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

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 ¬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

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 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™.

3.1 The supremum

Just as the real numbers are distinguished among all systems enjoying the

same algebraic properties (that is all ordered ¬elds) by the fundamental ax-

iom, so the integers are distinguished among all systems enjoying the same

algebraic properties (that is all ordered integral domains) by the statement

that every non-empty set of the integers bounded above has a maximum.

Well ordering of integers. 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.

(We proved this as a consequence of the fundamental axiom in Exer-

cise 1.4.4.)

The power of this principle is illustrated by the fact that it justi¬es the

method of mathematical induction.

Exercise 3.1.1. (i) State formally and prove the statement that every non-

empty set of integers bounded below has a minimum.

31

32 A COMPANION TO ANALYSIS

(ii) Suppose that P is some mathematical property which may be pos-

sessed by a positive integer n. Let P (n) be the statement that n possesses the

property P. Suppose that

(a) P (0) is true.

(b) If P (n) is true, then P (n + 1) is true.

By considering the set

E = {n ∈ Z : n ≥ 0 and P (n) is true},

and using (i), show that P (n) is true for all positive integers n.

(iii) Examine what happens to your proof of (ii) when P (n) is the state-

ment n ≥ 4, when P (n) is the statement n ¤ 4 and when P (n) is the

statement n = ’4.

Unfortunately, as the reader probably knows, a bounded non-empty set

of real numbers need not have a maximum.

Example 3.1.2. If E = {x ∈ R : 0 < x < 1}, then E is a non-empty

bounded set of real numbers with no maximum.

Proof. If a ≥ 1 or if a ¤ 0, then a ∈ E, so a is not a maximum. If 0 < a < 1,

/

then a < (a + 1)/2 ∈ E, so a is not a maximum.

However, we shall see in Theorem 3.1.7 that any non-empty bounded set

of real numbers does have a least upper bound (supremum).

De¬nition 3.1.3. Consider a non-empty set A of real numbers. We say

that ± is a least upper bound (or supremum) for A if the following two

conditions hold.

(i) ± ≥ a for all a ∈ A.

(ii) If β ≥ a for all a ∈ A, then β ≥ ±.

Lemma 3.1.4. If the supremum exists, it is unique.

Proof. The only problem is setting out the matter in the right way.

Suppose ± and ± least upper bounds for a non-empty set A of real num-

bers. Then

(i) ± ≥ a for all a ∈ A.

(ii) If β ≥ a for all a ∈ A, then β ≥ ±.

(ii ) ± ≥ a for all a ∈ A.

(ii ) If β ≥ a for all a ∈ A, then β ≥ ± .

By (i), ± ≥ a for all a ∈ A, so by (ii ), ± ≥ ± . Similarly, ± ≥ ±, so ± = ±

and we are done.

33

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

It is convenient to have the following alternative characterisation of the

supremum.

Lemma 3.1.5. Consider a non-empty set A of real numbers; ± is a supre-

mum for A if and only if the following two conditions hold.

(i) ± ≥ a for all a ∈ A.

(ii) Given > 0 there exists an a ∈ A such that a + ≥ ±.

Proof. Left to reader.

We write sup A or supa∈A a for the supremum of A, if it exists.

Exercise 3.1.6. Check that the discussion of the supremum given above car-

ries over to all ordered ¬elds. (There is no need to write anything unless you

wish to.)

Here is the promised theorem.

Theorem 3.1.7. (Supremum principle.) If A is a non-empty set of real

numbers which is bounded above (that is, there exists a K such that a ¤ K

for all a ∈ A), then A has a supremum.

Note that the result is false for the rationals.

Exercise 3.1.8. Let us work in Q. Using Exercise 1.3.5 (ii), or otherwise,

show that

{x ∈ Q : x2 < 2}

has no supremum.

We must thus use the fundamental axiom in the proof of Theorem 3.1.7.

One way to do this is to use ˜lion hunting™.

Exercise 3.1.9. (i) If A satis¬es the hypotheses of Theorem 3.1.7 show that

we can ¬nd a0 , b0 ∈ R with a0 < b0 such that a ¤ b0 for all a ∈ A but

[a0 , b0 ] © A = ….

(ii) Continuing with the discussion of (i) show that we can ¬nd a sequence

of pairs of points an and bn such that

a ¤ bn for all a ∈ A

[an , bn ] © A = …,

an’1 ¤ an ¤ bn ¤ bn’1 ,

and bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.

(iii) Deduce Theorem 3.1.7.

34 A COMPANION TO ANALYSIS

Here is another way of using the fundamental axiom to prove Theo-

rem 3.1.7. (However, if the reader is only going to do one of the two Exer-

cises 3.1.9 and 3.1.10 she should do Exercise 3.1.9.)

Exercise 3.1.10. (i) If A satis¬es the hypotheses of Theorem 3.1.7, explain

carefully why we can ¬nd an integer r(j) such that

r(j)2’j > a for all a ∈ A but

there exists an a(j) ∈ A with a(j) ≥ (r(j) ’ 1)2’j .

[Hint. Recall that every non-empty set of the integers bounded below has a

minimum.]

(ii) By applying the fundamental axiom to the sequence (r(j) ’ 1)2’j and

using the axiom of Archimedes, deduce Theorem 3.1.7.

We leave it to the reader to de¬ne the greatest lower bound or in¬mum

written inf A or inf a∈A a, when it exists.

Exercise 3.1.11. (i) De¬ne the greatest lower bound by modifying De¬ni-

tion 3.1.3. State and prove results corresponding to Lemmas 3.1.4 and 3.1.5.

(ii) Show that

inf a = ’ sup(’a),

a∈A a∈A

provided that either side of the equation exists.

(iii) State and prove a result on greatest lower bounds corresponding to

Theorem 3.1.7. (Part (ii) gives a short proof.)

As an example of how a ˜supremum argument™ can be used we reprove

the intermediate value theorem (Theorem 1.6.1).

Theorem 3.1.12. 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. Our proof will have three labelled main parts which may be compared

with those in the ˜lion hunting proof™ on page 15.

Part A Consider the set

E = {x ∈ [a, b] : f (x) ≥ 0}.

We observe that f (a) ≥ 0, so a ∈ E and E is non-empty. Since x ∈ E implies