ńņš. 1 |

A Second First and First Second Course in Analysis

T.W.KĀØrner

o

Trinity Hall

Cambridge

Note This is the ļ¬rst draft for a possible book. I would therefore be glad to receive

corrections at twk@dpmms.cam.ac.uk. Senders of substantial lists of errors or of lists of

substantial errors will receive small rewards and large thanks. General comments are also

welcome. Please refer to this as DRAFT F3 (note that Appendix K was reordered between

drafts E and F ). Please do not say ā˜I am sure someone else has noticed thisā™ or ā˜This

is too minor to matterā™. Everybody notices diļ¬erent things and no error is too small to

confuse somebody.

ii A COMPANION TO ANALYSIS

[Archimedes] concentrated his ambition exclusively upon those specula-

tions which are untainted by the claims of necessity. These studies, he be-

lieved, are incomparably superior to any others, since here the grandeur and

beauty of the subject matter vie for our admiration with the cogency and

precision of the methods of proof. Certainly in the whole science of geome-

try it is impossible to ļ¬nd more diļ¬cult and intricate problems handled in

simpler and purer terms than in his works. Some writers attribute it to his

natural genius. Others maintain that phenomenal industry lay behind the

apparently eļ¬ortless ease with which he obtained his results. The fact is that

no amount of mental eļ¬ort of his own would enable a man to hit upon the

proof of one of Archimedesā™ theorems, and yet as soon as it is explained to

him, he feels he might have discovered it himself, so smooth and rapid is the

path by which he leads us to the required conclusion.

Plutarch Life of Marcellus [Scott-Kilvertā™s translation]

It may be observed of mathematicians that they only meddle with such

things as are certain, passing by those that are doubtful and unknown. They

profess not to know all things, neither do they aļ¬ect to speak of all things.

What they know to be true, and can make good by invincible argument, that

they publish and insert among their theorems. Of other things they are silent

and pass no judgment at all, choosing rather to acknowledge their ignorance,

than aļ¬rm anything rashly.

Barrow Mathematical Lectures

For [A. N.] Kolmogorov mathematics always remained in part a sport.

But when . . . I compared him with a mountain climber who made ļ¬rst as-

cents, contrasting him with I. M. GelĀ“fand whose work I compared with the

building of highways, both men were oļ¬ended. ā˜ . . . Why, you donā™t think

I am capable of creating general theories?ā™ said AndreĖ˜ Nikolaevich. ā˜Why,

Ä±

you think I canā™t solve diļ¬cult problems?ā™ added I. M.

V. I. ArnolĀ“d in Kolmogorov in Perspective

Contents

Introduction vii

1 The real line 1

1.1 Why do we bother? . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 The fundamental axiom . . . . . . . . . . . . . . . . . . . . . 9

1.5 The axiom of Archimedes . . . . . . . . . . . . . . . . . . . . 10

1.6 Lion hunting . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.7 The mean value inequality . . . . . . . . . . . . . . . . . . . . 18

1.8 Full circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.9 Are the real numbers unique? . . . . . . . . . . . . . . . . . . 23

2 A ļ¬rst philosophical interlude ā™„ā™„ 25

2.1 Is the intermediate value theorem obvious? ā™„ā™„ . . . . . . . . 25

3 Other versions of the fundamental axiom 31

3.1 The supremum . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 The Bolzano-Weierstrass theorem . . . . . . . . . . . . . . . . 37

3.3 Some general remarks . . . . . . . . . . . . . . . . . . . . . . . 42

4 Higher dimensions 43

4.1 Bolzano-Weierstrass in higher dimensions . . . . . . . . . . . . 43

4.2 Open and closed sets . . . . . . . . . . . . . . . . . . . . . . . 48

4.3 A central theorem of analysis . . . . . . . . . . . . . . . . . . 56

4.4 The mean value theorem . . . . . . . . . . . . . . . . . . . . . 60

4.5 Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . 64

4.6 The general principle of convergence . . . . . . . . . . . . . . 66

5 Sums and suchlike ā™„ 75

5.1 Comparison tests ā™„ . . . . . . . . . . . . . . . . . . . . . . . . 75

iii

iv A COMPANION TO ANALYSIS

Conditional convergence ā™„ . . . . . . .

5.2 . . . . . . . . . . . . . 78

Interchanging limits ā™„ . . . . . . . . .

5.3 . . . . . . . . . . . . . 83

The exponential function ā™„ . . . . . .

5.4 . . . . . . . . . . . . . 91

The trigonometric functions ā™„ . . . . .

5.5 . . . . . . . . . . . . . 98

The logarithm ā™„ . . . . . . . . . . . .

5.6 . . . . . . . . . . . . . 102

Powers ā™„ . . . . . . . . . . . . . . . .

5.7 . . . . . . . . . . . . . 109

The fundamental theorem of algebra ā™„

5.8 . . . . . . . . . . . . . 113

6 Diļ¬erentiation 121

6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.2 The operator norm and the chain rule . . . . . . . . . . . . . . 127

6.3 The mean value inequality in higher dimensions . . . . . . . . 136

7 Local Taylor theorems 141

7.1 Some one dimensional Taylor theorems . . . . . . . . . . . . . 141

7.2 Some many dimensional local Taylor theorems . . . . . . . . . 146

7.3 Critical points . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8 The Riemann integral 169

8.1 Where is the problem ? . . . . . . . . . . . . . . . . . . . . . . 169

8.2 Riemann integration . . . . . . . . . . . . . . . . . . . . . . . 172

8.3 Integrals of continuous functions . . . . . . . . . . . . . . . . . 182

First steps in the calculus of variations ā™„

8.4 . . . . . . . . . . . . 190

8.5 Vector-valued integrals . . . . . . . . . . . . . . . . . . . . . . 202

9 Developments and limitations ā™„ 205

9.1 Why go further? . . . . . . . . . . . . . . . . . . . . . . . . . 205

9.2 Improper integrals ā™„ . . . . . . . . . . . . . . . . . . . . . . . 207

9.3 Integrals over areas ā™„ . . . . . . . . . . . . . . . . . . . . . . 212

9.4 The Riemann-Stieltjes integral ā™„ . . . . . . . . . . . . . . . . 217

9.5 How long is a piece of string? ā™„ . . . . . . . . . . . . . . . . . 224

10 Metric spaces 233

10.1 Sphere packing ā™„ . . . . . . . . . . . . . . . . . . . . . . . . . 233

10.2 Shannonā™s theorem ā™„ . . . . . . . . . . . . . . . . . . . . . . . 236

10.3 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

10.4 Norms, algebra and analysis . . . . . . . . . . . . . . . . . . . 246

10.5 Geodesics ā™„ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

v

Please send corrections however trivial to twk@dpmms.cam.ac.uk

11 Complete metric spaces 263

11.1 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

11.2 The Bolzano-Weierstrass property . . . . . . . . . . . . . . . . 272

11.3 The uniform norm . . . . . . . . . . . . . . . . . . . . . . . . 275

11.4 Uniform convergence . . . . . . . . . . . . . . . . . . . . . . . 279

11.5 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

11.6 Fourier series ā™„ . . . . . . . . . . . . . . . . . . . . . . . . . . 298

12 Contractions and diļ¬erential equations 303

12.1 Banachā™s contraction mapping theorem . . . . . . . . . . . . . 303

12.2 Solutions of diļ¬erential equations . . . . . . . . . . . . . . . . 305

12.3 Local to global ā™„ . . . . . . . . . . . . . . . . . . . . . . . . . 310

12.4 Greenā™s function solutions ā™„ . . . . . . . . . . . . . . . . . . . 318

13 Inverse and implicit functions 329

13.1 The inverse function theorem . . . . . . . . . . . . . . . . . . 329

13.2 The implicit function theorem ā™„ . . . . . . . . . . . . . . . . . 339

13.3 Lagrange multipliers ā™„ . . . . . . . . . . . . . . . . . . . . . . 347

14 Completion 355

14.1 What is the correct question? . . . . . . . . . . . . . . . . . . 355

14.2 The solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362

14.3 Why do we construct the reals? ā™„ . . . . . . . . . . . . . . . . 364

14.4 How do we construct the reals? ā™„ . . . . . . . . . . . . . . . . 369

14.5 Paradise lost? ā™„ā™„ . . . . . . . . . . . . . . . . . . . . . . . . 375

A The axioms for the real numbers 379

B Countability 383

C On counterexamples 387

D A more general view of limits 395

E Traditional partial derivatives 401

F Inverse functions done otherwise 407

G Completing ordered ļ¬elds 411

H Constructive analysis 415

I Miscellany 421

vi A COMPANION TO ANALYSIS

J Executive summary 427

K Exercises 431

Bibliography 603

Index 607

Introduction

In his autobiography [12], Winston Churchill remembered his struggles with

Latin at school. ā˜ . . . even as a schoolboy I questioned the aptness of the

Classics for the prime structure of our education. So they told me how Mr

Gladstone read Homer for fun, which I thought served him right.ā™ ā˜Naturallyā™

he says ā˜I am in favour of boys learning English. I would make them all learn

English; and then I would let the clever ones learn Latin as an honour, and

Greek as a treat.ā™

This book is intended for those students who might ļ¬nd rigorous analysis

a treat. The content of this book is summarised in Appendix J and corre-

sponds more or less (more rather than less) to a recap at a higher level of the

ļ¬rst course in analysis followed by the second course in analysis at Cambridge

in 2003 together with some material from various methods courses (and thus

corresponds to about 60 to 70 hours of lectures). Like those courses, it aims

to provide a foundation for later courses in functional analysis, diļ¬erential

geometry and measure theory. Like those courses also, it assumes comple-

mentary courses such as those in mathematical methods and in elementary

probability to show the practical uses of calculus and strengthen computa-

tional and manipulative skills. In theory, it starts more or less from scratch

but the reader who ļ¬nds the discussion of section 1.1 baļ¬„ing or the , Ī“

arguments of section 1.2 novel will probably ļ¬nd this book unrewarding.

This book is about mathematics for its own sake. It is a guided tour of a

great but empty Opera House. The guide is enthusiastic but interested only

in sight-lines, acoustics, lighting and stage machinery. If you wish to see the

stage ļ¬lled with spectacle and the air ļ¬lled with music you must come at

another time and with a diļ¬erent guide.

Although I hope this book may be useful to others, I wrote it for stu-

dents to read either before or after attending the appropriate lectures. For

this reason, I have tried to move as rapidly as possible to the points of dif-

ļ¬culty, show why they are points of diļ¬culty and explain clearly how they

are overcome. If you understand the hardest part of a course then, almost

automatically, you will understand the easiest. The converse is not true.

vii

viii A COMPANION TO ANALYSIS

In order to concentrate on the main matter in hand, some of the sim-

pler arguments have been relegated to exercises. The student reading this

book before taking the appropriate course may take these results on trust

and concentrate on the central arguments which are given in detail. The

student reading this book after taking the appropriate course should have

no diļ¬culty with these minor matters and can also concentrate on the cen-

tral arguments. I think that doing at least some of the exercises will help

students to ā˜internaliseā™ the material but I hope that even students who skip

most of the exercises can proļ¬t from the rest of the book.

I have included further exercises in Appendix K. Some are standard, some

form commentaries on the main text and others have been taken or adapted

from the Cambridge mathematics exams. None are just ā˜makeweightsā™, they

are all intended to have some point of interest. I have tried to keep to

standard notations but a couple of notational points are mentioned in the

index under the heading notation.

I have not tried to strip the subject down to its bare bones. A skeleton

is meaningless unless one has some idea of the being it supports and that

being in turn gains much of its signiļ¬cance from its interaction with other

beings, both of its own species and of other species. For this reason, I have

included several sections marked by a ā™„. These contain material which is

not necessary to the main argument but which sheds light on it. Ideally, the

student should read them but not study them with anything like the same

attention which she devotes to the unmarked sections. There are two sections

marked ā™„ā™„ which contain some, very simple, philosophical discussion. It is

entirely intentional that removing the appendices and the sections marked

with a ā™„ more than halves the length of the book.

My ļ¬rst glimpse of analysis was in Hardyā™s Pure Mathematics [23] read

when I was too young to really understand it. I learned elementary analysis

from Ferrarā™s A Textbook of Convergence [17] (an excellent book for those

making the transition from school to university, now, unfortunately, out of

print) and Burkillā™s A First Course in Mathematical Analysis [10]. The books

of Kolmogorov and Fomin [30] and, particularly, DieudonnĀ“ [13] showed me

e

that analysis is not a collection of theorems but a single coherent theory.

Strombergā™s book An Introduction to Classical Real Analysis [45] lies perma-

nently on my desk for browsing. The expert will easily be able to trace the

inļ¬‚uence of these books on the pages that follow. If, in turn, my book gives

any student half the pleasure that the ones just cited gave me, I will feel well

repaid.

Cauchy began the journey that led to the modern analysis course in his

Ā“

lectures at the Ecole Polytechnique in the 1820ā™s. The times were not propi-

tious. A reactionary government was determined to keep close control over

ix

Please send corrections however trivial to twk@dpmms.cam.ac.uk

the school. The faculty was divided along fault lines of politics, religion and

age whilst physicists, engineers and mathematicians fought over the contents

of the courses. The student body arrived insuļ¬ciently prepared and then

divided its time between radical politics and worrying about the job market

(grim for both staļ¬ and students). Cauchyā™s course was not popular1 .

Everybody can sympathise with Cauchyā™s students who just wanted to

pass their exams and with his colleagues who just wanted the standard ma-

terial taught in the standard way. Most people neither need nor want to

know about rigorous analysis. But there remains a small group for whom

the ideas and methods of rigorous analysis represent one of the most splen-

did triumphs of the human intellect. We echo Cauchyā™s deļ¬ant preface to his

printed lecture notes.

As to the methods [used here], I have sought to endow them

with all the rigour that is required in geometry and in such a

way that I have not had to have recourse to formal manipula-

tions. Such arguments, although commonly accepted . . . cannot

be considered, it seems to me, as anything other than [sugges-

tive] to be used sometimes in guessing the truth. Such reasons

[moreover] ill agree with the mathematical sciencesā™ much vaunted

claims of exactitude. It should also be observed that they tend to

attribute an indeļ¬nite extent to algebraic formulas when, in fact,

these formulas hold under certain conditions and for only certain

values of the variables involved. In determining these conditions

and these values and in settling in a precise manner the sense of

the notation and the symbols I use, I eliminate all uncertainty.

. . . It is true that in order to remain faithful to these principles,

I sometimes ļ¬nd myself forced to depend on several propositions

that perhaps seem a little hard on ļ¬rst encounter . . . . But, those

who will read them will ļ¬nd, I hope, that such propositions, im-

plying the pleasant necessity of endowing the theorems with a

greater degree of precision and restricting statements which have

become too broadly extended, will actually beneļ¬t analysis and

will also provide a number of topics for research, which are surely

not without importance.

1

Belhosteā™s splendid biography [4] gives the fascinating details.

Chapter 1

The real line

1.1 Why do we bother?

It is surprising how many people think that analysis consists in the diļ¬cult

proofs of obvious theorems. All we need know, they say, is what a limit is,

the deļ¬nition of continuity and the deļ¬nition of the derivative. All the rest

is ā˜intuitively clearā™1 .

If pressed they will agree that the deļ¬nition of continuity and the deļ¬ni-

tion of the derivative apply as much to the rationals Q as to the real numbers

R. If you disagree, take your favorite deļ¬nitions and examine them to see

where they require us to use R rather than Q. Let us examine the workings

of our ā˜clear intuitionā™ in a particular case.

What is the integral of t2 ? More precisely, what is the general solution of

the equation

g (t) = t2 ? (*)

We know that t3 /3 is a solution but, if we have been well taught, we know

that this is not the general solution since

t3

g(t) = + c, (**)

3

with c any constant is also a solution. Is (ā—ā—) the most general solution of

(ā—)?

If the reader thinks it is the most general solution then she should ask

herself why she thinks it is. Who told her and how did they explain it? If the

1

A good example of this view is given in the book [9]. The author cannot understand the

problems involved in proving results like the intermediate value theorem and has written

his book to share his lack of understanding with a wider audience.

1

2 A COMPANION TO ANALYSIS

reader thinks it is not the most general solution, then can she ļ¬nd another

solution?

After a little thought she may observe that if g(t) is a solution of (ā—) and

we set

t3

f (t) = g(t) ā’

3

then f (t) = 0 and the statement that (ā—ā—) is the most general solution of

(ā—) reduces to the following theorem.

Theorem 1.1.1. (Constant value theorem.) If f : R ā’ R is diļ¬eren-

tiable and f (t) = 0 for all t ā R, then f is constant.

If this theorem is ā˜intuitively clearā™ over R it ought to be intuitively clear

over Q. The same remark applies to another ā˜intuitively clearā™ theorem.

Theorem 1.1.2. (The intermediate value theorem.) If f : R ā’ R is

continuous, b > a and f (a) ā„ 0 ā„ f (b), then there exists a c with b ā„ c ā„ a.

such that f (c) = 0.

However, if we work over Q both putative theorems vanish in a puļ¬ of

smoke.

Example 1.1.3. If f : Q ā’ Q is given by

if x2 < 2,

f (x) = ā’1

f (x) = 1 otherwise,

then

(i) f is a continuous function with f (0) = ā’1, f (2) = 1, yet there does

not exist a c with f (c) = 0,

(ii) f is a diļ¬erentiable function with f (x) = 0 for all x, yet f is not

constant.

Sketch proof. We have not yet formally deļ¬ned what continuity and diļ¬eren-

tiability are to mean. However, if the reader believes that f is discontinuous,

she must ļ¬nd a point x ā Q at which f is discontinuous. Similarly, if she

believes that f is not everywhere diļ¬erentiable with derivative zero, she must

ļ¬nd a point x ā Q for which this statement is false. The reader will be in-

vited to give a full proof in Exercise 1.3.5 after continuity has been formally

deļ¬ned.

The question ā˜Is (ā—ā—) the most general solution of (ā—)?ā™ now takes on a

more urgent note. Of course, we work in R and not in Q but we are tempted

to echo Acton ([1], end of Chapter 7).

3

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This example is horrifying indeed. For if we have actually seen

one tiger, is not the jungle immediately ļ¬lled with tigers, and

who knows where the next one lurks.

Here is another closely related tiger.

Exercise 1.1.4. Continuing with Example 1.1.3, set g(t) = t + f (t) for all

t. Show that g (t) = 1 > 0 for all t but that g(ā’8/5) > g(ā’6/5).

Thus, if we work in Q, a function with strictly positive derivative need

not be increasing.

Any proof that there are no tigers in R must start by identifying the dif-

ference between R and Q which makes calculus work on one even though it

fails on the other. Both are ā˜ordered ļ¬eldsā™, that is, both support operations

of ā˜additionā™ and ā˜multiplicationā™ together with a relation ā˜greater thanā™ (ā˜or-

derā™) with the properties that we expect. I have listed the properties in the

appendix on page 379 but only to reassure the reader. We are not interested

in the properties of general ordered ļ¬elds but only in that particular prop-

erty (whatever it may be) which enables us to avoid the problems outlined

in Example 1.1.3 and so permits us to do analysis.

1.2 Limits

Many ways have been tried to make calculus rigorous and several have been

successful. We choose the ļ¬rst and most widely used path via the notion of

a limit. In theory, my account of this notion is complete in itself. However,

my treatment is unsuitable for beginners and I expect my readers to have

substantial experience with the use and manipulation of limits.

Throughout this section F will be an ordered ļ¬eld. The reader will miss

nothing if she simply considers the two cases F = R and F = Q. She will,

however, miss something if she fails to check that everything we say applies

to both cases equally.

Deļ¬nition 1.2.1. We work in an ordered ļ¬eld F. We say that a sequence

a1 , a2 , . . . tends to a limit a as n tends to inļ¬nity, or more brieļ¬‚y

an ā’ a as n ā’ ā

if, given any > 0, we can ļ¬nd an integer n0 ( ) [read ā˜n0 depending on ā™]

such that

|an ā’ a| < for all n ā„ n0 ( ).

4 A COMPANION TO ANALYSIS

The following properties of the limit are probably familiar to the reader.

Lemma 1.2.2. We work in an ordered ļ¬eld F.

(i) The limit is unique. That is, if an ā’ a and an ā’ b as n ā’ ā, then

a = b.

(ii) If an ā’ a as n ā’ ā and 1 ā¤ n(1) < n(2) < n(3) . . . , then an(j) ā’ a

as j ā’ ā.

(iii) If an = c for all n, then an ā’ c as n ā’ ā.

(iv) If an ā’ a and bn ā’ b as n ā’ ā, then an + bn ā’ a + b.

(v) If an ā’ a and bn ā’ b as n ā’ ā, then an bn ā’ ab.

(vi) Suppose that an ā’ a as n ā’ ā. If an = 0 for each n and a = 0,

then aā’1 ā’ aā’1 .

n

(vii) If an ā¤ A for each n and an ā’ a as n ā’ ā, then a ā¤ A. If bn ā„ B

for each n and bn ā’ b, as n ā’ ā then b ā„ B.

Proof. I shall give the proofs in detail but the reader is warned that similar

proofs will be left to her in the remainder of the book.

(i) By deļ¬nition:-

Given > 0 we can ļ¬nd an n1 ( ) such that |an ā’ a| < for all n ā„ n1 ( ).

Given > 0 we can ļ¬nd an n2 ( ) such that |an ā’ b| < for all n ā„ n2 ( ).

Suppose, if possible, that a = b. Then setting = |a ā’ b|/3 we have > 0. If

N = max(n1 ( ), n2 ( )) then

|a ā’ b| ā¤ |aN ā’ a| + |aN ā’ b| < + = 2|b ā’ a|/3

which is impossible. The result follows by reductio ad absurdum.

(ii) By deļ¬nition,

Given > 0 we can ļ¬nd an n1 ( ) such that |an ā’ a| < for all n ā„ n1 ( ),

()

Let > 0. Since n(j) ā„ j (proof by induction, if the reader demands a proof)

we have |an(j) ā’ a| < for all j ā„ n1 ( ). The result follows.

(iii) Let > 0. Taking n1 ( ) = 1 we have

|an ā’ c| = 0 <

for all n ā„ n1 ( ). The result follows.

(iv) By deļ¬nition, holds as does

Given > 0 we can ļ¬nd an n2 ( ) such that |bn ā’ b| < for all n ā„ n2 ( ).

( )

5

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Observe that

|(an + bn ) ā’ (a + b)| = |(an ā’ a) + (bn ā’ b)| ā¤ |an ā’ a| + |bn ā’ b|.

Thus if > 0 and n3 ( ) = max(n1 ( /2), n2 ( /2)) we have

|(an + bn ) ā’ (a + b)| ā¤ |an ā’ a| + |bn ā’ b| < /2 + /2 =

for all n ā„ n3 ( ). The result follows.

(v) By deļ¬nition, and hold. Let > 0. The key observation is

that

|an bn ā’ ab| ā¤ |an bn ā’ an b| + |an b ā’ ab| = |an ||bn ā’ b| + |b||an ā’ a| (1)

If n ā„ n1 (1) then |an ā’ a| < 1 so |an | < |a| + 1 and (1) gives

|an bn ā’ ab| ā¤ (|a| + 1)|bn ā’ b| + |b||an ā’ a|. (2)

Thus setting2 n3 ( ) = max(n1 (1), n1 ( /(2(|b| + 1)), n2 ( /(2(|a| + 1))) we see

from (2) that

|an bn ā’ ab| < /2 + /2 =

for all n ā„ n3 ( ). The result follows.

(vi) By deļ¬nition, holds. Let > 0. We observe that

|a ā’ an |

1 1

ā’ = . (3)

|a||an |

an a

Since a = 0 we have |a|/2 > 0. If n ā„ n1 (|a|/2) then |an ā’ a| < |a|/2 so

|an | > |a|/2 and (3) gives

2|a ā’ an |

1 1

ā’ ā¤ . (4)

|a|2

an a

Thus setting n3 ( ) = max(n1 (|a|/2), n1 (a2 /2)) we see from (4) that

1 1

ā’ <

an a

for all n ā„ n3 ( ). The result follows.

2

The reader may ask why we use n1 ( /(2(|b| + 1)) rather than n1 ( /(2|b|)). Observe

ļ¬rst that we have not excluded the possibility that b = 0. More importantly, observe that

all we are required to do is to ļ¬nd an n3 ( ) that works and is futile to seek a ā˜bestā™ n3 ( )

in these or similar circumstances.

6 A COMPANION TO ANALYSIS

(vii) The proof of the ļ¬rst sentence in the statement is rather similar to

that of (i). By deļ¬nition, holds. Suppose, if possible, that a > A, that is,

a ā’ A > 0. Setting N = n1 (a ā’ A) we have

aN = (aN ā’ a) + a ā„ a ā’ |aN ā’ a| > a ā’ (a ā’ A) = A,

contradicting our hypothesis. The result follows by reduction ad absurdum.

To prove the second sentence in the statement we can either give a similar

argument or set an = ā’bn , a = ā’b and A = ā’B and use the ļ¬rst sentence.

[Your attention is drawn to part (ii) of Exercise 1.2.4.]

Exercise 1.2.3. Prove that the ļ¬rst few terms of a sequence do not aļ¬ect

convergence. Formally, show that if there exists an N such that an = bn for

n ā„ N then, an ā’ a as n ā’ ā implies bn ā’ a as n ā’ ā.

Exercise 1.2.4. In this exercise we work within Q. (The reason for this will

appear in Section 1.5 which deals with the axiom of Archimedes.)

(i) Observe that if ā Q and > 0, then = m/N for some strictly posi-

tive integers m and N . Use this fact to show, directly from Deļ¬nition 1.2.1,

that (if we work in Q) 1/n ā’ 0 as n ā’ ā.

(ii) Show, by means of an example, that, if an ā’ a and an > b for all

n, it does not follow that a > b. (In other words, taking limits may destroy

strict inequality.)

Does it follow that a ā„ b? Give reasons.

Exercise 1.2.5. A more natural way of proving Lemma 1.2.2 (i) is to split

the argument in two

(i) Show that if |a ā’ b| < for all > 0, then a = b.

(ii) Show that if an ā’ a and an ā’ b as n ā’ ā, then |a ā’ b| < for all

> 0.

(iii) Deduce Lemma 1.2.2 (i).

(iv) Give a similar ā˜split proof ā™ for Lemma 1.2.2 (vii).

Exercise 1.2.6. Here is another way of proving Lemma 1.2.2 (v). I do not

claim that it is any simpler, but it introduces a useful idea.

(i) Show from ļ¬rst principles that, if an ā’ a, then can ā’ ca.

(ii) Show from ļ¬rst principles that, if an ā’ a as n ā’ ā, then a2 ā’ a2 .

n

2 2

(iii) Use the relation xy = ((x + y) ā’ (x ā’ y) )/4 together with (ii), (i)

and Lemma 1.2.2 (iv) to prove Lemma 1.2.2 (v).

The next result is sometimes called the sandwich lemma or the squeeze

lemma.

7

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 1.2.7. Suppose am ā„ cn ā„ bm for all m. Then, if an ā’ c and

bn ā’ c, it follows that cn ā’ c as n ā’ ā.

Suppose |am | ā„ |cm | ā„ |bm | for all m and that an ā’ c and bn ā’ c as

n ā’ ā. Does it follow that cn ā’ c? Give a proof or counterexample as

appropriate.

1.3 Continuity

Our deļ¬nition of continuity follows the same line of thought.

Deļ¬nition 1.3.1. We work in an ordered ļ¬eld F. Suppose that E is a subset

of F and that f is some function from E to F. We say that f is continuous

at x ā E if given any > 0 we can ļ¬nd Ī“0 ( , x) [read ā˜Ī“0 depending on and

xā™] with Ī“0 ( , x) > 0 such that

|f (x) ā’ f (y)| < for all y ā E such that |x ā’ y| < Ī“0 ( , x).

If f is continuous at every point of E we say that f : E ā’ F is a continuous

function.

The reader, who, I expect, has seen this deļ¬nition before, and is, in any

case, a mathematician, will be anxious to move on to see some theorems

and proofs. Non-mathematicians might object that our deļ¬nition does not

correspond to their idea of what continuous should mean. If we consult

the dictionary we ļ¬nd the deļ¬nition ā˜connected, unbroken; uninterrupted in

time or sequence: not discreteā™. A mathematician would object that this

merely deļ¬nes one vague concept in terms of other equally vague concepts.

However, if we rephrase our own deļ¬nition in words we see that it becomes

ā˜f is continuous if f (y) is close to f (x) whenever y is suļ¬ciently close to

xā™ and this clearly belongs to a diļ¬erent circle of ideas from the dictionary

deļ¬nition.

This will not be a problem when we come to deļ¬ne diļ¬erentiability since

there is no ā˜common senseā™ notion of diļ¬erentiability. In the same way the

existence of a ā˜common senseā™ notion of continuity need not trouble us pro-

vided that whenever we use the word ā˜continuousā™ we add under our breath

ā˜in the mathematical senseā™ and we make sure our arguments make no appeal

(open or disguised) to ā˜common senseā™ ideas of continuity.

Here are some simple properties of continuity.

Lemma 1.3.2. We work in an ordered ļ¬eld F. Suppose that E is a subset

of F, that x ā E, and that f and g are functions from E to F.

(i) If f (x) = c for all x ā E, then f is continuous on E.

8 A COMPANION TO ANALYSIS

(ii) If f and g are continuous at x, then so is f + g.

(iii) Let us deļ¬ne f Ć— g : E ā’ F by f Ć— g(t) = f (t)g(t) for all t ā E.

Then if f and g are continuous at x, so is f Ć— g.

(iv) Suppose that f (t) = 0 for all t ā E. If f is continuous at x so is

1/f .

Proof. Follow the proofs of parts (iii) to (vi) of Lemma 1.2.2.

By repeated use of parts (ii) and (iii) of Lemma 1.3.2 it is easy to show

that polynomials P (t) = n ar tr are continuous. The details are spelled

r=0

out in the next exercise.

Exercise 1.3.3. We work in an ordered ļ¬eld F. Prove the following results.

(i) Suppose that E is a subset of F and that f : E ā’ F is continuous at

x ā E. If x ā E ā‚ E then the restriction f |E of f to E is also continuous

at x.

(ii) If J : F ā’ F is deļ¬ned by J(x) = x for all x ā F, then J is continuous

on F.

(iii) Every polynomial P is continuous on F.

(iv) Suppose that P and Q are polynomials and that Q is never zero on

some subset E of F. Then the rational function P/Q is continuous on E (or,

more precisely, the restriction of P/Q to E is continuous.)

The following result is little more than an observation but will be very

useful.

Lemma 1.3.4. We work in an ordered ļ¬eld F. Suppose that E is a subset

of F, that x ā E, and that f is continuous at x. If xn ā E for all n and

xn ā’ x as n ā’ ā, then f (xn ) ā’ f (x) as n ā’ ā.

Proof. Left to reader.

We have now done quite a lot of what is called , Ī“ analysis but all we

have done is sharpened our proof of Example 1.1.3. The next exercise gives

the details.

Exercise 1.3.5. We work in Q. The function f is that deļ¬ned in Exam-

ple 1.1.3.

(i) Show that the equation x2 = 2 has no solution. (See any elementary

text on number theory or Exercise K.1.)

(ii) If |x| ā¤ 2 and |Ī·| ā¤ 1 show that |(x + Ī·)2 ā’ x2 | ā¤ 5|Ī·|.

(iii) If x2 < 2 and Ī“ = (2 ā’ x2 )/6 show that y 2 < 2 whenever |x ā’ y| < Ī“.

Conclude that f is continuous at x.

(iv) Show that if x2 > 2 then f is continuous at x.

(v) Conclude that f is a continuous function.

9

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Unless we can isolate the property that distinguishes the rationals from

the reals we can make no progress.

1.4 The fundamental axiom

The key property of the reals, the fundamental axiom which makes everything

work, can be stated as follows:

The fundamental axiom of analysis. If an ā R for each n ā„ 1, A ā R

and a1 ā¤ a2 ā¤ a3 ā¤ . . . and an < A for each n, then there exists an a ā R

such that an ā’ a as n ā’ ā.

Less ponderously, and just as rigorously, the fundamental axiom for the

real numbers says every increasing sequence bounded above tends to a limit.

Everything which depends on the fundamental axiom is analysis, every-

thing else is mere algebra.

I claim that all the theorems of classical analysis can be obtained from the

standard ā˜algebraicā™ properties of R together with the fundamental axiom. I

shall start by trying to prove the intermediate value theorem. (Here [a, b] is

the closed interval [a, b] = {x ā R : a ā¤ x ā¤ b}. )

Theorem 1.4.1. (The intermediate value theorem.) If f : [a, b] ā’ R

is continuous and f (a) ā„ 0 ā„ f (b) then there exists a c ā [a, b] such that

f (c) = 0.

(The proof will be given as Theorem 1.6.1.)

Exercise 1.4.2. Assuming Theorem 1.4.1 prove the apparently more general

result:ā“

If f : [a, b] ā’ R is continuous and f (a) ā„ t ā„ f (b) then there exists a

c ā [a, b] such that f (c) = t.

How might our programme of obtaining the intermediate value theorem

from the fundamental axiom fail?

(1) The reader has higher standards of rigour than I do but can ļ¬ll in

the gaps herself. For example, in the statement of Theorem 1.4.1, I do not

explicitly say that b ā„ a. Again, I talk about the ā˜algebraicā™ properties of R

when, strictly speaking, a set cannot have algebraic properties and I should

refer instead to the algebraic properties of (R, +, Ć—, >). Such problems may

annoy the reader but do not cause the programme to fail.

(2) As is almost certainly the case, my proofs contain errors or have gaps

which do not occur in other accounts of the material and which can thus be

10 A COMPANION TO ANALYSIS

corrected. In such a case, I apologise but it is I who have failed and not the

programme.

(3) My proofs contain a serious error or have a serious gap which occurs

in all accounts of this material. If this can be corrected then the programme

survives but looks a great deal more shaky. If a serious error has survived

for a century who knows what other errors may lurk.

(4) All accounts contain an error which cannot be corrected or a gap that

cannot be ļ¬lled. The programme has failed.

We start our attempt with a simple consequence of the fundamental ax-

iom.

Lemma 1.4.3. In R every decreasing sequence bounded below tends to a

limit.

Proof. Observe that if a1 , a2 , a3 , . . . is a decreasing sequence bounded below

then ā’a1 , ā’a2 , ā’a3 , . . . is an increasing sequence bounded above. We leave

the details to the reader as an exercise.

Exercise 1.4.4. (i) If m1 , m2 , . . . is an increasing sequence of integers

bounded above, show that there exists an N such that mj = mN for all j ā„ N .

(ii) Show that every non-empty set A ā Z bounded above has a maximum.

More formally, show that if A ā Z, A = ā… and there exists a K such that

K ā„ a whenever a ā A then there exists an a0 ā A with a0 ā„ a whenever

a ā A.

1.5 The axiom of Archimedes

Our ļ¬rst genuinely ā˜analysisā™ result may strike the reader as rather odd.

Theorem 1.5.1. (Axiom of Archimedes.)

1

ā’ 0 as n ā’ ā

n

Proof. Observe that the 1/n form a decreasing sequence bounded below.

Thus, by the fundamental axiom (in the form of Lemma 1.4.3), 1/n tends

to some limit l. To identify this limit we observe that since the limit of a

product is a product of the limits (Lemma 1.2.2 (v))

1 11 l

=Ć—ā’

2n 2n 2

and since the limit of a subsequence is the limit of the sequence (Lemma 1.2.2 (ii))

1

ā’l

2n

11

Please send corrections however trivial to twk@dpmms.cam.ac.uk

as n ā’ ā. Thus, by the uniqueness of limits (Lemma 1.2.2 (i)), l = l/2 so

l = 0 and 1/n ā’ 0 as required.

Exercise 1.5.2. [Exercise 1.2.4 (ii) concerned Q. We repeat that exercise

but this time we work in R.] Show, by means of an example, that, if an ā’ a

and an > b for all n, it does not follow that a > b. (In other words, taking

limits may destroy strict inequality.)

Does it follow that a ā„ b? Give reasons.

Theorem 1.5.1 shows that there is no ā˜exoticā™ real number say (to choose an

exotic symbol) with the property that 1/n > for all integers n ā„ 1 and yet

> 0 (that is is smaller than all strictly positive rationals and yet strictly

positive). There exist number systems with such exotic numbers (the famous

ā˜non-standard analysisā™ of Abraham Robinson and the ā˜surreal numbersā™ of

Conway constitute two such systems) but, just as the rationals are, in some

sense, too small a system for the standard theorems of analysis to hold so

these non-Archimedean systems are, in some sense, too big. Eudoxus and

Archimedes3 realised the need for an axiom to show that there is no exotic

number bigger than any integer (i.e. > n for all integers n ā„ 1; to see the

connection with our form of the axiom consider = 1/ ). However, in spite

of its name, what was an axiom for Eudoxus and Archimedes is a theorem

for us.

Exercise 1.5.3. (i) Show that there does not exist a K ā R with K > n for

all n ā Z by using Theorem 1.5.1.

(ii) Show the same result directly from the fundamental axiom.

Exercise 1.5.4. (i) Show that if a is real and 0 ā¤ a < 1 then an ā’ 0 as

n ā’ ā. Deduce that if a is real and |a| < 1 then an ā’ 0.

(ii) Suppose that a is real and a = ā’1. Discuss the behaviour of

1 ā’ an

1 + an

as n ā’ ā for the various possible values of a.

[ Hint (1 ā’ an )/(1 + an ) = (aā’n ā’ 1)/(aā’n + 1).]

Here is an important consequence of the axiom of Archimedes.

Exercise 1.5.5. (i) Use the fact that every non-empty set of integers bounded

above has a maximum (see Exercise 1.4.4) to show that, if x ā R, then there

exists an integer m such that m ā¤ x < m + 1. Show that |x ā’ m| < 1.

3

This is a simpliļ¬cation of a more complex story.

12 A COMPANION TO ANALYSIS

(ii) If x ā R and n is a strictly positive integer, show that there exists an

integer q such that |x ā’ q/n| < 1/n.

(iii) Deduce Lemma 1.5.6 below, using the axiom of Archimedes explicitly.

Lemma 1.5.6. If x ā R, then, given any > 0, there exists a y ā Q such

that |x ā’ y| < .

Thus the rationals form a kind of skeleton for the reals. (We say that the

rationals are dense in the reals.)

The reader will probably already be acquainted with the following deļ¬-

nition.

Deļ¬nition 1.5.7. If a1 , a2 , . . . is a sequence of real numbers, we say that

an ā’ ā as n ā’ ā if, given any real K, we can ļ¬nd an n0 (K) such that

an ā„ K for all n ā„ n0 (K).

Exercise 1.5.8. Using Exercise 1.5.3 show that n ā’ ā as n ā’ ā.

Exercise 1.5.8 shows that two uses of the words ā˜n tends to inļ¬nityā™ are

consistent. It is embarrassing to state the result but it would be still more

embarrassing if it were false. Here is a another simple exercise on Deļ¬ni-

tion 1.5.7.

Exercise 1.5.9. Let a1 , a2 , . . . be a sequence of non-zero real numbers.

Show that, if an ā’ ā, then 1/an ā’ 0. Is the converse true? Give a proof

or counterexample.

Exercise 1.5.10. It is worth noting explicitly that ordered ļ¬elds may satisfy

the axiom of Archimedes but not the fundamental axiom. Show in particular

that the rationals satisfy the axiom of Archimedes. (This is genuinely easy

so do not worry if your answer is brief.)

Exercise 1.5.11. The reader may be interested to see an ordered ļ¬eld con-

taining Z which does not satisfy the axiom of Archimedes. We start by con-

sidering polynomials P (X) = N an X n with real coeļ¬cients an and form

n=0

the set K of rational functions P (X)/Q(X) where P and Q are polynomials

and Q is not the zero polynomial (that is Q(X) = M bm X m with bM = 0

m=0

for some M ). Convince yourself that, if we use the usual standard formal

algebraic rules for manipulating rational functions, then K is a ļ¬eld (that is,

it satisļ¬es conditions (A1) to (D) as set out in the axioms on page 379).

To produce an order on K we deļ¬ne the set P to consist of all quotients

of the form

N n

n=0 an X

M m

m=0 bm X

13

Please send corrections however trivial to twk@dpmms.cam.ac.uk

with aN , bM = 0 and aN bM > 0. Convince yourself that this is a consistent

deļ¬nition (remember that the same quotient will have many diļ¬erent repre-

sentations; P (X)/Q(X) = R(X)P (X)/R(X)Q(X) whenever R(X) is a non-

zero polynomial) and that P satisļ¬es conditions (P1) to (P3). If we deļ¬ne

P1 (X)/Q1 (X) > P2 (X)/Q2 (X) whenever P1 (X)/Q1 (X)ā’P2 (X)/Q2 (X) ā P

condition (P4) is automatically satisļ¬ed and we have indeed got an ordered

ļ¬eld.

We note that the elements of K of the form a/1 with a ā R can be

identiļ¬ed in a natural way with R. If we make this natural identiļ¬cation, K

contains Z.

To see that the axiom of Archimedes fails, observe that 1/n > 1/X > 0

for all n ā Z, n ā„ 1.

Of course, since the axiom of Archimedes fails, the fundamental axiom

fails. By examining the proof of Theorem 1.5.1, show that the 1/n form a

decreasing sequence bounded below but not tending to any limit.

If the reader knows some modern algebra she will see that our presentation

can be sharpened in various ways. (It would be better to deļ¬ne K using equiv-

alence classes. We should take greater care over checking consistency. The

words ā˜identiļ¬ed in a natural wayā™ should be replaced by ā˜there is an isomor-

phism of (R, +, Ć—, >) with a subļ¬eld of Kā™.) Such readers should undertake

the sharpening as an instructive exercise.

Exercise 1.5.12. We shall not make any essential use of the decimal ex-

pansion of the real numbers but it is interesting to see how it can be obtained.

Let us write

D = {n ā Z : 9 ā„ n ā„ 0}.

(i) If xj ā D show that N xj 10ā’j ā¤ 1.

j=1

N

(ii) If xj ā D show that ā’j

j=1 xj 10 converges to a limit x, say, as

N ā’ ā. Show that 0 ā¤ x ā¤ 1 and that x = 1 if and only if xj = 9 for all j.

(iii) If y ā [0, 1] show that there exist yj ā D such that

N

y ā’ 10ā’N < yj 10ā’j ā¤ y

j=1

and that N yj 10ā’j ā’ y as N ā’ ā.

j=1

(iv) Identify explicitly the use of the axiom of Archimedes in the proof of

the last sentence of (ii) and in the proof of (iii).

(v) Suppose that aj , bj ā D, aj = bj for j < M and aM > bM . If

N

ā’ a and N bj 10ā’j ā’ b as N ā’ ā show that a ā„ b. Give

ā’j

j=1 aj 10 j=1

the precise necessary and suļ¬cient condition for equality and prove it.

14 A COMPANION TO ANALYSIS

Exercise 1.5.13. It seems, at ļ¬rst sight, that decimal expansion gives a nat-

ural way of treating real numbers. It is not impossible to do things in this

way, but there are problems. Here is one of them. Let 0 < a, b < 1 and

c = ab. If N aj 10ā’j ā’ a, N bj 10ā’j ā’ b, and N cj 10ā’j ā’ c ļ¬nd cj

j=1 j=1 j=1

in terms of the various ak and bk . (The reader is invited to reļ¬‚ect on this

problem rather than solve it. Indeed, one question is ā˜what would constitute

a nice solution?ā™)

Exercise 1.5.14. Here is a neat application of decimal expansion.

(i) Deļ¬ne f : [0, 1] ā’ [0, 1] as follows. Each x ā [0, 1] has a unique

non-terminating decimal expansion

ā

xj 10ā’j

x=

j=1

with the xj integers such that 0 ā¤ xj ā¤ 9. If there exists an integer N ā„ 2

such that x2j = 1 for all j ā„ N but x2N ā’2 = 1 we set

ā

x2(j+N )+1 10ā’j .

f (x) =

j=1

Otherwise we set f (x) = 0. Show that given any y ā [0, 1], any > 0 and

any t ā [0, 1] we can ļ¬nd an x ā [0, 1] with |x ā’ y| < such that f (x) = t.

In other words f takes every value in [0, 1] arbitrarily close to every point.

(ii) Show that if 0 ā¤ a < b ā¤ 1 then, given any t lying between f (a)

and f (b) (that is to say, t with f (a) ā¤ t ā¤ f (b) if f (a) ā¤ f (b) or with

f (b) ā¤ t ā¤ f (a) if f (b) ā¤ f (a)), there exists a c ā [a, b] such that f (c) = t.

(Thus the fact that a function satisļ¬es the conclusion of the intermediate

value theorem (Exercise 1.4.2) does not show that it is well behaved.)

(iii) Find a g : R ā’ [0, 1] such that, given any y ā R, any > 0 and any

t ā [0, 1], we can ļ¬nd an x with |x ā’ y| < with f (x) = t.

(iv) (This may require a little thought.) Find a g : R ā’ R such that given

any y ā R, any > 0 and any t ā R we can ļ¬nd an x with |x ā’ y| < such

that f (x) = t.

Although decimal expansion is a very useful way of representing numbers

it is not the only one. In Exercises K.13 and K.14 we discuss representation

by continued fractions.

1.6 Lion hunting

Having dealt with the axiom of Archimedes, we can go on at once to prove

the intermediate value theorem.

15

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Theorem 1.6.1. (The intermediate value theorem.) We work in R. If

f : [a, b] ā’ R is continuous and f (a) ā„ 0 ā„ f (b), then there exists a c ā [a, b]

such that f (c) = 0.

Proof. Since the method of proof is important to us, I shall label its three

main parts.

Part A Set a0 = a and b0 = b. We observe that f (a0 ) ā„ 0 ā„ f (b0 ). Now set

c0 = (a0 + b0 )/2. There are two possibilities. Either f (c0 ) ā„ 0, in which case

we set a1 = c0 and b1 = b0 , or f (c0 ) < 0, in which case we set a1 = a0 and

b1 = c0 . In either case, we have

f (a1 ) ā„ 0 ā„ f (b1 ),

a0 ā¤ a 1 ā¤ b 1 ā¤ b 0 ,

and b1 ā’ a1 = (b0 ā’ a0 )/2.

Continuing inductively we can ļ¬nd a sequence of pairs of points an and bn

such that

f (an ) ā„ 0 ā„ f (bn ),

anā’1 ā¤ an ā¤ bn ā¤ bnā’1 ,

and bn ā’ an = (bnā’1 ā’ anā’1 )/2,

for all n ā„ 1.

Part B We have a0 ā¤ a1 ā¤ Ā· Ā· Ā· ā¤ an ā¤ b0 so that the an form an increasing

sequence bounded above. By the fundamental axiom there is real number c,

say, such that an ā’ c as n ā’ ā. Since a = a0 ā¤ an ā¤ b0 we have a ā¤ c ā¤ b.

We note also that bn ā’ an = 2ā’n (b0 ā’ a0 ) so, by the axiom of Archimedes,

bn ā’ an ā’ 0 and thus

bn = an + (bn ā’ an ) ā’ c + 0 = c

as n ā’ ā.

Part C Since f is continuous at c and an ā’ c, it follows that f (an ) ā’ f (c)

as n ā’ ā. Since f (an ) ā„ 0 for all n, it follows that f (c) ā„ 0. A similar

argument applied to the bn shows that f (c) ā¤ 0. Since 0 ā¤ f (c) ā¤ 0, it

follows that f (c) = 0 and we are done.

Exercise 1.6.2. (i) Give the complete details in the inductive argument in

Part A of the proof of Theorem 1.6.1 above.

(ii) Give the details of the ā˜similar argument applied to the bn ā™ which

shows that f (c) ā¤ 0.

(iii) We use various parts of Lemma 1.2.2 in our Theorem 1.6.1. Identify

the points where we use Lemma 1.2.2.

16 A COMPANION TO ANALYSIS

Exercise 1.6.3. (i) Think how the argument used to prove Theorem 1.6.1

applies to [a, b] = [0, 1], f (x) = 2 ā’ 4x2 . (You are not asked to write anything

though you may well choose to draw a diagram.)

(ii) Think also how the argument used to prove Theorem 1.6.1 applies to

[a, b] = [0, 1], f (x) = (1 ā’ 5x)(2 ā’ 5x)(3 ā’ 5x).

The method used to prove Theorem 1.6.1 is called ā˜Lion huntingā™4 . The

method is also called ā˜successive bisectionā™, ā˜bisection searchā™ or simply ā˜bi-

sectionā™.

Let us summarise the proof. In Part A we have evidence of a lion in the

interval [anā’1 , bnā’1 ]. We split the interval into two halves [anā’1 , cnā’1 ] and

[cnā’1 , bnā’1 ] and show that, since there is evidence of a lion in the interval

[anā’1 , bnā’1 ], either there is evidence of a lion in [anā’1 , cnā’1 ], in which case we

take [an , bn ] = [anā’1 , cnā’1 ], or, if there is no evidence of a lion in [anā’1 , cnā’1 ]

(this does not mean that there are no lions in [anā’1 , cnā’1 ], simply that we do

not have evidence of one), then it follows that there must be evidence of a

lion in [cnā’1 , bnā’1 ] and we take [an , bn ] = [cnā’1 , bnā’1 ].

In Part B we use the fundamental axiom of analysis to show that these

successive bisections ā˜close inā™ on a point c which we strongly suspect of being

a lion. Finally in Part C we examine the point c to make sure that it really

is a lion. (It might be a wolf or a left handed corkscrew.)

Let us see what goes wrong if we omit parts of the hypotheses of Theo-

rem 1.6.1. If we omit the condition f (a) ā„ 0 ā„ f (b), then we cannot even

start Part A of the argument. The example [a, b] = [0, 1], f (x) = 1 shows

that the conclusion may indeed be false.

If we have f (a) ā„ 0 ā„ f (b) but replace R by another ordered ļ¬eld for

which the fundamental axiom does not hold, then Part A goes through per-

fectly but Part B fails. Example 1.1.3 with which we started shows that the

conclusion may indeed be false. (Take [a, b] = [ā’2, 0].)

If we have f (a) ā„ 0 ā„ f (b) and we work over R but we do not demand

f continuous then Part C fails. Working over R we may take [a, b] = [0, 1]

and deļ¬ne f (x) = 1 for x ā¤ 1/3 and f (x) = ā’1 for x > 1/3. Parts A and B

work perfectly to produce c = 1/3 but there is no lion (that is, no zero of f )

at c.

Exercises 1.6.4 to 1.6.6 are applications of the intermediate value theorem.

Exercise 1.6.4. Show that any real polynomial of odd degree has at least

one root. Is the result true for polynomials of even degree? Give a proof or

counterexample.

4

The name probably comes from A Contribution to the Mathematical Theory of Big

Game Hunting by H. PĀ“tard. This squib is reprinted in [8].

e

17

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 1.6.5. Suppose that g : [0, 1] ā’ [0, 1] is a continuous function. By

considering f (x) = g(x) ā’ x, or otherwise, show that there exists a c ā [0, 1]

with g(c) = c. (Thus every continuous map of [0, 1] into itself has a ļ¬xed

point.)

Give an example of a bijective continuous function k : (0, 1) ā’ (0, 1) such

that k(x) = x for all x ā (0, 1).

Give an example of a bijective (but, necessarily, non-continuous) function

h : [0, 1] ā’ [0, 1] such that h(x) = x for all x ā [0, 1].

[Hint: First ļ¬nd a function H : [0, 1] \ {0, 1, 1/2} ā’ [0, 1] \ {0, 1, 1/2}

such that H(x) = x.]

Exercise 1.6.6. Every mid-summer day at six oā™clock in the morning, the

youngest monk from the monastery of Damt starts to climb the narrow path

up Mount Dipmes. At six in the evening he reaches the small temple at the

peak where he spends the night in meditation. At six oā™clock in the morning

on the following day he starts downwards, arriving back at the monastery at

six in the evening. Of course, he does not always walk at the same speed.

Show that, none the less, there will be some time of day when he will be at

the same place on the path on both his upward and downward journeys.

Finally we give an example of lion hunting based on trisecting the interval

rather than bisecting it.

Exercise 1.6.7. Suppose that we have a sequence x1 , x2 , . . . of real num-

bers. Let [a0 , b0 ] be any closed interval. Show that we can ļ¬nd a sequence of

pairs of points an and bn such that

either xn ā„ bn + (bnā’1 ā’ anā’1 )/3 or xn ā¤ an ā’ (bnā’1 ā’ anā’1 )/3,

anā’1 ā¤ an ā¤ bn ā¤ bnā’1 ,

and bn ā’ an = (bnā’1 ā’ anā’1 )/3,

for all n ā„ 1.

Show that an and bn tend to some limit c ā [a0 , b0 ]. Show further that,

for each n ā„ 1, either xn ā„ c + (bnā’1 ā’ anā’1 )/3 or xn ā¤ c ā’ (bnā’1 ā’ anā’1 )/3

and so in particular xn = c.

Thus we cannot write the points of [a0 , b0 ] as a sequence. (We say that

[a0 , b0 ] is uncountable. The reader may know a proof of this via decimal

expansions.)

For more on countability and an important extension of this exercise see

Appendix B and Exercise B.7 within that appendix.

18 A COMPANION TO ANALYSIS

1.7 The mean value inequality

Having disposed of one of the three tigers with which we started, by proving

the intermediate value theorem, we now dispose of the other two by using

the ā˜mean value inequalityā™.

Theorem 1.7.1. (The mean value inequality.) Let U be the open inter-

val (Ī±, Ī²) on the real line R. Suppose that K ā„ 0 and that a, b ā U with

b > a. If f : U ā’ R is diļ¬erentiable with f (t) ā¤ K for all t ā U then

f (b) ā’ f (a) ā¤ (b ā’ a)K.

Before we can do this we must deļ¬ne diļ¬erentiability and the deriva-

tive. The reader will almost certainly be familiar with a deļ¬nition along the

following lines.

Deļ¬nition 1.7.2. Let U be an open set in R. We say that a function f :

U ā’ R is diļ¬erentiable at t ā U with derivative f (t) if, given > 0, we can

ļ¬nd a Ī“(t, ) > 0 such that (t ā’ Ī“(t, ), t + Ī“(t, )) ā U and

f (t + h) ā’ f (t)

ā’ f (t) <

h

whenever 0 < |h| < Ī“(t, ).

In Chapter 6 we shall deļ¬ne a more general notion of diļ¬erentiation and

derive many of its properties. For the moment all we need is Deļ¬nition 1.7.2.

Exercise 1.7.3. Let U be an open interval in R and suppose functions f, g :

U ā’ R are diļ¬erentiable at t ā U with derivatives f (t) and g (t). If Ī», Āµ ā

R, show that Ī»f + Āµg is diļ¬erentiable at t with derivative Ī»f (t) + Āµg (t).

To obtain Theorem 1.7.1 we prove an apparently weaker result.

Lemma 1.7.4. We use the notation and assumptions of Theorem 1.7.1. If

> 0, then f (b) ā’ f (a) ā¤ (K + )(b ā’ a).

Proof of Theorem 1.7.1 from Lemma 1.7.4. Since f (b)ā’f (a) ā¤ (K+ )(bā’a)

for all > 0, it follows that f (b) ā’ f (a) ā¤ K(b ā’ a).

Proof of Lemma 1.7.4. We suppose that f (b) ā’ f (a) > (K + )(b ā’ a) and

use lion-hunting to derive a contradiction. Set a0 = a, b0 = b. We observe

that f (b0 ) ā’ f (a0 ) > (K + )(b0 ā’ a0 ). Now set c0 = (a0 + b0 )/2. Since

f (c0 ) ā’ f (a0 ) ā’ (K + )(c0 ā’ a0 ) + f (b0 ) ā’ f (c0 ) ā’ (K + )(b0 ā’ c0 )

= f (b0 ) ā’ f (a0 ) ā’ (K + )(b0 ā’ a0 ) > 0,

19

Please send corrections however trivial to twk@dpmms.cam.ac.uk

at least one of the expressions f (c0 ) ā’ f (a0 ) ā’ (K + )(c0 ā’ a0 ) and (f (b0 ) ā’

f (c0 ) ā’ (K + )(b0 ā’ c0 ) must be strictly positive. If f (b0 ) ā’ f (c0 ) ā’ (K +

)(b0 ā’ c0 ) > 0, we set a1 = c0 and b1 = b0 . Otherwise, we set a1 = a0 and

b1 = c0 . In either case, we have

f (b1 ) ā’ f (a1 ) > (K + )(b1 ā’ a1 ),

a0 ā¤ a 1 ā¤ b 1 ā¤ b 0 ,

and b1 ā’ a1 = (b0 ā’ a0 )/2.

Continuing inductively, we can ļ¬nd a sequence of pairs of points an and

bn such that

f (bn ) ā’ f (an ) > (K + )(bn ā’ an ),

anā’1 ā¤ an ā¤ bn ā¤ bnā’1 ,

and bn ā’ an = (bnā’1 ā’ anā’1 )/2,

for all n ā„ 1.

We have a0 ā¤ a1 ā¤ Ā· Ā· Ā· ā¤ an ā¤ b0 so that the an form an increasing

sequence bounded above. By the fundamental axiom there is real number c,

say, such that an ā’ c as n ā’ ā. Since a = a0 ā¤ an ā¤ b0 we have a ā¤ c ā¤ b

and similarly aN ā¤ c ā¤ bN for all N . We note also that bn ā’an = 2ā’n (b0 ā’a0 )

so, by the axiom of Archimedes, bn ā’ an ā’ 0 and thus

bn = an + (bn ā’ an ) ā’ c + 0 = c

as n ā’ ā.

Since f is diļ¬erentiable at c, we can ļ¬nd a Ī“ > 0 such that (cā’Ī“, c+Ī“) ā U

and

f (c + h) ā’ f (c)

ā’ f (c) < /2

h

whenever 0 < |h| < Ī“. Thus

|f (c + h) ā’ f (c) ā’ f (c)h| ā¤ |h|/2

whenever |h| < Ī“, and so, since f (c) ā¤ K,

f (c + h) ā’ f (c) ā¤ (K + /2)h for 0 ā¤ h < Ī“

f (c) ā’ f (c + h) ā¤ ā’(K + /2)h for ā’Ī“ ā¤ h ā¤ 0

Since an ā’ c and bn ā’ c, we can ļ¬nd an N such that |aN ā’ c| < Ī“ and

|bN ā’ c| < Ī“. It follows, ļ¬rst taking h = aN ā’ c and then h = bN ā’ c, that

f (c) ā’ f (aN ) ā¤ (K + /2)(c ā’ aN )

and f (bN ) ā’ f (c) ā¤ (K + /2)(bN ā’ c).

20 A COMPANION TO ANALYSIS

Thus

f (bN ) ā’ f (aN ) = f (bN ) ā’ f (c) + f (c) ā’ f (aN )

ā¤ (K + /2)(bN ā’ c) + (K + /2)(c ā’ aN )

= (K + /2)(bN ā’ aN ),

contradicting our earlier assumption that f (bn ) ā’ f (an ) > (K + )(bn ā’ an )

for all n.

Thus our initial assumption must be wrong and the theorem is proved.

Theorem 1.7.1 immediately proves Theorem 1.1.1.

Theorem 1.7.5. (The constant value theorem.) Let U be the open in-

terval (Ī±, Ī²) or the real line R. If f : U ā’ R is diļ¬erentiable with f (t) = 0

for all t ā U , then f is constant.

Proof. Let b, a ā U with b ā„ a. By applying Theorem 1.7.1 to f with K = 0

we see that f (b) ā’ f (a) ā¤ 0. By applying Theorem 1.7.1 to ā’f with K = 0,

we see that f (a) ā’ f (b) ā¤ 0. Thus f (a) = f (b). But a and b were arbitrary,

so f is constant.

In section 1.1 we noted the importance of the following simple corollary.

Theorem 1.7.6. Let U be the open interval (Ī±, Ī²) or the real line R. If the

functions f, g : U ā’ R are diļ¬erentiable with f (t) = g (t) for all t ā U then

there exists a constant c such that f (t) = g(t) + c for all t ā U .

Proof. Apply Theorem 1.7.5 to f ā’ g.

In A Tour of the Calculus [5], Berlinski greets this theorem with a burst

of rhetoric.

. . . functions agreeing in their derivates, the theorem states,

diļ¬er on an interval only by a constant. It is the derivative of

a real-valued function that like some pulsing light illuminates

again the behaviour of the function, enforcing among otherwise

anarchic and wayward mathematical objects a stern uniformity

of behaviour. Such is the proximate burden of the mean value

theorem, which is now revealed to play a transcendental role in

the scheme of things.

21

Please send corrections however trivial to twk@dpmms.cam.ac.uk

To which the present author, constrained by the conventions of textbook

writing from such active expressions of enthusiasm, can only murmur ā˜Hear,

hearā™.

It is fairly easy to see that Theorem 1.7.1 is equivalent to the following

result.

Theorem 1.7.7. Let U be the open interval (Ī±, Ī²) or the real line R. Sup-

pose that a, b ā U and b > a. If g : U ā’ R is diļ¬erentiable with g (t) ā„ 0

for all t ā U then

g(b) ā’ g(a) ā„ 0.

Exercise 1.7.8. (i) By taking f = ā’g and K = 0, prove Theorem 1.7.7

from Theorem 1.7.1.

(ii) By taking g(t) = Ktā’f (t), prove Theorem 1.7.1 from Theorem 1.7.7.

Thus a function with positive derivative is increasing.

The converse result is ā˜purely algebraicā™ in the sense that it does not

involve the fundamental axiom.

Lemma 1.7.9. If g : (a, b) ā’ R is diļ¬erentiable and increasing on (a, b)

then g (t) ā„ 0 for all t ā (a, b).

Exercise 1.7.10. Use the deļ¬nition of the derivative to prove Lemma 1.7.9.

[Hint: Show ļ¬rst that given any > 0 we have g (t) > ā’ .]

Readers who know the mean value theorem (given as Theorem 4.4.1 later)

may wish to extend Theorem 1.7.1 as follows.

Exercise 1.7.11. Suppose that a, b ā R and b > a. If f : [a, b] ā’ R is

continuous and f is diļ¬erentiable on (a, b) with f (t) ā¤ K for all t ā (a, b),

use Theorem 1.7.1 on intervals (an , bn ) with a < an < bn < b and continuity

to show that

f (b) ā’ f (a) ā¤ (b ā’ a)K.

Experience shows that students do not fully realise the importance of

the mean value inequality. Readers should take note whenever they use

Theorem 1.7.6 or Theorem 1.7.7 since they are then using the mean value

inequality directly.

22 A COMPANION TO ANALYSIS

1.8 Full circle

We began this chapter with an example of an ordered ļ¬eld for which the

intermediate value theorem failed. A simple extension of that example shows

that just as the fundamental axiom implies the intermediate value theorem,

so the intermediate value theorem implies the fundamental axiom.

Theorem 1.8.1. Let F be an ordered ļ¬eld for which the intermediate value

theorem holds, that is to say:

Let a, b ā F with b > a and set [a, b] = {x ā F ; b ā„ x ā„ a}. If

f : [a, b] ā’ F is continuous and f (a) ā„ 0 ā„ f (b) then there exists a c ā [a, b]

such that f (c) = 0.

Then the fundamental axiom holds. That is to say:

If an ā F for each n ā„ 1, A ā F, a1 ā¤ a2 ā¤ a3 ā¤ . . . and an < A for each

n then there exists an c ā F such that an ā’ c as n ā’ ā.

Proof. Suppose a1 ā¤ a2 ā¤ a3 ā¤ . . . and an < A for all n. Choose a < a1 and

b > A. Deļ¬ne f : [a, b] ā’ F by

f (x) = 1 if x < an for some n,

f (x) = ā’1 otherwise.

Since f does not take the value 0, the intermediate value theorem tells us

that there must be a point c ā [a, b] at which f is discontinuous.

Suppose that y < aN for some N , so = aN ā’ y > 0. Then, whenever

|xā’y| < /2, we have x ā¤ aN ā’ /2, so f (x) = f (y) = 1 and |f (x)ā’f (y)| = 0.

Thus f is continuous at y. We have shown that c ā„ an for all n.

Suppose that there exists an > 0 such that y ā„ an + for all n. Then,

whenever |x ā’ y| < /2, we have x ā„ an + /2 for all n, so f (x) = f (y) = ā’1

and |f (x) ā’ f (y)| = 0. Thus f is continuous at y. We have shown that given

> 0 there exists an n0 ( ) such that c < an0 ( ) + .

Combining the results of the two previous paragraphs with the fact that

the an form an increasing sequence, we see that, given > 0, there exists an

n0 ( ) > 0 such that an ā¤ c < an + and so |c ā’ an | < for all n ā„ n0 ( ).

Thus an ā’ c as n ā’ ā and we are done.

Exercise 1.8.2. State and prove similar results to Theorem 1.8.1 for Theo-

rem 1.7.6 and Theorem 1.7.1. (Thus the constant value and the mean value

theorem are equivalent to the fundamental axiom.)

23

Please send corrections however trivial to twk@dpmms.cam.ac.uk

1.9 Are the real numbers unique?

Our statement of Theorem 1.8.1 raises the question as to whether there

may be more than one ordered ļ¬eld satisfying the fundamental axiom. The

answer, which is as good as we can hope for, is that all ordered ļ¬elds satisfying

the fundamental axiom are isomorphic. The formal statement is given in the

following theorem.

Theorem 1.9.1. If the ordered ļ¬eld (F, +, Ć—, >) satisļ¬es the fundamental

axiom of analysis, then there exists a bijective map Īø : R ā’ F such that, if

x, y ā R, then

Īø(x + y) = Īø(x) + Īø(y)

Īø(xy) = Īø(x)Īø(y)

Īø(x) > 0 whenever x > 0.

Exercise 1.9.2. Show that the conditions on Īø in Theorem 1.9.1 imply that

Īø(x) > Īø(y) whenever x > y.

We shall need neither Theorem 1.9.1 nor its method of proof. For com-

pleteness we sketch a proof in Exercises A.1 to A.5 starting on page 380,

but I suggest that the reader merely glance at them. Once the reader has

acquired suļ¬cient experience both in analysis and algebra she will ļ¬nd that

the proof of Theorem 1.9.1 writes itself. Until then it is not really worth the

time and eļ¬ort involved.

Chapter 2

A ļ¬rst philosophical interlude

ā™„ā™„

This book contains two philosophical interludes. The reader may omit both

on the grounds that mathematicians should do mathematics and not philosophise

about it1 . However, the reader who has heard Keynesā™ gibe that ā˜Practical

men who believe themselves to be exempt from any intellectual inļ¬‚uences,

are usually the slaves of some defunct economistā™ may wonder what kind of

ideas underlie the standard presentation of analysis given in this book.

2.1 Is the intermediate value theorem obvi-

ous? ā™„ā™„

It is clear from Example 1.1.3 that the intermediate value theorem is not

obvious to a well trained mathematician. Psychologists have established that

it is not obvious to very small children, since they express no surprise when

objects appear to move from one point to another without passing through

intermediate points. But most other human beings consider it obvious. Are

they right?

The Greek philosopher Zeno has made himself unpopular with ā˜plain hon-

est menā™ for over 2000 years by suggesting that the world may not be as

simple as ā˜plain honest menā™ believe. I shall borrow and modify some of his

arguments.

There are two ways in which the intermediate value theorem might be

obvious:- through observation and introspection. Is it obvious through ob-

1

My father on being asked by DieudonnĀ“ to name any mathematicians who had been

e

inļ¬‚uenced by any philosopher, instantly replied ā˜Descartes and Leibnizā™.

25

26 A COMPANION TO ANALYSIS

servation? Suppose we go to the cinema and watch the ļ¬lm of an arrowā™s

ļ¬‚ight. It certainly looks as though the arrow is in motion passing through

all the points of its ļ¬‚ight. But, if we examine the ļ¬lm, we see that it consists

of series of pictures of the arrow at rest in diļ¬erent positions. The arrow

takes up a ļ¬nite, though large, number of positions in its apparent ļ¬‚ight

and the tip of the arrow certainly does not pass through all the points of

the trajectory. Both the motion and the apparent truth of the intermediate

value theorem are illusions. If they are illusory in the cinema, might they

not be illusory in real life?

There is another problem connected with the empirical study of the in-

termediate value theorem. As Theorem 1.8.1 proves, the intermediate value

theorem is deeply linked with the structure of the real numbers. If the in-

termediate value theorem is physically obvious then the structure of the real

numbers should also be obvious. To give an example, the intermediate value

theorem implies that there ā a positive real number x satisfying x2 = 2.

is

We know that this number 2 is irrational. But the existence of irrational

numbers was so non-obvious that this discovery precipitated a crisis in Greek

mathematics2 .

Of course, it is possible that we are cleverer than our Greek forefathers, or

at least better educated, and what was not obvious to them may be obvious

ā

to us. Let us try and see whether the existence of 2 is physically obvious.

ā

One way of doing this would be to mark out a length of 2 metres on a

steel rod. We can easily mark out a length of 1.4 metres (with an error of less

than .05 metres). With a little work we can mark out a length of 1.41 metres

(with an error of less than .005 metres) or, indeed, a length of 1.414 metres

(with an error of less than .0005 metres). But it is hard to see why looking

at a length of 1.414 Ā± .0005 metres should convince us of the existence of a

ā

length 2. Of course we can imagine the process continued but few ā˜plain

honest menā™ would believe a craftsman who told them that because they

could work to an accuracy of Ā±.000 05 metres they could therefore work to

an accuracy of Ā±.000 000 005 metres ā˜and so onā™. Indeed if someone claimed

to have marked out a length of 1.414 213 562 373 095 metres to an accuracy

of Ā±5 Ć— 10ā’16 metres we might point out that the claimed error was less than

the radius of a proton.

If we try to construct such a length indirectly as the length of the hy-

potenuse of a right angled triangle with shorter sides both of length 1 metre

we simply transfer the problem to that of producing two lengths of 1 metre

2

Unfortunately, we have no contemporary record of how the discovery was viewed and

we can be certain that the issues were looked at in a very diļ¬erent way to that which we

see them today. Some historians even deny that there was a crisis, but the great majority

of commentators agree that there is ample indirect evidence of such a crisis.

27

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(we can produce one length of 1 metre by using the standard metre, but how

can we copy it exactly?) and an exact right angle. If we try to construct it

by graphing y = x2 ā’ 2 and looking at the intersection with the axis y = 0

close inspection of the so called intersection merely reveals a collection of

graphite blobs (or pixels or whatever). Once again, we learn that there are

many numbers which almost satisfy the equation x2 = 2 but not whether

there are any which satisfy the equation exactly.

Since the intermediate value theorem does not seem to be obvious by

observation, let us see whether it is obvious by introspection. Instead of

observing the ļ¬‚ight of an actual physical arrow, let us close our eyes and

imagine the ļ¬‚ight of an ideal arrow from A to B. Here a diļ¬culty presents

itself. We can easily picture the arrow at A and at B but, if we wish to

imagine the complete ļ¬‚ight, we must picture the arrow at the half way point

A1 from A to B. This is easy enough but, of course, the same argument shows

that we must picture the arrow at the half way point A2 from A to A1 , at the

half way point A3 from A to A2 and so on. Thus in order to imagine the ļ¬‚ight

of the arrow we must see it at each of the inļ¬nite sequence of points A1 , A2 ,

A3 , . . . . Since an electronic computer can only perform a ļ¬nite number of

operations in a given time, this presents a problem to those who believe that

the brain is a kind of computer. Even those who believe that, in some way,

the brain transcends the computer may feel some doubts about our ability to

picture the arrow at each of the uncountable set3 of points which according

to the ā˜obviousā™ intermediate value theorem (the intermediate value theorem

implies the fundamental axiom by Theorem 1.8.1 and the fundamental axiom

yields the result of Exercise 1.6.7) must be traversed by the tip of the arrow

on its way to the target.

There is another diļ¬culty when we try to picture the path of the arrow.

At ļ¬rst, it may seem to the reader to be the merest quibble but in my opinion

(and that of many cleverer people) it becomes more troubling as we reļ¬‚ect

on it. It is due to Zeno but, as with his other ā˜paradoxesā™ we do not have

his own words. Consider the ļ¬‚ying arrow. At every instant it has a position,

that is, it occupies a space equal to itself. But everything that occupies a

space equal to itself is at rest. Thus the arrow is at rest.

From the time of Zeno to the end of the 19th century, all those who argued

about Zenoā™s paradoxes whether they considered them ā˜funny little riddlesā™

or deep problems did not doubt that, in fact, the arrow did have a position

and velocity and did, indeed, travel along some path from A to B. Today

we are not so sure. In the theory of quantum mechanics it is impossible to

measure the position and momentum of a particle simultaneously to more

3

Plausible statement B.10 is relevant here.

28 A COMPANION TO ANALYSIS

than a certain accuracy. But ā˜plain honest menā™ are uninterested in what they

cannot measure. It is, of course, possible to believe that the particle has an

exact position and momentum which we can never know, just as it is possible

to believe that the earth is carried by invisible elephants standing on an

unobservable turtle, but it is surely more reasonable to say that particles do

not have position and momentum (and so do not have position and velocity)

in the sense that our too hasty view of the world attributed to them. Again

the simplest interpretation of experiments like the famous two slit experiment

which reveal the wavelike behaviour of particles is that particles do not travel

along one single path but along all possible paths.

A proper modesty should reduce our surprise that the real world and the

world of our thoughts should turn out to be more complicated than we ļ¬rst

expected.

Two things ļ¬ll the mind with ever-fresh admiration and rev-

erence, the more often and the more enduringly the mind is oc-

cupied with them: the starry heaven above me and the moral law

within me. [Kant, Critique of Practical Reason]

We cannot justify results like the intermediate value theorem by an appeal

to our fallible intuition or an imperfectly understood real world but we can

try to prove them from axioms. Those who wish may argue as to whether

and in what sense those axioms are ā˜trueā™ or ā˜a model for realityā™ but these

are not mathematical problems.

A note on Zeno We know practically nothing about Zeno except that he wrote

a book containing various paradoxes. The book itself has been lost and we

only know the paradoxes in the words of other Greek philosophers who tried

to refute them. Plato wrote an account of discussion between Socrates, Zeno

and Zenoā™s teacher Parmenides but it is probably ļ¬ctional. The most that we

can hope for is that, like one of those plays in which Einstein meets Marilyn

Monroe, it remains true to what was publicly known.

According to Plato:-

Parmenides was a man of distinguished appearance. By that time

he was well advanced in years with hair almost white; he may

have been sixty-ļ¬ve. Zeno was nearing forty, a tall and attractive

ļ¬gure. It was said that he had been Parmenidesā™ lover. They

were staying with Pythadorus . . . . Socrates and a few others

came there, anxious to hear a reading of Zenoā™s treatise, which

the two visitors had brought for the ļ¬rst time to Athens.

Parmenides taught that what is must be whole, complete, unchanging

29

Please send corrections however trivial to twk@dpmms.cam.ac.uk

and one. The world may appear to be made of many changing things but

change and plurality are illusions. Zeno says that his book is

. . . a defense of Parmenides argument against those who try to

make fun of it by showing that his supposition, that [only one

thing exists] leads to many absurdities and contradictions. This

book, then, is a retort against those who assert a plurality. It pays

them back in the same coin with something to spare, and aims

at showing that on a thorough examination, the assumption that

there is a plurality leads to even more absurd consequences than

the hypothesis of the one. It was written in that controversial

spirit in my young days . . . [40]

Many historians of mathematics believe that Zenoā™s paradoxes and the

discussion of the reasoning behind them were a major factor in the devel-

opment of the Greek method of mathematical proof which we use to this

day.

Chapter 3

Other versions of the

fundamental axiom

Since all of analysis depends on the fundamental axiom, it is not surprising

that mathematicians have developed a number of diļ¬erent methods of proof

to exploit it. We have already seen the method of ā˜lion huntingā™. In this chap-

ter we see two more: the ā˜supremum methodā™ and the ā˜Bolzano-Weierstrass

methodā™.

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