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, Sao 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 (II)
4
Set operations
5
Relations
6
Functions
7
Real numbers (I)
8
Principle of induction
9
Real numbers (II)
iot
Construction of the number systems
lit
Number theory
12
Cardinality
Book 2: Topological Ideas Introduction 13 13.1 13.3 13.5 13.9 13.11 13.14
Distance The space U" Length and angle in Un Some inequalities Modulus Distance Euclidean geometry and Un
xi 1 1 4 5 7 7 8
vi
Contents
13.17| 13.18 13.19| 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)
21 21 21 22 25 29
15 15.1 15.4 15.8| 15.11
Open and closed sets (II)
16 16.1 16.2 16.7 16.13
Continuity
16.17t 17 17.1 17.2 17.6 17.8 17.15 17.18 17.21| 17.25| 18 18.1 18.4| 18.9| 19 19.1
Introduction Boundary of a set Open balls Open and closed sets Open and closed sets in Un Interior and closure Closure properties Interior properties Contiguous sets Introduction Continuous functions The continuity of algebraic operations Rational functions Complex-valued functions Connected sets
Introduction Connected sets Connected sets in U1 Continuity and connected sets Curves Pathwise connected sets Components Structure of open sets in Un Cluster points
Cluster points Properties of cluster points The Cantor set Compact sets (I)
Introduction
31 31 32 33 34 39 39 39 41 44 46 47 47 47 49 50 52 53 56 57 60 60 62 63 66 66
|This 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
20f 20.lt 20.2t 20.4| 20.7f 20.15| 20.16| 20.20|
Compact sets (II) Introduction Open coverings Compact sets Compactness in Un 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 21.9| 21.15f 21.18f
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
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
Limits and continuity (II) Double limits Double limits (continued) Repeated limits Uniform convergence Distance between functions Uniform continuity
130 130 132 132 136 138 145
22.34t 23f 23.lt 23.3t 23.5t 23.11t 23.12f 23.20t
viii
Contents
24 24.1 24.2 24.3 24.4 24.5 24.8 24.111 24.12|
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 26.5| 26.11t
Oscillation Divergence Limit points Oscillating functions Lim sup and lim inf
181 181 182 184 186
27 27.1 27.2 27.8 27.13| 27.16| 27.18|
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| 28.15| 28.19| 28.26|
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
29|
Infinite sums Commutative and associative laws Infinite sums
218 218 220
29.lt 29.2t
Contents
ix
29.4| 29.9| 29.17| 29.23f
Infinite sums and series Complete spaces and the associative law Absolute sums Repeated series
221 223 226 230
30t
Separation in Un Introduction Separation Separating hyperplanes Norms and topologies in (1 Curves and continua Simple curves Simply connected regions
234 234 235 235 239 241 242 243
Notation
245
Index
246
30.lt 30.2| 30.4f 3O.8t 30.11f 30.12| 30.14|
The diagram on p. x illustrates the logical structure of the 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.
-»»-j 2. Logic (I)
Proofs
n
3. Logic (II)
±
-j4. Set operations
112. Cardinality L L J T
5. Relations
7. Real numbers (I)
6. Functions
8. Principle of induction
Book I Logic, Sets and Numbers
I
I 9. Real ] numbers ( i
T
1
110. Construction I of number systems]
—
I
I |
14. Open and closed sets (I)
11. Number theory
13. Distance
JL
1
15. Open and I closed sets (II)
16. Continuity
17. Connected sets
H
30. Separation I in W
18. Cluster points
i
Book 2 Topological
]20. Compact 1 I sets (II) !
P
19. Compact sets (I)
Ideas
—j 2 1 . T o p o l o g y (-*•
22. Limits and continuity (I)
24. Points at infinity
I
i 23. Limits and] j continuity (II) i
c
25. Sequences U H 26. Oscillation
- - T " ^ •427. Completeness!—*H I
I
I
28. Series
|—•{i 29. Infinite iI iI sums
I
i
,
I i
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 Un. 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 Un. 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 | ) 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 R"
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 Un are the rc-tuples
in which x19 x2,..., xn are real numbers. We usually use a single symbol x for the n-tuple and write
The real numbers xv x2,...,xn are called the co-ordinates or the components of x. It is often convenient to refer to an object x in R" as a vector. When doing so, ordinary real numbers are called scalars. If x = (x1, x 2 , . . . , xn) and y = (yu yii--) yn) a r e vectors and a is a scalar, we define 'vector addition' and 'scalar multiplication' by
ax = (ocxv OLX29. .., a x j .
These definitions have a simple geometric interpretation which we shall illustrate in the case n = 2. An object xe[R2 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 (x19 x2).
(xltx2)
x as a point
x as an arrow
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
ccx2
x2 y\
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
/ y / /
/io /
\
/ /
\
\\\
\
\\ /
^p
y
/
o Is
7/ / /
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
Distance
3
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 = (l, 2, 3) and y = (2, 0, 5). Then = ( l , 2 , 3) + (2,0,5) = (3,2, 8) 2x = 2(l,2, 3) = (2,4,6).
It is very easy to check U" 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,
The diagram below illustrates the vector y — x = (yx — xv y2 — x 2 , •. •, yn — xn) in the case n = 2.
y-x
---
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 U * with U. Nor is there a problem when n = 2 since we can then identify U 2 with C (§10.20). If n ^ 3 , however, there is no entirely satisfactory way of defining multiplication in Un. 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 R 3 , one can introduce the 'outer product' or 'vector product' of two vectors x and y. This is a vector denoted by x A y or x x y. Unfortunately, x A y = — y A x .
Distance
4
Multiplication is therefore something which does not work very well with vectors. Division is almost always meaningless. 13.3
Length and angle in Un The Euclidean norm of a vector x in Un is defined by
We think of ||x|| as the length of the vector x. This interpretation is justified in U2 by Pythagoras' theorem (13.15).
The inner product of two vectors x and y in R" is defined by <x,y> = x 1 >/ 1 +x 2 ); 2 + ... + xnyn. It is easy to check the following properties: (i) <x,x> = ||x||2 (ii) <x,y> =
(iii) The geometric significance of the inner product can be discussed using the cosine rule (i.e. c2 = a2 4- b2 — 2ab cos y) in the diagram below. y-x
O
Rewriting the cosine rule in terms of the vectors introduced in the right-
Distance hand diagram, we obtain that
But, llx -y|| 2 = <x - y , x - y > = <x, x - y > - = <x,x>-2<x,y> + = ||x||2 + ||y|| 2 -2<x,y> It follows that <x,y> = ||x||.||y||cosy. Of course, this argument does not prove anything. It simply indicates why it is helpful to think of
<x,y> IMNIyll 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 = (l, 2, 3) and y = (2, 0, 5) in R 3 . We have that
||y|| = {2 2 + 0 2 -f- 5 2 } 1 / 2 = V 2 9 ' <x, y> 1-2 + 2 - 0 + 3-5 17 ^14x29'
13.5
Some inequalities
In the previous section y was the angle between x and y. The fact that |cos y\ < 1 translates into the following theorem.
13.6
Theorem (Cauchy-Schwarz inequality) If xeR" and yeR", then
Proof Let aeR. Then ay||2 = < x - a y , x - a y > = ||x|| 2 -2a<x,y> + a2||y||2.
6
Distance
It follows that the quadratic equation ||x||2 —2a<x, y> + a 2 ||y|| 2 has at most one real root (§10.10). Hence ' b 2 - 4 a c g 0 ' - i.e. 4<x,y> 2 -4||x|| 2 ||y|| 2 ^0.
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
llyll
llxll
This geometric idea translates into the following theorem.
13.7
Theorem (Triangle inequality) If xeU" ajid yeR", then ||x + y||g||x|| + ||y||. Proof 2
13.8
, x + y> + 2<x, y> + ||y||2 + 2||x||. ||y|| + ||y||2
(theorem 13.6)
Corollary If xeR" and yeUn, then
Proof It follows from the triangle inequality that
Distance 13.9
7
Modulus
The modulus \x\ of a real number x coincides with the Euclidean norm of x thought of as a vector in R 1 . We have that
One has to remember that y1/2 represents the non-negative number whose square is y (§9.13). The modulus \z\ of a complex number z = x + iy is identified with the Euclidean norm of (x, y) thought of as a vector in R2. We have that = \\{x9 y)\\. 13.10 Then
Theorem Suppose that u and v are real or complex numbers. \uv\ = \u\-\v\.
Proof We need only consider the complex case. If u = a + ib and v = c + id, then \uv\2 - \u\2\v\2 = (ac - bd)2 + (be + ad)2 - (a2 + b2)(c2 + d2) = 0.
13.11
Distance The distance d(x, y) between two vectors x and y in R" is defined
by In interpreting this idea in Un, 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
8
Distance
13.12
Examples
(i) The distance between the vectors x = (l, 2, 3) and y = (2, 0, 5) in IR3 is
(ii) The distance between 3 and 7 (regarded as vectors in R1) is |3-7| = | - 4 | = 4 .
-,
13.13
i-
Exercise
(1) Let x = (0, 1, 0) and y = (1, 1, 0) be vectors in U3. Calculate the quantities (i) x + y (iv) ||x||
(ii) x - y (v) | | x - y | |
(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 + 4j (v) 4 - 3 i (vi) I (3) If a and b are any real numbers, prove that |a|0, prove that a = 0. (4) Let x and y be elements of Un. Prove that
[Hint: Use question 3.] (5) Let x and y be elements of Un. Prove that (6) Let \eUm and yeUn. Then (x, y) may be regarded as an element of Um+n. Prove that
||(x,y)||2H|x||
13.14
Euclidean geometry and Un
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 IR2 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 systematically 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 U2 is a model for Euclidean plane geometry one has to give a precise definition of each of these words in terms of IR2 and then prove each of Hilbert's axioms for Euclidean plane geometry as a theorem in U2. 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 Un. A point is simply an element xeUn. The line I through the distinct points a and b is the set This is easier to illustrate with a picture if we rewrite it in the form Z = {a + j8(b-a): 0eR} and think of / as the line through the point a in the direction of the vector b-a. /line
If, in the above, /? 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 a and /? 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
ray through b
half-line from a through b
line segment joining a and b
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 Un 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>=0. 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 [Rn. Then x and y are orthogonal if and only if
x+y
Distance
11
Proof We have that
and hence ||x + y||2 = ||x||2 + ||y||2 if and only if <x, y>=0. A circle in IR 2 with centre £ and radius r > 0 is a set of the form
circle in IR2
In IR3, the set S represents a sphere and in R" we call 5 a hyper sphere. The inside of a hypersphere - i.e. the set is called an open ball (We should perhaps really call it a hyperball.) In U3, a ball looks like a ball ought to look. In IR 2 a ball is a disc and in IR1 a ball is a bounded open interval (£ — r, ^ + r). A dosed box in R" in a set S of the form / x x / 2 x . . . x In where each of the sets Ik is a closed, bounded interval (i.e. an interval of the form [a, bj).
12
Distance
if a i d o n ! f
L
t
u
>n VL™
arC CXampleS
°f C°nVeX
Sets
- ^ set S in M" is convex
CVer X £ S a n d y e S il is t r u e t h a t
'
ocx + PyeS provided
hat a^O ^ 0 and a + ^ = 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 R3
through
piece of hyperplane in
13
Distance
In (Rn, we call the set P a hyperplane. A hyperplane in IR" with normal u / 0 is therefore a set of the form / / = {x:<x,u> = c}, where c is a constant. The hyperplane passes through any point £> which satisfies <£, u> = c. In R 3 a hyperplane is an ordinary plane. In U2 a hyperplane is a line. In IR * a hyperplane is just a point.
hyperplane is a point
hyperplane is a line
It is 'geometrically obvious' (at least in R 1 , U2 and IR3) 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 Un. Prove that
Interpret this result geometrically for n = 2 in terms of the diagonals of a certain parallelogram. (2) Let x and y be elements of Un. Prove that
l|x|| = ||yll~<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 IR2, a hyperplane is the same thing as a line. (4) The parallel postulate for Euclidean plane geometry asserts that given any line / and any point P not on /, there exists a unique line through P which does not meet /. Prove this for (R2. (5) Let us say that a closed box S = / x x I2 x . . . x In in IR" is proper if none
14
Distance
of the intervals Ik 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 Un. Prove that hyperspheres are not convex. t(7) Prove that the intersection of any collection of convex sets is convex. What about unions? |(8) A cone in IR" 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 U" into distinct 'half-spaces', explain why a line which enters a triangle in R2 through one of its sides must leave the triangle through one of the other sides (Pasch's theorem). |(10) A function / : Rn-^1R" is linear if and only if, for each x and y in U" and all real numbers a and /?,
Prove that, for each S in R", S convex =>f (S) convex. f(ll) Prove that a function/: IR"-^ 1 is linear if and only if there exists an eeR" such that/(x) = <x, e> for each xelR". t(12) Anticipating the definition of § 13.17, prove that a normed vector space % (with real scalars) admits an inner product satisfying (i), (ii) and (iii) of § 13.3 if and only if ||x + y||2 + ||x-y|| 2 = 2||x||2 + 2||y||2
for each xe % and each ye %. [Hint. Use exercises 13.16(1) and 13.13(5).]
13.17|
Normed vector spaces
The space Un is an example of a normed vector space. For a vector space, one first needs a commutative group % (§ 6.6) to serve as 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) (a + £)x = ax + fix (iii) (a0)x = a(£x) (iv) 0x = 0; lx = x. For a normed vector space, we usually insist that the field of scalars be R or C. A norm function from X to R is required which satisfies the following properties: (i)
(ii) ||x||=0~x = 0 (iii) ||ax|| = |a|||x|| (iv) ||x + y|| ^ ||xj| + ||y||
(triangle inequality)
Distance
15
for all xe X, ye Z and all scalars. (Here ||x|| 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 U" 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 % and a function d: %x Z-+U which
satisfies (i) (ii) (iii) (iv)
d(x, y ) ^ 0 d(x, y) = 0 ^ 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 Un. As we have seen in §13.11, the metric d2 :Un x Un'-+U in R" is defined by
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
Any subset £ of Un is also a metric space provided that we continue to use the Euclidean metric in £. We then say that £ is a metric subspace ofMn. For example, the interval (0, 1] is a metric subspace of IR1 provided that the metric d: (0, 1] x (0, 1]->R is defined by d(x9 y) = \x-y\.
d(x,y)
of More exotic examples of metric spaces can be obtained by starting with any normed vector space % and defining d: % x Z-+M by
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 U 2 (with the Euclidean metric) is a model for Euclidean geometry (§13.14), we shall give Poincare's model for a non-Euclidean geometry as an example of such a metric space.
13.19|
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 Ivanovich 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 U2 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 % in Poincare's metric space is the set 3T={(x, y):x2 + y2 <\) in U2. But the metric used in % is not the Euclidean metric.
Distance
17
Let P and Q be points of %. If P and Q lie on a diameter of the circle C which bounds £, 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 £ are as indicated in the diagram, we then define the Poincare distance between P and Q by d{P, Q) =
PB/PA
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 boundary and therefore, as far as he is concerned, X extends indefinitely in all directions.
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 L and M. With these definitions ft 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.
13.20
Distance between a point and a set
If £ is a point in a metric space X and S is a non-empty set in ft,
18
Distance
we define the distance between £ and S by
Obviously, d(£, S) is always non-negative.
13.21 Example Consider the point 3 and the set S = [0, 1] in R. We have that d(39 S) = d(3, 1) = 2.
s
13.22 Theorem Let ^ be a point in a metric space X and let S be a nonempty set in %. Then d& S) = 0 if and only if, for each e>0, there exists an xeS such that d(£, x)<£.
Distance
19
Proof Everything in the set D = {d(x^):xeS} is non-negative. Thus 0 is a lower bound for D. The statement dfe, S) = 0 is therefore equivalent to the assertion that no g>0 is a lower bound for D i.e. not 3e>0 VxeS (d( But this is equivalent to
as required (§3.10).
It is important to take note of the fact that d(£, S) = 0 does not imply that
13.23
Examples
(i) Consider the point 1 and the set S= (0, 1) in R. For each £>0, we can find an xeS such that x> 1 — s.
Since 1—e<x
(ii) Consider the point 0 and the set S = {1/n: ne N} in U. Given any e > 0, we can find an neN such that n> 1/e (because N is unbounded above). But then d(0, l/n) = l/n<e and hence d(0, S) = 0.
points of S
20
Distance
13.24
Exercise
(1) Find d(£, S) for the following points £ and sets S in (i) (ii) (iii) (iv) (v) (vi) (vii)
£ = 1; S = (0,2) £ = 0 ; S = [l,2] £=±; S=N £=0; S = (0,l) { = 1; S = (2,3) £ = 2; S = (0,l)u(3,4) £ = y/2; S = Q. [tfmr: Use theorem 9.20.]
(2) Find d($, S) for the following points $ and sets S in R 2 . (i) $=(1,1); S = {(x,y (ii) $ = (2,1); S = {(x,y (iii) $ = (0,0); S = {(x,>; (3) Let S be a non-empty set of real numbers which is bounded above. Prove that d(sup S, S) = 0. (4) Let $ be a point in a metric space X and let S and T be non-empty sets in X. Prove that (i) (ii) (iii) <*($, SuT) = minM$, S), d($, T)}. What can be said of d($, SnT) in terms of d($, S) and d($, T)? (5) Let $ and T| be points in a metric space X and let S be a non-empty set in X. Prove that
(6) Let Su S 2 , . . . , Sn be non-empty sets of real numbers. Write S = S 1 x S 2 x . . . x S n and suppose that x = ( x 1 , x 2 , . . . , xn)eUn. Prove that d(x, 5) = 0ifandonlyif d(xk, Sk) = 0 for eachfc= l, 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 iRn. However, with just an occasional exception, all examples will be of sets in Un. 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 0 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 endpoints of an interval. Our first priority will therefore be to give a precise mathematical definition of a boundary point of a set S in a metric space %. 14.2
Boundary of a set
Let S be a set in a metric space Z. A boundary point of the set S is a point \ of % such that (i) d& S) = 0 and
(ii) dft, 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 dS. Observe that S and CS have the same boundary - i.e. dS = d(CS).
21
22
Open and closed sets (I)
X
not boundary points
Some special consideration of the case S = 0 is necessary. In this case d(Z>9 S) = 0 is meaningless and hence cannot be true. Thus the empty set has no boundary points - i.e. d0 = 0 . Similarly, d% = 0.
14.3
Open balls
The open ball B with centre £, and radius r > 0 in a metric space % is defined by In U3, an open ball is the inside of a sphere. In U 2, an open ball is the inside of a circle. In U1, an open ball is an open interval.
%
14.4 Theorem Let S be a set in a metric space % and let £ be a point in %. 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 C S.
Open and closed sets (I) point of S
23 point of 6 S
Proof The fact that each open ball B with centre ^ and radius r > 0 contains a point of S and a point of CS is equivalent to the assertion that, for each r > 0 , there exists xeS and yeCS such that d(£, x ) < r and d(£, y)
14.5
Examples
(i) The boundary of the set F in !R2 defined by
is the set
Note that every boundary point of F belongs to F - i.e. dF c F. L
y
CF
(ii) The boundary of the set G in U2 defined by
is the set
Note that no boundary point of G belongs to G - i.e. dGczCG.
24
Open and closed sets (I)
(hi) The boundary of the set H in U 2 denned by
H = {(x9y): x2 + y2<\ andy^O} is the set dH = {(x,y):x2 + y2 = lm\d ^ 0 } u { ( . x , 0): - l ^ x ^ l } . Note that some boundary points of H belong to H and some do not. CH
^ _^
(iv) The boundary of the set S = {1/n: ne N} in R1 is
dS = Sv{0}. points of S
o
/f 4
\ 3
Each point of S is clearly a boundary point of S. For example,
)^d(i (i i))=o and thus \e dS. Similarly, 0 is a boundary point of S because d(0, S) = 0 (example 13.23(ii)) ) = d(0, 0) = 0. Finally, no other point of IRx can be a boundary point of S. For example, f is not a boundary point of S because
Open and closed sets (I) 14.6
25
Exercise
(1) Sketch the following sets in U2 and determine their boundary points (i) (ii) (iii) (iv)
P = {(x,y):0^x<,l andO
(2) Sketch the following sets in U * 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. f (3) Let S be afiniteset in Un. Prove that dS = S. (Note that this need not be true in a general metric space. For example, if % = S.) t(4) Show that the open ball B in Un with centre % and radius r>0 has boundary dB = {x: d(x, £) = r}. (Note that this need not be true in a general metric space. For example, if % is the metric subspace of 1R1 consisting simply of the two points £ and £ + r.) |(5) Find the boundary points of the set S in U2 defined by S = {(l/m9 1/n): meM and neN}. t(6) Let A and B be sets in a metric space %. Prove that d(A\jB)adAKjdB. [Hint: Use theorem 14.4.] Deduce that d{AnB)adAndB.
14.7
Open and closed sets
Let S be a set in a metric space %. The set S is open if and only if it contains none of its boundary points - i.e. dS^CS. The set S is closed if and only if it contains all of its boundary points - i.e. dSczS. 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 C S. 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.
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.
/
open
/
As with most city plots of land, sets are usually neither open nor closed.
14.8 (i) (ii) (iii) (iv)
14.9 closed.
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 Z is open if and only if CS is
Proof The theorem follows immediately from the fact that S and CS have the same boundary.
es
14.10 Theorem A set F in a metric space % is closed if and only if, for each xe X,
Proof (i) Suppose that d(x, F) = 0 => xeF. If £ is a boundary point of F, then dfe, F) = 0 and dfe, CF) = 0. Since d& F) = 0, it follows that Z,eF. Thus F contains its boundary points and hence is closed. (ii) Suppose that F is closed. If x£F, then xe CF and hence d(x, CF) = 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, x e F . This is a contradiction. Hence x<£F => d(x, F ) ^ 0 - i.e. d(x, F) = 0=>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) = 0 by exercise 13.24(3). Hence the result by theorem 14.10.
14.12 Theorem A set G in a metric space % is open if and only if each point £e G is the centre of an open ball B entirely contained in 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 CG (theorem 14.4). Thus G contains none of its boundary points and so is open. (ii) Suppose that G is open. Let ^eG. 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 CG - i.e. BczG.
14.13 (i) (ii) (iii) sets is
Theorem In a metric space X: The sets 0 and % are open. The union S of any collection W of open sets is open. The intersection T of any finite collection {G1? G 2 ,..., Gk} of open open.
Proof (i) The set 0 is open because d0 = 0ae0=Z. The set £ is open because dZ = 0aeX = 0. (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<=S. Since ^eS, there exists a GeW such that ^eG. Since G is open we can
28
Open and closed sets (I)
find an open ball B with centre % such that BczG. But GczS and so the result follows. open sets i n ^
(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 BczT.
Since ^ET, we have that Z>eGv £eG 2 ,..., ^eG k . Because the sets Gv G2, . . . , Gk are all open, we can find open balls Bv B2,..., Bk each with centre £ such that ^.czG y
0=1,2,.:.,*).
One of the open balls Bv B2,..., Bk has minimum radius. Let this open ball be B. Then B a Bj a Gj (j = 1, 2 , . . . ,fe)and hence BcT=n
G-.
Thus fi is an open ball with centre ^ such that BczT.
14.14 Theorem In a metric space X: (i) The sets 0 and % are closed (as well as open).
Open and closed sets (I)
29
(ii) The intersection S of any collection W of closed sets is closed, (iii) The union T of any finite collection {Fv F2,..., Fk] of closed sets is closed. Proof The theorem follows from theorems 14.9 and 14.13. (i) The sets 0 and X are closed because C0= % and C % = 0 are open, (ii) The set S is closed because
FeW
)
FeW
is open by theorem 14.13. (iii) The set Tis closed because J
Fj = n
eFj
is open by theorem 14.13.
14.15
Open and closed sets in Un
Of the various types of sets in R" 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:
and
single points lines (and half-lines) closed line segments hyper spheres closed boxes hyperplanes
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 Un are left as exercises.
14.16
Theorem Any hyperplane in IR" is closed.
Proof Consider the hyperplane H through the point £ with normal u # 0. This is given by H = {x: <x — £, u> = 0}. Suppose that d(y, H) = 0. We propose to use theorem 14.10 and therefore seek to prove that y e / / . By theorem 13.22, given any e>0, there exists an xeH such that d(y, x) = ||y —x||<e. But since xeH,
Hence, using the Cauchy-Schwarz inequality (13.6),
30
Open and closed sets (I)
Since this is true for every a>0, it follows that =0 and so yeH. Thus H is closed. 14.17
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 Un which are both open and closed are 0 and
or. (3) Let S be the set of exercise 14.6(5). Find a point ££S such that d(E>, S) = 0. Deduce that S is not closed. (4) Prove that open balls in a metric space % 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 % is closed. Deduce that any finite set in a metric space % is closed. (6) Let Sv S2,...,Sn be closed sets in U1. Prove that S = S1xS2x ... xSn is a closed set in Un. [Hint: Use exercise 13.24(6).] (7) Prove that the types of set in GT said in §14.15 to be closed actually are closed. (8) Aflat (or affine set) in Un 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 or are defined by (where u # 0 and c are constant). We call S a closed half-space of R" and T an open half-space. Prove that S is closed and T is open in the sense of §14.7. |(10) Give examples to show that a convex set in U" may be (i) open (ii) closed (iii) neither open nor closed (iv) both open and closed. |(11) Prove that any open set G in a metric space % is the union of the collection of all open balls it contains. |(12) Find a sequence of open intervals and a sequence <Jk> of closed intervals in U such that, if
then / is not open and J is not closed.
15
15.1
OPEN AND CLOSED SETS (II)
Interior and closure
Let £ be a set in a metric space %. The interior E of the set E is defined by E = E\dE. The closure E is defined by
The interior E is obtained from E by removing all the boundary points of E which happen to be elements of E. The closure E is obtained from E putting in all the boundary points of E which happen not to be elements of E.
15.2
Example Consider the set S in R 2 defined by
This set is neither open nor closed. Its interior S is the set $={(x,y):0
32
Open and closed sets (II)
Its closure S is the set
15.3
Exercise
(1) Prove that, for any set £ in a metric space X, (i) E is open <=>E = E (ii) E is closed <=>£ = £. (2) Determine the interior and closure of each of the sets of exercises 14.6(1) and (2). (3) Prove that, for any set £ in a metric space X,
(i) e(E)=eE (ii)
) )
_
[Hint: 6(E) = e(EnC(dE)) and C(E)=e(EudE)l (4) Let if be a hyperplane in Un. Prove that (i) dH = H (ii) i? = tf (iii) H = 0. [Hint: We have seen that H is closed (theorem 14.16). Hence dH^H. If ^EH and / / has normal u, show that d(^9 ^ + tu) = |t|-||u||. Deduce that HadHl (5) Let S = {x: <x, u> < c) and T = {x: <x, u> ^ c} (where u ^ 0) be half-spaces in Un. Let / / be the hyperplane H = {x: <x, u> =c}. Prove that (i) dS = H (iv) 8T=H
(ii) 5 = T (v) T = T
(iii) S = S (vi) f=S.
[Hint: For (iv), (v) and (vi), recall question (3).] t(6) Let B1 = {x: d& x)0) be balls in Un. Prove that (i)B 1 =B 1 (ii) ^7 = B2 (Hi) B^B, [Hint: Use exercise 14.6(4).]
(iv) B2 = B2.
15.4
Closure properties
15.5
Theorem For any set £ in a metric space X, xeE <=>d(x, E) = 0
Proof (i) If xeE = E\jdE, then x e £ or xedE. In the former case d(x, E) = d(x, x) = 0. In the latter case, we have by definition that d(x, E) = 0 and d(x, GE) = 0. Thus xeE =>d(x9 E) = 0. (ii) If d(x, E) = 0, then either xeEaE or else xeCE. In the latter case d(x, CE) = d(x9 x) = 0 and so xedEcE. Thus d(x, £) = 0 => xeE.
Open and closed sets (II) 15.6
33
Corollary Let S and T be sets in a metric space X. Then
Proof Let xeS. By theorem 15.5, d(x, S) = 0. But S c T implies , S)^d{\, T). Hence d(x, T) = 0. Thus x e T .
15.7f Theorem Let £ be a set in a metric space £. Then £ is the smallest closed set containing E. Proof To show that £ is closed, we use theorem 14.10. We therefore assume that dfe, £) = 0 and seek to deduce that £e£. By theorem 13.22, given any £>0, we ean find an xe£ such that d(£, x)<e. But xe£ <=>d(x, E) = 0 by theorem 15.5. Hence £) = d(§, x) (exercise 13.24(5)). Thus Since this is true for any e>0, we conclude that dfe, E) = 0 and hence ^eE by theorem 15.5. Now suppose that F is any closed set containing E. Since £czF, we have from corollary 15.6 that EaF. But F is closed and therefore F = F (exercise 15.3(lii)). It follows that EaF for each closed set F containing E. Thus E is the smallest closed set containing E.
15.8|
Interior properties From exercise 15.3(31) we know that
Thus any result about closures leads to a corresponding result about interiors. For example,
es^eT^Js^Jf=> e(es)c:e(ef) and so 5c=T => $czt. 15.9| Theorem Let £ be a set in a metric space X. Then £ is the largest open set contained in £. Proof Note first that E=C(CE) and hence is open because it is the complement of a closed set. If Gc£, then GczE. If G is also open it follows that
34
Open and closed sets (II)
GczE (because then G = (5). This argument shows that £ is the largest open set contained in E.
15.10|
Exercise
(1) Aflat (or affine set) F is the intersection of a collection of hyperplanes in U". Prove that (i) F = F
(ii) / = 0 .
Show that the same results hold if F is a half-line or a closed line segment. (2) Let £ be a set in a metric space %. Prove that xeE if and only if there exists an open ball B with centre £ such that BaE. (3) Let £ be a set in a metric space £. Prove that dE is closed. [Hint. d£ = £nC(£).] (4) Let £ be a set in a metric space %. Prove that (i) £ = H
£
(ii) E=\J
EczF F closed
G.
GczE G open
(5) Let A and £ be sets in a metric space %. Prove that (i)
0
)
o
(ii) AczB => AczS;AczB =>Aa3 o
(iii)
i^
Give examples to show that c cannot be replaced in general by = in (iii). (6) Let £ be a set in a metric space %. Prove that d(dE) = dE. Show also that: (i) d(E)adE (ii) d(£)czdE. Give examples to show that c cannot be replaced in general by = in either case. (7) Let C be a convex set in U" with a non-empty interior. Prove that (i) C=£
(ii) <5 = 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 U". Prove that C and C are also convex. (9) Let £ be a set in a metric space %. Let W denote the collection of all sets which can be obtained from £ by applying the operations C, ~ and ° to £ 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 % are contiguous (i.e. 'next to each other') if a point of the closure of one of the sets
Open and closed sets (II)
35
belongs to the other. Thus A and B are contiguous if and only if or
AnB=£0.
We say that the sets A and B are separated if and only if they are not contiguous.
AandB separated
A and B contiguous
To say that A and B are 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. 15.12 Theorem Two sets A and B are contiguous if and only if there 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
Similarly, beAnB 15.13
<=>beB and d(b, A) = 0.
Examples
(i) The intervals A = (0, 1) and J3 = [l, 2] in U1 are contiguous. We have that d(l, A) = 0. Alternatively one can note that
The diagram illustrates the owner of set A walking from his house aeA to visit his neighbour whose house is at beB. Observe that this visit does not involve crossing any point outside AKJB.
h
36
Open and closed sets (II)
(ii) The intervals ^4 = (0, 1) and B = (l, 2) in U1 are separated. The diagram illustrates the owner of A jumping over the point 1 in order to get to B.
—(
h •
H
•
h
0
a
1
b
2
A
B
1
In formal terms AnB = (0,
15.14 Theorem Two open sets G and H in a metric space % are contiguous if and only if they overlap.
/
1 1\ \
/
/ G
\
X \
i
\
H \
/
\
/
/
\
/
Proof Two sets which overlap are obviously contiguous. We therefore need only prove that, if G and H are contiguous open sets, then Suppose that GnH = 0. Then G<=CH. Hence GaeH = eH (corollary 15.6) because CH is closed. Thus GnH = 0. Similarly, Gntl = 0. Thus G and H are not contiguous.
15.15 Theorem Let A, B, G and H be sets in a metric space %. If AczG and B e / / , then ^4 and B contiguous => G and // contiguous.
Open and closed sets (II)
Proof Suppose that AnB^0.
37
Then GnH=>AnB^0.
Similarly,
15.16f Theorem Two sets A and B in a metric space % are separated if and only if there exist disjoint (i.e. non-overlapping) open sets G and H such that A a G and BaH. Proof (i) Suppose that G and H are disjoint open sets for which AaG and B<^H. Then A and B are separated by theorems 15.14 and 15.15. (ii) Suppose that A and B are separated. Define the sets G and H by G = {x:d(x, A)
Then the open ball with centre % and radius r > 0 lies entirely inside G. To prove this, we suppose that d(£, x)
Also
)-<%, x). d(x, A)^d(x, 5) + d(£, A).
Hence
)-d& A) B)-2r-d&
A)
= 0. Thus xeG and so G is open. Similarly, H is open. It remains to show that AaG and BaH. Since ^ and £ are separated, AnB Hence, given any aeA, d(a, B)>0 (theorem 15.12). Thus, for each B) and hence AaG. Similarly, BaH.
38
Open and closed sets (II)
15.17
Exercise
(1) Decide which of the following pairs of sets in (R2 are contiguous and which are separated. (i) A = {(x9 y): x>0}; B = {(x, y): x < 0 } (ii) A = {(x,y):x>0};B = {(x9y):y = 0andx£0} (hi) A = {(x, y): x2 +y2<1}; B = {(x, y): x2 + y2>0} (iv) 4 = {(x, y): x>0 and y ^ O } ; B = {(x, y): x^O and y<0} (v) X = { ( x , ) ; ) : x ^ 0 a n d ) ; ^ 0 } ; B = {(x,>;):x<0and);<0}. (2) Prove that two closed sets A and 5 in a metric space Z are contiguous if and only if they overlap. (3) Suppose that A and B are sets in a metric space % such that AKJB= %. If A and B are separated, prove that each set is both open and closed. What does this imply about A and B in the metric space 1R"?
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 % and y, be metric spaces and let S<= %. A function/: S-+% 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/: S-*% then f(f~\T)) =T for each Taf(S) (exercise 6.4(5)). It follows that an equivalent formulation of the definition is that/is continuous on S if and only if, for each pair C and D of subsets of f(S\ C and D separated => f~1(C) and f~l{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.
16.3 Theorem Let X, % and £ be metric spaces and let S c £ , Suppose that g: S^y, and / : T-»£ are continuous on S and T respectively and that g(S)czT. Then the composite function/°g: £-•£ is continuous on S.
Proof The diagram renders the proof obvious.
16.4 Theorem Let % and y, be metric spaces and let S c l Then / : S-+y, is continuous on the set S if and only if, for each xeS and each £ c S , d(x,£) = 0=>d(/(x), /(£)) = 0.
Continuity
41
Proof (i) Suppose that / is continuous on S. If d(x, E) = 0, then {x} and E are contiguous by theorem 15.12. Hence {/(x)} and f(E) are contiguous. Thus d(/(x), f(E)) = 0 by theorem 15.12 again. (ii) Suppose that d(\, E) = 0=>d(f(x)J(E)) = 0. Let A and B be contiguous subsets of S. Then a point of one set is at zero distance from the other by theorem 15.12. Suppose aeA and d(a, B) = 0. Then d(/(a), /(£)) = 0 and hence f(A) a n d / ( £ ) are contiguous by theorem 15.12 yet again.
If/: S->R", the formula
defines n real-valued functions/^ S->R,/ 2 : S—•IR,... ,/„: S-*R. We call these functions the component functions o f / a n d write
f=(fvf2,.. -./„)• 16.5 Theorem Let £ be a metric space and let Sa%. A function / : S^Un 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: tR"-*[R defined by Pk(xv
x 2 , . . . , xn) = xk
is continuous on Un. Proof The identity function / : [Rn->[Rn defined by /(x) = x is obviously continuous. But I = (PV P 2 , . . . , Pn) 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 % and y, be metric spaces and let Sa %. Then/: S-+y, is continuous on S provided that for each xeS there exists a y > 0 and a c > 0 such that
d(x,y)d(f(x)J(y))^cd(x,y) for each yeS. Proof Suppose that d(x, E) = 0. Given any ^>0, we shall prove that there exists a y e £ such that d(/(x),/(y))<e. Thus d(f(x),f{E))^=0 by theorem 13.22. Choose eo = min {sc'1, y}. Then eo>O and hence there exists yeE such that d(x, y)<e 0 by theorem 13.22. But so^y. It follows that , /(y)) S cd(x, y) < cs0 g s as required. 16.9 Theorem Let a and b be real numbers and let L: (R"xR"M: U x R ->R and D:Ux(U\ {0})^>U be defined by (i) L(x 1 ? x 2 ) = ax 1 (ii) M(x 1 ? x 2 ) = x 1 x 2
(iii) D(xv x2) = xjx2. Then L, M and D are continuous on their domains of definition. Proof (i) We have that || Xl - y ^ l K x ^ x 2 ) - ( y 1 , y2)|| and I|x 2 -y 2 ||^||(x 1 , x 2 )-(y 1 ? y2)|| because ||(x1? x 2 )-(y 1 ? Hence ||L(x1? x 2 )-L(y 1 ?
^{\a\ +
\b\}-\\(xvx2)-(y19y2)\\.
Thus Lis continuous on [R2" by lemma 16.8. (ii) We use lemma 16.8 with y = l. Assume that \\(xvx2)-(yvy2)\\
Continuity
43
Then l-Xj-yJ^IK*!, x2)-(yv
y2)\\
and
\x2-y2\^\\(xvx2)-(yvy2)\\
y2)\ = \x1x2£c\\(xl9x2)-{yl9y2)\\.
(iii) We use lemma 16.8. If (x l 9 x 2 ) e R xU\ y = i | x 2 | . Assume that
{0}, then x 2 ^ 0 and we take
\\(x19x2)-(y19y2)\\<±\x2\. Then and \x2-y2\<\\(xv
x2)-(yv
y2)\\<j\x2\.
In particular, I y2\ = I y2 ~ xi+x2\
^ |x 2 | - 1 y2 - x2\ >£|x 2 |.
2
Put c = 2|x 2 r (|x 1 | + |x2|). Then \D(xvx2)-D(yvy2)\
= x1y2-x1x2
+ xlx2-ylx2 x2 y2
<: 2|x 2 r 2{|x11 • | y2 - x2| -h |x2| • |
£c\\(x19x2)-(y19y2)\\ and the result follows from lemma 16.8. 16.10 Corollary Let X be a metric space and let S<= £. If/X: S->[Rn and / 2 : S-^R" are continuous on the set 5, then so is the function F:S->Mn
44
Continuity
defined by
Proof This follows immediately from theorems 16.3 and 16.9 since
= Lo(f1J2). 16.11 Corollary Let X be a metric space and let Scz X. lffx: S->R and f2: S^U are continuous on the set S, then so is the function F: S-^U defined by
F(x)=f1(x)f2(x). Proofs e have that F = Mo (/ l 9 / 2 ). 16.12 Corollary Let £ be a metric space and let 5 c X. If/i: S-»R and / 2 : S->IR \ { 0 } are continuous on the set S, then so is the function F: S-+M defined by F(x)=/i(x)// 2 (x). Proo/ We have that F = D o 16.13
(fl9f2).
Rational functions
Suppose that P(xv x 2 ,..., xw) is an expression obtained from the variables xv x2,..., xn and a finite number of real constants using a finite number of multiplications and additions. Then we say that P: Un-+M is a polynomial. If Q:Mn-*U is a polynomial and Z = {x:<2(x) = 0} then the function i ? : I R " \ Z ^ R defined by
is called a rational function.
16.14
Example
The function P : U 2 ->R defined by P(x, y) = x 2 + 3xy + x 2 y 3 + 1
is a polynomial. The function R: R 2 \ {(0, 0)} ->R 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/: R -»R, find non-empty contiguous sets A and B in R, for which f(A) and f(B) are not contiguous.
Deduce that in neither case is / continuous. (2) Let eeR". Prove that the functions/:R"-*R 1 and g-.W-^M1 defined by (i) /(x) = ||x|| (ii) (see §13.3) are continuous. (3) A function/: R"->Rm is said to be linear if and only if /(ax + jSy) = a/(x) + j8/(y) for all real a and /? and all x and y in U". Prove that a linear function f{. IR"-^ 1 satisfies / 1 (x) = <x, e> for some eeUn. Hence show that any linear function/: [R"->lRm is continuous. (4) Let % be a metric space and let S be a non-empty set in %. Prove that the function / : 5E-+R defined by /(x) = d(x, S) is continuous on %. [Hint: Begin by showing that \d(x, S) — d(y, S)\^d(x, y) using exercise 13.24(5).] (5) Let X and % be metric spaces and let ScTaZ. If/: T-^% and g: T->% satisfy f(x) = g(x) for each xeS, prove that g continuous on T implies / continuous on S. t(6) Prove theorem 16.9 with U replaced everywhere by C.
46
Continuity
16.171
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 U 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* is that it is 'all in one lump'. A little man standing on any point of an interval / would be able to walk to any other point of / without having to cross any points of the complement of /.
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 % is connected if and only if S 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 0 and X. Proof Observe that
(i) AKJB=% and AnB = 0 => AuB=X and AnB = 0 => B=CA => AuB=X. (ii) AuB=X and Ar\B = 0 => A\jB=X and Ar\B = 0 => B=6A Taking these two results together we obtain that AKJB=%
and AnB = 0 o B = B and B=6A o B is closed and A = CB.
It follows that X can be split into two disjoint, separated sets A and B if and only if A = C B and B is simultaneously open and closed. The theorem is an immediate consequence.
17.4
Corollary The metric space R" 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 j;. Then
r=U s Sew
is connected. Proo/ Let A and 5 be non-empty subsets of T such that AuB= T. Either £e,4 or ^eB, Suppose that ^ e A Then all of the sets SnA with SeW are non-empty. Suppose that SnB = 0 for all S e ^ Then
S = BnT=B Seiu
Seut
because BaT. This is a contradiction because B^0.
Hence there exists an
Connected sets SxeW such that S^nB^^.
49 Also S1nAj=0
and
Because S1 is connected, it follows that S1nA and S1nB are contiguous. Thus A and B are contiguous by theorem 15.15.
17.6
Connected sets in U1
An interval I in U* is a set with the property that, if ael and then
acel. a
c
b
It seems 'geometrically obvious' that a connected set in U1 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 1R1 is connected if and only if it is an interval.
Proof (i) Suppose E is connected. If aeE, beE and a
= (_oo, c)
points in B
H= (C,
Connected sets (ii) Suppose that E is an interval. Split E into two non-empty subsets A and B with AKJB = E. We need to show that A and B are contiguous. Let aeA and beB. We shall assume that a^b. (If not, reverse the roles of A and B.) Let S = Ar\\_a, b~\ and T=5n[a, b]. If S and Tare contiguous, then so are A and B by theorem 15.15. Since E is an interval, [a, b]aE and thus SuT=[a, b]. In particular, c = sup S satisfies a^c^b and hence ceS or ceT. points of S
By exercise 13.24(3), d(c, S) = 0. Hence, if ceT, then 5 and Tare contiguous by theorem 15.12. If c$T, 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 T are contiguous.
17.8
Continuity and connected sets
17.9 Theorem Let % and y, be metric spaces and let 5c: Z. If / : S-+y, is continuous on the set 5, then S connected =>/(S) connected.
Proo/ Let C and D be non-empty subsets of f(S) such that CuD=/(S). We need to show that C and D are contiguous. If C and D are separated, then so a r e / ' ^ C ) a n d / " 1 ^ ) (§16.2). But this contradicts the hypothesis that S is connected because f~1(C)vf~1(D) = S and neither of the sets f~1{C) or / " 1 (D) is empty.
51
Connected sets
17.10 Theorem Suppose that / is an interval in U and that f:I-+M is continuous on /. Then/(/) is an interval. Proof This is an immediate consequence of the previous two theorems. 17.11 Corollary (Intermediate value theorem) Suppose that/:[<2, b]->R is continuous on [a, b~\ and X lies between f(a) and f(b). Then there exists a [a, b] such that
f(a) <-
17.12
Example Show that the equation 17x7-19x5-l=0
has a solution £ which satisfies — 1 < ^ < 0.
1
I 1
1 1 1 r
V \
L
0 X
i
\
52
Connected sets
Proof The function/: R-+R defined by/(x) = 17x 7 -19x 5 -l is a polynomial and hence is continuous on R. Thus it is continuous on [ - 1 , 0] cR, Now/(-1)= 1 and/(0) = - 1 . Since - 1 < 0 < 1, it follows from the intermediate value theorem that/(£) = O for some £ between — 1 and 0.
17.13 Example Let/:[a, ft]-*[a, ft] be continuous on [a, ft]. Then, for some £e[a, ft], /(£) = £. Thus a continuous function from [a, ft] to [a, ft] 'fixes' some point of [a, ft].
{=/«)<
1-
Proo/ The image of [a, ft] under / is a subset of [a, ft]. Thus ^a and/(ft) ^ ft. The function #: [a, b]->R defined by #(x) = /(x) — x is continuous on [a, ft] by corollary 16.10. But g(a)7±0 and g(b)^O. By the intermediate value theorem, it follows that there exists a £e[a, ft] such that
17.14 Note The previous example is the one-dimensional version of Brouwefs fixed point theorem. This asserts that, if B is any closed ball in R" and/:£-•£ is continuous on £, then there exists a ^eB such that/(^) = ^. Thus, if a spherical tank full of water is shaken around, one tiny drop of water will end up where it started. This is by no means 'obvious' and the proof in the general case is quite difficult. It is, however, a very important result with many applications. 17.15
Curves What is a curve? A physicist might say that a curve is the path
Connected sets
53
traced out by a particle as it changes its position with time. This idea leads us to the following definition. Let % be any metric space. We shall say that a path in % is a function / : [0, l ] - > £ which is continuous on / = [0, 1]. One can think of x=/(£) as the position in % of a particle at time t. The diagram illustrates the case
z=n2.
We shall call the set / ( / ) 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/(0) = a a n d / ( l ) = b, we shall say that the curve /(I) joins a and b.
17.16
Example The closed line segment in Un given by
is a curve which joins the points a and b. We have that L = / ( / ) w h e r e / : I-+M" is defined by
17.17
Theorem A curve in a metric space % is connected. Proof This is an immediate consequence of theorems 17.7 and
17.9.
17.18
Path wise 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
pathwise connected set
curve joining a and b
17.19 Theorem A pathwise connected set £ in a metric space X is connected. Proof Let Z^eE and let W be the collection of all curves which contain £ and are subsets of E. The hypothesis of the theorem ensures that £=U
S.
Sew
Hence E is connected by theorems 17.5 and 17.17.
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 U 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
Connected sets
55
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 U2 defined by
For each point iieS, it is true that the closed line segment joining 0 and u lies in S. Hence S is connected.
17.21f
Components
17.22f Theorem Let C and D be contiguous sets in a metric space X. If C and D are connected, then so is CuD. Proof Suppose that A and B are non-empty subsets of CuD such that A\JB = C\JD. We must show that A and B are contiguous. If AnC^0 and BnC^0, then AnC and BnC are contiguous because C is connected. Hence A and B are contiguous by theorem 15,15. If BnC = 0, then either (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 £ be a set in a metric space %. If £e£, then the component of £ containing £ is the largest connected subset S of £ which contains £ - i.e. if U is the collection of all connected subsets of £ containing £, then
If one thinks of £ as a country entirely surrounded by water, then it is an archipelago of which the constituent islands are the components of £. components of£
17.23f Theorem Any set £ in a metric space X is the union of its components. Any two distinct components of £ are separated. Proof The first sentence follows immediately from the definition because each £eS 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 £ containing y. But, if C and D are contiguous, then CuD is a connected subset of £ larger than C. Hence C and D are separated. 17.24
Examples
(i) The components of the set £ in U 2 defined by £ = {(*, >>):x^0 or x > l } are
and
57
Connected sets
(ii) The components of the set S = {l/n: ne N} in U1 are the sets {1}, {£}, {£},.... Each point of S therefore constitutes a separate component of S. We say that such a set is 'totally disconnected*. (hi) The components of the set T = {0}u5 are the sets {0}, {1}, {^}, {£},.... Notice that any two distinct components are separated but that {0} is not separated from the union of the other components.
17.25|
Structure of open sets in Un
Theorem 14.13 asserts that the union of any collection of open sets is open. It follows that the set 4, oo)
is an open set in
We shall prove that all open sets in Ui are of this general type. We begin with the following theorem.
17.26f
Theorem Any component of an open set G in R" is itself an open set.
Proof Let S be a component of G. Let ^eS. Since S a G, £eG. Because G is open there exists an open ball B with centre £ such that B cz G (theorem 14.12). Observe that it must also be true that BaS. 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
components of G
17.27f Theorem Any open set G in U1 is the union of a countable collection of disjoint open intervals. Proof We know from theorem 17.23 that Ieia
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->Q which has the property that f(l)el for each IeW.
The function / is injective because the intervals I in W are disjoint. Since O is countable, it follows from theorem 12.7 that W is countable. 17.28
Exercise
(1) Which of the following sets in U2 are connected? (i) {(x, y (ii) {(x,y
Connected sets
59
(iii) {(x,y):x^l orx<0} (iv) {{x,y):l<x2+y2<2}
(v) {(i,±):nelM} (vi) {(x, y): x ^ y a n d x ^ -y} (vii) {{x,y):xyl} (ix) {(X,)/);XJ; = 1 }
(x) {(x,;y):x), = O}. (2) Prove that any convex set in i " is connected. (See §13.14.) Deduce that open balls, closed boxes, lines, half-lines and hyperplanes are all connected sets in Mn. [Hint Use exercise 13.16(6).] Show also that halfspaces (exercise 14.17(9)) are connected. (3) Let / be an interval in U1 and suppose that/: I-+U1 is an injection which is continuous on /. Prove that/is either strictly decreasing on / or else strictly increasing on /. [Hint: Intermediate value theorem.] (4) Let S be the set in U2 defined by
Give an example of a continuous function/: S-+S which has no fixed point. (5) The sets A and B in U 2 are defined by Explain why a function/: R 2 ->R 2 cannot be continuous on the set A if t(6) Let if be a hyperplane in Un. Prove that Un\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 if. (See §13.14.)
18
18.1
CLUSTER POINTS
Cluster points A cluster point of a set £ in a metric space % is a point £e X such
that
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.
isolated points ofE
cluster points ofE
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
Cluster points TABLE I
(i) ^ is a cluster point of E. (ii) \eE but is not an isolated point of E. (iii) dfe, E \{£}) = 0. (iv) Each open ball with centre % contains a point of E other than £. (v) Each open ball with centre ^ contains an infinite subset of E.
61 TABLE II
(i) £ is an isolated point of £. (ii) £e£ but is not a cluster point of E.
P) .
-
_
(iv) There exists an open ball containing no point of E except £.
To deduce Iv 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: neN} in U* has a single cluster point, namely 0. Note that 0£S. Every point of S is an isolated point of S. (ii) The set N in U * has no cluster points. All its points are isolated, (iii) The cluster points of the set Tin U2 defined by are the points of {(x, y): x2 + y2 ^ 1}. The point (2, 2) is an isolated point of T. (2,2)
18.3
Exercise
(1) Find the cluster points and the isolated points of the following sets in
62
Cluster points (i) (0,1) (ii) [0,1) (iii) ( - o o , 0 ] (iv) N (vi) U (vii) 0 (viii) {1, 2, 3} (ix) [0, l]u{2}
(v) Q (x) [0, l ] u ( l , 2]
(2) Repeat the above question for the following sets in U2. (i) (ii) (iii) (iv) (v) (vi) (vii)
A = {(x9y): B = [(x,y):x^0} C = {(x,y):x2 + y2l}.
+ y2^l
(3) Prove the equivalences given in tables I and II of §18.1.
18.4|
Properties of cluster points
18.5f
Theorem
Any isolated point £ of a set £ in IR" is a boundary point of E.
Proof By §18.1 (Iliv), there exists an open ball B with centre £ all of whose points are in Q, E except for ^ 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
18.6f Theorem Let £ be a set in a metric space X. A boundary point r\ of E which is not a cluster point of E is an isolated point of E. Proof If r\ is not a cluster point of £, then Iiv does not hold and hence there exists an open ball B with centre r\ containing no point of E except (possibly) t|. Since r\ is a boundary point, any open ball B with centre r\ contains a point of E. Hence r\eE. Thus r\ is an isolated point of £ by §18.1 (Ilii).
18.7| Theorem A set £ in a metric space X is closed if and only if it contains all its cluster points. Proof (i) Suppose £ is closed and £ is a cluster point of £. Then (ii) Suppose that £ contains all its cluster points. Since it contains all its isolated points (by definition), it follows from theorem 18.6 that £ contains all its boundary points and hence is closed.
Cluster points 18.8|
63
Exercise
(1) Let £ 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 £, prove that E has no cluster points. (2) Show that any set £ 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 N is a closed set in U1 and that any finite set in a metric space X is closed.
18.9|
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 Ui with no isolated points. The fact that perfect sets exist is not particularly surprising. Any closed interval in U1 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 (^, §). This leaves the set
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, f] and [§, 1] which make up Fv This leaves the set
which is closed and satisfies F2<=zFv Now delete the open middle thirds of each of these intervals and so on. We thus obtain a sequence of closed sets which satisfy
The Cantor set is defined by 00
F=n Fn. Observe that F is closed by theorem 14.14. Obviously F contains all the endpoints of the closed intervals which make up the sets Fv F 2 , F 3 , Thus F contains all the points r> i 1 2 1 2 7 8 1 ' x» 3 ' 3 ' 9 ' 9 ' 9 ' 9 ' 2 7 ' ' ' * '
u
64
Cluster
points
1 3 v/<'////////////////////////////A
0
1 9
0
2 9
1 '////*
3
l 3
7 9
2 3
Fi
8 9 '////A
0
1 27 'X
2 1 27 9 Z//A
2 7 9 27
V
j
Y/ /&
8 27
1 3
2 3
1920 7 2727 9 y//\ V//\ Yy/A
3
A 'A
8 25 9 27
26 , 27
V/\
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 exercises, (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 Iv / 2 , ... exist such that F c ^ u ^ u ... and
I Kh)
where 1(1k) denotes the length of the interval Ik.
18.10t
Exercise
(1) Let F be the Cantor set as constructed in §18.9. Prove that xeF if and only if
where each of the coefficients an 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 FN are the numbers of the form
where each an is 0 or 2. Moreover, each of these closed intervals is of length 3 ~ N and ^
(2) Prove assertions (i) - (vi) of §18.9. [Hint F=FcFH(n
1
= l, 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 [R1, perhaps the most important in analysis are those of the form
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/: [a, b]->U attains a maximum and a minimum value on [a, &]. 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 / is compact if and only if it is closed and bounded. Should we therefore define a set S in a metric space % 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 % is bounded if and only if there exists an open ball B of which S is a subset.
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 Un. 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 Un is a set S of the form Ix x 12 x ... x 7n, where each of the sets Ik is a compact interval in U *. The diagram illustrates a nested sequence of closed boxes in U 2.
A sequence <5fe) 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 Un. Then
n fc=i
Proof We begin with the proof for M1. Suppose therefore that <7fc> is a nested sequence of compact intervals lk = \_ak, bk~]. Put A = {ak: ke N} and B = {bk: ke N}. Since is nested, each element of B is an upper bound for A.
68
Compact sets (I) points of A
points of B
Let £ = sup A. Then ak^£^bk for each keN. Hence £e/ k for all keN. This proves the theorem in the case n = l . Now let <5fc> be any nested sequence of closed boxes of the form Sk = Ik{1)x Ik{2) x ... x Ik{n) in Un. The first part of the proof demonstrates the existence of a real number <^e/fc(0 for each keN. But then
for each keN. Hence there exists a Z>eSk for each keN.
19.4 Example We give another proof of the fact that R is uncountable using the Chinese box theorem. Suppose that [0, 1] is countable and t h a t / : N-*[0, 1] is a surjection. Split [0, 1] into three congruent compact subintervals [0, | ] , [|, §], [f, 1], At least one of these subintervals does not contain/(I). Call this subinterval Iv Split the compact interval Ix into three congruent compact subintervals in the same way. At least one of these subintervals does not contain /(2). Call this subinterval I2.
69
Compact sets (7) /(I)
\
In this way we construct inductively a nested sequence <7k> of compact intervals such that
f(k)eeik for each keN. It follows that
k=l
This contradicts the assumption that / : N ->[0, 1] is a surjection because the intersection of the sequence <7k> 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 {Kn} of non-empty compact sets in a metric space X has a non-empty intersection.
ClK,
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 (I)
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 % 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 % is compact if and only if each infinite subset E of S has a cluster point
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 'V£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 R 1 is compact in the sense defined above. For this purpose we require the following important theorem. 19.6 Theorem (Bolzano-Weierstrass theorem) Every bounded infinite set E in R" has a cluster point. Proof Since E is bounded, it lies in some closed box S (exercise 13.16(5)). The box S can 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 Et of E. Let Sx be the sub-box containing Ev
Compact sets (I)
71
We can now repeat the process with S 1 replacing S and Ex 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 closed box SK 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(Iv)).
• •
• •
• • • •
•
•
•
*
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 %. This point is taken up again in chapter 20. 19.8 Theorem (Heine-Borel theorem) A set X in R" is compact if and only if it is closed and bounded. Proof The proof that a compact set in 1R" 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 tR" is compact. Suppose that K is a closed, bounded set in Un 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 Thus K is compact.
19.9 Theorem (Cantor intersection theorem) Let be a nested sequence of non-empty closed subsets of a compact set K in a metric
72
Compact sets (I)
space X. Then
n 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 F . points ofE
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 nef^J. Thus
19.10 Example The sequence 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?
Compact sets (I)
73
(3) Give examples of a nested sequence of non-empty closed sets in U* and a nested sequence of non-empty bounded sets in Ul which have empty intersections.
19.12
Compactness and continuity
19.13 Theorem Let X and % be metric spaces and let Scz X. If/: S^y, 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 that/(gf(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 Now let TJ =/(£). Then r\ef(S). Also since/is continuous on <S, d& D \ &}) = 0 => d(i,, f(D \ ft})) = 0 by theorem 16.4. B u t / ( D \ { ^ } ) = £ \{i|} and so i| 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 / : R-+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 % and t h a t / : K-+M is continuous on the set K. T h e n / achieves a maximum and a minimum value on the set K - i.e. there exists ^eK and H\GK such that, for any xeK,
/(I)
By theorem 19.13, f(K) is compact. A compact set in R 1 is closed and bounded by the Heine-Borel theorem (19.8). Because/(X) is a closed, non-empty set of real numbers which is bounded above it has a maximum (corollary 14.11). For similar reasons, f(K) has a minimum.
19.15 Corollary Suppose that K is a non-empty compact set in a metric space %. and that ye %. Then there exists a ^ e K such that
Compact sets (I)
75
Proof The function/: Z-+U defined by/(x) = d(y, x) is continuous on % 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 Un and n that yeU . Then there exists a ^ e F such that
Proof Let B be the closed ball with centre y and radius d(y, F ) + 1 . Then BnF is a non-empty, closed and bounded set in Mn. It follows from the Heine-Borel theorem (19.8) that BnF 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 B in R 2 are defined by Explain why a function / : R 2 -»R 2 cannot be continuous on A if (2) Let / be an interval in (R1 and l e t / : I R 1 - ^ 1 be continuous on /. Give counter-examples to each of the following (false) propositions: (i) / closed =>/(/) closed (ii) / closed =>/(/) bounded (iii) / open =>/(/) open (iv) / bounded =>/(/) bounded (v) / open and bounded => / ( / ) has no maximum. (3) Which of the propositions above are necessarily true when/: [ R 1 - ^ 1 is continuous on a compact interval J containing /. t(4) The distance d(S, T) between two non-empty sets S and Tin a metric space % is defined by d(S, T) = inf {d(x, y):xeS and ye'T}.
76
Compact sets (I)
Show that, if S and Tare compact, then there exist ssS and t e T s u c h that d(S, T) = d(s, t). Show that the same result holds when X = U" if 5 is compact but Tis only closed. Give examples of closed sets S and Tin !R 2 for which the result is false. f(5) Let SczU and suppose t h a t / : S-+M is continuous on the set S. The graph of/is the set Tin 1R2 defined by
Prove that (i) S connected => T connected (ii) S compact => T compact. Give examples of non-continuous functions/: S-+M for which (i) and (ii) are true. t(6) L e t / : U^-U be a bounded function whose graph is a closed set in R2. Prove t h a t / is continuous on U.
20f
20. If
COMPACT SETS (II)
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 Z. 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
SeU
sets in U
20.3
Example
Let S: [0,1] ->(0, oo). Then the collection
covers the set £ = [0, 1].
77
78
Compact sets (II)
20.4|
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 / m t e subcollection 5 cz U which covers X.
20.5 by
Example
Suppose the function <5:[0, l]->(0, oo) of example 20.3 is given
*
(
x
)
.
x+2 Then a finite subcollection 5 of U which covers [0, 1] is
- i.e. one needs only the intervals (x — S{x% x + d(x)) with x = 0 and x = l.
20.6|
Exercise
(1) A function/: R-*R has the property that, for each xe[0, 1], there exists a <5>0 such t h a t / is either increasing or else decreasing on (x —<5, x + d). Prove t h a t / 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 Un
In this section we shall show that both our definitions for compactness are equivalent in the space Un. 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 Un. As we know, the Bolzano-Weierstrass theorem is false in a general metric space. This is
Compact sets (II)
79
Kis compact
I | Lindelofs J theorem
\7
I J_
ins closed
A: is countably compact
K has the Chinese box property
K has the BolzanoWeierstrass property BolzanoWeierstrass theorem
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 % has the Bolzano-Weierstrass property if and only if each infinite subset E of K has a cluster point £. A set K in a metric space % has the Chinese box property if and only if any nested sequence 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 % is said to be countably compact if and only if each countable collection U of open sets which covers K has a finite subcovering § 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 (II)
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 £ be a set in a metric space X. We say that D is dense in E if and only if EczD - i.e. for each xeE, d(x, D) = 0.
points of D
20.8 Example The set Q is dense in Hence Q" is dense in R" by exercise 13.24(6).
This follows from theorem 9.20.
20.9| Theorem Let £ 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 J8 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 58 is countable. (Consider D x O and apply theorem 12.10.)
rational radius
points of D
Compact sets (II) If GeU, then each
81
lies in an open ball BeS. Hence G c U B. GeU
Be®
The first sentence of this paragraph is proved in the following way. Since G is open, there exists an open ball Bo with centre £ and radius r > 0 such that Bo a G (theorem 14.12). Since D is dense in £, there exists a deD such that d(£, d)
and so yeBoczG. Now let/: $-•% be chosen so that Baf(B). Then/($) is a countable subcollection of U and EcU
f(B)= *<J G. Gej\9i)
other sets C€t t which contain B We have shown that/(S) is a countable subcollection of %L which covers £.
20.10| Corollary (Lindelofs theorem) Any set £ in H" has the Lindelof property. Proof The set Q" is countable by exercise 12.13(3). Since Q" is dense in U" (example 20.8), On is dense in E. The result therefore follows from theorem 20.9. 20.1 If
Corollary Any countably compact set K in Rn is compact.
82
Compact sets (II) Proof This follows immediately from Lindelof s theorem.
20.12f
Theorem
Any countably compact set K in a metric space Z is bounded.
Proof Let £ be any point of Z and let U be the nested collection of open balls of the form
where neN. Then U is countable. Since K is countably compact, a finite subcollection $ oi U covers K. Let B denote the open ball of maximum radius in S. Then and so K is bounded.
20.13|
Theorem
Any countably compact set K in a metric space £ is closed.
Proof Let £ be any point not in K and let % be the nested collection of all open sets of the form
where neN. Then % is countable and covers K. Since K is countably compact, a finite subcollection 5 of U covers K. Let £ = Z \ S be the ball of minimum radius such that Se-S. Then B contains no point of K. Hence £ is not a cluster point of X. Thus K is closed.
20.14| Theorem Any closed set K with the Chinese box property in a metric space Z 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 X is countably compact. Let the sets in U be the terms of the sequence . Since {Gl9 G2, . . . , Gk} does not cover K,
is non-empty.
Compact sets (II)
83
But is a nested sequence of closed subsets (theorem 14.14) of K. Since K has the Chinese box property, has a non-empty intersection. Thus
and so % does rcot cover K.
This theorem completes the proof of the implications of the diagram of §20.7.
20.15|
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:||x|| = l} 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 / > 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 /. 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 £, 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 definite collection §r of open balls of radius r > 0 which covers S.
It is obvious that any totally bounded set S in a metric space X is bounded. (If
Compact sets (II)
84
contains N elements, 5 is a subset of any open ball B with radius IN provided that the centre of B lies in S.) In R" the converse is also true because, by the Heine-Borel theorem, if S is bounded, then S is compact. But S is covered by the collection of all open balls of radius r. Hence S and therefore S is covered by a finite number of open balls of radius r. In a general metric space ft, 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 ft 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 ft, 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.
20.16|
Compactness in general metric spaces The diagram below summarises the results of this chapter.
compact
K has the Lindelof property
ATis closed
Kis countably compact
There exists a countable set which is dense inK
K has the Chinese box property
K has the BolzanoWeierstrass property
Kis totally 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 (II) —*-
K
— +~
85
True in any metric space
Not yet proved
True in a complete metric space
Proved already
True in Un
20.17| Theorem A set S in a metric space % which has the BolzanoWeierstrass property is totally bounded. Proof If S is not totally bounded, then for some r > 0 no finite collection of open balls of radius r covers S. If \ v x 2 , . . . , xn_1 are points of 5 one can therefore always find a point x n e S such that d(\n, \k) ^ r (k = 1, 2 , . . . , n -1).
The set S therefore contains an infinite subset with no cluster point (exercise 18.8(1).)
20.18f Theorem If 5 is a totally bounded set in a metric space %, there exists a countable set D which is dense in S. Proof Let Cn 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, Cn)< 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,
86
Compact sets (II)
Since this is true for all neN, d(x, D) = 0.
20.19")" Theorem Any countably compact set K in a metric space % 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 of F
For each x e F we can find an open ball which contains no point of F except x (§18.1(IIiv)). The collection of all such open balls together with the open set CF 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.
20.20|
A spherical cube
In this section, we give one of our rare examples of a metric space other than or. We begin by observing that one can think of a sequence <xfc> of real numbers as the 'infinite-dimensional' vector x = (x1? x 2 , x3, ...) in which the co-ordinates of the vector x are just the terms of the sequence <xfc). Vector addition and scalar multiplication may then be defined just as we did in §13.1 for vectors in Un. We shall consider the normed vector space /°° consisting of all bounded sequences of real numbers. The norm on /°° is defined by ||x|| = sup |x k |. The distance between points x and y of /°° is therefore given by d(x9 y) = ||x-y|| = sup keN
\xk-yk\.
Compact sets (II)
87
Thefirstthing to notice is that the proof that we gave for the Chinese box theorem (19.3) works equally well for / °°. The same is therefore true of the Bolzano- Weierstrass theorem provided that 'bounded' is replaced by 'totally bounded'. Thus /°° is a complete metric space. Consider the set S = {x: ||x|| ^ 1} in /°°. This can also be written in the form 5 = {x:|x f c |^l for each keN}. The first expression would describe a ball in U3 and the second would describe a cube in R3. Hence the title of this section. The set S is clearly a closed and bounded set in /°°. 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 Un see exercise 20.21(6).
20.21|
Exercise
(1) Prove that Z00 as described in §20.20 is a metric space. (2) The normed vector space I2 consists of all sequences x = <xfc> of real numbers for which
converges. An inner product is defined on I2 by 00
k=l
Prove a version of the Cauchy-Schwarz inequality (13.6) and deduce that I2 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 5' = / 1 x I2 x / 3 x ... where is a sequence of compact intervals. If <Sk> is a nested sequence of such closed boxes, prove that its intersection is not empty. Deduce from this version of the Chinese box theorem that both /°° and I2 are complete. (4) Prove that I2 contains a countable dense subset. [Hint: Look at the set D of all points of the form (ql9 q2, . . . , qn, 0, 0, ...) where ql9 q2, . . . , qn are rational.] Deduce that S = {x: ||x||<;i} has the Lindelof property in I2.
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Compact sets (II)
(5) Prove that the set 5 = {x:||x||^l} is closed and bounded in I2 but not totally bounded. Deduce that 5 is not compact in I2. [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:\xk\^l/k for each keN} is closed and totally bounded in I2. [Hint: Given r>0, choose n so that 11/2
and then use the fact that closed boxes are totally bounded in (R".] Deduce that S is compact in I2.
21
21.1
TOPOLOGY
Topological equivalence
Suppose that X and % are metric spaces and that/: X -+ y, is a bijection with the property that bothf: X-+% a n d / " 1 : ^-•fC 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 homeomorphism/: X-*% exists, we say that X and % 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 U3) are therefore topologically equivalent. The topological property that they have
ball
89
doughnut
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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.
Sweden
Denmark Poland West Germany
Norway Czechoslovakia USSR Austria
France
•
•
Spain Italy Greece
n 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
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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 U*). 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 of the 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 U. Suppose t h a t / : (—1, l]->( —1, 1) is a bijection. L e t / ( 1 ) = ^. Then / ( ( — I, 1)) = ( —1, £)u(R defined by /(*) =
1-X
2'
This is a homeomorphism between ( — 1,1) and U. (For a proof of this result which avoids the need for any calculation, see theorem 22.22.) It follows that ( — 1, 1) and U are topologically equivalent.
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21.4
Circles and spheres
It is easy to see that a circle C and a line segment L (regarded as metric subspaces of U1) cannot be topologically equivalent. If / : 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. B u t / " 1 : L^C then maps the connected set L \ {A, B) onto the disconnected set C\ {A', B'\ and hence is not continuous.
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.
Similarly, a 'punctured sphere' is R 3 (i.e. a sphere with one point removed) is topologically equivalent to a plane. Again, the stereographic projection provides a suitable homeomorphism.
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Let S be the sphere in R 3 with centre (0, 0, j) and radius j . Take as the 'north pole' the point N = (0, 0, 1) and let P be the plane {(x, y, z): z = 0}. Then the stereographic projection/: P-+S \ {N} is given by
fix, y, OH
-
a n d / " 1 : S \ {iV}->P is given by
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) Let 5 be any set in R" 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 $eS such that/(§) = £. (3) Let 3 denote the collection of intervals in R1 which have at least two points. Let U denote the collection of all open intervals in 5 , let V denote the collection of all compact intervals in 3 and let W denote the remaining intervals in 3 .Show that U, V andW are the equivalence classes into which 3 is split by the relation of topological equivalence. (4) Let L be a closed line segment in U2 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 sphere S\{N}. t(6) A function F: C->S\{N} is denned by 3mz
\z\2 \
W'TW/ 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 subspace of IR3). Show that
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21.6
Continuous functions and open sets
21.7 Theorem Let X and % be metric spaces and let/: X-+%. Then the following statements are all equivalent. (i) / is continuous on X. (ii) For any xe X and any S<= X,
(iii) For any SczX,
(iv) For any F c ^ , F closed =>f"1(F) closed, (v) For any G c ^, G open =>/~ 1 (G) open. Proo/ We shall show that (i)=>(ii)=>(iii)=>(iv)=^(v)=>(i). (i)=>(ii). This is just theorem 16.4. (ii) => (Hi). Assume (ii) and suppose that yef(S). Then y =/(x) where xeS. Since d(x, S) = 0 it follows that d(f(x), f(S)) = 0. Thus y=/(x)e/(S). (iii)=>(iv). Assume (iii) and suppose that Fczy, is closed. Since
Hence f~\F)czf-\F). T h e r e f o r e / " 1 ^ ) is closed. (iv)=>(v). Assume (iv) and suppose that G<^% is open. Then CG is closed and hence
ef-1(G)=f-\eG) is closed. T h u s / - 1 ( G ) is open. (v)=>(i). Assume (v) and suppose that C and D are separated sets in y,. Then disjoint, open sets G and H exist with CaG and DczH by theorem 15.16. The s e t s / ' ^ G ) a n d / - 1 ( i J ) are then disjoint, open sets in £ by (v). But then/" 1 (C)cz/- 1 (G) and/'~ 1 (Z))c:/- 1 (/J) are separated sets in 2 . Thus C and D separated implies / - 1 ( Q a n d f~1{D) separated. Hence / is continuous (§16.2).
The most significant of the equivalences in the above theorem is that which asserts that a function/: X-*% is continuous if and only if, for each
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G c ^ , G open implies/ 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 y are topologically equivalent if and only if there exists a bijection/: X^y such that, for each Gay, G open <*> f~\G)
open.
(1)
Thus, if we wish to determine whether or not X and y are topologically equivalent, the only question we need to ask about X and y is: what are their open sets? Any bijection/: X-^y can then be tested to see whether it satisfies (1). Other information about X and y 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.
21.9|
Relative topologies
Notice that theorem 21.7 refers to a function/: Z-^y, where % and y, are metric spaces. Various equivalent conditions are then given for/ to be continuous on the whole space X. In contrast, the theorems of chapter 16 always involved a subset £ of X and were concerned with the continuity of a function/: Z^% on the set Z.
If £ 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 %. We say that £ is a metric subspace of %.
In chapter 16, we carefully chose a definition for a function /: Z^% to be continuous on the set £ which makes it quite irrelevant whether we regard our
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underlying metric space to be X or whether we simply throw out all the points of X which are not in £ and take our underlying metric space to be £. This is because two subsets A and B of £ are contiguous in the metric space X if and only if they are contiguous in the metric space £. It follows that theorem 21.7 applies equally well in respect of a function/: Z-+% which is continuous on a subset £ of the metric space X. One simply takes note of the fact that £ is a metric subspace of X and replaces X at each occurrence by £. 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 X to £. 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 X is a boundary point of S if and only if <*(£, S) = 0 and dfe, <2S) = 0. But, if ^ £ , it ceases to be eligible as a boundary point when one switches the underlying space from X to £. Even if £ e £ it may still cease to be a boundary point when we switch from X to £ because the nature of CS depends on whether we use X or £ as the universal set. If £ replaces X as the universal set, CS will in general become a smaller set and hence d(£, CS) may cease to be zero. Since a subset S of £ has fewer boundary points relative to the metric space £ than it does relative to the larger metric space X, it is easier for S to satisfy the criteria for being open or closed relative to £ than it is to satisfy the criteria for being open or closed relative to X. Thus the subsets of £ which are not open or closed relative to X may very well become open or closed when the underlying metric space X is replaced by the metric subspace £. Returning to items (iii), (iv) and (v) of theorem 21.7, this observation simply reflects the fact that it is easier for a function/ to be continuous on a subset £ of X than it is f o r / to be continuous on the whole of X. If £ is a metric subspace of the metric space X, we define the relative topology of £ to be the collection of subsets of £ which are open relative to £. As explained above, the relative topology of £ is NOT just the collection of all subsets of £ which are open relative to X. It is a larger collection than this.
21.10
Example Suppose that
X=U2.
Let Zt = {(x, y): x>l}
and let
£ 2 = £ 1 u{(l,0)}.Take Note that 5 1 c £ 1 c £ 2 . The set Sx is neither open nor closed relative to the metric space 5£ = (R2. The boundary points on its straight edge belong to ft \ 5 X while those on the curved edge belong to Sv The set Sx is closed relative to the metric space Zv The boundary points of Sx relative to the metric space £ x are those on the curved edge of Sx and these all belong to Sj. Notice that the points on the straight edge of 5j do not belong to £j and hence are not eligible as boundary points relative to the metric space £ j . The set S^ is not closed relative to the metric space £ 2 . We have that (1, 0) is a boundary point of 5X relative to the metric space £ 2 which does not belong to S^
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The set S 2 = 5 1 u{(l, 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 Zx since it is not a subset of Zv It is closed relative to the metric space £ 2 . Observe that although (1, 0)6 £ 2 and is a boundary point of S 2 relative to the metric space X, it is not a boundary point of S2 relative to the metric space £ 2 . We have that Of course £ j is both open and closed relative to the metric space £ l (theorem 14.14). Similarly, £ 2 is both open and closed relative to the metric space £ 2 . Observe finally that Zx is an open set relative to the metric space £ 2 . Its only boundary point relative to the metric space £ 2 is (1, 0).
The next theorem makes it quite easy to find the relative topology of a metric subspace £ of a metric space X provided that one knows the topology of X to begin with.
21.1 If Theorem Let £ be a metric subspace of a metric space X. Then H is an open set relative to the metric space £ if and only if
for some set G which is open relative to the metric space X.
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Topology
Proof Recall that an open ball B with centre ^ e £ and radius r > 0 in the metric space £ is defined by B = {z: d(z, £)0 in the metric space Z. Thus B' = {x: d(x, £)
I = \J B. Beta
Let
Beat
Then G is open relative to the metric space Z because it is the union of sets which are open relative to the metric space Z (theorem 14.14). Also H = GnZ.
(ii) Suppose that H = GnZ where G is open relative to the metric space Z. denote the collection of all open balls B' (relative to the metric space Z) which are contained in G. Then B. But then B B'ew'
and hence is the union of sets which are open relative to the metric space £.
21.12|
Corollary
Let £ be a metric subspace of a metric space Z. Then H is a
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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 £ \ # is open relative to the metric space £ 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=6GnZ and the result follows.
21.13 Example The set [0, 1) is neither open nor closed relative to the metric space U1. However, [0, 1) is open relative to the metric subspace [0, 2] because [0, 1) = ( —1, l)n[0, 2] and (—1, 1) is an open set relative to the metric space (R1. 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 IR1. metric subspace of IR1
metric subspace of IR1
J^ r
A ^
r
2
-1
y
m 0 f open
21.14|
1
3-
0 closed
Exercise
(1) Let A, B, C and D be the sets in IR2 given by { =
{(x,y):y2<2},
Explain why the function/: R2->[R2 cannot be continuous on IR2 if any of the following properties hold: (i)/" 1 (B) = C (w)f(A) = C
iii) f~\A) = D (iv)/(£) = /).
(2) Let 1 be a compact metric space and let f\%-*% be a bijection which is continuous on Z. Prove that / is a homeomorphism. [Hint: Use theorem 21.7iv and exercise 20.6(2).] (3) Let S be a set in U2 defined by
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 R 1 . [Hint. Use theorem 17.27.]
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Topology
(5) Prove that every subset of N is both open and closed relative to N regarded as a metric subspace ofU1. (The relative topology of N is said to be discrete.) (6) Let A and B be subsets of a metric subspace £ of a metric space X. Show that A and B are contiguous relative to the metric space £ if and only if they are contiguous relative to the metric space X. (7) Let £ be a metric subspace of a metric space X. Show that a subset C of Z is connected relative to Z if and only if it is connected relative to X. Give an example of a connected set D inU2 such that DnZ is not connected relative to £ = {(x,y):x2-hy2^l}. (8) Let £ be a metric subspace of a metric space X. Prove that a subset K of £ is compact relative to £ if and only if it is compact relative to X. Give an example of a compact set L in U1 such that L n £ is not compact relative to £ = (0, 1). (9) Let £ be a metric subspace of X. Prove that £ is a connected subset of X if and only if the only subsets of £ which are both open and closed relative to £ are 0 and £.
21.15f
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 3 of sets in X. The collection 3 will serve as the topology of X. In order that it be sensible to regard the sets in 3 as the 'open sets' of X we require that 3 satisfy the conclusions of theorem 14.13 - i.e.
(i) 0e5,
Xe3.
(ii) Any union of sets in 3 is in 3. (iii) Any finite intersection of sets in 3 is in 3. 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 U". 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 U" 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.
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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 £ of a set £ in a topological space to be the intersection of all closed sets which contain E. Similarly, the interior L 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) (ii) (iii) (See exercise 15.10(5).) We can now define the boundary dE of a set £ in a topological space by
If X and y, are topological spaces, we use theorem 21.7(v) to provide a definition for the continuity of a function/: X-+%. We say t h a t / : X-^% is continuous (on £)ifandonlyif, for each G c ^ , G open => / " 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 y, are topological spaces, a homeomorphism/: X^>% is a continuous bijection with a continuous inverse/" 1 : y,-+%. If a homeomorphism/: X^% exists, we say that the topological spaces X and y, 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 £ be a subset of a topological space X with topology S. Then
is a collection of subsets of £ which satisfies the conditions for a topology on £. We call V the topology on £ relative to X. Thus theorem 21.11 for a metric space becomes a definition for a topological space. The set £ 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 £ of a topological space X. We use the definition given above but with £ (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 0 and X. Theorem 17.3 for a metric space therefore becomes a
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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.
21.16f Theorem Let X and y be topological spaces and let f:X^y continuous surjection. Then
be a
X connected => y connected. Proof Suppose that E is a set in y which is both open and closed. Since f:X-+y is c o n t i n u o u s , / " 1 ^ ) is both open and closed in X. But % is connected. Thus, / " 1 ( £ ) = 0 or f~1(E)= X. Because / is a surjection, it follows that E = 0 or E= X. Thus y, is connected.
A compact topological space X is one with the property that any collection U of open sets which covers X has a finite subcollection which covers X. 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.
21.17f Theorem Let X and y be topological spaces and let f\X-+y continuous surjection. Then
be a
X compact =>y compact. Proof Let V be a collection of open sets which covers y. Then
is a collection of open sets which covers X. Since X is compact, a finite subcollection £ covers X. Let
$={f(G):Ge£}. Then 5 is a finite subcollection of V which covers y. Hence y is compact.
21.18t
Product topologies In the space U2 the Euclidean metric d: U2-^U is defined by
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 R 2 . The function m: [R2—>IR defined by m(x, y) = max{|x 1 - t y 1 |,
\x2-y2\}
is an example of an alternative metric. So is the function /:R2->[R defined by /(x, y) = | x 1 -
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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 / and m looks very different from a Euclidean open ball as the diagrams below illustrate.
£, x)
= {x:d(e,x)
L={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 U2. Now suppose that X^ and X2 are topological spaces. What is the appropriate topology for I j x I 2 ? If X1 and X2 a r e metric spaces with metrics dx\ Xx->M and d2: X2-+U, w e c a n proceed as with the space U2 = U x IR. In particular, the functions d: Xx x X2->U a n d m: Xx x £ 2 ->R defined by d(x, y) =
{(d1(xl9
x2i y2)):, 211/2
(1)
and m(x, y) =
2,
y2)}
(2)
are metrics on. X^ x X2 which generate the same topology on Xi x X2. Why should this topology be more useful than the various other topologies which one might impose on Xx x £ 2 ? The significant fact about the metrics d and m is that they generate a topology on XxxX2 which makes the projection functions Pl: Xxx X2^-X1 and P2: X1 x X2^X2 continuous. For example, if we use the metric d in Xi x X2, then
and hence P1 is continuous on X±x X2 by lemma 16.8.
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P 2 (x)
X = (Xj, X 2 )
•X,
This observation makes it natural to define the product topology on Zl x X2 to be the weakest topology (i.e. the topology with the fewest open sets) with respect to which the projection functions P x : %l x X2~*Xl and P2:Xlx X2-+X2 are continuous. It follows that, if Gx is an open set in Xu then Pi~1(G1) must be an open set in the product topology of X± x X 2. Similarly, if G2 is an open set in X2, then P2~ 1{G2) must be an open set in the product topology of Xx x X2.
Since the intersection of a finite collection of open sets must be open, it follows that
must be an open set in the product topology of X^ x X2. Finally, the union of any collection of open sets must be open. Thus, if W is any collection of sets of the form Gx x G2 (where Gj is open in Xx and G2 is open in X2\ then
5= KJ
(3)
must be an open set in the product topology of Xi x X2. The collection 5 of all sets 5 of the form (3) satisfies the requirements for a topology on Xtx X2 given in §21.15. It follows that 3 is the product topology on Xi x X2.
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S=
Now return to the case in which %1 and £ 2 a r e metric spaces with metrics dx and d2 respectively. If we use the metric m: Xx x £ 2 ->R defined by (2), then the open balls B in %l x ft2 are of the form B = BX xB2 where JBJ is an open ball in %x and 5 2 is an open ball in %2. It is therefore apparent from the preceding discussion, that the open sets in %x x %2 generated by the metric m are precisely those which lie in the product topology 3'. The same is therefore true of the metric d: %x x SC2-+U 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 x %2 since they all generate the same topology. When discussing topological matters, we therefore work with the metric in %1 x %2 which happens to be most convenient for the problem in hand. In U2 (or U") this is usually the Euclidean metric. When Zx and %2 are more general metric spaces, however, the metric m is often much less cumbersome.
22
22.1
LIMITS AND CONTINUITY (I)
Introduction
->^ and suppose that £eX and \\e%. In this chapter, we shall study the meaning of the statement as Sometimes this is written in the equivalent form
We say that 4/(x) tends to r\ as x tends to £' or that '/(x) converges to the limit r\ as x approaches £'. The diagrams below illustrate the idea we are trying to capture. The first diagram is of a function/: R 2 ->R 2 for which/(x)->77 as x-»£ It shows x approaching £ along a path. As x describes this path,/(x) approaches r\. We shall of course want the same to be true however x approaches £.
The next diagram shows the graph of a function/: as x-ȣ.
for which f(x)-+r]
Limits and continuity (I)
107
y
Note that in both diagrams it is false that r\ =f(Z>). It is not the case that one can always find the value of
by replacing x in the formula for /(x) by £. This point is of some importance in calculus. The derivative of a function f:M->U at the point £ is defined by
/'(«) = li /<*)-/«) (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 = £. Sometimes, of course, it will be true that as We then say t h a t / i s continuous at the point . The diagram below illustrates a function / : U -+U which is continuous at
y=f(x)
108
Limits and continuity (I)
Observe that, in drawing the graph of/, 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 f(x)-+r\
as
x-+£,
through the set S.
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 'near9
We are seeking a precise mathematical definition of the statement /(x)-*t|
as
x->£.
The intuitive notion which we wish to capture is that/(x) gets 'nearer and nearer' to r\ 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/(x) is near to r\.9 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 G in 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 ^.
\ G
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 y, be metric (or topological) spaces and l e t / : X where S is a set in X. Suppose that Z>eX and x\ey,. Then we say that as through the set S if and only if For any open set G containing t|, there exists an open set H containing £ such that
xeHnS=>f(x)eG provided
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Limits and continuity (I)
One thinks of the choice of G as specifying how 'near' we want/(x) to be to r\. 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. x e H - then/(x) will be as 'near' to r\ as we specified - i.e. /(x)eG. We now define the statement as
This should mean that /(x) approaches r\ however x approaches appropriate definition is therefore that:
The
For any open set G containing r\, there exists an open set H containing £ such that xeH =>/(x)eG provided In order that this definition makes sense, it is necessary that/be defined on some open set S containing £ (except possibly at £ itself). Such an open set S contains all points of Z which are 'sufficiently near' to £. Indeed, our definition of the statement '/(x)-n] as x->^' is equivalent to the assertion that/(x)^i| asx-^^ through the set S for some open set S containing £. 22.4
Limits and continuity
If/(x)--•/(£) as x->^, we say that/is continuous at the point ^. The next theorem relates this terminology to our previous work on continuity. 22.5
Theorem Let % and y. be metric (or topological) spaces and let
Limits and continuity (I)
111
f: Z-*y. where S is a set in Z. Then/ is continuous on the set S if and only if, for each as 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 / is continuous on the set S. Let \ be any point of S and let G be an open set containing/^). Then/~ 1 (G) is an open set relative to S by theorem 21.7. From theorem 21.11, it follows that there exists an open set if in Z such that/" 1 (G) = if nS. We then have that xeHnS=>/(x)eG. This shows that/(x)->/(!;) as x->^ through the set S. (ii) Suppose that / is not continuous on the set S. Then there exists an open set G in y, such that/ - 1 (G) is not open relative to S (theorem 21.7). Let iief~1{G) be a boundary point of/ - 1 (G) relative to S. Then any open set H in Z which contains \ has the property that HnS contains a point
Since zeHnS but/(z)<£G, it follows that it is false that/(x)-*/(£) as through the set S. 22.6 Corollary Let P: Un- M and Q: Un-+U be polynomials and suppose that g(^)^0. Then P(x) as
112
Limits and continuity (I) Proof This follows immediately from theorems 16.15 and 22.5.
Note that the same result holds with R replaced throughout by C. 22.7 Examples (i) Let S = R 2 \ { ( 0 , 0)} and consider the rational function/: S->R defined by
By corollary 22.6, f(x, y ) - 3
as
(x, y)->(l, 1).
(ii) Let S = U \ { 0 } and consider the rational function F: S-+U defined by
We know from corollary 22.6 that F(x)->1 as x-> — 1 but corollary 22.6 does not help immediately in evaluating lim F(x) x->0
because F(0) is not defined. Note, however, that for x # 0, _ . l + 2x + x 2 - l F(x) = x If G: IR->R is the polynomial defined by G(x) = 2 + x, it follows that F(x) = G(x) unless x = 0. Since we explicitly ignore what happens when x equals 0 when evaluating a limit as x approaches 0, it follows that lim F(x) = lim G(x) = lim (2 + x) = 2. x->0
x~*0
x-*0
22.8 Exercise (1) Evaluate the following limits
(i)
lim
y2-x2 -= 2
l.D y
(ii)
^
m
v2-x2
(2) Let K = {(x, y): O ^ x ^ l and O^ygl} and let/: [R2-IR be defined by fl
(x,y)eK
Limits and continuity (I)
113
Prove that/(x, y)-+\ as (x, y)->(l, 1) through the set K and/(x, y)-+0 as (x, y)->(l, 1) through the set e K. (3) Suppose that 2; is not a cluster point of S. Explain why it is true that f(x)-+r\ as x-»£ through the set 5 for all r\. If £ is a cluster point of S, prove that there exists at most one r\ such that/(x)->Tj as x->£ through the set S provided that, for each pair of distinct points r\1 and r\2 in yy there exist disjoint open sets Gx and G2 such that r\ieG1 and r\2eG2. Show that such disjoint open sets always exist when y is a metric space. (4) Suppose that S ^ L ^ ^ . Prove that/(x)-nj as x->£ through 5 if and only if/(x)-*Tj as x->^ through Sx and through S2. Deduce that, if/ is the function of question 2, then there is no r\eU for which/(x, y)->r\ as (x, JO->(1, 1).
(5) Suppose that T is a subset of 5. If f(x)-+r\ as x-ȣ through the set S9 prove that/(x)-^i| as x->^ through the set T. (6) Suppose t h a t / : R-+R and g: U-^U are continuous on the set U and /(x) = gf(x) for each xeO. Prove that f = g. [Hint: Let J be irrational. Then f is a cluster point of Q (why?). Also/(x)-^/((^) a s i ^ through Q and g(x)-^g(^) as x ^ ^ through O.]
22.9
Limits and distance
In a metric space % a set G is open if and only if each xeG is the centre of an open ball B which is entirely contained in G. This fact means that, if X and y, 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 % and y. be metric spaces and let/: S^y where S is a set in X. Then /(x)->i| a s x ^ through the set S if and only if, for each open ball E with centre t|, there exists an open ball A with centre \ such that xeAnS =>/(X)G£
provided that
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Limits and continuity (I)
If we take the radius of the open ball E to be e and that of A to be 3, the definition assumes the following familiar form: For any e>0, there exists a 3>0 such that for each xeS,
It is sometimes helpful to think of/(x) as an approximation to r\. The quantity d(r\, /(x)) is then the error in approximating to r\ by /(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 \ {^} are to be ignored.) 22.10 Example Consider the function/: [R2->[R2 defined by (u,v)=f(x,y) where u=x+y) v = x-y.) The fact that/(x, y)->(2, 0) as (x, y)->(l, 1) follows from theorem 22.5 because/is continuous on U2. 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 d > 0 such that We begin by observing that \\(x,y)-(l
l)\\ = {(x-l)2
and y, x-y)-(2,
Given e > 0 we therefore have to choose S > 0 such that 0<||(x, y)-(l, 1)|| <5 =» J2\\(x, y)-(l, l)\\<e. The choice of S=s/y/2 (or any smaller positive number) therefore completes the proof.
Limits and continuity
22.11
115
(I)
Example Consider the function/: R 2 - ^ 1 defined by
0
(x, y) = (09 0).
Let La be the line La={(x, y): y = ax}. For points (x, y)eLa, we have that oc2x2 — x2
a 2 —1
(x^O).
Hence a2-!
as
(x,yH(0,0)
feg /me 3; = ax - i.e. through the set La. It follows that/(x, y) cannot tend to a limit as (x, ^)->(0, 0) through 1R2. If it did, it would have to tend to the same limit through each of the sets La (see exercise 22.8(4)). The diagram indicates how the function tends to different limits along different straight lines through (0, 0).
The part of the surface z=/(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 / : R -*R we have that f(x)-+rj as x-+£ if and only if for any e > 0, there exists a 3 > 0 such that 0<\x-£\<3=>\f(x)-ri\<e. This definition can be illustrated graphically as below. y=f(x)
Y\ + E
Y\-E
In this diagram /^ is a rectangle with centre (£, ?y) of height 2e and width 23. 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 of/ above (f - 3, £ + 3) lies within R (except possibly for the point (£, /(£))). Now suppose that I = (a, b) is an open interval in U and that / : (a, b)-*M. We write f(x)-+rj
as x-»a +
and say that f(x) tends to ^ as x tends to a /ram the right if and only if f(x)-+rj a s x - * a through the set (a,fc).Similarly, we write f{x)-+r\ if and only if f(x)->rj as
as x->b —
->fc through the set (a, b).
/ ti + E -
Limits and continuity (I)
117
It is useful to note that f(x)^rj as x-»a + if and only if, for any 8>0, there exists a 8>0 such that a<x \f(x) — rj\<8. Also f(x)-+rj as x-+b— if and only if, for any £>0, there exists 3. S>0 such that b-S<x \f(x)-rj\<e. 22.13 Theorem Let / be an open interval containing £ and suppose that /:/\{£}-»R. Then /(x)->77 as x-+f if and only if/(x)->77 as x-»£— andf(x)^rj as x-xi; + . Proo/ This follows immediately from exercise 22.8(4). 22.14
Example Let /: U ->R be denned by 2x
Then/(*)-• 0 as x->l— and/(x)->2 as x->l+. It follows from theorem 22.13 that lim f(x) does not exist. In particular, / is not continuous at the point 1.
y=fW
Note that it is not necessary to use an argument involving e and 6 to show that/(*)-> 2 as x->l +. The function g: U -*[R defined by g(x) = 2x is a polynomial and hence is continuous everywhere. A\sof(x) = g(x) when x> 1.
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Limits and continuity (I)
In calculating a limit as x-*l + we ignore what happens when x ^ 1. Hence lim f(x)= lim g(x)= lim 2x = 2. X-*\+
X-+1+
X-+1 +
Similarly for/(x)—•() as x->l —. 22.15
Some notation
When the limits exist, the notation - ) = lim/(x);
= lim f(x)
can often be useful. However, it is important not to confuse /(£ —) or/(£ + ) with/(£). These three quantities are equal only when/is continuous at the point £. If / ( £ - ) = /(£) we say that / is continuous on the left at £. If/(
/«)=/«+) neither / ( £ —) nor + ) is equal to
119
Limits and continuity (I) 22.16
Monotone functions
Suppose that S is a non-empty subset of U. We say that / : Sincreases on S if and only if
for each x and y in S. Similarly, / : S->R decreases on S if and only if
for each x and y in 5. We say t h a t / i s strictly increasing on S if and only if
xf(x)
xf(x)>f(y).
strictly increasing function
decreasing function
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 U and that / : I-+R is increasing on /. (i) If/is bounded above with supremum L on /, then/(x)-»L as x^b — . (ii) If / is bounded below with infimum / on /, then /(x)->/ as x-+a +. Proof We prove only (i). From §22.12, we know that, given any e > 0, we have to demonstrate the existence of a 8 > 0 such that
b-5<x\f(x)-L\<e. But \f(x) — L\<s
— 8
+ s (exercise 13.13(3)). The inequality
120
Limits and continuity (I)
I -
/(x)L —e. But/increases on (a, b). Therefore, for any x satisfying y<x
22.18 Corollary Suppose that I = (a, b) is an open interval in U and that / : /->(R is increasing on /. If £e (a, b), then /(
a<x<£
Limits and continuity (I)
121
Proof The function / is bounded above on the interval (a, £) by By theorem 22.17, the smallest upper bound of/on (a, £) is/(£ — ). It follows that, for any xe(a, £), A similar argument for the interval (£, b) yields the other inequalities. 22.19
Inverse functions
Recall that a function /: S-»T admits an inverse function f'1: T^>S if and only if/ 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(x1) = f(x2) <=>xx =x2.
In this section, we are interested in the case when S and Tare intervals in U and / is continuous on S. It then seems 'intuitively obvious' that / is bijective if and only if T=f(S) and / is strictly increasing or strictly decreasing on S. Furthermore, it seems equally clear that the inverse function/" 1 : T^S will be continuous on T.
x=fl(y)
*~ y
122
Limits and continuity (I)
But, like so many 'intuitively obvious' theorems, the proof requires some deep results. 22.20 Proposition Let / be an interval in R and suppose that/: J->R is continuous on L Then/is injective if and only if/is either strictly increasing* or strictly decreasing on /. 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/: I-+R which is continuous on / but neither increasing nor decreasing on /. If a
22.21 Theorem Let / be an interval in R and let J be a set of real numbers. Suppose that / : /-> J is strictly increasing or strictly decreasing on / and that J =/(/). Then / is continuous on / if and only if J is an interval. Proof (i) Suppose that / is continuous on /. Then J =fij) is an interval by theorem 17.10. (ii) Suppose that / is not continuous on L Then, by theorem 22.5, for some £e/, it is false that/(x)->/(£) as x->£ through the set /. From theorem 22.17 it follows that at least one of the quantities/^ — ) or/(£ + ) exists but is not equal to /(£). But corollary 22.18 then implies than J is not an interval.
Limits and continuity (I)
123
22.22 Theorem Let / and J be intervals in U and suppose t h a t / : I-+J is surjective, continuous and strictly increasing or decreasing on /. Then / : I-+ J is a homeomorphism - i.e. / is bijective and / " *: J-+I is continuous on J. Proof The fact that / is bijective follows from the two preceding results. Since f~l: J-+I 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 x ^ O such that y = xn. We call x the nth root of y and write x = y1/n.
A more elegant proof of this result uses the theorems of the preceding section. Let neN and define / : [0, oo)->[0, oo) by
The function / 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 = / ( [ 0 , oo)) is an interval. But/(0) = 0 and J is clearly unbounded above. It follows that J = [0, oo) and s o / i s surjective. By theorem 22.22,/admits an inverse function/" 1 : [0, oo)—•[(), oo). We have that
provided x^O and y^O. T h u s / *: [0, oo)->[0, oo) is the nth root function - i.e. f~\y)=ylln
124
Limits and continuity (I)
Note that theorem 22.22 supplies the extra information that this function is continuous on [0, oo).
22.24 Exercise (1) Let/: U2^>U2 be defined by/(x, y) = (u9 v) where u=
1+x
l+y {y<x)
v=
What can be said about the limits of/as (x, y)->(0, 0) through (i) the set S = {(x, y):x<0 and y>0} and (ii) the set T={(x, y):x>0 and (2) Let #: R 2 - ^ 1 be defined by
Prove that g(x, y)-+0 as (x, y)-+{0, 0) along each of the lines y = ocx. But show that g(x, y)A0 as (x, y)-+(0, 0) along the parabola x = ay2. (3) A function/: U2\ {(0, 0)}-^U is defined by
If
^, show that |/(x, y)|<e if and only if or
<2e{l+v/(l-4e2)}-1.
If ^(x, y)-^0 as (x, y)-(0, 0), prove that/(x, j/)->0 as (x, through the set
9
0, 0)
y)}.
(4) Let % and ^ be metric spaces and let/: S-+y, where S is a subset of X. If £e £ and r\ey., prove that/(x)-H| as x ^ ^ through the set S if and only if, for each subset E of S not containing £,
Use this result to deduce theorem 22.5 from theorem 16.4. (5) Let / be an interval in 1R and let/: I-+U be increasing on /. Prove that the set of points in / at which / is not continuous is countable. [Hint Recall the proof of theorem 17.27.] (6) Let / : U -»R be defined by /(x)= 1 + x + x3. Show that / has an inverse function/" 1 : U->M and that this is continuous on U.
Limits and continuity (I) 22.25
125
Combining limits
22.26 Theorem Suppose that X, % and £ are metric (or topological) spaces and that / : T->£ and g: 5->T where Scz X and Ta%. Let £e X and let i)€T . If/is continuous on the set Tand g(x)^r\ as x->£ through the set S, then /(0()H/(I)
as
through the set 5.
Proof Let G be any open set in Z which contains f(r\). Since / is continuous on the set T, it follows from theorem 22.5 that there exists an open set H in y, containing r\ such that y6tfnr=>/(y)eG.
(1)
(Note that we do not need to exclude the possibility that y = T| because f(r\)eG by assumption.) Also, since H is an open set containing r\, there exists an open set J containing £ such that xeJnS =>g(\)eHnT (2) provided x # £ . (Note that g(S)c:T and so xeS=>g(x)eT.) Combining (1) and (2), we obtain that
xeJnS=>f(g(x))eG provided x # £. It follows that /(0(x)H/(n)
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)^rj as x-*^ through the set S, then {)y^nlln through the set S for any neN.
as
126
Limits and continuity (I) Proof This follows immediately from §22.23 and theorem 22.26.
22.28 Theorem Suppose that X is a metric (or topological) space and t h a t / : S-+Mn where S is a set in X. Let ^e % and write
Then/(x)-»Ti as x->^ through the set S if and only if, for each k = 1, 2,..., n, fk(*)-+rik
as
x
~^
through the set 5. Proo/ This is most conveniently deduced from exercise 22.24(4) and exercise 13.24(6).
22.29
Example Consider the function/: G32->[R2 denned by
where g: R 2 - ^ 1 and h: [ R 2 - ^ 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 /(x,yH(0,0)
as
(x, y)->(0,0)
along the line x = y. On the other hand, g(x, y) -/> 0 as (x, y)->(0, 0) along the parabola x = y2 and hence /(x,y)>(0,0)
as
(x, 3>)->(0,0)
2
along x = y .
22.30 Theorem Suppose that % is a metric (or topological) space and t h a t / i : 5^1Rn and/ 2 : S->Rn where 5 is a set in %. Let ^e £ and suppose that /i(x)->ih
and
x
f2( )-*t\2
as
x-^^
as
x
^^
through the set S. Then for any real numbers a and b, af1(x)-\-bf2(x)-^ar\1-{-br\2 through the set S.
as
x-^^
Limits and continuity (I)
127
Proof Define F: S-^Un by/(x) = af^x) + 6/2(x). From theorem 22.28, we have that ,/ 2 (x))-*(i| l5 i|2)
as
x-+^
through the set S. But, with the notation of theorem 16.9,
F = Lo(fl9f2). Since Lis continuous on !R"x[Rn (theorem 16.9), it follows from theorem 22.26 that /(x)->L(th, TJ2) as
x-^
through the set S.
22.31 Theorem Suppose that % is a metric (or topological) space and that f{. S^U and f2: S-^U where S is a set in %. Let ^e % and suppose that
and
f1(x)-^rj1
as
x
as
/2( )^^2
x^^ x
~^^
through the set S. Then
f1(x)f2(x)^n1n2
as \-*§
through the set S. Define F: S-+U by F(x)=/ 1 (x)/ 2 (x). From theorem 22.28, we have that aS
through the set S. But, with the notation of theorem 16.9, F = M o (/ 1? / 2 ). Since M is continuous on U xU (theorem 16.9), it follows from theorem 22.26 that F{x)^nln2 as x ^ ^ through the set S.
22.32
Theorem With the same hypotheses as in the previous theorem,
7TT-
as
^
through the set S provided that rj2¥"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, i f / ( x ) - ^ 2 as x-^^ through the set S and rj2^0, then there exists an open set J containing £ such t h a t / ( x ) / 0 for any xeJnS.)
128
Limits and continuity (I)
22.33 Exercise (1) Suppose that g: U2-^Ul and h: IR2-*^1 have the property that flf(x, y)-+l as (x, y)-»(0, 0) and h(x, y)^m as (x, jO->(0, 0). Find (i)
lim
/i(x, y) (ii)
(x, y)->(0, 0)
lim
/2(x, y)
(x, y)-+{0, 0)
when/ji IR 2 -^ 1 and/ 2 : R 2 ^ ^ 1 are defined by
Also find lim
J{x, y)
(x, y)-(0, 0)
where / : R 2 ^R 2 is given by f = (fv f2). (2) Let/: R-*R and t/: U^U be defined by 1 ( , = 3) 2
,(x) =
Prove that/(j/)-»2 as y-^3 and ^f(x)^3 as x-+4 but show that it is false that/(0(x))->2 as x-»4. (3) Suppose that SaZ and that g: S-+y, has the property that g(x)-^r\ as x->!; through the set S while/: %-+Z has the property that/(y)->£ as y-^rj. Prove that as through the set S provided that either of the conditions (i) / is continuous at the point r\ or (ii) ntg(S) is satisfied. 22.34|
Complex functions
Let S be a set of complex numbers (§10.20) and consider a function / : S-*C. If oo and f are complex numbers, we are interested in the statement /(z)->a>
as
z-C
(1)
through the set 5. As we know from §13.9, the complex number z = x + iy can be identified with the point (x, y) in IR2 in which case ||z|| = {x2 + )>2}1/2 = ||(x, y)\\. There is therefore no difficulty in interpreting (1) which simply means that: V£>O3<5>OVZGS,
Limits and continuity (I)
129
It is useful to note that theorems 22.30, 22.31 and 22.32 apply equally well if U is replaced throughout by C. 22.35 Example Consider the function/: C-+C defined by/(z) = z3. We shall prove that Observe that i3 = i2i = — i. Hence
(see exercise 10.24(5)). We have that |i| = l and, if | z - i | < l , then |z|^|i| + l=2. Hence, if \z — i| 0 is given and choose S = min{e/7, 1}. Then, if \z — i\ < S, \z3-(-i)\<7\z-i\
23f
23.lt
LIMITS AND CONTINUITY (II)
Double limits
We begin with some notation. Let X, % and £ be metric spaces and let ^e %, x\ey, and £e£. Suppose that A is a set in X and B is a set in % and write S = AxB. We shall consider a function
In mathematical applications one often comes across statements of the type '/(x, y)-»£ as x-»£ through A and y->T| 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 /(x,y)->£
as
(x, y)->& i|)
through the set S = AxB. However, in order to use the definition of a limit given in §22.3 we need to know what sets are open in Xxfy. We therefore augment our definition of a double limit in this case by observing that the open sets of X x y, are to be those in the product topology of X xy, (§21.18).
130
Limits and continuity (II)
131
As explained in §21.18, when % and y are metric spaces with metrics dx and d2 respectively, there are various metrics which we can introduce into % x y which generate the product topology of % x y and the choice of which of these metrics to use in % x y is largely a matter of convenience. When % = y. = M and so % x y = U2, we usually use the Euclidean metric but, in many cases, it is easier to use the metric m: Zxy-*M defined by 1,
), (x 2 ,
x 2 ), d2(y1, y2)}.
The use of this metric 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 s>0, there exists a 3>0 such that for each (x, y)eS = Ax B, ),
(x, y))<3 =
x, y)) <e.
This can in turn be written in the following less clumsy form: For any g>0, there exists a 3>0 such that for any xeA and any yeB, 9 x)<3
and d2(r\,
x, y))<e
provided that (x, y) # (£,
23.2 Example Suppose t h a t / : [ R 2 - ^ 1 and that/(x, y)->£ as (x, y)->(£, rj). If we use the Euclidean metric in U 2, the definition of this statement can be expressed in the form: For any g>0, there exists a 3>0 such that 0<{(x — £)2+(_y — Y])2}1'2 <3 => |/(x, y) — Q <e.
(1)
2
Alternatively, we can use the metric m in U and obtain the definition in the form: For any £>0, there exists a A > 0 such that
\y-r\\ \f(x, y)-(\<s
(2)
provided (x, y)#(^, rj). The first form of the definition says that \f(x, y) — Q <E provided that (x, y)^ (^, rj) is in a sufficiently small disc with centre (£, rj). The second says that |/(x, y) — Q<£ provided that (x, y)#(<^, rj) is in a sufficiently small box with centre (£, rj). Both definitions are plainly equivalent because, if 3 = 3^^ works in (1), then A = Sl/y/2 works in (2). On the other hand, if A works in (2), then 3 = A works in (1).
132
Limits and continuity (II)
23.3f
Double limits (continued) We have not yet finished with the statement '/(x, y)->£ as x->£ through A and y-n] 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 Zxy, with the property that (x, y) not in R are to be disregarded. Our double limit statement then means that /(x,y)->§ as (x,"y)->&ii) through the set Rr\S (where S — AxB topology is used in Z x y).
and it is understood that the product
23A Example Consider the function/: [R2->[R of exercise 22.8(2). If one is asked to consider the double limit £ defined by the statement/(x, y)->C as x-»l and y->l where x + y>l, then one should interpret this as meaning that f(x9 JO-C
as (JC, y M l , 1)
through the set J = {(x, y):x + y>l}. Since J<=CK, it follows from exercise 22.8(2) that C = 0. 23.5|
Repeated limits The notation lim /(x, y)
is available for a double limit as discussed in §23.1. We shall however prefer the less clumsy notation lim
/(x, y)
(x, y)->& I])
although the latter notation is somewhat less precise in that it takes for granted the use of the product topology in Zxy,. It is important not to confuse either of these pieces of notation with repeated limits lim I lim /(x, y)); lim (lim /(x, y) I. x-*
\y-ii
/
y-i
\x-*
/
The reasons why one needs to be careful in dealing with repeated limits are best explained with the help of some examples.
Limits and continuity (II) 23.6
133
Example Observe that r
x2
-y2
hm —-z
1 T
= 1
and hence lim (lim-T-S-1 = 1. 2 2 x^o\y-,o x + y j On the other hand, a similar argument shows that lim (lim —z
T|
= — 1.
The order in which the limits appear cannot therefore be reversed in general without changing the result. 23.7
Example Observe that, for each xeU,
It follows that (
*y
lim I lim —
\ 1 = 0.
Similarly, lim f lim —z. j I = 0. However, the double limit nm
z
IT
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: U2^>M defined by h(x,y) = x
We have that \h(x, y)\ = \x\£\\(x, y)\\
134
Limits and continuity (II)
and so it is easily shown that lim
h(x, y) = 0.
(x, y)-(0, 0)
On the other hand, we have that h{x, y)-*x as y-»0 + and /i(x, y)—• — x as y—•0 —. It follows that lim h(x9 y)
(1)
y-0
does not exist unless x = 0 and hence that the repeated limit lim (lim h(x, y) J 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 /(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 Aim f(x, y)\
x - + 0 \ v -*•()
I
(x.y)
—»
t
1
(0,0)
(0,0)
lim | lim fix, 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)
(0,0)
(1,0)
135
Limits and continuity (II)
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.
23.9| Theorem Let %, fy and £ be metric spaces and let ^e %, \\e% and Suppose that A is a set in X with cluster point £, and that B is a set in y. Let / : S\(£, !))-•£ where S = A x B. Suppose that /(x,y)->£
as
(x,
(2)
/(x, y)->Z(y) as
(3)
through the set S and that, for each yeB,
through the set A where /: £->£. Then /(y)-»£
as y
(4)
through the set B.
B4
Proof Let e>0 be given. By (2) there exists a £ > 0 such that for each xeA and each yeB & x) < d and d2(t!, y) < 5 => ^ , /(x, y)) < e/2 provided (x, xeA,
x\). By (3), for each yeB there exists a A y >0 such that, for each ), f(x, y))<e/2.
136
Limits and continuity (II)
For each ye£, choose xyeA so that d^, 0
\y)<3 y,
y
and d^,
x y )
y), /(y))
<e/2 + e/2 = £ and hence (4) holds.
It is sometimes important to know whether or not it is true that lim (lim / ( x , y)) = lim (lim /(x, y) j . x-S \y-ii
/
y-n \x-$
(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. lOf Theorem Let %, y and £ be metric spaces and let S be an open set in X x y containing (£, r\). If/: S->£ is continuous on S, then lim ( lim /(x, y) j = lim (lim /(x, y) 1. x-*% \y->ti
J
y-*ti \x->£
/
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.
23.1 If
Uniform convergence
Suppose that % and £ are metric spaces and that/:X xB^Z where A is a set in %. Let £E % and suppose that /:B->£ has the property that, for each yeB, /(x,yH/(y)
as
x^£
through the set A. Expressed in full detail, this assertion means that Vyefl V £ >0 3(5>O VxeX, (1) We know from chapter 3 that the order in which the quantifiers V and 3 occur in a statement is significant to its meaning. In particular, if P(u, v) is a predicate, it is in
Limits and continuity (II)
137
general false that Vw
3vP(u9v)o3vVuP(u9v).
It follows that, if we move the term 'Vyel?' 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 V£>0
VxeA VyeB, (2)
does not mean the same as (1).
pointwise convergence
uniform convergence
138
Limits and continuity (II)
The difference in the meaning of the two statements is that, in statement (1), the value of S 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 S depends only on the value of e - i.e. the same value of S works uniformly for all values of yeB. If (2) holds, we say that /(x,y)->/(y)
as
x-+^
through the set A uniformly for yeB. It is sometimes necessary to emphasise that this statement is stronger than (1). When (1) holds, we therefore say /(x,yH/(y)
as
x->^
through the set A pointwise for yeB. Note that it is obvious that uniform convergence implies pointwise convergence. But fhe converse is very definitely false.
23.12f
Distance between functions With the notation of the previous section, suppose that /(x,y)-/(y)
as
x->$
(1)
through the set A uniformly for yeB. Suppose that £ > 0 is given. Then e/2 > 0. Hence there exists a S > 0 such that for each xeA and yeB 0 < dft, x) < d => d(l(yl /(x, y)) < e/2. Since the final inequality holds for all yeB we may conclude that, for any e > 0 there exists a d>0 such that, for each xeA, 0 < d& x)<S=> sup d(l(y\ f{x, y)) < e yeB
i.e. sup <*(/(y),/(x, y ) ) - 0
as
x->$
(2)
yeB
through the set A. We have shown that (1)=>(2; and it is even easier to show that (2)=>(1). This observe ion suggests the following definition. Suppose that F: B^Z G: B->£. Then we define the uniform distance between F and G by II(F, G) = supd(F(y), G(y)) yeB
where this quantity exists.
and
(3)
Limits and continuity (II)
139
23.13 Example Consider the function F: [0,1]->R defined by F(y) = $y2 + \ and the function G: [0, 1]->R defined by G(y) = 6_y. The uniform distance between F and G is given by u(F,G)= sup
= max \Sy2-6y + l\. The maximum of H(y) = \8y2 — 6y + 1| is either attained at y = 0 or y = 1 or else at a point 7/e(0, 1). In the latter case, differentiation yields that 16/7-6 = 0
Since tf(0) = l,
= 3 and if(|) = i, it follows that M(F,
G) = 3.
140
Limits and continuity (II)
In terms of the uniform distance notation, statement (2) assumes the less clumsy form, u(/,/(x,#)H0
as x ^
(4)
through the set A, where /(x,»): B-+Z is the function whose value at yeB is /(x, y).
B
If F and G are bounded on B then w(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: as
(5)
through the set A. It is often helpful to 'forget' that/(x,«) and / are functions and to think of them instead as 'points' in the metric space of bounded functions on B.
space of bounded functions on B
Limits and continuity (II)
141
It is important, however, not to forget that the distance between the 'points' /(x,») and / is defined by p(yJ(x,y)).
(6)
yeB
The above discussion is intended to indicate why uniform convergence is in some respects a more straightforward concept 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 ^ = R \ { 0 } and define g: AxU->U by
We begin by observing that, for each yeU, g(x, y)-+y as x->0 - i.e. g(x, y)->y as x-»0 pointwise for yeU. It follows that, if g(x, y)-*l(y) as x->0 uniformly for yeU, then it must be the case that l{y) = y. Consider u(l, g(x,m)) = sup \l(y)-g(x, y)\. yen
Taking l(y) = y, we obtain that, for each xeA,
\Ky)-g(x,y)\ =
y— |x|3
Hence, for each xeA, Ixl 3 yeU X
\y
Since |x|—>0 as x-^0, it follows that g(x, y)-*y as x-»0 uniformly for yeU. The diagram over illustrates the set of (x, y) for which it is true that I y — g(x, y)\ < £• Observe that the choice 5 = s ensures that o < |x| < d =>| y — g(x, y)\ < e for all yeU.
142
Limits and continuity (II)
23.15
Example Let A = R\{0} and define h: AxU-+U by h{x, y)=~2
xv
j•
xz + y2 We begin by observing that, for each yeU, h(x, y)-*0 as x-+0 - i.e. h(x, y)->0 as x->0 pointwise for yeU. It follows that, if h(x, y)—>l{y) as x->0 uniformly for yelR, then /(y) = 0. Observe that, for each xeA, 0-yen
(The geometric-arithmetic mean inequality gives \xy\i^\{x2 + y2) and equality is attained when x = y) Since i/(0, h(x9 •))-f>0 as x^>0, it follows that it is not true that h(x, y)^>0 as x->0 uniformly for yeU. The diagram below illustrates the set of (x, y) for which it is true that |/z(x, y)\ <s when 0 < £ < ^ . (See exercise 22.24(3).)
143
Limits and continuity (II)
-4s2)}'1
\h(x,y)\<8
It is clear that, i
y
, then there is no value of d>0 such that for all xeA and all
yen If y 7^0, the largest value of d>0 for which it is true that 0<|x|<(5=>/z(x, y)<£ is
23.16| Theorem Let X, % and Z be metric spaces and let \e X, x\e% and Suppose that A is a set in X and that £ is a set in ^/. Let/: S\(£, il)->£ where S = i x 5 and let /: £ ^ £ . Suppose that /(x, y)-W(y) as x-^^
(7)
through the set X uniformly for y e 5 and that /(y)-5
as y->ri
(8)
through the set B. Then
/(x,y)-S as (x,yHfen) through the set S. Proof Let £ > 0 be given. By (7) there exists a ^ > 0 such that, for each xeA and each yeB,
By (8), there exists a S2 >0 such that for each y
144
Limits and continuity (II)
Choose ^ = min{^1? S2}. Then, if 0 < ^ 1 ( ^ , x)<<5 and 0
y))£d& l(y)) + d(l(y)J(x, y)) <s/2 + s/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.
23.171 Theorem Let X, y and £ be metric spaces and let ^e X, x\ey and i Suppose that A is a set in X for which £ is a cluster point and that B is a set in y. Let / : S \ ( £ , t | ) - + £ whereS = AxB. Let /: £ - • £ and let m: 4->£. Suppose that (i) / ( x , y)-»/(y) as x->^ through the set A uniformly for yeB, (ii) /(y)-»£ as y-*ri through the set B, (iii) /(x, y)->m(x) as y - n j through the set B pointwise for xeA. 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 # be defined as in example 23.14. We have that
(i) g(x, y)->y as x->0 uniformly for yeU. (ii) y-»0 as y-»0. (iii) #(x, y)->x as y^O for xelR \ { 0 } . From theorem 23.17 we may conclude that lim lim g(x, y) = \im lim g{x, y). y-*0 x->0
x->0 >;-+0
This is easily verified directly.
23.19*}"
Exercise
(1) Consider the function/: U2 \{(0, 0)}->[R defined by J \x-> y)—~~i
2'
Prove the following: (i) / ( x , y)^l as (x, y)->(0, 0) along the line y = Q. (ii) / ( x , y)-> — 1 as (x, y)-*(0, 0) along the line x = 0. (iii) / ( x , y)-*0 as (x, y)->(0, 0) along the line x = y.
Limits and continuity (II)
145
Deduce that/(x, y) does not tend to a limit when x->0 and y->0 independently. (2) Let / be as in question 1. If 0 < e < l , prove that |/(x, y)\<s if and only if 1-e 1+e
<
(y\2 \x/
<
1+e 1-e
Prove that/(x, y)->0 provided that x-+0 and y-+0 in such a manner that
(3) For each of the functions f:(U\ {0}) x U ->R given below, decide whether or not it is true that lim (lim f(x, y) I = lim ( lim/(x, y) I. 2
2
(i) f(x, y)= * ~[
2 4-
2
(ii) /(x, y) = 4 ^ T
l + x z + y2
x2 + y
(4) In each case considered in question 3, find a function /: IR ->R such that /(x, y)^/(y)
as x->0
pointwise for (5) In each case considered in question 3, decide whether or not it is true that /(x, y)-*l(y)
as x->0
uniformly for (a) yeU and (b) ye[ — l, 1]. (6) A function/: R2-»[R1 has the property that (x1^x2
and y i ^ ^ ) ^ / ^ ! . y i ) ^ / f e , y2)-
Prove that lim I lim /(x, y) 1= lim ( lim /(x, y) 1. £ \ } \Z J
23.20|
Uniform continuity
Let % and y, be metric spaces and suppose that/:S->^ where 5 is a set in %. To say that / is continuous on S is equivalent to the assertion that, for each / ( x ) - , / © as through the set S. This means, in turn, that
x ^
146
Limits and continuity (II)
In this statement the value of 3>0 which is asserted to exist may depend both on the value of s>0 and on the value of ^eS. If, given e>0, a value of S>0 can be found which works for all ^eS simultaneously, then we say that / is uniformly continuous on the set S. Thus / is uniformly continuous on the set S if and only if V £ >0 35>O VxeS
23.21
Example The function /: [0, oo)->[0, oo) defined by
is uniformly continuous on [0, oo). Proof It is evident from the diagram that \y/x — y/^\ is largest for x^O, £^0 and \x-£\^d when £ = 0 and x = S. (This is easily checked analytically by proving that Jb-^Ja^{b-a)112 when Ogagfr.)
=
It follows that, given any £>0, there exists a any x^O and any y^O,
23.22
y/x
(namely S = s2) such that, for
Example The function/: (0, oo)->(0, oo) defined by
is not uniformly continuous on (0, oo).
Limits and continuity (II)
147
y
Proof Given e > 0, the largest value of d > 0 for which
satisfies
1
1
-S
I
1
1
x
t
= e.
Hence Since this expression tends to zero as £—>0 + it follows that no d>0 can work for all
23.23| Theorem Let X and y be metric spaces and suppose that / : K^y, where K is a compact set in X. If / is continuous on K, then / is uniformly continuous on K. Proof Since/is continuous on K,/(x)-»/(y) as x->y through K for each yeK. Let e > 0 be given. Then ^e>0 and so, for each yeK, there exists a (5(y)>0 such that, for each xeK, d(y9 x)«5(y) => d(f(y),
(1)
Let 1L denote the collection of all open balls with centre yeK and radius ^S(y). Then U covers K. From the definition of a compact set (§20.4) it follows that a finite subcollection § of It covers K. Let d>0 be the radius of the open ball of minimum radius in the subcollection $. Now suppose that x and £, are any points of K which satisfy d(£, x)
148
Limits and continuity
Also, d(y, x)^d(y,
(II)
£) + <*& x)<|<5(y) + <5g<5(y). Thus, by (1) again, d(f(y),/(*))<&.
(3)
Combining (2) and (3), we obtain that
<*(/©,/(x))<e. We have therefore shown that, for any & > 0, there exists a 3 > 0 such that, for each and each
and so the proof of the theorem is complete.
23.24|
Exercise
(1) Determine which of the following functions / : (0, 1)->R are uniformly continuous on the interval (0, 1).
(2) Let X and y, be metric spaces and suppose that/:S->^ where S is a set in X. Let Qv [0» ! ] - ^ s a n d ^2- [0, l]->5 and suppose that dig^t), g2(t))-*0 as t->0 + . If/ is uniformly continuous on S, prove that as Interpret this result geometrically. Show that the result need not be true if / i s only continuous on S. (3) Let {a0, a l 9 . . . , an} and {bl9 b2, •. •, ^n} be two sets of real numbers and suppose ... [R which satisfies that 0 = ao
=> g(x) = bk
(/c = l, 2 , . . . , n)
is said to be a step function. Prove that any continuous function /:[0, 1]->IR 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]->R such that
u(f,g)= sup
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 00 + 00 = 00
.'. 2 x oo = 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 X 0 , X l9 X 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. N of Q) has cardinality X 0 (aleph-zero). The next largest type of infinite set has cardinality X 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 U 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.
Jline at infinity' parallel lines
In this chapter we shall be concerned with what might be called 'topological infinities'. These are the infinities which appear in statements like /(x)-> + oo
as
xH—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 U 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 U is not compact because it is not bounded (theorem 20.12). What can be done about this?
Points at infinity
151
The natural mathematical response to such a problem is to ask whether it is possible to fit R 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 % which has only one more point than R. This is called the one-point compactification of R. For most purposes, however, it is very much more useful to employ a two-point compactification of R. (See §24.4.) We begin by considering the one-point compactification. This consists of a compact space R # which contains R as a subspace but has only one point more than R. For reasons which will soon be apparent we denote this extra point by oo. Recall from §21.4 that R is topologically equivalent to a circle C in U2 with one point N deleted. The stereographic projection / : R indicated below provides an appropriate homeomorphism.
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 lR # =Ru{oo}. The function / : U -+C\{N} is extended to be a function g: R # ->C by defining f{x) ixeU) N ix = oo)
152
Points at infinity
We want R * 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: R #->(R for the space R * by defining c(x9y) = d(g(x)9g(y)).
(1)
Thus the distance between x and y regarded as elements of R * is the Euclidean distance between the points g(x) and g(y) in U2. If x and y are real numbers,
If z is a real number, c(z, oo) = c(oo, z) =
(1+z 2 ) 1 ' 2 '
In view of (1), the function g: R # ->C is a homeomorphism. Since C is compact it therefore follows that [R# is compact (theorem 19.13). It may be helpful to think of R * 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 U#
c(x,y)
Notice that the distance c(x, y) between x and y regarded as elements of 1R# is not the same as the Euclidean distance between x and y - i.e. c(x, y)¥z\x — y\. 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' U # ? The answer is that IR is a topological subspace of IR *. In fact, a subset of R 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 / : R ^ C \ { N } is a homeomorphism and hence both / a n d / " 1 preserve open sets (theorem 21.7). The fact that IR fits inside R * as a topological structure but not as a
Points at infinity
153
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 U * - 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 % is open if and only if, for each £eG, there exists an open ball B with centre % such that BaG (theorem 14.12). This criterion remains valid for an open set G in U * if we think of an 'open ball' in IR * as an interval of the form (£ — S, £ + 8) when its centre is a real number £ and as a set of the form {oo}u{x:xelR and |x|>X} when its centre is oo. (The abuse lies in using the Euclidean metric when the centre is a real number and the U* metric when the centre is oo. When indulging in this abuse we shall write 'open ball' in inverted commas.)
'open balls'
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
are those with the property that, for each £e G, there exists an 'open ball' B with centre .£ such that J5cG. Here an 'open ball' is to be understood as a Euclidean ball when its centre ^eUn - i.e.
and as a set of the form B =
{X:\\X\\>X}U{M}
when its centre is oo. The case n = 2 is particularly important since (R2 can be identified with
154
Points at infinity
the system C of complex numbers (§10.20). The one-point compactification of U2 is called the Riemann sphere or the Gaussian plane depending on whether one prefers to think of U 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 before, the stereographic projection. (See §21.4.) 'open balls' on Riemann sphere 'open balls' on Gaussian plane
We mentioned in §24.2 that the one-point compactification of U is not the most useful possible compactification of !R. But this ceases to be true for Un when n ^ 2 . In particular, one always uses the one-point compactification of U 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 U and C \ {JV}. A somewhat more natural topological equivalence, however, is that which holds between IR and the open interval (—1, 1). (See §21.3.) The diagram below illustrates a homeomorphism which can be used to demonstrate this topological equivalence. .fix)
Points at infinity
155
It is natural here to augment (R 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 [—00, + o o ] = R u { — oo, + oo}. We then define an open set in [ — oo, + oo] to be a set G with the property that, for each £eG, there exists an 'open ball' B with centre £ such that B a G. Here an 'open ball' is an interval of the form (£ — 6, £ + S) when its centre is a real number £ but is a set of the form (X, +oo)u{ + oo} when its centre is + oo and a set of the form (-ao, Y)u(-cx)} when its centre is — oo.
r////////// V//////////
'open balls' in [—°°, + °°]
This definition for an open set in [— oo, +oo] makes the function g\\_— oo, +oo]->[—1, 1] illustrated below a homeomorphism. Thus [ — oo, + oo] is topologically equivalent to [ — 1,1]. In particular, [ — oo, + oo] is compact.
• -f oo
Moreover, U is a topological subspace of [— oo, + oo]. In fact, a subset of U is open relative to [ — oo, + oo] if and only if it is open in the usual sense. When discussing the one-point compactification of IR, we made no attempt to extend anything but the topological structure of U to U *. In particular, there is no way in which the order structure of IR can sensibly be extended to IR # . But this is not true of the two-point compactification of IR
156
Points at infinity
and the order structure of U extends in a very natural way to [— oo, + oo]. An ordering ^ on [—oo, +00] is obtained by requiring that, for all xe[—00, +00], — oorgxrg + 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 [ — 00, +00] all non-empty sets have suprema and infima. If S is a nonempty set of real numbers which is bounded above in U (i.e. which has a real number as an upper bound), then sup S has its usual meaning. If S is unbounded above in R, then its smallest upper bound in [—00, +00] is + 00 and hence sup S= + 00. Thus sup S = + 00 is a shorthand way of saying that S is unbounded above in R. Note also that sup 0 = —00. The reason is that all elements of [ — 00, + 00] are upper bounds for 0 (because xe0 is always false and therefore xe0=>x^y is always true). With + 00 and — 00 available, one can therefore use the notation sup S 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 U to [— 00, + 00]. One cannot, of course, extend all of the algebraic structure of U to [—00, +00] since then [—00, +00] would be a continuum ordered field and we know from §9.21 that U is the only such object. However, one can extend some of the algebraic structure of IR to [—00, +00]. We make the following definitions. (i) x + (+oo) (+oo) + x + o o \ . V (provided x^ - c o ) x - ( - o o ) = - ( - o o ) + x = + ooj (ii) x + (-oo) = ( o o ) + x o o ) > (provided x^ +00); x - ( + o o ) = - ( + o o ) + x = - o o j VF (iii) x(+oo) = (+oo)x=+oo) } (provided x>0) x(-oo) = (-oo)x= - o o j (iv) x / + o o = 0 ) ' > (provided x ^ ± 00). x/ - 00 = 0 j Perhaps the most significant feature of these definitions are those items which are omitted. Observe that (+00) + ( — 00), ( + 00) —(+00), 0(+oo),
Points at infinity
157
0(— oo), +00/ +00 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 + 00 and — 00. 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 5 and T of real numbers and two non-empty sets P and Q of positive real numbers. We then have that (i) sup ( — x)= —I inf x xeS
(ii)
V xeS
sup (x + y) = ( sup x l + l sup y (x,y)eSxT
(iii)
\ xeS
)
\ yeT
sup (xy) = I sup x 11 sup y {x,y)ePxQ
\xeP
) \
yeQ
without any need for assumptions about the boundedness of the sets concerned provided that the suprema and infima are allowed the values + 00 or —00 (See exercise 9.10(9).) Note, however, that some vigilance is necessary if one or more of the sets is empty. For example, if S — 0 and T is unbounded above, the right-hand side of (ii) becomes meaningless. The system [—00, +00] 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 [— 00, + 00]. Since other authors use U * 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 fallows, we shall wish to regard S as a set in the space [—00, +00]. Next consider a function / : S-*[— 00, +00] for which f(S)aM. Again, one would normally regard / as taking values in IR but here it is convenient to regard / as taking values in [— 00, + 00].
158
Points at infinity •
+00
+00
If ^e[— 00, + 00] and ne\_— 00, + 00], it makes sense to ask whether or not/(x)->77 as x->£ through the set S. The definition is the same as always. (See §22.3.) For any open set G in [—00, +00] containing rj, there must exist an open set H in [—00, +00] containing £ such that xeHnS =>f(x)eG provided that x # £ . As we found in §22.4, it is more helpful for technical purposes to rephrase this definition in terms of 'open balls' in [—00, +00]. We obtain that f(x)-+n as x-+£ through 5 if and only if: For each 'open ball' E in [— 00, + 00] with centre n, there exists an 'open ball' A i n [ — 0 0 , +00] with centre £ such that xeAnS =>f(x)eE provided x ^ £ . If £ and n are real numbers, this definition is exactly the same as the familiar definition of §22.4. But if £ or n are + 00 or — 00 then something new is obtained. If we insert the relevant definition for an 'open ball' given in §24.4 and bear in mind that/takes only real values, we obtain the criteria listed in the table below. In this table / 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 '/(x)-W as x-> + 00'. One can think of f(x) as an approximation to /. Then |/(x) —/| is the error involved in using
159
Points at infinity
f(x) as an approximation for /. The statement '/(x)-W 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, '/(x)-» —oo as x-> —oo' means that we can make/(x) as negative as we choose (i.e. less than Y) by making x sufficiently negative (i.e. less than X). /(x)->/ as n through S
V £ >0 3^>0 VxeS, 0<|x — a\<S => 1./
/(x)-W as x-i through S
V£>0 3X VxeS,
/(x)->/ as x-» — oo through S /(x)-» + oo as x->a through S
VX 3^>0 VxeS,
/(x)—• — oo as x->a through S
VY 3(S>0 VxeS,
/(x)-» + oo as x-> 4- oo through S
VX 3YVxeS, x VX
/(x)-* + ooas x ^ — oo through S /(x)-> — oo as x-»- + oo through S
VY3X VxeS,
f(x)-> — oo as x-> — oo through S
VY3X VxeS, >/(x)
x>x
24.6 Example Consider the /(x) = l/x. We have that (i) l/x->0 as x ^ + oo (ii) 1/x-^ + oo as x->0 +
=*/(x)
function
/ : R\{O}-^IR
defined
by
160
Points at infinity
(iii) l/x-> — oo as x-+0 — (iv) l/x->0 a s x - ^ - oo. To prove (i) we have to show that, for each e > 0, there exists an X such that x>X
<£.
Clearly the choice X = l/e suffices. To prove (iii), we have to show that, for each 7, there exists a S > 0 such that x Any value of S>0 is adequate for this purpose when Y^0. When Y<0, the choice S = —1/7 suffices.
Note that in spite of (ii) and (iii) we say that/(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 I for which/(x)->/ 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 R, it is therefore untrue to say that f(x) converges asx->0+ or x->0 —. We next consider a set S in Un# and a function/: S->Rm#. (Recall that U is the one-point compactification of Un.) We are interested in the case when SczM" and f(S)cRm. If ^eUn# and r\eMm#, we have that/(x)->Ti as x->£ through S if and only if: For each 'open ball' E in R m# with centre TJ, there exists an 'open ball' A in R" # with centre \ such that n#
xeAnS=>/(x)e£ provided x ^ ^ .
Points at infinity
161
If we insert the relevant definition for an 'open ball' given in §24.2 and bear in mind that/takes values only in Um, we obtain the criteria listed in the table below. In this table, I and a are to be understood as vectors in Um and Un respectively. /(x)->lasx->a through S /(x)-»lasx->oo through 5 /(x)->oo as x->a through S /(x)->oo as x->oo through S
24.7
Ve>0 3S>0 VxeS 0<||x-a||<<5=>||/(x)-I||<e Va>0 3X VxeS ||x||>X=>||/(x)-l||<e VX 3<5>O VxeS 0<||x-a||<(3=>||/(x)||>X VX 37 VxeS ||x||>y=>||/(x)||>X
Example Consider the function/: [R2\{(0, 0)}->[R2 defined by /
x
—y
We have that as
as
Proof To prove (i), we have to show that, for any e > 0, there exists an X such that
x 1 The choice X = 1/e clearly suffices. To prove (ii) we have to show that, for any X, there exists a 5>0 such that If X ^ 0 any value of d > 0 is adequate. Otherwise we require a 8 > 0 such that 2 0<x 2 +y 2 <(5 2 => / 2>X x +y
162
Points at infinity
and the choice 6 = 1/X clearly suffices. The diagram illustrates (i). The function value /(x, y) lies in the disc with centre (0, 0) and radius s>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 1 x + iy
1 (x — iy) x — iy (x + iy) (x — iy) x2 + y2'
we can identify the given function with the function/: C\{0}-»C defined b y / ( z ) = l / z . To prove (i) we then have to show that, for any e>0, there exists an X such that
When working with U1 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, co) = {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 U it is therefore as well to employ notation of the type as |x| >oo X) as or in order to avoid any possible confusion. Similar notation is also sometimes useful when working in R" 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 / : R"->[Rm can be replaced by convergence problems by replacing U by [—00, +00] and Un by Un#. 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 [— 00, + 00] or R" # replaces the metric space X in these theorems. Thus, for example, theorem 22.30 shows that, if /j(*)->/! as x-> — 00 and/ 2 (x)-»/ 2 as x-> — 00, then as
x-> — 00.
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 j \ : S-*[— 00, +00] and/ 2 : S-*[—00, +00], where S is a set in X. Let £e X and suppose that Mx^rji x
and
/2( )-^2
as
x->£
as
x
~ȣ
through the set S. Then, if a and b are real numbers (i) a (ii) (iii) 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 00 for xeS and n2 = 0, then /i(x)// 2 (xH + oo
as
x->^
through S. Similarly, if — 00 ^nx < 0 , / 2 ( x ) > 0 for xeS and n2 = 0, then /2W// 2 (x)-^-co
as
x-ȣ
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, if/^x)-* + 00 as x->0+ and /2(x)--> + oo as x->0 + , then fi(x)/f2(x)-+l as x->0 + . However, as the examples below indicate, this result is false in general.
(i)f1(x) = x2, f2(x) = x
164
Points at infinity
(ii)/ 1 (x) = 2x, f2(x) = x (iii)/ 1 (x) = x, / 2 (x) = x 2 .
24.10 Exercise (1) Prove the following results: (i) sup (2x + 3) = + oo (ii) sup (2x + 3) = — oo x 2 <0
x>0
(iii)
inf
(xy) = — oo (iv)
x>0,y<0
sup
(y — x)= + oo.
x>0,y>0
(2) L e t / : U \{1}-+R be denned by
Prove the following results: (i) /(x)->l (ii) /(x)->l (iii) /(x)->l (iv) /(x)-> + oo (v)/(x)-> —oo (vi) |/(x)|->oo (vii) x/(x)-> + oo (viii) x/(x)-> — oo (ix) |x/(x)|-»oo
as x-> + oo as x - > - o o as |x|->oo as x-»l + asx-»l — as x->l as x-> + oo as x-> — oo as |x|-^oo.
(3) (i) Let / : R2->R be defined by / ( x , y)=(l + x 2 + ^ 2 ) " 1 . Prove that /(x, y ) ^ 0 as (x, y)^co. Consider also the function g: U2^U defined by g(x, y) = (l+(x — y)2)~1. Prove that g(x, y)-+l 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)->R2 given by h(t) = (l/t, t) defines a curve in U2. Sketch this curve. Prove that (a) h(t)^oo as £-•() + (b) h(t)-+co as r-> + oo. (4) A function f:M-+M has the property that f{x)->l as x-> + oo and f(x)^m as x-* — oo where / and m are real numbers. Define g: [^-oo, +oo]->[R by
(f(x) g(x)=\ I [ m
(xeU) (x=+oo) ( x = — oo).
I f / i s continuous on M, explain why g is continuous on [— oo, oo]. Deduce the following results: (i) The set /(R)u{/, m} is compact, (ii) The function / is bounded on U. (iii) If there exists anxeIR such that /(x) ^ / and /(x) ^ m,
Points at infinity
165
then / achieves a maximum on U. t(iv) The function / i s uniformly continuous on U (§23.20). (5) The metric space R # with metric c: U*-*M was introduced in §24.2. Explain why a set G in U * is open if and only if, for each £eG, there exists an 'open ball' B with centre £ such that BaG. Here 'open ball' has the meaning assigned in §24.2. (6) Prove proposition 24.9 Suppose that /i(x)-> + oo as x-» — oo and / 2 (x)-^ + oo as x-+ —oo. Give examples to show that neither of the following statements need be true. (i)/i(x)-/2(x)->0 (ii) / i (x)//2(x) -»1
24.11|
as
as
*->-oo x - • - oo.
Complex functions
We have studied the one-point compactification of U 2 and observed that it should always be assumed that this compactification is in use when U2 is identified with the system C of complex numbers. Since C has a rich algebraic structure (i.e. C is a field), it is natural to seek to extend some of this structure to C * just as we extended some of the algebraic structure of U to [— oo, + oo]. There is no point, of course, in seeking to extend the order structure of C since C has no order structure. (See §10.20.) In this context, we make the following definitions: (i) (ii) x • oo = oo • x = oo (iii) x/oo=0 (iv) x/0=oo
(x^O) (x^oo) (x#0).
With these definitions proposition 24.9 remains valid when C * replaces [ — oo, +oo]. Note that oo4-oo, 0*oo, oo/oo and 0/0 are not defined. In particular, we do not define 00 + 00 = 00. 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.
24.12|
Product spaces
The one-point compactification of R" is not the only possible compactification of Un. An obvious and important alternative is to regard Un as sitting inside the space [—00, +oo] n . In the case of IR2 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 oo, + oo)
<
(x*y) o,0)
(0,0)
u2 9
O,
-oo)
\-°°)
(X, - o
Schematic diagram of [— °°, + °°] 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 % = (£i, £2, •••» £n) m [ — oo, + oo]" to be a set of the form
where each set Bt(i = 1, 2,..., n) is an 'open ball' in [ — oo, +00] with centre £-. A set G is then open in [ — 00, + 00]" if and only if each %EG is the centre of an 'open ball' B such that BcG. Note that Un is a topological subspace of [— 00, + 00]". In fact a subset of R" is open relative to [—00, + 00]" if and only if it is open in the usual sense. Next consider a set S in [— 00, + oo]" and a function/: £-•[— 00, + oo]m. We are interested in the case when SczR" and f(S) c Um. If^e[ — 00, +00]" and Tje[ — 00, +oo]m, we have that f(x)-+r\ as x->^ through S if and only if: For each 'open ball' E in [— 00, + oo]"1, with centre n there exists an 'open ball' A in [—00, +00]" with centre \ such that xeAnS=>f(x)eE provided x ^ . The table below gives some sample criteria in the case m—\ and n = 2. In this table /, ax and a2 are to be understood as real numbers. Ve>0 3£>O VxeS,
/ as x-+ through S o, a2)
\x1—a1\<3 and \x2 — a2\<S=>\f(x) — provided x ^ a. V £ >0 3X 35>0 VxeS,
xx >X and \x2 — a2\<5 => | / ( x ) — / | < e through S /(x)-> + oo as x-»(a p - oo) through S \x1 — ax\<5 and x2f(x)>X /(x)-*-oo asx-^(+oo, +00) VY IX VxeS through S xx > X and x2 > X =>f(\)< Y
As we know from §23.1, the first criterion in this table is equivalent to the usual definition of the statement '/(x)-+l as x-»a through S\
Points at infinity 24.13
Example
167
Consider the function/: [R2->!R defined by (xy)2 (*, y) =
\+(xy)2
This does not tend to a limit as (x, y)->oo. We have t h a t / ( x , y)-+j as (x, y)-*oo along the curve xy = \ but/(x, y)-+0 as (x, y)->ao along the line x = 0. However, (xy)2
lim
(1)
(x, y)->(+ oo, + oo)
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, (xy)2
l+(xy) 2
<£.
But (xy)2 r-1 \+(xy)2
(xy)2
and hence the choice X = s 1/4 clearly suffices. It is worth noting that, although (1) exists, (2)
lim
does wot 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 lim
1
=1.
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
24.14f
Exercise
(1) Prove that (i) lim I lim (ii) lim lim Discuss the existence of (iii)
lim (x,y)-»(+», +00)
(2) L e t / : R2->IR2 and consider the following statements (i) lim ( lim / ( x , y)) = l (ii) lim (iii)
lim
lim
f(x,y))=\ f(x, y)=\
(x, y)-*(+oo, +oo)
(iv)
lim f(x, y) = l. (x, 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/: U2^U defined by
Prove that/(x, y)^y as * - • + oo uniformly for Orgy^ 1 - i.e. prove that, for any £>0, there exists an X such that, for any x>X and any y satisfying Ofgy^ 1,
\f(x, y)-y\<e. (The point is that the same X must work for all y satisfying 0 ^ y g 1. Thus X must depend only on £.)
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 % is a function/: N -+y,. We call/(/c) the /cth term of the sequence. The notation
will be used to denote the sequence whose /cth term is yk. Thus denotes the function f:N-+y. defined by f(k) = yk (fcefol).
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 for example are 2,5, 10,17,26,37,... 169
170
Sequences
25.2
Convergence of sequences
Suppose that {yk} is a sequence of points in a metric space y, and that x\ey,. We say that {yk} converges to r\ and write • 00
if and only if V e > 0 3K, k>K =>d{r\,yk)<8. This definition is notning new. If/: N-+y. satisfies f(h) = yfc, the definition simply asserts that/(/c)-m as fc-» + 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 % = [ — oo, + oo], S = N and ^ = + oo. Some of these special results are of sufficient importance to make it worthwhile listing them below.
25.3 Theorem Suppose that <xk> is a sequence of points in R" and that £ is a point in Un. We write xk = (xlk9 x2ki..., xnk) and § = (^, <J2,..., £n). Then x k ^ ^ as /c-»oo if and only if as for each7 = 1, 2,..., n. Proo/ Theorem 22.28.
/c->oo
171
Sequences 25.4
Example Consider the sequence ^cos k sin k
of points in U2. We have that cos k
as fc-+oo
1 k
sin k <—>0 k
and
as fc->oo.
Hence, by theorem 25.3, xfc-•((), 0) as
^^"^ X
r X
\ \ \
/
\
I \ \
\ X
c
5
"\
N
\ i /
y /
3
25.5 Theorem Suppose that <xfe> and are sequences of points in Un and that and
xfc->^ as /c->oo yk—>i| as fc->oo.
Then, for any real numbers a and b, as /c-^oo. Proo/ Theorem 22.30.
25.6 Theorem Suppose that and that
and are sequences of real numbers as k-+oo
and Then;
as
/c->oo.
172
Sequences
(i) xkyk-+£rj as /c-»oo and, provided that x
Theorems 22.31 and 22.32.
This theorem remains true if U is replaced throughout by C.
25.7
Convergence of functions and sequences
25.8 Theorem Let % and y, be metric spaces and suppose t h a t / : where S is a^ set in %. Then the statements and
(1)/(x)->T| as x->£ through S; (2) For each sequence <xfc> of points o
are equivalent. Proof (i) (1)=>(2). Suppose that -(1) holds. Then, given any e>0, there exists a £ > 0 such that for any xeS \{^} (3) Now suppose that <xk> is a sequence of points of S \ { ^ } such that
>^ as
k-KX>. Then there exists a K such that (4)
Sequences
173
Combining (3) and (4), we obtain that and hence /(x k )-n| as k-+co. (ii) not (1) => not (2) (see §2.10). Suppose that it is false that f(x)-+x\ as x->^ through S. If it is false that Ve>0 then it is true that 3 £ >0 V(5>0 3xeS\{£}, (d($, x)<5 and d(r\, This latter expression is true for some fixed e > 0 and all S>0. Given 8 = l//c (fc = 1, 2, 3,...), we can therefore find x k eS\{^} such that d& x k )
and d(i|,/(x k ))^e.
From the first of these inequalities it follows that xk-+^ asfc->ooand, from the second, it follows that/(xk)-Ar| as k-*co. 25.9 Corollary Let % and y, be metric spaces and suppose that/: S->^ where 5 is a set in %. Then / is continuous on S if and only if, for each sequence <xk) of points of S which converges to a point of S,
limf(xk)=fUmx\ 25.10 Example In §9.2 we explained how Archimedes trapped the area of a quarter-circle of radius 1 between two sequences {An} and <£„> of rational numbers. The terms of these sequences satisfy A1=^, B1 = l and (ii)
B;+\=2{A;+\+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 From this it follows that and <£„> ^AnSAn+t^Bn+1g>BnR defined by f(x) = y/x is continuous on [0, oo). We
174
Sequences
have that AnBn-+AB as rc->oo and hence, by corollary 25.9, as
Since An + l=^/(AnBn), is obtained again.
25.11
it follows that A = J(AB) and hence the result A = B
Exercise
(1) Determine whether or not the sequence <xfc> converges in U2 in each of the following cases. 'fc + 1 fc + 2\
(iii) xfc = f-, k) \k )
_
/
1
(iv) xk = (cos /c, sin k)>
(2) Write down definitions for the statements 'xfc-» + oo as /c-*oo' and '**-* — oo as /c->oo' in the case when <xfc> is a sequence of real numbers. Show that /c2-> + oo as /c->oo and that — ^Jk-+ — oo as /c->oo. Discuss the sequences {{ — If} and . (3) Let <xk> be a sequence of points in 1R". Write down a definition for the statement cxk—XX) as fe->oo' based on the one-point compactification of Un. For which of the sequences of question 1 is it true that xfc->oo a? /c->oo? (4) A sequence <xk> of real numbers is said to increase if and only if x x k^ k+i f° r e a c n keM. Prove that an increasing sequence <xk> of real numbers is unbounded above if and only if x k -* + oo as /c->oo. Prove that an increasing sequence of <xk> of real numbers is bounded above if and only if xfc->x as /c->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 x1 = 1 and xk + 1 =f{xk) where/:[0, oo)->lR is defined by J \x) —
TTT •
Prove that 0<xk+1^xk (fc = l, 2, ...) and explain why it follows that <xk> converges. If xk->x as /c-^oo, justify the conclusion that x=f(x) and hence show that
Sequences
175
(6) Prove that the sequence <xk> of rational numbers defined by Xj = 2 and (
2
decreases and is bounded below. Deduce that <xfc> converges and prove that the limit is yjl.
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 % and let ^G Z. Then d(^, S) = 0 if and only if there exists a sequence {x^) of points of S such that xfc->£ as /c-+oo.
Proof (i) Suppose that d(^ S) = 0. By theorem 13.22, given any e>0, there exists an xeS such that d{£,, x)0, it follows that there exists an xkeS such that d(£, x k )
as
/c^oo.
Hence x f c - ^ as /c-»oo. (ii) Suppose that there exists a sequence <xfc> of points of S such that x f e -^ as /C-KX). Then, given any a>0, there exists a K such that k>K=> dfe, xk)<s. In particular, dfe, xK+1)<s. It follows from theorem 13.22 that
25.14 Theorem Let 5 be a set in a metric space X and let ^ e £ . Then ^eS if and only if there exists a sequence <xfc) of points of S such that xfc—•I; as fc-»oo.
176
Sequences
ProofBy theorem 15.5, ^eSd(^9 S) = 0 and so the theorem follows immediately from lemma 25.13.
25.15 Corollary A set S in a metric space X is closed if and only if each convergent sequence of points of 5 converges to a point of S. Proof By exercise 15.3(1), S is closed if and only if S = S.
25.16"}* 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 Rn such that xfc->£ as /c-*oo, explain why (i) <xfc, u>-><£, u>
(ii) llxJMISH as
as
/C->GO
fc-oo.
[Hint: Use corollary 25.9.] Deduce from corollary 25.15 that the sets S = {x: <x, u>^c} and T={x: ||x||^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, if 5 is a connected set in X and /: S-+y, 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 ^e X. Prove that £ is a cluster point of S if and only if there exists a sequence <xk> of distinct points of <S such that xk-+Z> as /c->oo. (Distinct means that x} = xk <=>j = /c.)
25.18
Subsequences
Suppose that g: N-*N is strictly increasing. Then the sequence fOg:N -+y. is said to be a subsequence of the sequence/: N -+y,. If we think of the subsequence illustrated in the diagram as a list, then its first few terms are
177
Sequences 1
1
2
2
3
3
4
4
Z7^ —
5
V
6 7
Note that this is obtained from the original sequence
by omitting some of the terms. If (yky is a sequence and is a strictly increasing sequence of natural numbers, then
denotes a subsequence of . (Iff(k) = y a n d g(l) = kh then For example, if = and = <2/>, then (yki} is <4/2 +1>. This subsequence is obtained from the sequence by deleting terms as indicated below: /, 5, jtf, 17, ^ , 37, jtf, 65, ^ , 101, ...
25.19 Theorem Let be a sequence of points in a metric space y, with the property that y k - m as k-+oo. If is any subsequence of , then as
yjt - ^
/->oo.
Proof Let / : f^J->^ be defined by f(k) = yk and let gf: M-^l^l be defined by g{l) = kt. We have that /(/c)-ni as /c-> + oo through N and that 6f(/)^ + oo as /-> + oo through l\l (see exercise 25.22(1)). It follows from exercise 22.33(3ii) that ii
as
/-^ + oo
through N - i.e. yk -*r\ as /->oo.
25.20
Example We have that l/k->0 as k-*oo. From theorem 25.19 it
178
Sequences
follows without further calculation that 1/2*^0
as /
because is a subsequence of . 25.21 Example Consider the sequence <( —l)fc> of real numbers. If k ( - l ) ^ / a s fc->oo, then (-1) 2 *-Was /c-^oo and ( - l ) 2 k + 1-W as /c->oo. But (-l) 2 / c = l and ( - l ) 2 k + 1 = - l . Hence / = 1 = - 1 . From this contradiction we deduce that <( —l)fc> diverges.
25.22
Exercise
(1) Suppose that is a strictly increasing sequence of natural numbers. Prove by induction that kt ^ / for all le N and deduce that fcj-> + oo
as
/->oo.
(2) Show that, for any mef^J, the sequence of real numbers
is increasing and bounded above. Deduce that the sequence converges to a real number (m). By considering an appropriate subsequence, prove that
(3) Let be a sequence of points in a metric space y.. If r\ is a cluster point of £ = {yk: /CGN}, prove that with the property that as (4) Let be a sequence of points in a metric space % and let £ = |y k : /ce N }.Ifyk-^>T] as h-> oo, prove that £U{TJ} is closed. If has no convergent subsequences, prove that E is closed. (5) Let (xky 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 U2. Prove that <(xk, j/k)> has a convergent subsequence. [Hint. Begin with a convergent subsequence <xk > of <xk> and consider
Sequences 25.23
179
Sequences and compactness
The Bolzano-Weierstrass theorem (19.6) was central to our discussion of compact sets in chapter 19. We therefore begin by giving a version of the Bolzano-Weierstrass theorem for sequences.
25.24 Theorem {Bolzano-Weierstrass theorem) Any bounded sequence <xfc) of points in Un has a convergent subsequence. Proof This may be deduced from theorem 19.6 as follows. If E = {xk: he N} is infinite, then theorem 19.6 asserts that E has a cluster point £. This cluster point must be the limit of a subsequence of <xfe> (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 K in a metric space % 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 X 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 % is compact if and only if each infinite subset E of S has a cluster point ^eS. (i) Let K be compact and let <xfc> be a sequence of points of K. If E = [\k: he N} is finite, then <xk> has a subsequence <xk ) all of whose terms are equal to one of the terms of <xfc>. Hence <xkj>" converges to a point of K. If E is infinite, then E has a cluster point ^eX. By exercise 25.22(3), <xfc> 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.
180
Sequences
25.26| 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.
25.27 Example Suppose that X and % are metric spaces and that S c l If/: S-+y, is continuous on the set S, 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 be any sequence of points in/(S). We seek to show that has a subsequence which converges to a point of f(S). Write yk =/(x k ). Since S is compact, <x k ) has a subsequence which converges to a point §eS. Suppose xk/-+£ as /->oo. By corollary 25.9, /(xki)->/(!;) as /->oo. It follows that the subsequence converges to the point/(£)e/(S).
25.28
Exercise
(1) Let be a sequence of points in a metric space % and let E = {yk: keN}. If yk->il, prove that Eu{r\] is compact. t(2) Let % and y. be metric spaces and let/: K->y, be continuous on the compact set K in %. Use theorem 25.25 to show that/ is uniformly continuous on K (see §23.20). [Hint: The contradictory of the statement that / is uniformly continuous on K begins with ' 3s > 0 V S > 0\ Use this assertion with d = l//c for each ke N.] t(3) Given an example of a bounded sequence in Z00 (see §20.20) which has no convergent subsequence.
26
26.1
OSCILLATION
Divergence
Suppose that % and y, are metric spaces and t h a t / : S-+y. where S is a set in %. Let tje^/ and let ^ be a cluster point of S.
g3 oscillates 'finitely' as x->-0+
181
I g4 oscillates 'infinitely' as x->0 +
182
Oscillation
If/(x)->i| as x->£, through the set S, 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 g1 and g2 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 U, they do converge when regarded as mappings to the space [— oo, + oo]. The functions g3 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 y, are metric spaces and t h a t / : S^y, where S is a set in X. Let x\e% and let £ be a cluster point of S. Suppose it is true that, for some as 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 o f / a s x approaches £ through S.
Observe that there is a distinction between a 'limit' and a 'limit point'. A limit exists only when /(x) converges as x approaches £ through S. But limit points may exist even though /(x) diverges as x approaches \ through S.
Oscillation 26.3
183
Examples (i) Consider the function/: U->U defined by 1
(x^O)
This function diverges as x approaches 0 but /(x)-»l as x-»0 + and /(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/: [ R 2 - ^ 1 of example 22.11. We have seen that as
along the line y = ax. Examining each aeU in turn, we find that every point of [0, 1) is a limit point of /(x, y) as (x, y) approaches (0, 0). Also Ax,y)->1
as
(XJH0,0)
along the line x = 0. Hence 1 is also a limit point. Since — l:g/(x, y ) ^ l for all (x,'y)eU2, the function can have no other limit points. It follows that the set of limit points of/(x, y) as (x, y)->(0, 0) is equal to [—1, 1].
26.4 Exercise (1) Suppose that % and y are metric spaces and t h a t / : S^y where 5 is a set in % for which ^ is a cluster point. Prove that 'key, is a limit point o f / a s x approaches ^ through the set S if and only if there exists a sequence <xfc) of points of S such that x f c 7^ (/c = 1, 2,...) and x k -^^ as /c->oo for which /(x k )-»i,
as
/C-KX).
[Hint: Use theorem 25.8]. (2) Find all real limit points of the following sequences of real numbers: (i) (iv) <(-!)*>
(ii) <(-l) fe /c> (V)
(iii) (Vi)
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+ / : (0, oo)-*R:
of the following functions
(i) fix) = sin — x
(ii) fix) = x sin — x
(iii) /(x) = - sin -
(iv) /(x) = 1 - sin -
XX
X
184
Oscillation
(v) f(x) =
sin X
X
(vi) /(x) = - j — sin - . X
X
X
X
These functions may also be regarded as taking values in [ — oo, -f oo]. Find all limit points from [—00, + 00] of the given functions as x->0 + . (4) Find all real limit points as (x, y) approaches (0, 0) of the function / of exercise 22.24(3). (5) Give examples of functions/: (0, oo)->[R for which the set L of real limit points as x->0 + is as indicated below: (i) L = 0 (iv) L = [ 0 , 00) (vii) L = {0} but
(ii) L = U (iii) L = [ 0 , 1] ( v ) L = { - l , 1} (vi) L=N f(x)±0 as x->0 + .
Show that in the case of each of your examples except (iii) and (v), + 00 or — 00 is also a limit point. (6) Suppose t h a t / : (0, oo)->IR is continuous on (0, 00). Let L be the set of real limit points as x->0 + . Prove that L is connected.
26.5f
Oscillating functions
26.61 Theorem Suppose that % and y, are metric spaces and let/: S-*y. where S is a set in X for which \ is a cluster point. Then the set L of limit points of/ as x approaches \ through S is given by L = O f(SnAk) where the set Ak is obtained by removing £ from the open ball with centre £ and radius l//c.
Proof (i) Suppose that AeL. Then there exists a sequence <xfc> of points of 5 \{£} satisfying x k ->^ as /c->oo such that/(x k )-*A as /c->oo (exercise 26.4(1)). Given any J, there exists a X such that xkeAjnS => f(xk)ef(AjnS). Hence Xef(AjnS). It follows that f(AknS). (ii) Suppose that Aef(AknS) for each /ce^J. Then for each fceN, there exists a sequence <xM> of points of AknS such that f{xkj)->'k as /->oo. For each keN, choose lk so that d(X,f(xktlJ£ as /c->oo and/(xMfc)->X, as fc->oo.
185
Oscillation Thus XeL and so f(AknS)czL.
26.7| Theorem Let ft and ^ be metric spaces and suppose that/: S^y. where 5 is a set in % for which £ is a cluster point. Then the set L of limit points of/ as x approaches £ through S is closed. Proof By the previous theorem, L is the intersection of a collection of closed sets. Thus L is closed by theorem 14.14. 26.8| Theorem In addition to the assumptions of the previous theorem, suppose that/(S) lies in a compact subset K of y. Then the set L of limit points of/ 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 that/(SnAfc) is non-empty for each keN. Thus (f(SnAk)} is a nested sequence of non-empty, closed subsets of a compact set K. It follows from the Cantor intersection theorem (19.9) that the sequence has a non-empty intersection i.e. 26.9f
Theorem With the assumptions of the previous theorem, /(x)-*t| as
through S if and only if L= {r\}:
x-^
186
Oscillation
Proof If /(X)-»TI as x->£ through S, then it is trivial to prove that L={TI}. If it is false that / ( x ) - n j as x->E, through S, then there exists a sequence <xfc> of points of S \ { £ } such that xk->£ as /c->oo but/(x k ) -/m as /c->oo. We deduce the existence of an so>0 and a subsequence <xki> for which d(r\, f(\k())^e0 But has a convergent subsequence by theorem 25.25. Hence L l )
The assumption that f(S) be a subset of a compact set Kay which is made in theorems 26.8 and 26.9 is not so restrictive as it may seem at first sight. For example, if y. = U, one can always begin by replacing U by the compact space [— oo, + oo].
26.10 that
Example
Consider the sequence <( — l)kk — k} of real numbers. We have (-l) 2 k 2/e-2/e-»0 2
as
/c->oo --oo
as
k-+ao.
The sequence therefore has only one real 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 U. 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 t h a t / : S->R 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^0. If L consists of a single point, then either/converges as x approaches § through S or else /diverges to + oo or diverges to - oo as x approaches. E, through S. If L consists of more than one point, we say that / oscillates as x approaches % through S. If/oscillates as x approaches £ through S and L is a compact subset of R, it is customary to say that / 'oscillates finitely'. (Note that, since L is closed, it follows that L is a compact subset of U if and only if L is a subset of R which is bounded in U.) Otherwise/is said to 'oscillate infinitely'.
26.1 It
Lim sup and lim inf
Let X be a metric space and suppose t h a t / : S-+M where S is a set in X for which £ is a cluster point. The set L of limit points from [ — oo, + o o ] a s 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 X (corollary 14.11). We call A the limit superior (or lim sup) o f / a s x approaches % through S. We call k the limit inferior (or lim inf). Sometimes A is referred to as the upper limit and X as the lower limit.
Oscillation
187
The function g3 illustrated in §26.1 is an example of a function for which both X and A are finite (i.e. X and A are elements of U). The function g 4 of §26.1 is an example of a function for which X is finite but A = + oo.
26.12| Theorem Let % be a metric space and suppose t h a t / : S->R where S is a set in % for which ^ is a cluster point. Then the function / oscillates as x approaches £, through S if and only if X # A. Moreover, given any ^G[—oo, +oo], f(x)-+rj
as
x-+£
through iS if and only if A = rj = A. Proof This is an immediate consequence of theorem 26.9.
26.13| Proposition With the assumptions of the preceding theorem, A is the limit superior o f / a s x approaches \ through S if and only if (i) V/>A 3(5>OVxeS, o/(x)0 3xeS,
0«/(§,x)<<5
and
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 L is closed, this means that AeL. Similar criteria, of course, hold for the limit inferior X. The notation A = lim sup / (x);
X = lim inf / (x)
x-$
x-$
is often used. The reason for the choice of this notation will be evident from the statement of the following result.
26.14| Proposition Let % be a metric space and suppose t h a t / : 5->R where S is a set in % for which £ is a cluster point. Let A8 denote the set obtained by removing \ from the open ball with centre \ and radius b > 0. Then (i) A = lim < sup /(x) <5->0+
lxeSn/4,
(ii) X= lim < inf
/(x)j>.
188
Oscillation
The proof of (i) above depends on the observation that the function F: (0, oo)->[— oo, + oo] defined by F(d)= sup /(x) increases on (0, oo). This guarantees the existence of the limit as (5->0 + . The identification of the limit with A is then achieved with the help of proposition 26.13. 26.15| 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 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 of , ak -+rj as /—• oo => rj ^ A and
(ii) There exists a subsequence of such that ak —>A as /-•oo.
What is the corresponding result for the lim inf of a sequence? (3) Suppose that is any sequence of real numbers. Prove that A e [ — oo, + oo] is the lim sup of the sequence if and only if and
(i) VL>A, {k:ak^L) is finite (ii) V / < A, {k: ak ^ /} 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 + bk) ^ (lim sup ak I + (lim sup bk I fc->oo
\
fc-oo
/
\
/e-+oo
/
whenever the right-hand side makes sense. Give an example for which the lefthand side is — oo and the right-hand side is H-oo. Show that the two sides are equal whenever 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) ^ (lim inf ak I + ( lim sup bk j ^ lim sup (ak + bk). k~*oo
\
/c-»oo
/
\
k-+ao
J
k-*oo
Show also that lim sup (— ak) = — ( lim inf ak I. Deduce that, for any bounded sequence of real numbers, lim inf (ck — c f e + 1 )^0^1im sup (ck — /c-+oo
k-+oo
Oscillation
189
(6) Suppose that is any sequence of positive real numbers. If there exists a K such that ak + 1^rak for any /c>K, prove that there exists an H such that ak < Hrk for all keN. Deduce that lim sup allk-^\\m sup ——.
27
27.1
COMPLETENESS
Cauchy sequences
A Cauchy sequence <xfc> of points in a metric space % 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 k>K
x
>xkeE.
i X
2
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:
(k>K and 1>K) => d(xk, \t)<e.
27.2
Completeness
A complete metric space % 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
Completeness
191
27.3")" Theorem A metric space % 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 S and let <xfc> 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 <xfc> 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 <xfc> be a Cauchy sequence. Then its range is totally bounded and hence is either finite or else possesses a cluster point. In either case <xfc> 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.
27.4f Theorem A set K in a complete metric space X is compact if and only if it is closed and totally bounded. Proof See §20.16.
27.5f Theorem Let be a nested sequence of non-empty closed balls in a complete metric space X whose radii tend to zero. Then 00
P i Bk k=l
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 x k e£ k . Given any e>0, let BK be the first ball with radius less than e. Then
k>K=>xkeBK and hence <xk> is a Cauchy sequence. Since % 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 £ is a complete metric space, this theorem takes a more satisfactory form.
192
Completeness
27.6f Theorem Let X, % and £ be metric spaces and suppose that £ is complete. Let £e X and Tje^. Suppose that A is a set in X for which £ is a cluster point and that B is a set in %. Let/: S \(£, f|)->£ where S = A x £. Let l:B-+Z and letm:,4->£. Suppose that (i) /(x, y)->/(y) as x->£ through the set A uniformly for yeB, and (ii) /(x, y)->m(x) as y—m through the set B pointwise for xeA. Then there exists a £e£ such that (iii) /(y)-»£ as y->n through the set B, and (iv) m(x)-+£ as x->^ through the set A.
Proof Let s>0 be given. From (i) we have that there exists a, d>0 such that for each xeA and each yeB,
From (ii) we have that, for each xeB there exists a Ax such that 0«i(y, 11) < AX => d(m(x)J(x, y))<e/4. It follows that if xeA satisfies 0)<S and 0 is any sequence of points of A \{£} such that xfc-+£ as /c-*oo, then <m(xfc)> is a Cauchy sequence in X. Item (iv) then follows from
Completeness
193
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, x k + 1 ) g 2 ~ k (JceN) is a Cauchy sequence. Prove that any Cauchy sequence <xfc> in a metric space X has a subsequence <xki> which satisfies.
d(xki,xkiJS2-1
(ZeN).
(3) Prove that any Cauchy sequence in a metric space X which has a convergent subsequence is itself convergent. t(4) Let y. be a set in a metric space X. Then y may be regarded as a metric subspace of X. Prove the following: (i) If X is complete, then y is complete if and only if y is closed in X. (ii) If y is compact in X, then y 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 a < 1 such that
for each xe X and each yeX. Let/: X-+X be a contraction. Prove that: (i) / is continuous on X. (ii) Any sequence <xfc> of points of X which satisfies xk + j =/(x fc ) is a Cauchy sequence. (iii) If X is complete, t h e n / has a fixed point - i.e. for some £e5C,/(£) = £. (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 Un9 it follows from theorem 27.3 that R" is complete. The proof of the next theorem establishes this result in a more direct fashion.
27.9
Theorem The metric space U" is complete.
194
Completeness
Proof Let <xfc> be a Cauchy sequence in Un. By exercise 27.7(1), <xk> is bounded and hence, by theorem 25.24, <xfc> has a convergent subsequence. The conclusion therefore follows from exercise 27.7(3).
Let S be a non-empty set and let y be a metric space. We shall denote the set of* all bounded functions/: S-^y by 53 (S, %\ The set $ (S, %) becomes a metric space if we define the distance between two 'points' / and goffB (S, y) by u(fg)= sup d(f(x\g(x)). xeS
We call u the uniform metric on & (S, %).
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 /°° = 5
27.10|
Theorem If y is complete, then so is the metric space $ (S, %).
Proof Let be a Cauchy sequence in 3 (S, ^). Then, for each xeS, is a Cauchy sequence in y,. Since ^ is complete, it follows that, for each xeS, converges - i.e. there exists a function/: S-+y, such that/fc(x)->/(x) as k-^oo. In the language of §23.11 and §23.12, we have shown that converges pointwise for xeS. What we need to show is that converges in the space $ {S, %) - i.e. converges uniformly for xeS. Given any e>0, there exists a K such that for any k>K and any
It follows that, for each xeS, But, for each xeS,/fc(x)->/(x) as /c->oo. Thus, for any rf(/(x),/(x))^/2. (See exercise 16.16(4).) We deduce that, for any 1>K u(fbf)^s/2<s. It follows that/->/ as /->oo in the space S (S, ^). 27.1 If Theorem Suppose that y, is a complete metric space. Then the set C of all continuous functions in $ (5, ^) is closed in 3? (S, ^). Proof Suppose that is a sequence of functions in C such that/ k ->/ as /c-»oo. To prove that C is closed, we need to show that feC (corollary 25.15). Let £>0 be given. Then there exists a K such that k>K=>u(fJk)<s/3.
Completeness
195
Let ^e 5 and let J > K. Since/; is continuous on S, there exists a S > 0 such that for each xeS
It follows that, if d(&, x) < S, then
fMUx)) + d(Ux)J(x)) and so / is continuous on S - i.e. fe C. We shall use the notation C (5, y) to denote the space of all continuous functions / : S^-y.. Such functions need not in general be bounded. However, if S is a compact set in a metric space %, then a continuous function / : S-> ^/ is bounded and, moreover,
xeS
To obtain this result we apply theorem 19.14 to the continuous function F:
defined by F = do{f,g). 27.12"}" Theorem Suppose that ^ is a complete metric space and S is compact. Then d (S, %) is complete. Proof This follows from theorem 27.11 and exercise 27.7(4i).
27.13|
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 U). 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 U (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 yt=j
Examples
Consider the sequences <xfc> and defined by
x1=2,
and
1/
2\
1/
1
These are both sequences of rational numbers. The sequence <xk> decreases and converges (in the space U) to ^2. (See exercise 25.11(6).) Similarly, the sequence increases and converges (in the space U) to ^J2.
196
Completeness
1—v
V2
(i) Since <xfc> converges in the space R, it is a Cauchy sequence in IR (exercise 27.7(1)). It follows that <xfc> is a Cauchy sequence in the space Q. But <xfc> does not converge if it is regarded as a sequence of points in Q. There is no rational number £ such that xk—•£ as k—>oo. In fact, xk-*^/2 as k-+oo and yjl is irrational. (ii) Let Ik = {r: reQ and xk^r^yk}. Then Ik = \_xk, > J n Q and hence is closed in the metric space Q (corollary 21.12). It follows that is a nested sequence of closed boxes in Q (§ 19.2) but 00
n ik=o - i.e. the Chinese box theorem fails in the space Q. (iii) Consider the set S = {r: reQ and O ^ r ^ ^ / 2 } . This set is closed in the metric space Q. It is also totally bounded. But it is not compact in O. For example, the continuous function/: Q-+R defined by/(x) = x does not achieve a maximum on S.
27.15f
Exercise
(1) Find the uniform distance between the functions/: R -•R and g: R -•R defined by l+x2'
"
w
l+x2
where these functions are regarded as points in the metric space 3 (IR, IR). (2) Let % be the set of all continuous functions/: [«,fr]-*(R.Let the metric in Z be defined by
/, g)= f
d(f,g)=\ \f(x)-g(x)\dx. Ja
Prove that X is not complete. (3) Let S be a set in a metric space % and let £e Z. Let ^ be a complete metric space and suppose that L^(S, y) is the subset of 9$ (S, y) consisting of all bounded functions / : S->y. such that there exists an x\ey. for which /(x)->i] as x-+£ through S. Prove that L^(5, y) is closed in SB (S, y). [Hmt: 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/: R xR-»(0, 1) be defined by/(x, y) = xy. Show that/(x, y)-*y as x->l through (0, 1) uniformly for ye(0, 1) and that/(x, y)-+x as y-+l through (0, 1) pointwise for xe (0, 1). Explain the relevance of this result to the assumption in theorem 27.6 that £ is complete. (5) Let Pn: [0, 1]->R be defined by P()
l
- + - + ...+-.
Completeness
197
Prove that converges in the space S([0, 1], U). (6) Let 9 denote the metric space of all real polynomials where the metric is defined by
u(P, Q)=max \P(x)-Q(x)\. 0£x£l
Prove that 9 is not complete. [Hint. See the previous question.]
27.16|
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 snace? 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 %*. 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 satisfy d(\k, yk)->0
as
/c-^oo
(1)
then we shall want both sequences to converge to the same limit. We therefore begin by introducing an equivalence relation ~ on the set 5 of Cauchy sequences of points of X by writing
if and only if (1) holds. The set ft* is defined to be the set of all equivalence classes of S defined by this equivalence relation. We shall use the notation [<xfc>] to denote the equivalence class containing the Cauchy sequence <xfc>. The original space X is fitted inside the new space X* by identifying an element x of X with the element [<x>] of Z*. 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: £*->R'by <*([<x*>], [])=lim
d(xk,yk).
k-+oo
Note that the limit on the right-hand side exists because and are Cauchy sequences in X. (See exercise 27.20(1).) But every Cauchy sequence of real numbers converges because U is complete (theorem 27.9).
198 27.17|
Completeness 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 <Xy> is a sequence in X* satisfying d(Xp Xj+1)<2-«
+1
\
(2)
then there exists an Xe X* such that X^X as7-+00. 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
where <xk0)> is a sequence in X which satisfies
and hence d(xku\ x ^ ) < 2 - ( / c + 1)
(3)
provided that l^k. Now (2) asserts that lim d(xku\ x ^ + We may therefore deduce the existence of a strictly increasing sequence of natural numbers such that, for any l^kp d{xt{j\ x ^ + 1 ) ) < 2 - ( j + 1).
(4)
From (3) we also have that d(xkjij+1\
xkj+u+1))<2-ikJ+1)^2-u+1\
(5)
We are now in a position to define XeX* by
For this to be an acceptable definition, it is necessary that <x k 0) > is a Cauchy sequence in X. This fact follows from exercise 27.7(2) because
by (4) and (5). It remains to show that X^X
as /->oo. From (4) we have that, for each 7 ^ / ,
Completeness
199
But Xj = [<xk.(Z)>] and thus it follows that d(Xl9 X)<>2~~l-*0
27.18|
as
/->oo.
Completeness and the continuum axiom
After the previous discussion it is natural to consider first the completion Q * of the rational number system. Is Q * the same as the real number system R? Before seeking to answer this question, we should remind ourselves that Q and U are not just metric spaces. Both are also ordered fields. The question is therefore only meaningful if we can extend the algebraic structure of Q 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 (ii)
(iii) l<xkyi>L(ykyi *>3K(k>K => xk>yk). With these definitions Q* becomes an ordered field and ], C< yk>]) = l To prove this is a somewhat tiresome but essentially trivial task and so we shall omit the details. Recall that Q is the 'smallest' ordered field. Thus O* is the 'smallest' complete ordered field. We know from theorem 27.9 that U is complete and thus K*c:[R. (These remarks, of course, take for granted that structures isomorphic to Q, Q * or U are to be identified with Q, Q* or U respectively - see §9.21.) The next theorem shows that the system Q * and the system U are the same - i.e. the real number system is the completion of the rational number system. Recall that U 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.
27.19|
Theorem
The system Q* satisfies the continuum axiom.
Proof Let S be a non-empty set in Q* which is bounded above. Let axeS and let b1 be an upper bound of S. We construct an increasing sequence of points of S and a decreasing sequence (bk} of upper bounds of S such that
The construction is inductive. If ck=^{ak + bk) is an upper bound of 5, we take bk + 1=ck and ak + 1=ak. If ck is not an upper bound we take bk + 1=bk and choose ak + i G S so that ak +1 ^ ck. The sequences <%> and
200
Completeness
hence converge. They have a common limit £, which is easily shown to be the smallest upper bound of the set S.
points of S
27.20|
Exercise
(1) Let <xfc> and be Cauchy sequences in a metric space X. Prove that is a Cauchy sequence of real numbers and hence converges. (2) Check that Q * 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 U is Archimedean (theorem 9.17) and hence so are Q and Q*. Prove that an Archimedean ordered field X is complete if and only if it satisfies the continuum axiom (i.e. %=M). (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
28
28.1
SERIES
Convergence of series
Suppose that is a sequence of points in a normed vector space Z. (See §13.17.) The Kth partial sum of the sequence is defined by
If the limit of the sequence <sK> exists, then we say that the series 00
converges and we write 00
s= I ak if and only if sK->s as X->oo. If the sequence of partial sums does not converge, we say that the series diverges.
28.2
Example If |x| < 1, then
To prove this result we observe that K
k=0
=
l-xK+1 1-x
•
1 1-x
as
K-»oo.
If |x| ^ 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 % is a complete normed vector space and that
(i)
£ K is a convergent series of non-negative real numbers. If Hak||^fck
(fceN),
then the series
converges. Proof The partial sums of the sequence (1) are a Cauchy sequence (exercise 27.7(1)). Given any e>0, it follows that there exists an N such that, for any K>J>N, K
where tK denotes the Kth partial sum of the series (1). It follows that, for
any K>J>N, K
bk<s k=J+\
k=J+l
and so <sk> is a Cauchy sequence. Since % 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 denote a sequence of positive real numbers. Then the series
k=l
Series
203
converges if (i)
I
i
fc->oo
^ bk
and diverges if (ii) liminf ^ ^ > 1 . k-+oo
Ok
Proof If (i) holds, there exists a p < 1 and an H such that bk < Hpk (fceN). The convergence of the series then follows from the comparison test. If (ii) holds, then bfe-/»0 as /c-»oo and so the series diverges.
28.5 Proposition (Root test) Let denote a sequence of nonnegative real numbers. Then the series
k=l
converges if (i) limsup
bk1/k<\
and diverges if (ii) lim sup
bk1/k>l.
fc^oo
Proof The same remarks apply as in the proof of proposition 28.4.
28.6
Example Consider the case bk = k~a. Then l i m - 4 ±1 = l; k-+cc
Dk
lim bki/k = l. fc^oo
It follows that neither the ratio test nor the root test are helpful in determining the convergence or divergence of the series
I1i—i La k= 1 K
As it happens, the series converges for a > l and diverges for a ^ l . (See exercise 28.10(3).)
204
Series
28.7
Absolute convergence A series GO
fc=l
converges absolutely if and only if
t Ikll fc=l
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 withfcjt= ||ak|| 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 is a sequence of real (or complex) numbers and £ is a real (or complex) number, then the series
k=l
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 \ak\1/k = fc^oo
Then
If 0 < p < + o o , it follows that the power series converges when \z — £| < 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 converges when \z — £ | < # and diverges when |z — £|>K. It follows that the set S of values of z for which the power series converges satisfies
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 series.
////////////A 7777777777777
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 \z — (|<JR). If it converges at a boundary point of 5, however, there is no reason why the convergence should necessarily be absolute.
28.9
Example The power series oo
I (has radius of convergence 1. We have that lim
1/fc
(-1)*
= 1.
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 is a sequence of points in a normed vector space % and that the series 00
k= 1
converges. Prove that ak—•() as fc->oo. Give an example of a sequence
206
Series (bky of real numbers for which bk^0
as /c->oo but the series
k=l
diverges. (2) Suppose that {ak} is a sequence of non-negative real numbers. Prove that the infinite series
exists if and only if the sequence of partial sums is bounded above. (3) Prove that
K
k=l
diverges by demonstrating that
where sK denotes the Kth partial sum. Prove that, if <x> 1, then "
1
1
1
converges by demonstrating that
where tK denotes the Kth partial sum. (4) If is any decreasing sequence of non-negative real numbers, prove that K+L
aK-aK+1S\
I (-l) fc ak \^aK. k=K
Show that, if also afc->0 as /c-»oo, then
converges. [Hint. Show that the sequence of partial sums is Cauchy.] Deduce that the series - (-If'1
111
Series
207
converges and that j ^ s r g l . Prove also that 3
1 1 1 1 1 1 1 1
S
2~
3 2
5 7 4
9
11 6
[/fmt: If rK denotes the Xth partial sum of the second series, then 1 1
1
1
1
(5) Suppose that
converges if and only if t = sup is finite and that, in this case, s = t. (6) Suppose that A and B are non-empty sets and that/: A x B-+M. Prove that s u p / s u p / ( a , b))= as A \ beB
sup
I
f(a, b).
(a,b)eAxB
Let (ajky be a 'double sequence' of non-negative real numbers and let 3 be the collection of all finite subsets of N. Prove that o o / o o
exists if and only if t= sup Ge5 x 5
Z
a
jk
(j,/c)eG
is finite and that, in this case, s = t. What can be said of 00
/
«= Z ( (
(See exercise 23.19(6).)
GO
208
Series
28.11|
Uniform convergence of series Under what circumstances is it true that 00
00
lim X fk(x) = X lim/t(x)? *-*$
fc=l
fc=l
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->y, where y,is a. normed vector space. We say that I /k(x)
(1)
fc=i
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. 28.12*1* Theorem Suppose that % is a metric space and that y. is a complete normed vector space. Let SaX and suppose .that, for each keN, the functions fk:S-^% satisfy /*(*)-•% as x->£ through the set S and that
k=l
converges uniformly for xeS. Then 00
00
k=\
k=l
£ AW-* E % as x -^ through the set S. Proof This is an immediate consequence of theorem 27.6. 28.13f Corollary Suppose that % is a metric space and that y. is a complete normed vector space. Let Scz% and suppose that, for each keN, the function fk: S-+y, is continuous on S. Then the function/: S-+y. defined by
/(x)= I Mx) k=l
is continuous on S provided the series converges uniformly for xeS.
Series 28.14
Example
209 The series
converges pointwise for each xe[0, 00). For x > 0 , / ( x ) may be evaluated using the formula for a geometric progression. We obtain that
l-^r1
(x>0)
Observe that/(x)->l as x->0 + but/(0) = 0. Thus / is not continuous on [0, oo) and hence the series does not converge uniformly for xe[0, OO).
28.15f
Series in function spaces
Recall from §27.8 that $ (5, %) denotes the space of all bounded functions F : S-+y with metric u defined by u(F, G) = supd(F(x), G(x)). xeS
If y, is a normed vector space, then we may regard $ (S, y) as a normed vector space by defining vector addition and scalar multiplication in $ (S, y) by (F + G)(x) = F(x) + G(x) (aF)(x) = aF(x). The norm on fS (S, y) is, of course, defined by ||F|| = sup||F(x)|| xeS
so that u(F, G) = \\F-G\\. We proved in theorem 27.10 that, if y is complete, then so is $ (S, y). In particular, if y is a complete normed vector space, then so is $ (S, y). Suppose that y is a normed vector space and that, for each keN, fk:S-+y is bounded - i.e. is a sequence of 'points' in the normed vector space ® (S, y). Then, as explained in §23.12, to say that the series
converges uniformly for xeS is the same as saying that the series OO
converges in the space 3$ (S, y). 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. 28.16| Theorem (Weierstrass 'M test) Suppose that, for each keN,fk: S-+y., where y, is a complete normed vector space. Suppose also that is a sequence of non-negative real numbers for which 00
IK k=l
converges and that, for each keN and each xeS,
IL/iMII^*.
(i)
Then the series
I k=l
converges uniformly for xeS. Proof Condition (1) implies that \\fk\\£bk Hence the series 00
I/* converges in the space $ (S, %) by the comparison test. 28.17 Example Consider the function fk: [0, oo)-»R defined by fk(x) = x2e~kx. This achieves a maximum at x = 2k~1. Hence, for each xe[0, oo) and each keN,
It follows from the Weierstrass 'M test' that the series
k=\
converges uniformly for xe[0, oo). From corollary 28.13 we may conclude that / is continuous on [0, oo). In fact, for each xe[0, oo), we may evaluate f(x) as in example 28.14 to obtain
l-rr1
(x>0) (x = 0).
Observe that /(x)-*/(0) as x->0 +.
Series 28.18J
211
Exercise
(1) Prove that the series
converges point wise for x^O. Show that / is not continuous on [0, oo) and deduce that the series does not converge uniformly for x^O. Does the series converge uniformly for (a) O g x ^ l , (b) l ^ x ^ 2 ? (2) Prove that the series
/<*)=! *i£ converges uniformly for xeR. Explain w h y / i s continuous on R. (3) Suppose that <wk> and are sequences of real numbers and let K
U = y
uk
(K>J).
Prove that K k=J+l
K-l k=J+l
(Abel's 'partial summation' formula). (4) Suppose that (ak) is a decreasing sequence of positive real numbers which tends to zero and that 4>k: S-+M satisfies K
£ (t>k(x) ^H lc=l
for all KeN and all XES. Prove that
converges uniformly for xeS. [Hint: Use question 3.] (5) Establish the identity . ^ . cos \x-cos{n+j)x sin x + sin 2x + ... + sin nx= :—'-. 2 sin ^x and deduce that the series
/w= Z k=i
sin kx
kK
converges uniformly for xe[(5, 2n — S'] provided that <5>0. [Hint: Use question 4.] Deduce that / is continuous on (0, n).
212
Series
(6) Suppose that the real power series
f(x)= £ akxk k=O
has interval of convergence / and let J be any compact subinterval of /. Prove that the series converges uniformly for xel. [Hint: To show that the series converges uniformly on [0, X]ciJ, take uk = akXk and vk = (xX~1)k in question 3.] Deduce that the function F: [ — 1, 1)—•IR defined by oo
k
is continuous on [ — 1, 1).
28.19|
Continuous operators
Suppose that % and y. are metric spaces and that T: %-*% is continuous on %. Then we know from corollary 25.9 that r / l i m XfcJ = lim T{Xk) \k-*co
I
k-*ao
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 % is a function space. We then usually call the continuous function T: Z-^y, a. continuous operator. We shall illustrate the usefulness of the result by examining the case in which % is the space of all continuous functions/: [a, b]-*R (usually denoted by C[a, bj) and T is the 'integration operator' on C[a, b~\.
28.20| Theorem Let C[a, fr] denote the space of all continuous functions / : [a, b~\-+U and define the operator T: C[a, &]-*R by
Then T is continuous on C[a, b~\. Proof We have that
\T(f)-T(g)\=
- | | g(t)dt
213
Series
S\ \f(t)-g(t)\dt £(b-a)
max \f(t)-g(t)\
= (b-a)\\f-g\\. It follows that T(/)->T(g) as f-*g and hence that T is a continuous operator on Cla, b].
28.21"]" Corollary suppose that
For each /ceN, l e t / t : [a, £>]->IR be continuous on [a, b~\ and
(1) converges uniformly for xe[a, 6]. Then (2) Proof To say that (1) converges uniformly for xe[a, i>] is the same as saying that
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 Fk: [0, 1]->R illustrated below.
We have that, for each xe[0, 1], Ffe(x)-»0 as /c->oo - i.e. converges pointwise to
214
Series
the zero function. But
f
Fk(x)dx = l
It follows that the series whose sequence of partial sums is (Fk(x)} fails to satisfy (2) of corollary 28.21.
28.23f Theorem Suppose that X and y, are metric spaces and that T: X-+y, is continuous and bijective. Then 00
\
=i
/
00
) k=i
provided that the right-hand side exists. Proof Since T is continuous
I yt. k=l
Hence 00
X r- 1 (yj=r- 1 k=l
We illustrate this result by taking X = C[a, b] and % = C£a, b] where the latter notation denotes the set of all F: [a, b]-+U which can be written in the form
F(x) =
f{t)dt
{a^x^b)
where/is a function in C[a, b]. The role of the operator T:X-*y, will be assumed by the 'integration operator' /: C[a, b] -> C\[a, b] defined by / ( / ) = F. The fundamental theorem of calculus asserts that the 'differentiation operator' D:C£a, b]->C[a, b~\ defined by D ( F ) = / where
f(x) = F(x)
(a<x
is inverse to the integration operator - i.e.
It is easily shown that / is continuous on C[a, b~] and we therefore obtain the following theorem in which C'[a, b~] denotes the set of all/:[tf, b]->M which have a continuous derivative on [a, b\.
215
Series 28.24f
Theorem
Suppose that, for each fceN, gkeC[_a, b~] and that
I gk'ix) k=l
converges uniformly for xe[a, b~] and
converges for some {e [a, b~\. Then, for each xe(a, b) (3) P r o o / Apply theorem 28.23 with yfc=/fc, where/ k : [a, fr]->R is defined by
28.25
Example
Consider the function Gk: [0,1]->R illustrated below.
We have that GkeC[a, b] and that converges uniformly to the zero function for xe[0, 1]. But
It follows that the series whose sequence of partial sums is fails to satisfy (3) of theorem 28.24.
28.26t
Applications to power series
A trivial consequence of the Weierstrass 'M test' (28.16) is that a real (or complex) power series
converges uniformly for \z — Q ^ 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.
28.27| Proposition (Abel's theorem) The sum of any real power series is continuous on its interval of convergence.
28.28*f" Proposition real power series
If c and d are any points in the interval of convergence of a
/(*)= £ ak(x-Z)\ then fd
00
fd
f(x)dx= X flk (x-tfdx.
28.29| Proposition real power series
If y is any interior point of the interval of convergence of a oo
f(x)= £ ak(x-0\ k=0
then/is differentiate at y and 00
f\y)= Z
ka
k(y-Ok~l-
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 Cl but it is in general false that/(z)->/(>>) as z->y along 'tangential' curves like C 2 .
Series 28.30
217
Example If |x| < 1, we have that 00
1
k=o
1+x
By proposition 28.28, if |y| < 1, then fy
fy / °°
A
\
=Jo T^=\ ( I (-l)VW 1 + x Jo \k = o J i)fc+1
00
= Z (-D
+1
The latter power series has interval of convergence ( — 1, 1] and, by Abel's theorem (28.27), its sum/(y) is therefore continuous on ( — 1, 1]. But/(^) = log(l +y) for ye(-l, 1). It follows that/(I) = log 2 - i.e. log 2 = 1 1 1 * 2 3 4 5
28.31|
...
Exercise
(1) Prove that, for |x| < 1,
(-i)kx2k
1+x 2
Hence obtain a power series expansion for arctan x which is valid for - 1 <xS 1. Deduce that n
1 1 1 1 1
+
(2) Obtain the general binomial theorem by differentiating
(3) The power series
(z)=% bk(z-ok k=0
both converge for \z — Q
29f
29.1|
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 Xy x + y = y + x. The associative law asserts that for any xe X, ye X and ze X,
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, = 6 + 4=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 on a set A is simply a bijective function $: A-+A. We shall be concerned with permutations (p o n the set N of natural numbers. Given such a permutation 0 : N-+N and a sequence , we call the 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 <\> on N,
provided that one side or the other exists. Unfortunately, this result is false in
218
Infinite sums
219
general. Exercise 28.10(4) provides a counterexample. If we write
then
fc= 1
where : N -+ N is the permutation defined by
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 + 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
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
29.2|
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 Z is a normed vector space and that a: I-+Z is a function. If se Z. there is, of course, no difficulty in defining (1) iel
in the case when the 'index set' / is finite. In the case when / is infinite, we shall define (1) to mean that
For any e>0, there exists a. finite set Fez I such that, for each finite set G,
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 /, it is clear that infinite sums must satisfy the commutative law. The next theorem puts the proposition in formal terms.
29.3| Theorem Suppose that Z is a normed vector space and that a: I- Z. If 0 is a permutation on /, then
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 and (j> by (f)'1.) Let £ > 0 be given. Then there exists a finite set F a I such that, for any finite set G,
iel
Hence, for any finite set H,
Infinite sums
221 Hl£a(O-
iel
Z fl(OII
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.
29.4|
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 28.10(1)).
29.5| Theorem suppose that
Suppose that % is a normed vector space. Let a: 1->Z and
Z isl
exists. Then, given any £>0, there exists a finite set Fez I such that, for e a c h ; e / ,
JtF=>\\aU)\\<e. Proof Let £ > 0 be given. Then E / 2 > 0 and hence there exists a finite set Fez I such that for any finite set G
LQtjeI\F. Then
)II = II Z a ( 0 -
2 >
ieFv{j}
ieF
ieF
222
Infinite
29.6| Corollary suppose that
sums Suppose that Z is a normed vector space. Let a: I-+Z and
iel
exists. Then the set H = {i: a(i) # 0} is countable. Proof Let fceN. By the previous theorem the set Gk = {i: ||fl(0|| = l/fc} is finite. But
and so if 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 / is countable. Indeed, it will often be convenient to assume that I = N. This assumption can be made without loss of generality when / is countable because there then exists a bijection <j>: N-+I. Thus / may be replaced by N provided that a: /-> Z is replaced by the sequence b:N-+% where b = a O <j>.
29.7| Theorem Suppose that ft is a normed vector space and that (bky is a sequence of points in Z. Then
Z K= £ K keN
(l)
k=l
provided that the left-hand side exists. Proof Let e > 0 be given. If the left-hand side of (1) exists, then there exists a finite set Fez I such that for any finite set G,
2> fceN
keG
Let X 0 = max F. Then, provided that K>K0, the set G = {1, 2, 3, . . . , K} satisfies ^J. Hence
and the equation (1) follows.
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.
29.8| Proposition bers. Then
Suppose that (bk} is a sequence of non-negative real num-
keN
k= 1
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.
29.9|
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 % 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 Un.
29.10| Theorem Suppose that Z is a complete normed vector space and that is the collection of all finite subsets of/. Let a: I-+Z and suppose that
exists. Then the set
is totally bounded and hence its closure is compact.
224
Infinite sums
Proof We have to show that, for each £>0, there exists a. finite collection of open balls of radius £ which covers S. Let £ > 0 be given. Then there exists a finite set Fez I such that, for each finite set
Define the set C by
Then C is a finite set. The collection of all open balls of radius s whose centres are elements of C is therefore finite. It covers S because, for each He 5,
||2>(0-S+ X 4011 = 11 I 0(O-S||<£. ieH
ieF\H
ieHvF
29.1 If Theorem Suppose that X is a complete normed vector space and that a :I-+Z. Then the infinite sum (1) iel
exists if and only if, for any e > 0, there exists a finite set F a I such that, for any finite set if, e.
(2)
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
£ 40: He 5.and £c= ieH
Then {SE: Ee 5} is a collection of closed subsets of the compact set S. 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 S. Let £ > 0 be given. Then, by (2), there exists a finite set Foa:I such that, for any finite set H,
ieH
Also, since SGS FQ , there exists a finite set F satisfying F0<^F^I
such that
Infinite sums
225
ieF
Now suppose that G=>F. Then G\Fcil\F0
and so
||s-£a(OI|:£||s-£ 4011 + 11 I 40 ieG
ieF
ieG\F
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.
29.12"}* Corollary Suppose that X is a complete normed vector space and that a: / - • X. If J c /, then the existence of the infinite sum
£40 iel
implies the existence of
140. ieJ
29.13f Corollary Suppose that X is a complete normed vector space and that a: J-* X. If J and K are disjoint sets with union /, then
£40 = 140+140 iel
ieJ
ieK
provided that one side or other of the equation exists.
29.14*}* Corollary Suppose that X is a complete normed vector space and that a: I->Z. Then the existence of the infinite sum
iel
implies that, for any £>0, there exists a finite set Fez I such that for any set G,
||£fl(0-£fl(0ll = || E 40ll<e. iel
ieG
ieI\G
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.
29.15| Theorem Suppose that % is a complete normed vector space and that a : I-+X. Let W denote any collection of disjoint sets with union /. Then
I«(0= I jz««} iel
JeW- lieJ
(3)
)
provided the left-hand side exists. Proof Let e > 0 be given. Since the left-hand side of (3) exists, it follows from corollary 29.14 that there exists a finite set Fal such that, for any set G,
iel
ieG
Let 5 — {J:JeW and FnJ^ 0} and suppose that Q is a finite collection of sets satisfying S a QaW. Then the set G=U J JeQ
satisfies Fa Gal
and therefore, using corollary 29.13,
iel
JeQ LieJ
)
iel
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
10 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.
Infinite sums
227
29.16f Proposition Suppose that
keN
Jem ikeJ
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.
29.17f
Absolute sums
We met the notion of an absolutely convergent series in §28.7. A similar notion can be defined for infinite sums. If % is a complete normed vector space and a: / - + £ , we say that
exists absolutely if and only if
11WOII
(1)
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
converges.
29.18| Proposition (Comparison test) Suppose that X is a complete normed vector space and that a: / - • £ and b:I->U. Suppose that the infinite sum
I MO exists and that \\a(i)\\^b(i)
(iel).
228
Infinite
sums
Then the infinite sum 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.
29.19"}* Corollary Suppose that % is a complete normed vector space and that a: I-^Z. Then the existence of
El won iel
implies the existence of iel
Proof Take b(i) = \\a(i)\\ in proposition 29.18.
29.20 Corollary Suppose that % is a complete normed vector space and that a: I-^Z. Let W denote any collection of disjoint sets with union /. Then the existence of
(2)
I il Jem
U
J
implies the existence of the infinite sum
(3)
Z«(0iel
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.
29.211 Corollary Suppose that % is a complete normed vector space and that a: I^Z. Let W denote any collection of disjoint sets with union /. Then
2X0= £ {z«(»)} iel
Jew. UeJ
provided the ng/zt-hand side exists absolutely.
)
Infinite sums
229
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.
29.22|
Exercise
(1) Suppose that % is any normed vector space and that a: I-+%. If J and K are disjoint sets with union /, prove that
iel
ieJ
provided the right-hand side exists. (2) Show that the equation of question 1 holds when % is a complete normed vector space provided the left-hand side exists (i.e. prove corollaries 29.12 and 29.13). (3) Suppose that % is a complete normed vector space and that a: / - » £ . Show that.
iel
exists if and only if, for any e > 0, there exists a finite set G such that for any set
HczI\G ieH
(4) Let be a sequence of real numbers. Write a k + = m a x ak~ = max{ — ak, 0}. Prove that the infinite sum
{ak, 0} and
exists absolutely if and only if any two of the series 00
00
(i) £ ak (ii) £ ak+ k=l
00
(iii) £ ak~
k=l
k=l
converge. If (i) converges but neither (ii) nor (iii) converge, then the series (i) is said to converge conditionally. Prove that, in this case, given any £eM there exists a permutation <j)\ N -• N such that
(The latter result is called Riemann's theorem.) (5) If a: /->[R, prove that the infinite sum
230
Infinite sums
exists if and only if it exists absolutely. [Hint: Use the previous question.] Obtain the same result when a: /->R". Let be the sequence in Z00 (see §20.20) whose terms are all zero except for the /cth which is equal to l//c. 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 Z = Un). (6) Prove that a normed vector space % is complete if and only if, for every a: / - • £, the existence of
I IMOI iel
implies the existence of
I a(Q. iel
29.23|
Repeated series One sometimes has to deal with repeated series of the form
I
I *A-
(1)
It follows from corollary 29.21 that if 00
I
f
00
)
I IM
7=1 U=i
(2) J
exists, then o o f o o
0=1
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 ( l - z ) - 2 = l + 2 z + 3z2 + 4z3 + ...
(3)
provided that |z| < 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 (l-z)-1
which is valid for | z | < l . Thus
= l + z + z2 + z3 + ...
(4)
Infinite sums
231
and so
a-*r2=E i ^
(5)
provided that | z | < l . But we know that the power series (4) exists absolutely for \z\ < 1. Thus (5) exists absolutely and so we can reverse the order of summation and obtain oo
I
oo
I
(1-*)-*= E I z ' = I z < I 1. 1=0 j=0
Z=0
j=0
Thus
provided that \z\ < 1. Some explanation of these calculations may be helpful. Given that oo
j=0
oo
k=0
exists, the same result will be obtained regardless of whether the rows or columns in the array below are summed first.
+ r2
= a20
+ a21 +
a22
+ r
i
=
a
io
+
a
n
+
II
a2l
a22
^23
+
+
4
+ al3
a
n
+
4
r0
^20
+
= a 00 + fl01 + a 02 II
c
o
+
+
^02
fl
n
II
4 c.
+ C2
O3 II
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
("11
"00
+ "01
(733
0
0
•f
"23
0
0
+ "12 "f
"13
0
("n
+
"02
•f
"12
+
"ool
"03
0
"01
"02
Then 00
00
= Z rj= Z
00
Z
s
j=0
00
a
j= 0 1 = j
j= 0 1 = 0
00
00
1=0
00
00
Ji= Z Z aji 00
i
1 = 0 j=0
1 = 0 j=0
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.
a
30\.
5
"31
=
"32
"33
"22
"23
d{
Observe that
This method is particularly suitable for multiplying power series. We have that
= f Z «,vi+* [=o
j+fc=i
00
= Z *' Z «A
Infinite sums
233 00
I
Zl
= Z
a
Z m=0
1=0
mbl-m
provided that |z|
(1-*)-*=( I z'Vf
29.25| Exercise (1) Use the method of example 29.24 to show that, for each natural number n, -!)! k - — z* kl(n-l)\
k%
provided |z|
^ j=l
00
^ k=l
I
,2 J
k2 K
00
00
^
^
k=l
j=l
l
1
-J
i2-k2' J
^
(3) Prove that 1
+ 1! (1+x)
1
1
.
7
(x provided that
XG(R\Z.
2! (2 + x)
30f
30.11
SEPARATION IN
Introduction
In this chapter we return to the study of the geometry of Un 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 U2, then U2\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 Un, the half-spaces S1 = {x: <x, %> >c} are convex, open and disjoint. It follows that Si and S2 are the components of U" \H. (See theorem 15.14 and §17.21.) If xeSi and yeS 2 , t n e n t n e n n e segment L joining x and y must cut H. Otherwise, LczS1uS2 and so S1uS2 is connected (theorem 17.19). An analogous argument shows that any hyperplane (§13.14) splits U" 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 of 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 Un do not extend to infinitedimensional spaces. For example, a hyperplane H in an infinite-dimensional space. % need not be closed. But it can be shown that, in this case, H = X and % \ H is connected! Only when H is closed does %\H split nicely into two components.
234
235
Separation in 30.2|
Separation
The word 'separate' is used with various meanings in different branches of mathematics. In §15.11, we said that two sets A and B are 'separated' if and only if they are not contiguous. This usage is sometimes extended by saying that a topology 5 'separates' points of Z if and only if, for each xe Z and ye Z with x ^ y , there exist disjoint open sets S and T for which xeS 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 Z separates Z into two components if and only if Z \ H has two components S and T. If A and B are two sets in Z, we shall say that H separates A and B if and only if H separates Z into two components S and T for which A a S and BaT.
30.3
line and
Example The
= {{x, 0): *, y): (x +
xe
separates inU2.
the
sets
H
30.4|
Separating hyperplanes
We begin with the observation that, if Y is a non-empty, closed set in Un and ^eU", then there exists a point zeYwhich is nearest to \ — i.e. ||£ — z|| = J(^, Y). (See corollary 19.16.)
236
Separation
in Un
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 6 is obtuse for all yeY. This expectation is confirmed by the following theorem.
30.5| Theorem (Projection theorem) Let Y be a closed, convex, non-empty set in Un. Then for any ^eUn, there exists a unique zeYwhich satisfies ||£ —z|| = d(£, S). This value of z is characterised by the fact that (1) for all yeY Proof Recall from our discussion of the 'cosine rule' in §13.3 that 2
2
-z|!2-2<^-z, y-z>.
(2)
It follows that, if (1) holds, then | | £ - y | | ^ | | £ - z | | for all yeY and 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.) If yeY, then u = ay + (1 — a)z = z + a(y — z)eY provided that 0 ^ a ^ 1 because Y is convex. Replacing y in (2) by u, we obtain that
If <^ — z, y — z>>0, then we can ensure that ||£ — u||<||£ — z|| by taking a sufficiently small and positive. Hence <^ — z, y — z>fgO for all y e Y - i.e. (1) holds. It remains to be shown that z is unique. Observe that, if yeY and y # z , then it follows from (1) and (2) that
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 Un. In this case, the theoFem tells us that the nearest point zeYto £ is characterised by the fact that \ — z is orthogonal to Y.
Separation in
237
We may then define a function P: R"-»Y by writing P(Z>) = z. For obvious reasons, P is said to be the orthogonal projection of R" onto Y Suppose that Y is a convex, non-empty set in Un. Then Y is closed and convex. (See exercise 15.10(8).) Suppose that £<£Yand let z be the nearest point in Yto £. We know from theorem 30.5 that, for all yeY,
Taking u = (£-z)/||£-z||, it follows that
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 ||u|| = l. (See §13.14.)
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
30.6f
Theorem
Let Y be a non-empty, convex set in Un and suppose that
238
Separation in Un
Then there exists a u / O such that ^0 for all yeY. Then £fc->£ as /c->oo and has a convergent subsequence because the unit sphere is closed and bounded and thus is compact by the Heine-Borel theorem. Suppose that uw->u as /->oo. Then, from the continuity of the inner product,
as required. 30.7| Theorem (Theorem of the separating hyperplane) Suppose that A and B are non-empty, convex sets in 1R" for which (i) A^0 and (ii) AnB = 0. Then there exists au^O such that ^ for all ae A and all beB. Proof We apply the previous theorem to the convex set C = {a-b:ae^andbe5}. Conditions (i) and (ii) ensure that 0 <£ C We obtain that there exists a u # 0 for which ^ 0 for all a — beC. Hence ^ for each aeA and each beB. If the real number c is chosen so that sup ^ c ^ then theorem 30.7 tells us that the hyperplane
u, x> = c
Separation
in Un
239
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 Un.
30.8|
Norms and topologies in Un
In §13.17 we introduced the idea of a normed vector space. The natural norm to use in the space Un is the Euclidean norm
In §21.18, however, we came across some other norms which are compatible with the vector space structure of Un. We shall use the notation ||x||m = max{|x1|, |x 2 |,..., |xB|> and 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 1R" 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 R" and it is natural to ask whether all these different norms generate the same topology on Un. The next proposition answers this question in the affirmative. Note, however, that the corresponding result is very definitely false in infinite-dimensional spaces.
30.9|
Proposition
All norms on Un generate the same topology on Un.
Proof We shall use the usual notation for the Euclidean norm on Un. We shall suppose that we are also given an alternative norm on Un and use the notation |||x||| to denote this norm. As indicated in §21.18, it is enough if we can show that the set
contains an open Euclidean ball Br = {\: ||x|| 0 while aX<£S for large a > 0 . First suppose that S is unbounded (with respect to the Euclidean metric). Each unbounded convex set in Rn contains a half-line (proof?). Let X^O be a point on this half-line. Then ocXeS for all a > 0 and this is a contradiction. Hence S is bounded and so there exists an R for which
240
Separation
in Un
Next suppose that 0 is a boundary point of S (with respect to the Euclidean metric). Since S is convex, it follows from theorem 30.6 that there exists an X^O such that <X, s > ^ 0 for all seS. But, if aXeS and a > 0 , then <X, aX> = a||X|| 2 >0. This is a contradiction. We conclude that 0 is an interior point of S (with respect to the Euclidean metric) and hence there exists an r > 0 such that Br(^S.
<X,x> =
30.1 Of
Exercise
(1) Prove that any hypersphere (§13.14) separates U" into two components. (2) Prove that any ball, box or line segment S (§13.14) has the property that U" \S is connected. (3) Find hyperplanes which separate the following sets in U2: (i) A = {(x, y): x2 + y 2 g l } ; B = {(x, y): y^l} (ii) A = {{x, y): x>0 and xy> 1}; B = {(x, y): x>0 and xy< - 1 } . (4) Find two non-empty, convex sets A and B in R 2 with AnB = 0 which cannot be separated by a hyperplane. (5) Let S be a convex set in (RM. (i) If ^edS, prove that ^ed(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 Z00 for which this result is false. (6) Prove that in a Hausdorff space (§30.2) every compact set is closed. Prove that Rn is a Hausdorff space.
Separation in Rn 30.1 If
241
Curves and continua
In the remainder of this chapter, we focus our attention on !R2. We have seen that lines and circles separate !R2 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 R 2 to be the image/(/) of a compact interval / under a continuous function / : I-+U2. 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 U 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 / = [0, 1] and let S be the square [0, 1] x [0, 1]. For each neN, Hilbert constructed a continuous function fn: I->S. The curves ft(I), j'2(I) and / 3 (/) are illustrated in the diagram below.
I i —U i
The sequence of functions converges uniformly to a continuous function / : / - » £ . As is evident from the construction, the curve/(/) passes through every point of S i.e. / ( / ) = 5. We say that / ( / ) is a 'space-filling curve'.
242
Separation in Un
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/: 7-+S has the property that it is continuous on I but differentiate at no point of /.) 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 U2 is locally. connected if for each eeE and each £>0 there exists a <5>0 such that for each xeE satisfying ||e — x\\<3 there exists a connected subset of E containing both e and x which has diameter at most e.) If we are to discuss separation by curves in U2, it is therefore necessary to begin by restricting the class of curves which we are to study.
30.12|
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 / : [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 U2.
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 U2 \ A should be connected while if J is a Jordan curve, then U 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.
Separation
in Un
243
3O.13| Proposition Let J be a Jordan curve in U2. Then U2\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: U2-^U2 such that F(U) = J. But we are only given that there exists a homeomorphism/: U->J. It is a result of Schonflies that each such homeomorphism/: U-*J can be extended to a homeomorphism F: IR2->R2 but the proof of this useful result is harder than the proof of the Jordan curve theorem itself.
30.14J
Simply connected regions
Recall that the Riemann sphere (or Gaussian plane) U 2 * is the one-point compactification of M2. This notion is particularly important in those cases when one is identifying U 2 with the system C of complex numbers. We begin by observing that the Jordan curve theorem remains true with U2 replaced by U 2 * (except that, if oo e J, neither component is bounded in U 2). In complex analysis, one is often concerned with open, connected subsets of IR2#. 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 of IR 2 * \ J. In particular, if oo ^R, then R contains the bounded component of IR 2 * \ J.
244
Separation
in Un
\
30.15f Exercise (1) Explain why Brouwer's set E (§30.11) separates U2 into two components. Explain also why E is not a locally connected continuum. (2) Consider the set R = {(x, y):y>x2} in U2. Explain why R is simply connected. (3) Suppose that % and y. are topological spaces and that y, is Hausdorff (§ 30.2 and exercise 30.10(6)). If X is compact and / : %-+% is continuous and bijective, prove that / is a homeomorphism. (4) If / : /->R 2 and g: U-+M2 (where / = [0, 1] and U is the unit circle in U2) are continuous and bijective, prove t h a t / ( / ) 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 IR2 and that ^eJ. Let S and T denote the inside and the outside of J respectively. Let a e S and be T. 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
{x:P(x)}
0
V 3 AaB
es
AuB AnB A\B
(§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)
7 9 14 14 15 15 16 16 21 22 23 23 23
(§7.13) 52 (§7.13) 52 52 (§7.13) 52 (§7.13) 52 (§7.13) 52 (§7.13) 54 (§8.2) 56 (§8.7) 60 (§8.13) 60 (§8-14) (§9.7, §24.4) 68,156 (§9.7, §24.4) 65,156 69 (§9.10)
(-oo, b)
fob]
fo oo) (-oo, b\
fob)
]a,bl N
I Q
sup S inf S
sup f(x) xeS
U S
(§4.10)
24
Se WL
n s
(§4.10)
24
Se W
(a,b) AxB A2 Un
9 (A) f:A->B AS)
f~\T) f-u.B->A <xky fog
u uU_+ (a,b) (a, oo)
inf/W
(§9.10)
(§5.1, §7.13) 28,52 28 (§5.2) 29 (§5.2) (§5.2, §13.1) 29 32 (§5.9) 33 (§6.1) 35 (§6.2) 35 (§6.2) 36 (§6.2) (§6.3, §25.1) 59,169 (§6.5) 39 (§7.2) 46 50 (§7.10) 50 (§7.10) (§7.13, §5.1) 52,28 52 (§7.13)
max S min 5 00
Q
+
C i AB X
(xvx2, ..., 0
INI <x,y> |x|
d(x, y) d&S) dS
t
E
(§9.7) 68 (§9.7) 68 (§9.16, §24.2) 73,151 84 (§10.11) (§10.20) 92 93 (§10.20) (§12.22) 122 1 (§13.1) 1 •*„) (§13.1) 3 (§13.1) (§13.3, §13.17) 4, 14 4 (§13.3) 7 (§13.9) (§13.9, §13.18) 7, 15 18 (§13.20) 21 (§14.2) 31 (§15.1) 31 (§15.1)
* Numbers in italics refer to pages in Book 1: Logic, Sets and Numbers.
245
69
xeS
246
Notation
r I2 m(x, y) /(x, y)
(§20.20) (§20.21) (§21.18). (§21.18)
86 87 102 102
[ - o o , -hoc] - o o , +oo [-oo, +oo]" ( + oo, +oo)
(§24.4) (§24.4) (§24.12) (§24.12)
155 155 165 166
/ ( X ) - T I as x-+§
(§22.1)
106
lim/(x)
(§22.1)
106
lim sup/(x)
(§26.11)
187
(§22.12) (§22.12) (§22.15)
116 116 118
liminf/(x)
(§26.11)
187
%(S, %)
(§27.8, §28.15)
(§23.1)
130
(§23.5)
132
6(S, y.) Z*
(§27.8) (§27.16)
194, 209 195 197
(§28.1)
201
(§23.5)
132 (§28.19) (§28.19) (§28.19)
212 214 214
(§29.2)
220
/(x)-» I as x-+a + /(x)-» •/ as x-+bf(i+) • , / ( £ - ) /(x,y; and y lim
/(x, y)
I ak ', y) y-^n 1
c , G) * * *
(§23.12) 138 (§24.2) 151 (§24.3) 153 (§24.3, §24.11) 154,165
C[fl, C[«, b] C [a,
Z 40 ie/
fc] fe]
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-Schwarz 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 (§27.1.6) 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) }A 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 convergence (§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 endpoint, 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 Hausdorff 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, 100 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,
249
Index 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) 16
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
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
