple arc, then IR 2 \A should be connected while if J is a Jordan curve, then IR 2 \ J should have precisely two components. Both results are true but neither is easily proved. The latter result is the more important and we quote it as the Jordan curve theorem.

243

Separation in 1R"

30.13t Proposition Let J be a Jordan curve in IR 2 . Then IR 2 \J has two components and J is the boundary of both components. One of the components is bounded and the other is unbounded. (The bounded component is called the 'inside' of J and the unbounded component is called the 'outside' of J.)

outside

Note that the Jordan curve theorem would be trivial if it were known that there exists a homeomorphism F: IR 2 --+IR 2 such that F(V)=J. But we are only given that there exists a homeomorphism f: V --+J. It is a result of Schonflies that each such homeomorphism f: V --+J can be extended to a homeomorphism F: IR 2 --+IR 2 but the proof of this useful result is harder than the proof of the Jordan curve theorem itself.

30.14t

Simply connected regions

Recall that the Riemann sphere (or Gaussian plane) IR 2 # is the one-point compactification of IR 2 • This notion is particularly important in those cases when one is identifying IR 2 with the system IC of complex numbers. We begin by observing that the Jordan curve theorem remains true with IR 2 replaced by IR2# (except that, if OOEJ, neither component is bounded in IR 2 ). In complex analysis, one is often concerned with open, connected subsets of IR 2 #. Such sets are called regions (or domains). Of particular importance, are simply connected regions. These are regions which have no 'holes'. More precisely, a region R in IR 2 # is simply connected if and only if IR 2 # \ R is connected.

simply connected region

multiply connected region

An important alternative characterisation of a simply connected region R is the fact that if R contains a Jordan curve J, then R contains one of the two components ofiR 2 # \ J. In particular, if oo rtR, then R contains the bounded component ofiR 2 # \ J.

244

Separation in 1R"

30.15t

Exercise

(1) Explain why Brouwer's set E (§30.11) separates IR 2 into two components. Explain also why E is not a locally connected continuum. (2) Consider the set R={(x, y):y>x 2 } in IR 2 • Explain why R is simply connected. (3) Suppose that X and 'IJ are topological spaces and that 'IJ is Hausdorff (§30.2 and exercise 30.10(6)). If X is compact and f: X _.'IJ is continuous and bijective, prove that f is a homeomorphism. (4) If f: I _.[R 2 and g: U _.[R 2 (where I= [0, 1] and U is the unit circle in IR 2 ) are continuous and bijective, prove that f(J) and g(U) are respectively a simple arc and a Jordan curve. Explain why Hilbert's space-filling curve is not simple. (5) Suppose that J is a Jordan curve in IR 2 and that ~EJ. Let S and T denote the inside and the outside of J respectively. Let aE S and bET. Prove that there exists a simple arc joining a and ~ which lies entirely inside S and that there exists a simple arc joining b and ~ which lies entirely inside T. [Hint: Use the result of Schonflies quoted in §30.12.]. (6) Explain why the result of the previous question is false when J is replaced by Brouwer's set E.

NOTATION*

=

E

rf {x:P(x)}

0 'r/

3 AcB

es

AuB AnB A\B

s

u Se

( -00,

[a, b] [a, oo)

15 15 16 16 21 22 23 23 23

(§4.10)

24

14 14

IJ\Ii" .9 (A)

f:A-+B f(S) f- 1(T) f-I:B->A <xk> fog IJ\Ii IJ\Ii+ IJ\Ii (a, b) (a, oo)

(-oo,b] [a, b) ]a, b[ N

7!.. 10

supS inf S sup f(x)

inf f(x)

(§9.10)

(§4.10)

24

(§5.1, §7.13) 28, 52 28 (§5.2) (§5.2) 29 29 (§5.2, §13.1) 32 (§5.9) 33 (§6.1) (§6.2) 35 35 (§6.2) (§6.2) 36 (§6.3, §25.1) 39,169 (§6.5) 39 (§7.2) 46 (§7.10) 50 (§7.1 0) 50 (§7.13, §5.1) 52, 28 (§7.13) 52

max S min S 00

10+ I[

i AB X

(x 1 , x 2 , 0 llxll (x, y) lxl d(x, y) d(~, S)

as

£ E

••• ,

(§9.7) 68 (§9.7) 68 (§9.16, §24.2) 73,151 (§10.11) 84 (§10.20) 92 (§10.20) 93 (§12.22) 122 (§13.1) 1 x.) (§13.1) 1 (§13.1) 3 (§13.3, §13.17) 4, 14 (§13.3) 4 7 (§13.9) (§13.9, §13.18) 7, 15 (§13.20) 18 (§14.2) 21 (§15.1) 31 31 (§15.1)

*Numbers in italics refer to pages in Book 1: Logic, Sets and Numbers.

245

69

xeS

tu

(a, b) AxB Az

(§7.13) 52 (§7.13) 52 (§7.13) 52 (§7.13) 52 (§7.13) 52 (§7.13) 52 (§8.2) 54 (§8.7) 56 (§8.13) 60 (§8.14) 60 (§9.7, §24.4) 68,156 (§9.7, §24.4) 68,156 69 (§9.10)

b)

xeS

tu

n s Se

7

9

(§2.4) (§2.10) (§3.1) (§3.1) (§3.1) (§3.1) (§3.4) (§3.4) (§4.1) (§4.4) (§4.7) (§4.7) (§4.9)

Notation

246

I"' J2 m(x, y) /(x, y) f(x)-+TJ as X-+~ limf(x) ·-~ f(x)--+1 as x-+a + f(x)--+1 as x-+bf(~ + ),f(~-) f(x, y)-+~ as X-+~ and y-+T) lim

f(x, y)

[- oo, + oo] - 00, + 00 [- oo, + oo]" (+oo, +oo)

(§24.4) (§24.4) (§24.12) (§24.12)

155 155 165 166

lim sup f(x) ·-~

(§26.11)

187

116 116 118

lim inf f(x) ·-~ YJ(S, '/))

(§26.11)

187

(§27.8, §28.15)

130

C(S, 'lj)

(§27.8) (§27.16)

194, 209 195 197

(§28.1)

201

(§28.19) (§28.19) (§28.19)

212 214 214

(§29.2)

220

(§20.20) (§20.21) (§21.18) (§21.18) (§22.1) (§22.1)

86 87 102 102 106 106

(§22.12) (§22.12) (§22.15) (§23.1) (§23.5)

132

(x,y)-(~.'1)

limf(x, y) ·-~ Y-'1 u(F, G) IR# IR"# i[;#

X* 00

L

(§23.5)

132

(§23. 12) 138 (§24.2) 151 (§24.3) 153 (§24.3, §24.11) 154,165

ak k=! C[a, b] C'[a, b] c~ [a, b]

L a(i) iel

INDEX*

Abel's theorem (28.27) 216 absolute convergence (§28.7) 204 absolute sums (§29.17) 226 accumulation point (§18.1) 60 affine set (14.17) 30 angle (§13.3) 4 Appel (§21.2) 91 Archimedean ordered field (27.20) 200 Archimedes (§9.2, §9.16) 63, 74 associative law (§6.6) 41 axiom of the continuum (§9.5, §27.18) 67, 199 ball (§13.14, §14.3) 11, 22 bijection (§6.2) 36 binomial theorem (28.31) 217 Bolyai (§13.19) 16 Bolzano-Weierstrass property (§20.7, 27.3) 78, 191 Bolzano-Weierstrass theorem (19.6, 25.24, 27.3) 70, 179, 191 boundary (§14.2) 21 boundary point (§14.2) 21 bounded sets (§9.3, 19.1) 66, 66 Brouwer (17.14, §17.18) 52, 53 Cantor (§12.20, §12.23, §24.1) 121, 123, 150 Cantor intersection theorem (19.9, 20.6) 71, 78 Cantor set (§18.9) 63 cardinality (§12.2) 110 cardinal numbers (§12.25) 126 Cartesian product (§5.2) 28 Cauchy-Schwiuz inequality (13.6) 5 Cauchy sequence (§27.1) 190 Chinese box property (§20. 7) 78 Chinese box theorem (19.3, 27.5) 67, 191 circle (§13.14) 8 closed interval (§7.13) 57 closed set (§14.7, §21.15) 25, 100 closure (§15.1, §21.15) 31, 100 cluster point (§18.1) 60

commutative law (§6.6) 41 compactification (§24.2, §24.4) 150, 154 compact interval (§7.13) 57 compact set (§19.5, §20.4, §21.15, §25.23) 69, 78, 100, 179 comparison test (28.3) 202 complement (§4.4) 22 complete metric space (§20.15, §27.2) 83, 190 completion of a metric space (§27.16) 197 complex number (§10.20) 92 component function (§16.2) 41 components of a set (§17.21) 56 components of a vector (§13.1) 1 composition (§6.5) 39 co-ordinates (§13.1) 1 cone (13.16) 14 connected set (§17.2, §21.15) 47, 101 consistent (§10.1, §13.14) 78, 9 contiguous sets (§15.11) ;34 continuous (§16.2, §21.15, §22.4) 39, 101, 110 continuous at a point (§22.4) 110 continuous on the left (§22.15) 118 continuous on the right (§22.15) 118 continuous operator (§28.19) 212 continuum (§8.14, §30.11) 62, 241 continuum axiom (§9.5, §2.18) 67, 199 continuum ordered field (§9.5) 66 contraction.(27.7) 193 contradiction (§2.9) 9 contradictory (§2.9) 9 cosine rule (§13.3) 4 countable set (§12.4) 112 countably compact set (§20.7) 78 con vergence (§22.1) 106 convex set (§13.14) 12 cover (§20.2) 77 curve (§17.15) 52 decreasing function (§22.16) dense (§9.19, §20.7) 75, 80 disc (§13.14) 11

*Numbers in italics refer to pages in Book 1: Logic, Sets and Numbers.

247

119

248

Index

disc of convergence (§28.8) 205 discrete topology (21.14) 100 disjoint (§4.9) 23 distance (§13.11, §13.18, §23:12) 7, 15, 138 divergence (§26.1) 181 domain (§6.2, §30.14) 35, 243 double limit (§23.1) 130 element (§3.1) 14 empty set (§3.1) 15 end point, of interval (§7.13) 52 equivalence class (§5.5) 31 equivalence relation (§5.5) 30 Euclid (§1.8, §7.2, §8.14, §10.1, §11.1, §13.4) 5, 45, 61, 78, 99, 9 Euclidean geometry (§13.14) 8 Euclidean metric (§ 13.18) 15 Euclidean norm (§13.3) 4 extended real number system (§24.4) 154 extension, of a function (§6.4) 39 field (§7 .3) 46 finite (§12.1) 107 finite intersection property (20.6) four colour problem (§21.2) 90 function (§6.1) 33 Gauss (§13.19) 16 Gaussian plane (§24.3) graph (§6.1) 34

interval (§7.13, §17.6) 57, 49 interval of convergence (§28.8) 205 inverse function (§6.2) 36 irrational number (§8.15, §11.1) 61, 98 isolated point (§18.1) 60 isomorphic (§9.21) 75 Jordan curve (§30.12) 242 Jordan curve theorem (30.13)

243

left-hand limit (§22.12) 116 length (§13.3) 4 limit (§22.3) 109 limit inferior (§26.11) 186 limit point (§ 18.1, §26.2) 60, 182 limit superior (§26.11) 186 Lindelof property (§20.7) 79 Lindelof theorem (20.10) 81 line (§13.14) 9 linear (13.16) 14 Lobachevski (§13.19) 16 locally connected set (§30.11) 242 lower limit (§26.11) 186

78

154

Haken (§21.2) 91 half-line (§13.14) 10 Hausdorlf space (§30.2) 235 Heine-Borel theorem (19.8, 27.4) 71, 191 Hilbert (§13.14) 9 homeomorphic spaces (§21.15) 101 homeomorphism (§21.1, §21.15) 89, 101 homomorphism (§21.1) 89 hyperbolic geometry (§13.19) 16 hyperplane (§13.14) 13 hypersphere (§13.14) 11 identity function (§6.5) 39 image (§6.2) 34 implies (§2.10) 10 increasing function (§22.16) 119 independent (§12.26, §13.19) 127, 16 index set (§29.2) 220 inductive definition (§8.7) 56 infinite set (§12.1) 107 infinite sum (§29.2) 220 infinity (§9.16, §24.1) 73, 149 injection (§6.2) 36 inside (§30.12) 242 integers (§8.13, §11.2) 60, 100 interior (§15.1, §21.15) 31, 101 intersection (§4. 7) 23

mapping (§6.1) 34 Mazurkiewicz-Moore theorem (§30.11) metric (§13.18) 15 metric space (§13.18) 15 metric subspace (§13.18, §21.9) 15, 95 monotone function (§22.16) 119

242

natural number (§8.2, §10.3) 54, 79 nested sequence (§19.2) 67 non-Euclidean geometry (§13.19) 16 norm (§13.17) 14 normal (§13.14) 12 normed vector space (§13.17) 14 north pole (§21.4) 92 nowhere dense set (§18.9) 64 one-point compactification (§24.2) 150 open ball (§14.3, §24.2) 22, 153 open covering (§20.2) 77 open interval (§7.13) 52 open set (§14.7, §21.15) 25, lOO operator (§28.19) 212 ordered field (§8.1) 54 orthogonal (§13.14) 10 orthogonal projection (§30.4) 237 oscillating function (§26.5) 186 outer product (§13.1) 3 outside (§30.12) 242 parallelogram (§13.1) 2 parallel postulate (§10.1, 13.16, §13.19) 13, 16 partial sum (§28.1) 201 Pasch's theorem (13.16) 14

78,

Index

249

path (§17.15) 53 pathological (§18.9) 63 pathwise connected set (§17.18) 53 Peano (§30.11) 241 perfect set (§18.9) 63 permutation (§29.1) 218 perpendicular (§13.14) 10 plane (§13.14) 12 Poincare (§13.19) 16 point (§13.14) 9 point of accumulation (§18.1) 60 pointwise convergence (23.11) 138 polynomial (§10.23, §16.13) 96, 44 power series (§28.8) 204 product topology (§21.18) 104 projection function (16.6, §21.18) 41, 103 projection theorem (30.5) 236 punctured sphere (§21.4) 92 Pythagoras' theorem (13.15) 10

separating hyperplane (§30.4) 238 separation (§30.2) 235 sequence (§6.3, §25.1) 36, 169 sequentially compact set (§25.23) 179 series (§28.1) 201 set (§3.1) 14 simple arc (§30.12) 242 simple curve (§30.12) 242 simpy connected (§30.14) 243 space-filling curve (§30.11) 241 sphere (§13.14) 11 stereographic projection (§21.4, §24.2) 92, 151 strictly decreasing function (§22.16) 119 strictly increasing function (§22.16) 119 subsequence (§25.18) 176 subset (§4.1) 21 supporting hyperplane (§30.4) 237 surjection (§6.2) 36

quantifier (§3.4)

term, of a sequence (§25.1) 169 topological equivalnce (§21.1, §21.15) 89, 101 topological space (§21.15) 100 topological subspace (§21.15) 101 topology (§21.8, §21,15) 95, 100 totally bounded set (§20.15) 83 totally disconnected set (17.24) 57 triangle inequality (13.7, §13.18) 6, 15 two-point compactification (§24.4) 154

16

radius of convergence (§28.8) 204 range (§6.2) 35 rational function (§16.13) 44 rational number (§8.14, §11.13) 60, 102 ratio test (28.4) 202 ray (§13.14) 10 real number (§7.2) 44 re-arrangement (§29.1) 218 region (§30.14) 243 relative topology (§21.9, §21.15) 96, 101 repeated limit (§23.5) 132 repeated series (§29.23) 230 Riemann sphere (§24.3) 154 Riemann's theorem (29.22) 229 right angle (§13.14) 10 right-hand limit (§22.12) 116 root (§9.13, §22.23) 72, 123 root test (28.5) 203 scalar (§13.1, §13.17) 1, 14 scalar multiplication (§13.1, §13.17) Schonflies (§30.12) 243 separated sets (§15.11) 35

1, 14

uncountable set (§12.14) 118 uniform continuity (§23.20) 145 uniform convergence (§23.11, §28.11) 138, 208 uniform distance (§23.12) 138 uniform metric (§23.12, §27.8) 138, 194 union (§4.7) 23 upper limit (§26.11) 186 vector (§13.1, §13.17) 1 vector addition (§13.1, §13.17) vector space (§13.17) 14 Weierstrass M test (28.16)

210

1, 14

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