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A Companion to Analysis
A Second First and First Second Course in Analysis


T.W.K¨rner
o
Trinity Hall
Cambridge


Note This is the ¬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 ˜I am sure someone else has noticed this™ or ˜This
is too minor to matter™. Everybody notices di¬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 ¬nd more di¬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 e¬ortless ease with which he obtained his results. The fact is that
no amount of mental e¬ort of his own would enable a man to hit upon the
proof of one of Archimedes™ 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-Kilvert™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 a¬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 a¬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 ¬rst as-
cents, contrasting him with I. M. Gel´fand whose work I compared with the
building of highways, both men were o¬ended. ˜ . . . Why, you don™t think
I am capable of creating general theories?™ said Andre˜ Nikolaevich. ˜Why,
±
you think I can™t solve di¬cult problems?™ added I. M.
V. I. Arnol´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 ¬rst philosophical interlude ™™ 25
2.1 Is the intermediate value theorem obvious? ™™ . . . . . . . . 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 ™ 75
5.1 Comparison tests ™ . . . . . . . . . . . . . . . . . . . . . . . . 75

iii
iv A COMPANION TO ANALYSIS

Conditional convergence ™ . . . . . . .
5.2 . . . . . . . . . . . . . 78
Interchanging limits ™ . . . . . . . . .
5.3 . . . . . . . . . . . . . 83
The exponential function ™ . . . . . .
5.4 . . . . . . . . . . . . . 91
The trigonometric functions ™ . . . . .
5.5 . . . . . . . . . . . . . 98
The logarithm ™ . . . . . . . . . . . .
5.6 . . . . . . . . . . . . . 102
Powers ™ . . . . . . . . . . . . . . . .
5.7 . . . . . . . . . . . . . 109
The fundamental theorem of algebra ™
5.8 . . . . . . . . . . . . . 113

6 Di¬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 ™
8.4 . . . . . . . . . . . . 190
8.5 Vector-valued integrals . . . . . . . . . . . . . . . . . . . . . . 202

9 Developments and limitations ™ 205
9.1 Why go further? . . . . . . . . . . . . . . . . . . . . . . . . . 205
9.2 Improper integrals ™ . . . . . . . . . . . . . . . . . . . . . . . 207
9.3 Integrals over areas ™ . . . . . . . . . . . . . . . . . . . . . . 212
9.4 The Riemann-Stieltjes integral ™ . . . . . . . . . . . . . . . . 217
9.5 How long is a piece of string? ™ . . . . . . . . . . . . . . . . . 224

10 Metric spaces 233
10.1 Sphere packing ™ . . . . . . . . . . . . . . . . . . . . . . . . . 233
10.2 Shannon™s theorem ™ . . . . . . . . . . . . . . . . . . . . . . . 236
10.3 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
10.4 Norms, algebra and analysis . . . . . . . . . . . . . . . . . . . 246
10.5 Geodesics ™ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 ™ . . . . . . . . . . . . . . . . . . . . . . . . . . 298

12 Contractions and di¬erential equations 303
12.1 Banach™s contraction mapping theorem . . . . . . . . . . . . . 303
12.2 Solutions of di¬erential equations . . . . . . . . . . . . . . . . 305
12.3 Local to global ™ . . . . . . . . . . . . . . . . . . . . . . . . . 310
12.4 Green™s function solutions ™ . . . . . . . . . . . . . . . . . . . 318

13 Inverse and implicit functions 329
13.1 The inverse function theorem . . . . . . . . . . . . . . . . . . 329
13.2 The implicit function theorem ™ . . . . . . . . . . . . . . . . . 339
13.3 Lagrange multipliers ™ . . . . . . . . . . . . . . . . . . . . . . 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? ™ . . . . . . . . . . . . . . . . 364
14.4 How do we construct the reals? ™ . . . . . . . . . . . . . . . . 369
14.5 Paradise lost? ™™ . . . . . . . . . . . . . . . . . . . . . . . . 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 ¬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. ˜ . . . 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.™ ˜Naturally™
he says ˜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.™
This book is intended for those students who might ¬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
¬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, di¬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 ¬nds the discussion of section 1.1 ba¬„ing or the , δ
arguments of section 1.2 novel will probably ¬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 ¬lled with spectacle and the air ¬lled with music you must come at
another time and with a di¬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-
¬culty, show why they are points of di¬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 di¬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 ˜internalise™ the material but I hope that even students who skip
most of the exercises can pro¬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 ˜makeweights™, 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 signi¬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 ™. 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 ™™ which contain some, very simple, philosophical discussion. It is
entirely intentional that removing the appendices and the sections marked
with a ™ more than halves the length of the book.
My ¬rst glimpse of analysis was in Hardy™s Pure Mathematics [23] read
when I was too young to really understand it. I learned elementary analysis
from Ferrar™s A Textbook of Convergence [17] (an excellent book for those
making the transition from school to university, now, unfortunately, out of
print) and Burkill™s A First Course in Mathematical Analysis [10]. The books
of Kolmogorov and Fomin [30] and, particularly, Dieudonn´ [13] showed me
e
that analysis is not a collection of theorems but a single coherent theory.
Stromberg™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
in¬‚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
´
lectures at the Ecole Polytechnique in the 1820™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 insu¬ciently prepared and then
divided its time between radical politics and worrying about the job market
(grim for both sta¬ and students). Cauchy™s course was not popular1 .
Everybody can sympathise with Cauchy™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 Cauchy™s de¬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 sciences™ much vaunted
claims of exactitude. It should also be observed that they tend to
attribute an inde¬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 ¬nd myself forced to depend on several propositions
that perhaps seem a little hard on ¬rst encounter . . . . But, those
who will read them will ¬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 bene¬t analysis and
will also provide a number of topics for research, which are surely
not without importance.




1
Belhoste™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 di¬cult
proofs of obvious theorems. All we need know, they say, is what a limit is,
the de¬nition of continuity and the de¬nition of the derivative. All the rest
is ˜intuitively clear™1 .
If pressed they will agree that the de¬nition of continuity and the de¬ni-
tion of the derivative apply as much to the rationals Q as to the real numbers
R. If you disagree, take your favorite de¬nitions and examine them to see
where they require us to use R rather than Q. Let us examine the workings
of our ˜clear intuition™ 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 (——) the most general solution of
(—)?
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 ¬nd another
solution?
After a little thought she may observe that if g(t) is a solution of (—) and
we set
t3
f (t) = g(t) ’
3
then f (t) = 0 and the statement that (——) is the most general solution of
(—) reduces to the following theorem.
Theorem 1.1.1. (Constant value theorem.) If f : R ’ R is di¬eren-
tiable and f (t) = 0 for all t ∈ R, then f is constant.
If this theorem is ˜intuitively clear™ over R it ought to be intuitively clear
over Q. The same remark applies to another ˜intuitively clear™ theorem.
Theorem 1.1.2. (The intermediate value theorem.) If f : R ’ R is
continuous, b > a and f (a) ≥ 0 ≥ f (b), then there exists a c with b ≥ c ≥ a.
such that f (c) = 0.
However, if we work over Q both putative theorems vanish in a pu¬ of
smoke.
Example 1.1.3. If f : Q ’ Q is given by
if x2 < 2,
f (x) = ’1
f (x) = 1 otherwise,
then
(i) f is a continuous function with f (0) = ’1, f (2) = 1, yet there does
not exist a c with f (c) = 0,
(ii) f is a di¬erentiable function with f (x) = 0 for all x, yet f is not
constant.
Sketch proof. We have not yet formally de¬ned what continuity and di¬eren-
tiability are to mean. However, if the reader believes that f is discontinuous,
she must ¬nd a point x ∈ Q at which f is discontinuous. Similarly, if she
believes that f is not everywhere di¬erentiable with derivative zero, she must
¬nd a point x ∈ 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
de¬ned.
The question ˜Is (——) the most general solution of (—)?™ 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
Please send corrections however trivial to twk@dpmms.cam.ac.uk

This example is horrifying indeed. For if we have actually seen
one tiger, is not the jungle immediately ¬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(’8/5) > g(’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 ˜ordered ¬elds™, that is, both support operations
of ˜addition™ and ˜multiplication™ together with a relation ˜greater than™ (˜or-
der™) 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 ¬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 ¬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 ¬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.

De¬nition 1.2.1. We work in an ordered ¬eld F. We say that a sequence
a1 , a2 , . . . tends to a limit a as n tends to in¬nity, or more brie¬‚y

an ’ a as n ’ ∞

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

|an ’ a| < for all n ≥ 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 ¬eld F.
(i) The limit is unique. That is, if an ’ a and an ’ b as n ’ ∞, then
a = b.
(ii) If an ’ a as n ’ ∞ and 1 ¤ n(1) < n(2) < n(3) . . . , then an(j) ’ a
as j ’ ∞.
(iii) If an = c for all n, then an ’ c as n ’ ∞.
(iv) If an ’ a and bn ’ b as n ’ ∞, then an + bn ’ a + b.
(v) If an ’ a and bn ’ b as n ’ ∞, then an bn ’ ab.
(vi) Suppose that an ’ a as n ’ ∞. If an = 0 for each n and a = 0,
then a’1 ’ a’1 .
n
(vii) If an ¤ A for each n and an ’ a as n ’ ∞, then a ¤ A. If bn ≥ B
for each n and bn ’ b, as n ’ ∞ then b ≥ 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 de¬nition:-
Given > 0 we can ¬nd an n1 ( ) such that |an ’ a| < for all n ≥ n1 ( ).
Given > 0 we can ¬nd an n2 ( ) such that |an ’ b| < for all n ≥ n2 ( ).
Suppose, if possible, that a = b. Then setting = |a ’ b|/3 we have > 0. If
N = max(n1 ( ), n2 ( )) then

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

which is impossible. The result follows by reductio ad absurdum.
(ii) By de¬nition,

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

Let > 0. Since n(j) ≥ j (proof by induction, if the reader demands a proof)
we have |an(j) ’ a| < for all j ≥ n1 ( ). The result follows.
(iii) Let > 0. Taking n1 ( ) = 1 we have

|an ’ c| = 0 <

for all n ≥ n1 ( ). The result follows.
(iv) By de¬nition, holds as does

Given > 0 we can ¬nd an n2 ( ) such that |bn ’ b| < for all n ≥ n2 ( ).
( )
5
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Observe that

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

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

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

for all n ≥ n3 ( ). The result follows.
(v) By de¬nition, and hold. Let > 0. The key observation is
that

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

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

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

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

|an bn ’ ab| < /2 + /2 =

for all n ≥ n3 ( ). The result follows.
(vi) By de¬nition, holds. Let > 0. We observe that

|a ’ an |
1 1
’ = . (3)
|a||an |
an a

Since a = 0 we have |a|/2 > 0. If n ≥ n1 (|a|/2) then |an ’ a| < |a|/2 so
|an | > |a|/2 and (3) gives

2|a ’ an |
1 1
’ ¤ . (4)
|a|2
an a

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

1 1
’ <
an a

for all n ≥ n3 ( ). The result follows.
2
The reader may ask why we use n1 ( /(2(|b| + 1)) rather than n1 ( /(2|b|)). Observe
¬rst that we have not excluded the possibility that b = 0. More importantly, observe that
all we are required to do is to ¬nd an n3 ( ) that works and is futile to seek a ˜best™ n3 ( )
in these or similar circumstances.
6 A COMPANION TO ANALYSIS

(vii) The proof of the ¬rst sentence in the statement is rather similar to
that of (i). By de¬nition, holds. Suppose, if possible, that a > A, that is,
a ’ A > 0. Setting N = n1 (a ’ A) we have

aN = (aN ’ a) + a ≥ a ’ |aN ’ a| > a ’ (a ’ 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 = ’bn , a = ’b and A = ’B and use the ¬rst sentence.
[Your attention is drawn to part (ii) of Exercise 1.2.4.]

Exercise 1.2.3. Prove that the ¬rst few terms of a sequence do not a¬ect
convergence. Formally, show that if there exists an N such that an = bn for
n ≥ N then, an ’ a as n ’ ∞ implies bn ’ a as n ’ ∞.

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 ∈ Q and > 0, then = m/N for some strictly posi-
tive integers m and N . Use this fact to show, directly from De¬nition 1.2.1,
that (if we work in Q) 1/n ’ 0 as n ’ ∞.
(ii) Show, by means of an example, that, if an ’ 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 ≥ 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 ’ b| < for all > 0, then a = b.
(ii) Show that if an ’ a and an ’ b as n ’ ∞, then |a ’ b| < for all
> 0.
(iii) Deduce Lemma 1.2.2 (i).
(iv) Give a similar ˜split proof ™ 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 ¬rst principles that, if an ’ a, then can ’ ca.
(ii) Show from ¬rst principles that, if an ’ a as n ’ ∞, then a2 ’ a2 .
n
2 2
(iii) Use the relation xy = ((x + y) ’ (x ’ 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 ≥ cn ≥ bm for all m. Then, if an ’ c and
bn ’ c, it follows that cn ’ c as n ’ ∞.
Suppose |am | ≥ |cm | ≥ |bm | for all m and that an ’ c and bn ’ c as
n ’ ∞. Does it follow that cn ’ c? Give a proof or counterexample as
appropriate.


1.3 Continuity
Our de¬nition of continuity follows the same line of thought.
De¬nition 1.3.1. We work in an ordered ¬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 ∈ E if given any > 0 we can ¬nd δ0 ( , x) [read ˜δ0 depending on and
x™] with δ0 ( , x) > 0 such that

|f (x) ’ f (y)| < for all y ∈ E such that |x ’ y| < δ0 ( , x).

If f is continuous at every point of E we say that f : E ’ F is a continuous
function.
The reader, who, I expect, has seen this de¬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 de¬nition does not
correspond to their idea of what continuous should mean. If we consult
the dictionary we ¬nd the de¬nition ˜connected, unbroken; uninterrupted in
time or sequence: not discrete™. A mathematician would object that this
merely de¬nes one vague concept in terms of other equally vague concepts.
However, if we rephrase our own de¬nition in words we see that it becomes
˜f is continuous if f (y) is close to f (x) whenever y is su¬ciently close to
x™ and this clearly belongs to a di¬erent circle of ideas from the dictionary
de¬nition.
This will not be a problem when we come to de¬ne di¬erentiability since
there is no ˜common sense™ notion of di¬erentiability. In the same way the
existence of a ˜common sense™ notion of continuity need not trouble us pro-
vided that whenever we use the word ˜continuous™ we add under our breath
˜in the mathematical sense™ and we make sure our arguments make no appeal
(open or disguised) to ˜common sense™ ideas of continuity.
Here are some simple properties of continuity.
Lemma 1.3.2. We work in an ordered ¬eld F. Suppose that E is a subset
of F, that x ∈ E, and that f and g are functions from E to F.
(i) If f (x) = c for all x ∈ 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 de¬ne f — g : E ’ F by f — g(t) = f (t)g(t) for all t ∈ E.
Then if f and g are continuous at x, so is f — g.
(iv) Suppose that f (t) = 0 for all t ∈ 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 ¬eld F. Prove the following results.
(i) Suppose that E is a subset of F and that f : E ’ F is continuous at
x ∈ E. If x ∈ E ‚ E then the restriction f |E of f to E is also continuous
at x.
(ii) If J : F ’ F is de¬ned by J(x) = x for all x ∈ 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 ¬eld F. Suppose that E is a subset
of F, that x ∈ E, and that f is continuous at x. If xn ∈ E for all n and
xn ’ x as n ’ ∞, then f (xn ) ’ f (x) as n ’ ∞.
Proof. Left to reader.
We have now done quite a lot of what is called , δ 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 de¬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| ¤ 2 and |·| ¤ 1 show that |(x + ·)2 ’ x2 | ¤ 5|·|.
(iii) If x2 < 2 and δ = (2 ’ x2 )/6 show that y 2 < 2 whenever |x ’ y| < δ.
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 ∈ R for each n ≥ 1, A ∈ R
and a1 ¤ a2 ¤ a3 ¤ . . . and an < A for each n, then there exists an a ∈ R
such that an ’ a as n ’ ∞.

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 ˜algebraic™ 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 ∈ R : a ¤ x ¤ b}. )

Theorem 1.4.1. (The intermediate value theorem.) If f : [a, b] ’ R
is continuous and f (a) ≥ 0 ≥ f (b) then there exists a c ∈ [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:“
If f : [a, b] ’ R is continuous and f (a) ≥ t ≥ f (b) then there exists a
c ∈ [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 ¬ll in
the gaps herself. For example, in the statement of Theorem 1.4.1, I do not
explicitly say that b ≥ a. Again, I talk about the ˜algebraic™ properties of R
when, strictly speaking, a set cannot have algebraic properties and I should
refer instead to the algebraic properties of (R, +, —, >). 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 ¬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 ’a1 , ’a2 , ’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 ≥ N .
(ii) Show that every non-empty set A ⊆ Z bounded above has a maximum.
More formally, show that if A ⊆ Z, A = … and there exists a K such that
K ≥ a whenever a ∈ A then there exists an a0 ∈ A with a0 ≥ a whenever
a ∈ A.


1.5 The axiom of Archimedes
Our ¬rst genuinely ˜analysis™ result may strike the reader as rather odd.
Theorem 1.5.1. (Axiom of Archimedes.)
1
’ 0 as n ’ ∞
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
=—’
2n 2n 2
and since the limit of a subsequence is the limit of the sequence (Lemma 1.2.2 (ii))
1
’l
2n
11
Please send corrections however trivial to twk@dpmms.cam.ac.uk

as n ’ ∞. Thus, by the uniqueness of limits (Lemma 1.2.2 (i)), l = l/2 so
l = 0 and 1/n ’ 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 ’ 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 ≥ b? Give reasons.

Theorem 1.5.1 shows that there is no ˜exotic™ real number say (to choose an
exotic symbol) with the property that 1/n > for all integers n ≥ 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
˜non-standard analysis™ of Abraham Robinson and the ˜surreal numbers™ 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 ≥ 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 ∈ R with K > n for
all n ∈ 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 ¤ a < 1 then an ’ 0 as
n ’ ∞. Deduce that if a is real and |a| < 1 then an ’ 0.
(ii) Suppose that a is real and a = ’1. Discuss the behaviour of

1 ’ an
1 + an
as n ’ ∞ for the various possible values of a.
[ Hint (1 ’ an )/(1 + an ) = (a’n ’ 1)/(a’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 ∈ R, then there
exists an integer m such that m ¤ x < m + 1. Show that |x ’ m| < 1.
3
This is a simpli¬cation of a more complex story.
12 A COMPANION TO ANALYSIS

(ii) If x ∈ R and n is a strictly positive integer, show that there exists an
integer q such that |x ’ q/n| < 1/n.
(iii) Deduce Lemma 1.5.6 below, using the axiom of Archimedes explicitly.
Lemma 1.5.6. If x ∈ R, then, given any > 0, there exists a y ∈ Q such
that |x ’ 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 de¬-
nition.
De¬nition 1.5.7. If a1 , a2 , . . . is a sequence of real numbers, we say that
an ’ ∞ as n ’ ∞ if, given any real K, we can ¬nd an n0 (K) such that
an ≥ K for all n ≥ n0 (K).
Exercise 1.5.8. Using Exercise 1.5.3 show that n ’ ∞ as n ’ ∞.
Exercise 1.5.8 shows that two uses of the words ˜n tends to in¬nity™ 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 De¬ni-
tion 1.5.7.
Exercise 1.5.9. Let a1 , a2 , . . . be a sequence of non-zero real numbers.
Show that, if an ’ ∞, then 1/an ’ 0. Is the converse true? Give a proof
or counterexample.
Exercise 1.5.10. It is worth noting explicitly that ordered ¬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 ¬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 coe¬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 ¬eld (that is,
it satis¬es conditions (A1) to (D) as set out in the axioms on page 379).
To produce an order on K we de¬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
de¬nition (remember that the same quotient will have many di¬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 satis¬es conditions (P1) to (P3). If we de¬ne
P1 (X)/Q1 (X) > P2 (X)/Q2 (X) whenever P1 (X)/Q1 (X)’P2 (X)/Q2 (X) ∈ P
condition (P4) is automatically satis¬ed and we have indeed got an ordered
¬eld.
We note that the elements of K of the form a/1 with a ∈ R can be
identi¬ed in a natural way with R. If we make this natural identi¬cation, K
contains Z.
To see that the axiom of Archimedes fails, observe that 1/n > 1/X > 0
for all n ∈ Z, n ≥ 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 de¬ne K using equiv-
alence classes. We should take greater care over checking consistency. The
words ˜identi¬ed in a natural way™ should be replaced by ˜there is an isomor-
phism of (R, +, —, >) with a sub¬eld of K™.) 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 ∈ Z : 9 ≥ n ≥ 0}.
(i) If xj ∈ D show that N xj 10’j ¤ 1.
j=1
N
(ii) If xj ∈ D show that ’j
j=1 xj 10 converges to a limit x, say, as
N ’ ∞. Show that 0 ¤ x ¤ 1 and that x = 1 if and only if xj = 9 for all j.
(iii) If y ∈ [0, 1] show that there exist yj ∈ D such that
N
y ’ 10’N < yj 10’j ¤ y
j=1

and that N yj 10’j ’ y as N ’ ∞.
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 ∈ D, aj = bj for j < M and aM > bM . If
N
’ a and N bj 10’j ’ b as N ’ ∞ show that a ≥ b. Give
’j
j=1 aj 10 j=1
the precise necessary and su¬cient condition for equality and prove it.
14 A COMPANION TO ANALYSIS

Exercise 1.5.13. It seems, at ¬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 10’j ’ a, N bj 10’j ’ b, and N cj 10’j ’ c ¬nd cj
j=1 j=1 j=1
in terms of the various ak and bk . (The reader is invited to re¬‚ect on this
problem rather than solve it. Indeed, one question is ˜what would constitute
a nice solution?™)
Exercise 1.5.14. Here is a neat application of decimal expansion.
(i) De¬ne f : [0, 1] ’ [0, 1] as follows. Each x ∈ [0, 1] has a unique
non-terminating decimal expansion

xj 10’j
x=
j=1

with the xj integers such that 0 ¤ xj ¤ 9. If there exists an integer N ≥ 2
such that x2j = 1 for all j ≥ N but x2N ’2 = 1 we set

x2(j+N )+1 10’j .
f (x) =
j=1

Otherwise we set f (x) = 0. Show that given any y ∈ [0, 1], any > 0 and
any t ∈ [0, 1] we can ¬nd an x ∈ [0, 1] with |x ’ 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 ¤ a < b ¤ 1 then, given any t lying between f (a)
and f (b) (that is to say, t with f (a) ¤ t ¤ f (b) if f (a) ¤ f (b) or with
f (b) ¤ t ¤ f (a) if f (b) ¤ f (a)), there exists a c ∈ [a, b] such that f (c) = t.
(Thus the fact that a function satis¬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 ’ [0, 1] such that, given any y ∈ R, any > 0 and any
t ∈ [0, 1], we can ¬nd an x with |x ’ y| < with f (x) = t.
(iv) (This may require a little thought.) Find a g : R ’ R such that given
any y ∈ R, any > 0 and any t ∈ R we can ¬nd an x with |x ’ 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] ’ R is continuous and f (a) ≥ 0 ≥ f (b), then there exists a c ∈ [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 ) ≥ 0 ≥ f (b0 ). Now set
c0 = (a0 + b0 )/2. There are two possibilities. Either f (c0 ) ≥ 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 ) ≥ 0 ≥ f (b1 ),
a0 ¤ a 1 ¤ b 1 ¤ b 0 ,
and b1 ’ a1 = (b0 ’ a0 )/2.

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

f (an ) ≥ 0 ≥ f (bn ),
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.
Part B We have a0 ¤ a1 ¤ · · · ¤ an ¤ b0 so that the an form an increasing
sequence bounded above. By the fundamental axiom there is real number c,
say, such that an ’ c as n ’ ∞. Since a = a0 ¤ an ¤ b0 we have a ¤ c ¤ b.
We note also that bn ’ an = 2’n (b0 ’ a0 ) so, by the axiom of Archimedes,
bn ’ an ’ 0 and thus

bn = an + (bn ’ an ) ’ c + 0 = c

as n ’ ∞.
Part C Since f is continuous at c and an ’ c, it follows that f (an ) ’ f (c)
as n ’ ∞. Since f (an ) ≥ 0 for all n, it follows that f (c) ≥ 0. A similar
argument applied to the bn shows that f (c) ¤ 0. Since 0 ¤ f (c) ¤ 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 ˜similar argument applied to the bn ™ which
shows that f (c) ¤ 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 ’ 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 ’ 5x)(2 ’ 5x)(3 ’ 5x).

The method used to prove Theorem 1.6.1 is called ˜Lion hunting™4 . The
method is also called ˜successive bisection™, ˜bisection search™ or simply ˜bi-
section™.
Let us summarise the proof. In Part A we have evidence of a lion in the
interval [an’1 , bn’1 ]. We split the interval into two halves [an’1 , cn’1 ] and
[cn’1 , bn’1 ] and show that, since there is evidence of a lion in the interval
[an’1 , bn’1 ], either there is evidence of a lion in [an’1 , cn’1 ], in which case we
take [an , bn ] = [an’1 , cn’1 ], or, if there is no evidence of a lion in [an’1 , cn’1 ]
(this does not mean that there are no lions in [an’1 , cn’1 ], simply that we do
not have evidence of one), then it follows that there must be evidence of a
lion in [cn’1 , bn’1 ] and we take [an , bn ] = [cn’1 , bn’1 ].
In Part B we use the fundamental axiom of analysis to show that these
successive bisections ˜close in™ 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) ≥ 0 ≥ 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) ≥ 0 ≥ f (b) but replace R by another ordered ¬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] = [’2, 0].)
If we have f (a) ≥ 0 ≥ 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 de¬ne f (x) = 1 for x ¤ 1/3 and f (x) = ’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. P´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] ’ [0, 1] is a continuous function. By
considering f (x) = g(x) ’ x, or otherwise, show that there exists a c ∈ [0, 1]
with g(c) = c. (Thus every continuous map of [0, 1] into itself has a ¬xed
point.)
Give an example of a bijective continuous function k : (0, 1) ’ (0, 1) such
that k(x) = x for all x ∈ (0, 1).
Give an example of a bijective (but, necessarily, non-continuous) function
h : [0, 1] ’ [0, 1] such that h(x) = x for all x ∈ [0, 1].
[Hint: First ¬nd a function H : [0, 1] \ {0, 1, 1/2} ’ [0, 1] \ {0, 1, 1/2}
such that H(x) = x.]

Exercise 1.6.6. Every mid-summer day at six o™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 o™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 ¬nd a sequence of
pairs of points an and bn such that

either xn ≥ bn + (bn’1 ’ an’1 )/3 or xn ¤ an ’ (bn’1 ’ an’1 )/3,
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/3,

for all n ≥ 1.
Show that an and bn tend to some limit c ∈ [a0 , b0 ]. Show further that,
for each n ≥ 1, either xn ≥ c + (bn’1 ’ an’1 )/3 or xn ¤ c ’ (bn’1 ’ an’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 ˜mean value inequality™.
Theorem 1.7.1. (The mean value inequality.) Let U be the open inter-
val (±, β) on the real line R. Suppose that K ≥ 0 and that a, b ∈ U with
b > a. If f : U ’ R is di¬erentiable with f (t) ¤ K for all t ∈ U then

f (b) ’ f (a) ¤ (b ’ a)K.

Before we can do this we must de¬ne di¬erentiability and the deriva-
tive. The reader will almost certainly be familiar with a de¬nition along the
following lines.
De¬nition 1.7.2. Let U be an open set in R. We say that a function f :
U ’ R is di¬erentiable at t ∈ U with derivative f (t) if, given > 0, we can
¬nd a δ(t, ) > 0 such that (t ’ δ(t, ), t + δ(t, )) ⊆ U and
f (t + h) ’ f (t)
’ f (t) <
h
whenever 0 < |h| < δ(t, ).
In Chapter 6 we shall de¬ne a more general notion of di¬erentiation and
derive many of its properties. For the moment all we need is De¬nition 1.7.2.
Exercise 1.7.3. Let U be an open interval in R and suppose functions f, g :
U ’ R are di¬erentiable at t ∈ U with derivatives f (t) and g (t). If », µ ∈
R, show that »f + µg is di¬erentiable at t with derivative »f (t) + µg (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) ’ f (a) ¤ (K + )(b ’ a).
Proof of Theorem 1.7.1 from Lemma 1.7.4. Since f (b)’f (a) ¤ (K+ )(b’a)
for all > 0, it follows that f (b) ’ f (a) ¤ K(b ’ a).
Proof of Lemma 1.7.4. We suppose that f (b) ’ f (a) > (K + )(b ’ a) and
use lion-hunting to derive a contradiction. Set a0 = a, b0 = b. We observe
that f (b0 ) ’ f (a0 ) > (K + )(b0 ’ a0 ). Now set c0 = (a0 + b0 )/2. Since

f (c0 ) ’ f (a0 ) ’ (K + )(c0 ’ a0 ) + f (b0 ) ’ f (c0 ) ’ (K + )(b0 ’ c0 )
= f (b0 ) ’ f (a0 ) ’ (K + )(b0 ’ a0 ) > 0,
19
Please send corrections however trivial to twk@dpmms.cam.ac.uk

at least one of the expressions f (c0 ) ’ f (a0 ) ’ (K + )(c0 ’ a0 ) and (f (b0 ) ’
f (c0 ) ’ (K + )(b0 ’ c0 ) must be strictly positive. If f (b0 ) ’ f (c0 ) ’ (K +
)(b0 ’ c0 ) > 0, we set a1 = c0 and b1 = b0 . Otherwise, we set a1 = a0 and
b1 = c0 . In either case, we have
f (b1 ) ’ f (a1 ) > (K + )(b1 ’ a1 ),
a0 ¤ a 1 ¤ b 1 ¤ b 0 ,
and b1 ’ a1 = (b0 ’ a0 )/2.
Continuing inductively, we can ¬nd a sequence of pairs of points an and
bn such that
f (bn ) ’ f (an ) > (K + )(bn ’ an ),
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/2,
for all n ≥ 1.
We have a0 ¤ a1 ¤ · · · ¤ an ¤ b0 so that the an form an increasing
sequence bounded above. By the fundamental axiom there is real number c,
say, such that an ’ c as n ’ ∞. Since a = a0 ¤ an ¤ b0 we have a ¤ c ¤ b
and similarly aN ¤ c ¤ bN for all N . We note also that bn ’an = 2’n (b0 ’a0 )
so, by the axiom of Archimedes, bn ’ an ’ 0 and thus
bn = an + (bn ’ an ) ’ c + 0 = c
as n ’ ∞.
Since f is di¬erentiable at c, we can ¬nd a δ > 0 such that (c’δ, c+δ) ⊆ U
and
f (c + h) ’ f (c)
’ f (c) < /2
h
whenever 0 < |h| < δ. Thus
|f (c + h) ’ f (c) ’ f (c)h| ¤ |h|/2
whenever |h| < δ, and so, since f (c) ¤ K,
f (c + h) ’ f (c) ¤ (K + /2)h for 0 ¤ h < δ
f (c) ’ f (c + h) ¤ ’(K + /2)h for ’δ ¤ h ¤ 0
Since an ’ c and bn ’ c, we can ¬nd an N such that |aN ’ c| < δ and
|bN ’ c| < δ. It follows, ¬rst taking h = aN ’ c and then h = bN ’ c, that
f (c) ’ f (aN ) ¤ (K + /2)(c ’ aN )
and f (bN ) ’ f (c) ¤ (K + /2)(bN ’ c).
20 A COMPANION TO ANALYSIS

Thus

f (bN ) ’ f (aN ) = f (bN ) ’ f (c) + f (c) ’ f (aN )
¤ (K + /2)(bN ’ c) + (K + /2)(c ’ aN )
= (K + /2)(bN ’ aN ),

contradicting our earlier assumption that f (bn ) ’ f (an ) > (K + )(bn ’ 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 (±, β) or the real line R. If f : U ’ R is di¬erentiable with f (t) = 0
for all t ∈ U , then f is constant.

Proof. Let b, a ∈ U with b ≥ a. By applying Theorem 1.7.1 to f with K = 0
we see that f (b) ’ f (a) ¤ 0. By applying Theorem 1.7.1 to ’f with K = 0,
we see that f (a) ’ f (b) ¤ 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 (±, β) or the real line R. If the
functions f, g : U ’ R are di¬erentiable with f (t) = g (t) for all t ∈ U then
there exists a constant c such that f (t) = g(t) + c for all t ∈ U .

Proof. Apply Theorem 1.7.5 to f ’ 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,
di¬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 ˜Hear,
hear™.
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 (±, β) or the real line R. Sup-
pose that a, b ∈ U and b > a. If g : U ’ R is di¬erentiable with g (t) ≥ 0
for all t ∈ U then

g(b) ’ g(a) ≥ 0.

Exercise 1.7.8. (i) By taking f = ’g and K = 0, prove Theorem 1.7.7
from Theorem 1.7.1.
(ii) By taking g(t) = Kt’f (t), prove Theorem 1.7.1 from Theorem 1.7.7.

Thus a function with positive derivative is increasing.
The converse result is ˜purely algebraic™ in the sense that it does not
involve the fundamental axiom.

Lemma 1.7.9. If g : (a, b) ’ R is di¬erentiable and increasing on (a, b)
then g (t) ≥ 0 for all t ∈ (a, b).

Exercise 1.7.10. Use the de¬nition of the derivative to prove Lemma 1.7.9.
[Hint: Show ¬rst that given any > 0 we have g (t) > ’ .]

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 ∈ R and b > a. If f : [a, b] ’ R is
continuous and f is di¬erentiable on (a, b) with f (t) ¤ K for all t ∈ (a, b),
use Theorem 1.7.1 on intervals (an , bn ) with a < an < bn < b and continuity
to show that

f (b) ’ f (a) ¤ (b ’ 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 ¬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 ¬eld for which the intermediate value
theorem holds, that is to say:
Let a, b ∈ F with b > a and set [a, b] = {x ∈ F ; b ≥ x ≥ a}. If
f : [a, b] ’ F is continuous and f (a) ≥ 0 ≥ f (b) then there exists a c ∈ [a, b]
such that f (c) = 0.
Then the fundamental axiom holds. That is to say:
If an ∈ F for each n ≥ 1, A ∈ F, a1 ¤ a2 ¤ a3 ¤ . . . and an < A for each
n then there exists an c ∈ F such that an ’ c as n ’ ∞.

Proof. Suppose a1 ¤ a2 ¤ a3 ¤ . . . and an < A for all n. Choose a < a1 and
b > A. De¬ne f : [a, b] ’ F by

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

Since f does not take the value 0, the intermediate value theorem tells us
that there must be a point c ∈ [a, b] at which f is discontinuous.
Suppose that y < aN for some N , so = aN ’ y > 0. Then, whenever
|x’y| < /2, we have x ¤ aN ’ /2, so f (x) = f (y) = 1 and |f (x)’f (y)| = 0.
Thus f is continuous at y. We have shown that c ≥ an for all n.
Suppose that there exists an > 0 such that y ≥ an + for all n. Then,
whenever |x ’ y| < /2, we have x ≥ an + /2 for all n, so f (x) = f (y) = ’1
and |f (x) ’ 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 ¤ c < an + and so |c ’ an | < for all n ≥ n0 ( ).
Thus an ’ c as n ’ ∞ 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
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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 ¬eld satisfying the fundamental axiom. The
answer, which is as good as we can hope for, is that all ordered ¬elds satisfying
the fundamental axiom are isomorphic. The formal statement is given in the
following theorem.

Theorem 1.9.1. If the ordered ¬eld (F, +, —, >) satis¬es the fundamental
axiom of analysis, then there exists a bijective map θ : R ’ F such that, if
x, y ∈ R, then

θ(x + y) = θ(x) + θ(y)
θ(xy) = θ(x)θ(y)
θ(x) > 0 whenever x > 0.

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

θ(x) > θ(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 su¬cient experience both in analysis and algebra she will ¬nd that
the proof of Theorem 1.9.1 writes itself. Until then it is not really worth the
time and e¬ort involved.
Chapter 2

A ¬rst philosophical interlude
™™

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 Keynes™ gibe that ˜Practical
men who believe themselves to be exempt from any intellectual in¬‚uences,
are usually the slaves of some defunct economist™ 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? ™™
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 ˜plain hon-
est men™ for over 2000 years by suggesting that the world may not be as
simple as ˜plain honest men™ 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 Dieudonn´ to name any mathematicians who had been
e
in¬‚uenced by any philosopher, instantly replied ˜Descartes and Leibniz™.


25
26 A COMPANION TO ANALYSIS

servation? Suppose we go to the cinema and watch the ¬lm of an arrow™s
¬‚ight. It certainly looks as though the arrow is in motion passing through
all the points of its ¬‚ight. But, if we examine the ¬lm, we see that it consists
of series of pictures of the arrow at rest in di¬erent positions. The arrow
takes up a ¬nite, though large, number of positions in its apparent ¬‚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 √ 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

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

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 ± .0005 metres should convince us of the existence of a

length 2. Of course we can imagine the process continued but few ˜plain
honest men™ would believe a craftsman who told them that because they
could work to an accuracy of ±.000 05 metres they could therefore work to
an accuracy of ±.000 000 005 metres ˜and so on™. Indeed if someone claimed
to have marked out a length of 1.414 213 562 373 095 metres to an accuracy
of ±5 — 10’16 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 di¬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 ’ 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 ¬‚ight of an actual physical arrow, let us close our eyes and
imagine the ¬‚ight of an ideal arrow from A to B. Here a di¬culty presents
itself. We can easily picture the arrow at A and at B but, if we wish to
imagine the complete ¬‚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 ¬‚ight
of the arrow we must see it at each of the in¬nite sequence of points A1 , A2 ,
A3 , . . . . Since an electronic computer can only perform a ¬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 ˜obvious™ 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 di¬culty when we try to picture the path of the arrow.
At ¬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 re¬‚ect
on it. It is due to Zeno but, as with his other ˜paradoxes™ we do not have
his own words. Consider the ¬‚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 Zeno™s paradoxes whether they considered them ˜funny little riddles™
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 ˜plain honest men™ 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 ¬rst
expected.

Two things ¬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 ˜true™ or ˜a model for reality™ 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 Zeno™s teacher Parmenides but it is probably ¬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-¬ve. Zeno was nearing forty, a tall and attractive
¬gure. It was said that he had been Parmenides™ lover. They
were staying with Pythadorus . . . . Socrates and a few others
came there, anxious to hear a reading of Zeno™s treatise, which
the two visitors had brought for the ¬rst time to Athens.

Parmenides taught that what is must be whole, complete, unchanging
29
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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 Zeno™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 di¬erent methods of proof
to exploit it. We have already seen the method of ˜lion hunting™. In this chap-
ter we see two more: the ˜supremum method™ and the ˜Bolzano-Weierstrass
method™.


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