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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.)
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
There are two ways in which the intermediate value theorem might be
obvious:- through observation and introspection. Is it obvious through ob-
My father on being asked by Dieudonn´ to name any mathematicians who had been
in¬‚uenced by any philosopher, instantly replied ˜Descartes and Leibniz™.


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.
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
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.
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
Plausible statement B.10 is relevant here.

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

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
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
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

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.


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

It is convenient to have the following alternative characterisation of the
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.

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
(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

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