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

x ¤ b, the set E is automatically bounded above.

Part B Since every non-empty set bounded above has a supremum, E has a

supremum, call it c.

35

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

Part C Let > 0. Since f is continuous at c we can ¬nd a δ > 0 such that

if x ∈ [a, b] and |x ’ c| < δ then |f (x) ’ f (c)| < . We must consider three

possible cases according as a < c < b, c = b or c = a.

If a < c < b, we proceed as follows. Since c = sup E we can ¬nd x0 ∈ E

such that 0 ¤ c ’ x0 < δ and so |f (x0 ) ’ f (c)| < . Since x0 ∈ E, f (x0 ) ≥ 0

and so f (c) ≥ ’ . On the other hand, choosing y0 = c + min(b ’ c, δ)/2 we

know that 0 ¤ y0 ’ c < δ and so |f (x0 ) ’ f (c)| < . Since y0 > c it follows

that y0 ∈ E so f (y0 ) < 0 and f (c) < . We have shown that |f (c)| ¤ .

/

If c = b, we proceed as follows. Since c = sup E, we can ¬nd x0 ∈ E such

that 0 ¤ c ’ x0 < δ and so |f (x0 ) ’ f (c)| < . Since x0 ∈ E, f (x0 ) ≥ 0 and

so f (c) ≥ ’ . By hypothesis f (b) ¤ 0 so |f (c)| ¤ .

If c = a we proceed as follows. Since c = a we know that f (c) ≥ 0. On

the other hand, choosing y0 = c + min(b ’ c, δ)/2 we know that 0 ¤ y0 ’ c < δ

and so |f (y0 ) ’ f (c)| < . Since y0 > c it follows that y0 ∈ E so f (y0 ) < 0

/

and f (c) < . We have shown that |f (c)| < .

In all three cases we have shown that |f (c)| ¤ for all > 0 so f (c) =

0.

Exercise 3.1.13. In both the ˜lion hunting™ and the ˜supremum argument™

proofs we end up with a point c where f (c) = 0, that is a ˜zero of f ™. Give an

example of a function f : [a, b] ’ R satisfying the hypotheses of the interme-

diate value theorem for which ˜lion hunting™ and the ˜supremum argument™

give di¬erent zeros.

Let us summarise the proof just given. In Part A we construct a set E

on which f has a certain kind of behaviour and show that E is bounded and

non-empty. In Part B we use the fact that E is bounded and non-empty to

give us a point c = sup E which is in some sense a ˜transition point™. Finally

in Part C we examine the point c to make sure that it really has the desired

property. Note that we had to examine the cases c = a and c = b separately.

This often happens when we use the ˜supremum argument™.

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

orem 3.1.12. If we omit the condition f (a) ≥ 0, then we cannot even start

Part A of the argument.

If we have f (a) ≥ 0 but replace R by another ordered ¬eld for which it is

not necessarily true that every non-empty bounded subset has a supremum,

Part A goes through perfectly but Part B fails.

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

continuous then Part C fails. Part C also fails if f (a) ≥ 0 and f is continuous

but we do not know that 0 ≥ f (b).

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

reprove Lemma 1.7.4 from which the mean value inequality (Theorem 1.7.1)

36 A COMPANION TO ANALYSIS

follows.

Lemma 3.1.14. Let U be the open interval (±, β) on the real line R. Suppose

that a, b ∈ U and b > a. If f : U ’ R is di¬erentiable with f (t) ¤ K for

all t ∈ U and > 0 then

f (b) ’ f (a) ¤ (b ’ a)(K + ).

Proof. Consider the set

E = {x ∈ [a, b] : f (t) ’ f (a) ¤ (K + )(t ’ a) for all a ¤ t ¤ x}.

We observe that f (a) ’ f (a) = 0 = (K + )(a ’ a) ≥ 0 so a ∈ E and E is

non-empty. Since x ∈ E implies x ¤ b, the set E is automatically bounded

above.

Since every non-empty set bounded above has a supremum, E has a

supremum, call it c.

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

We must consider three possible cases according as a < c < b, c = b or c = a.

If a < c < b we proceed as follows. If a ¤ t < c, then, by the de¬nition of

the supremum, we can ¬nd an x ∈ E with t < x ¤ c and so, by the de¬nition

of E,

f (t) ’ f (a) ¤ (K + )(t ’ a).

If c ¤ t < c + δ, then, choosing a t0 ≥ a with c > t0 > c ’ δ, we have

f (t0 ) ’ f (a) ¤ (K + )(t0 ’ a)

whilst, by the result of the previous paragraph,

f (c) ’ f (t0 ) ¤ (K + /2)(c ’ t0 )

f (t) ’ f (c) ¤ (K + /2)(t ’ c),

37

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

so, adding the last three inequalities, we obtain

f (t) ’ f (a) ¤ (K + )(t ’ a).

Thus f (t) ’ f (a) ¤ (K + )(t ’ a) for all a ¤ t < c + δ, contradicting the

de¬nition of c.

If c = a, then we know that f (t) ’ f (a) ¤ (K + /2)(t ’ a) for all

a ¤ t < a + δ, again contradicting the de¬nition of c. Since we have shown

that it is impossible that c = a or a < c < b, it follows that c = b. Since the

supremum of a set need not belong to that set we must still prove that b ∈ E.

However, if we choose a t0 ≥ a with b > t0 > b ’ δ the arguments of the

previous paragraph give f (b) ’ f (t0 ) ¤ (K + /2)(b ’ t0 ) and f (t0 ) ’ f (a) ¤

(K + )(t0 ’ a), so f (c) ¤ (K + )(c ’ a). The arguments of the previous

paragraph also give f (t) ’ f (a) ¤ (K + )(t ’ a) for a ¤ t < c, so c ∈ E and

the theorem follows.

We now discuss the relation between the fundamental axiom and the

supremum principle. In the proof of Theorem 3.1.7 we saw that the funda-

mental axiom together with the usual rules of algebra implies the supremum

principle. In the proof of Theorem 3.1.12 we saw that the supremum principle

together with the usual rules of algebra implies the intermediate value the-

orem. However, Theorem 1.8.1 tells us that the intermediate value theorem

together with the usual rules of algebra implies the fundamental axiom.

Although our argument has been a bit informal the reader should be

satis¬ed that the following is true. (If she is not satis¬ed she may write out

the details for herself.)

Theorem 3.1.15. Let (F, +, —, >) be an ordered ¬eld. Then the following

two statements are equivalent.

(i) Every increasing sequence bounded above has a limit.

(ii) Every non-empty set bounded above has a supremum.

The next exercise sketches a much more direct proof that the supremum

principle implies the fundamental axiom.

Exercise 3.1.16. Let (F, +, —, >) be an ordered ¬eld such that every non-

empty set bounded above has a supremum. Suppose that an ∈ F for each

n ≥ 1, A ∈ F, a1 ¤ a2 ¤ a3 ¤ . . . and an < A for each n. Write E = {an :

n ≥ 1}. Show that E has a supremum, a say, and that an ’ a.

3.2 The Bolzano-Weierstrass theorem

This section is devoted to the following important result.

38 A COMPANION TO ANALYSIS

Theorem 3.2.1. (Bolzano-Weierstrass.) If xn ∈ R and there exists a K

such that |xn | ¤ K for all n, then we can ¬nd n(1) < n(2) < . . . and x ∈ R

such that xn(j) ’ x as j ’ ∞.

Mathematicians say a sequence converges if it tends to a limit. The

Bolzano-Weierstrass theorem thus says that every bounded sequence of reals

has a convergent subsequence. Notice that we say nothing about uniqueness;

if xn = (’1)n then x2n ’ 1 but x2n+1 ’ ’1 as n ’ ∞.

Exercise 3.2.2. (i) Find a sequence xn ∈ [0, 1] such that, given any x ∈

[0, 1], we can ¬nd n(1) < n(2) < . . . such that xn(j) ’ x as j ’ ∞.

(ii) Is it possible to ¬nd a sequence xn ∈ [0, 1] such that, given any x ∈

[0, 1] with x = 1/2, we can ¬nd n(1) < n(2) < . . . and x ∈ R such that

xn(j) ’ x as j ’ ∞ but we cannot ¬nd m(1) < m(2) < . . . such that

xm(j) ’ 1/2 as j ’ ∞? Give reasons for your answer.

The proof of the Bolzano-Weierstrass theorem given in modern textbooks

depends on an ingenious combinatorial observation.

Lemma 3.2.3. If xn ∈ R, then at least one of the following two statements

must be true.

(A) There exist m(1) < m(2) < . . . such that xm(j) ≥ xm(j+1) for each

j ≥ 1.

(B) There exist n(1) < n(2) < . . . such that xn(j) ¤ xn(j+1) for each

j ≥ 1.

Exercise 3.2.4. Prove Theorem 3.2.1 from Lemma 3.2.3 and the fundamen-

tal axiom.

Proof of Lemma 3.2.3. Call an integer m ≥ 1 a ˜far seeing integer™ if xm ≥ xn

for all n ≥ m. (The term ˜far seeing™ is invented for use in this particular

proof and is not standard terminology.) There are two possibilities:

(A) There exist in¬nitely many far seeing integers. Thus we can can ¬nd

m(1) < m(2) < . . . such that each m(j) is far seeing and so xm(j) ≥ xm(j+1)

for each j ≥ 1.

(B) There are only ¬nitely many far seeing integers. Thus there exists

an N such that, if n ≥ N , there exists an n > n with xn > xn and so, in

particular, xn ≥ xn . Thus, given n(j) ≥ N , we can ¬nd n(j + 1) > n(j)

with xn(j) ¤ xn(j+1) . Proceeding inductively, we obtain n(1) < n(2) < . . .

with xn(j) ¤ xn(j+1) for each j ≥ 1.

I admit that the proof above is very clever but I feel that clever proofs

should only be used when routine proofs do not work. Here is a proof of the

Bolzano-Weierstrass theorem by lion hunting.

39

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

Exercise 3.2.5. We assume the hypotheses of Theorem 3.2.1. Set [a0 , b0 ] =

[’K, K]. Show that we can ¬nd a sequence of pairs of points an and bn such

that

xm ∈ [an , bn ] for in¬nitely many values of m,

an’1 ¤ an ¤ bn ¤ bn’1 ,

and bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.

Show that an ’ c as n ’ ∞ for some c ∈ [a0 , b0 ]. Show further that

we can ¬nd m(j) with m(j + 1) > m(j) and xm(j) ∈ [aj , bj ] for each j ≥ 1.

Deduce the Bolzano-Weierstrass theorem.

Exercise 3.2.5 links directly with the origins of the theorem. Proof by

successive bisection (our ˜lion hunting™) was invented by Bolzano and used

by Weierstrass to prove Theorem 3.2.1.

Here is another natural proof of Theorem 3.2.1 this time using a supre-

mum argument.

Exercise 3.2.6. We assume the hypotheses of Theorem 3.2.1. Explain why

yn = sup{xm : m ≥ n}

is well de¬ned. Show that the yn form a decreasing sequence bounded below

and conclude that yn tends to a limit y.

Show carefully that we can ¬nd n(j) with n(j + 1) > n(j) such that

|y ’ xn(j) | < j ’1 . Deduce the Bolzano-Weierstrass theorem.

The y of Exercise 3.2.6 is called lim supn’∞ xn . More formally we have

the following de¬nition.

De¬nition 3.2.7. If xn is a sequence of real numbers which is bounded above

we write

lim sup xn = lim sup{xm : m ≥ n}.

n’∞

n’∞

Exercise 3.2.8. Let xn be a bounded sequence of real numbers.

(i) De¬ne lim inf n’∞ xn by analogy with lim sup, showing that lim inf n’∞ xn

exists.

(ii) Show that lim inf n’∞ xn = ’ lim supn’∞ (’xn ).

(iii) Show that lim inf n’∞ xn ¤ lim supn’∞ xn .

(iv) Show that lim inf n’∞ xn = lim supn’∞ xn if and only if xn tends to

a limit.

40 A COMPANION TO ANALYSIS

(v) If n(1) < n(2) < . . . and xn(j) ’ x as j ’ ∞ show that

lim inf xn ¤ x ¤ lim sup xn .

n’∞ n’∞

(vi) If lim inf n’∞ xn ¤ x ¤ lim supn’∞ xn , does it follow that there

exist n(1) < n(2) < . . . such that xn(j) ’ x as j ’ ∞? Give a proof or

counterexample.

(vii) Show that y = lim supn’∞ xn if and only if both the following con-

ditions hold.

(A) Given > 0 we can ¬nd an N ( ) such that xn < y + for all

n ≥ N ( ).

(B) Given > 0 and N we can ¬nd n(N, ) ≥ N such that xn(N, ) > y’ .

State and prove a similar result for lim inf.

Although we mention lim sup from time to time, we shall not make great

use of the concept.

The reader will not be surprised to learn that the Bolzano-Weierstrass

theorem is precisely equivalent to the fundamental axiom.

Exercise 3.2.9. Suppose that F is an ordered ¬eld for which the Bolzano-

Weierstrass theorem holds (that is, every bounded sequence has a convergent

subsequence). Suppose that an is an increasing sequence bounded above. Use

the Bolzano-Weierstrass theorem to show that there exists an a ∈ F and

n(1) < n(2) < . . . such that an(j) ’ a as j ’ ∞. Show that an ’ a as

n ’ ∞ and so F obeys the fundamental axiom.

We illustrate the use of the Bolzano-Weierstrass theorem by applying it

to the familiar example of the intermediate value theorem.

Theorem 3.2.10. 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 ˜lion hunting proof™ on page 15 and the ˜supremum proof™ on

page 34.

Part A Without loss of generality, we may suppose a = 0 and b = 1. Consider

the real numbers f (0), f (1/n), f (2/n), . . . , f (1). Since f (0) ≥ 0 and f (1) ¤

0 there must exist an integer r with 0 ¤ r ¤ n ’ 1 such that f (r/n) ≥ 0 ≥

f ((r + 1)/n). Set xn = r/n.

Part B Since the xn ∈ [0, 1], they form a bounded sequence and so, by the

Bolzano-Weierstrass theorem, we can ¬nd a c ∈ [0, 1] and n(1) < n(2) < . . .

such that xn(j) ’ c as j ’ ∞.

41

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

Part C Since f is continuous and xn(j) ’ c, it follows that f (xn(j) ) ’ f (c)

as j ’ ∞. Since f (xn(j) ) ≥ 0, it follows that f (c) ≥ 0.

By the axiom of Archimedes, xn(j) + n(j)’1 ’ c so f (xn(j) + n(j)’1 ) ’

f (c) as j ’ ∞. Since f (xn(j) + n(j)’1 ) ¤ 0 it follows that f (c) ¤ 0.

Combining this result with that of the paragraph above, we obtain f (c) = 0

as required.

Exercise 3.2.11. Produce a version of the proof just given in which we do

not assume that a and b take particular values.

Exercise 3.2.12. Extend Exercise 3.1.13 to cover the ˜Bolzano-Weierstrass™

argument as well.

Let us summarise the proof just given. In Part A we construct a sequence

of points xn which look ˜more and more promising™. In Part B we use the fact

that every bounded sequence has a convergent subsequence to give a point

which ˜ought to be very promising indeed™. Finally in Part C we examine the

point c to make sure that it really has the desired property.

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

rem 3.2.10. If we omit the condition f (a) ≥ 0, f (b) ¤ 0 then we cannot even

start Part A of the argument.

If we have f (a) ≥ 0 ≥ f (b) but replace R by another ordered ¬eld

for which the Bolzano-Weierstrass theorem does not hold then Part A goes

through perfectly but Part B fails. As usual this is shown by Example 1.1.3.

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

continuous then Part C fails.

Exercise 3.2.13. Let f : [0, 1] ’ R Show that, if f (1)’f (0) ≥ L, then there

must exist an integer r with 0 ¤ r ¤ n ’ 1 such that f ((r + 1)/n) ’ f (r/n) ≥

L/n.

By imitating the proof of Theorem 3.2.10, give a Bolzano-Weierstrass

proof of Lemma 1.7.4 and thus of Theorem 1.7.1 (the mean value inequality).

Exercise 3.2.14. In this exercise you may use the axiom of Archimedes and

the fact that any non-empty bounded set of integers has a maximum.

(i) Let E be a non-empty set of real numbers which is bounded above

(that is there exists a K such that K ≥ x whenever x ∈ E). If n is a strictly

positive integer show that there exists an integer r such that

there exists an e ∈ E with e ≥ r/n

but (r + 1)/n ≥ x whenever x ∈ E.

(ii) Arguing in the manner of the proof of Theorem 3.2.10, show, using

the Bolzano-Weierstrass theorem, that E has a supremum.

42 A COMPANION TO ANALYSIS

3.3 Some general remarks

We have now obtained three di¬erent but equivalent forms of the fundamental

axiom (the fundamental axiom itself, the existence of a supremum for a non-

empty bounded set, and the Bolzano-Weierstrass theorem) and used methods

based on these three forms to prove the intermediate value theorem and the

mean value inequality (themselves equivalent to the fundamental axiom). I

make no excuse for the time we have spent on this programme. All of analysis

rests like an inverted pyramid on the fundamental axiom so it makes sense

to study it closely.

For reasons which will become clear in the next chapter, we will rely most

strongly on ˜Bolzano-Weierstrass™ techniques. However, there will be several

places where we prefer ˜supremum methods™. Exercises K.118 to K.121 show

that lion hunting is useful in theory of integration and, although it lies outside

the scope of this book, it should be remarked that the standard proof of

Cauchy™s theorem, on which complex analysis is based, relies on lion hunting.

There is a fourth method of exploiting the fundamental axiom based on

˜Heine-Borel™ or ˜compactness™ which is not discussed here (see Exercises K.29

to K.36, which are intended to be done after reading Chapter 4) but which,

when she does meet it, the reader should consider in the context of this

chapter.

Whenever we use one of these techniques it is instructive to see how the

others could have been used. (Often this is easy but occasionally it is not.)

It is also worth bearing in mind that, whenever we genuinely need to use one

of these methods or some theorem based on them, we are using the basic

property of the reals. Everything that depends on the fundamental axiom is

analysis ” the rest is mere algebra.

However, the reader should also remember most of the di¬culties in anal-

ysis are resolved not by the precise manipulation of axioms but by the clear

understanding of concepts.

Chapter 4

Higher dimensions

4.1 Bolzano-Weierstrass in higher dimensions

In 1908, Hardy wrote a textbook to introduce the new rigorous analysis (or

˜continental analysis™ as it was known in a Cambridge more insular than

today) to ˜¬rst year students at the Universities whose abilities approach

something like what is usually described as “scholarship standard” ™. Apart

from the fact that even the most hardened analyst would now hesitate to

call an introduction to analysis A Course of Pure Mathematics [23], it is

striking how close the book is in both content and feel to a modern ¬rst

course in analysis. (And, where there are changes, it is often not clear that

the modern course1 has the advantage.) One major di¬erence is that Hardy

only studies the real line but later advances in mathematics mean that we

must now study analysis in Rm as well.

We start with some algebra which is probably very familiar to most of

my readers. If x, y ∈ Rm , we de¬ne the inner product (or dot product) x · y

of the two vectors by

m

x·y = xj yj .

j=1

(We shall sometimes use the alternative notation x, y = x · y. Many texts

use the notation x.y = x · y.)

Lemma 4.1.1. If x, y, z ∈ Rm and » ∈ R then

(i) x · x ≥ 0 with equality if and only if x = 0,

1

Indeed, anyone who works through the exercises in Hardy as a ¬rst course and the

exercises in Whittaker and Watson™s even older A Course of Modern Analysis [47] as a

second will have had a splendid education in analysis.

43

44 A COMPANION TO ANALYSIS

(ii) x · y = y · x,

(iii) (»x) · y = »(x · y),

(iv) (x + y) · z = x · z + y · z.

Proof. Direct calculation which is left to the reader.

Since x · x ≥ 0 we may de¬ne the ˜Euclidean norm of x™ by

x = (x · x)1/2

where we take the positive square root.

Lemma 4.1.2. (The Cauchy-Schwarz inequality.) If x, y ∈ Rm then

|x · y| ¤ x y .

Proof. If x = 0, then x = 0, so x · y = 0 and the inequality is trivial. If

not, we observe that

0 ¤ (»x + y) · (»x + y)

= »2 x 2 + 2»x · y + y 2

2

(x · y)2

x·y 2

’

= »x + +y .

x2

x

If we now set » = ’(x · y)/ x 2 , this gives us

(x · y)2

2

0¤ y ’ ,

x2

which, after rearrangement and taking square roots, gives the desired result 2 .

Exercise 4.1.3. Although the proof just given is fairly detailed, it is a worth-

while exercise to extend it so that all the steps are directly justi¬ed by reference

to the properties of the inner product given in Lemma 4.1.1.

We can now obtain the standard properties of the Euclidean norm.

2

The reader may ask why we do not ˜simply™ say that x · y = x y cos θ where θ is

the angle between the vectors x and y. The Cauchy-Schwarz inequality is then ˜simply™

the statement that | cos θ| ¤ 1. However, our program is to deduce all of analysis from a

limited set of statements about R. We have not yet discussed what ˜cos™ is to be and, even

more importantly what the ˜angle between two vectors™ is to mean. When we ¬nally reach

a de¬nition of angle in Exercise 5.5.6, the reader will see that we have actually reversed

the suggested argument of the ¬rst two sentences of this footnote. This way of proceeding

greatly amuses mathematicians and greatly annoys educational theorists.

45

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

is the Euclidean norm on Rm , then the following re-

Lemma 4.1.4. If

sults hold.

(i) x ≥ 0 for all x ∈ Rm .

(ii) If x = 0, then x = 0.

(iii) If » ∈ R and x ∈ Rm , then »x = |»| x .

(iv) (The triangle inequality.) If x, y ∈ Rm then x + y ¤ x + y .

Proof. The triangle inequality can be deduced from the Cauchy-Schwarz in-

equality as follows.

2

=(x + y) · (x + y) = x 2 + 2x · y + y 2

x+y

¤ x 2 + 2 x y + y 2 = ( x + y )2 .

The remaining veri¬cations are left to the reader.

Exercise 4.1.5. (i) By carefully examining the proof of Lemma 4.1.2, or

otherwise, show that we have equality in the Cauchy-Schwarz inequality (that

is, we have |x · y| = x y ) if and only if x and y are linearly dependent

(that is, we can ¬nd real » and µ, not both zero, such that »x = µy).

(ii) Show that we have equality in the triangle inequality (that is x +

y = x + y ) if and only if we can ¬nd positive » and µ, not both zero,

such that »x = µy.

We observe that

1/2

m

(xj ’ yj )2

x’y =

i=1

which (at least if m = 1, m = 2 or m = 3) is recognisably the distance (more

properly, the Euclidean distance) between x and y. If we set x = a ’ b and

y = b ’ c, then the triangle inequality of Lemma 4.1.4 (iv) becomes

a’c ¤ a’b + b’c

which is usually read as saying that the length of one side of a triangle is less

than or equal to the sum of the lengths of the other two sides.

Exercise 4.1.6. If x = (x1 , x2 , . . . , xm ) ∈ Rm show that

m

max |xi | ¤ x ¤ |xi | ¤ m max |xi |.

1¤i¤m 1¤i¤m

i=1

Looking at each of the 3 inequalities in turn, ¬nd necessary and su¬cient

conditions for equality.

46 A COMPANION TO ANALYSIS

Exercise 4.1.7. We proved the triangle inequality by going via the Cauchy-

Schwarz inequality. Try and prove the inequality

1/2 1/2 1/2

3 3 3

x2 2

(xj + yj )2

≥

+ yj

j

i=1 i=1 i=1

directly.

Although we shall not take the matter to extremes, we shall have a strong

preference for coordinate free methods and statements. So far as I am aware,

no one has found a set of labelled axes (perhaps carved in stone or beau-

tifully cast in bronze) bearing an attestation from some higher power that

these are ˜nature™s coordinate axes™. Coordinate free statements and methods

encourage geometric intuition and generalise more readily.

Maxwell who played a crucial role in the development of vector methods

wrote in the ¬rst chapter of his great Treatise on Electricity and Magnetism

For many purposes of physical reasoning, as distinguished from

calculation, it is desirable to avoid explicitly introducing the Carte-

sian coordinates, and to ¬x the mind at once on a point of space

instead of its three coordinates, and on the magnitude and di-

rection of a force instead of its three components. This mode of

contemplating geometrical and physical quantities is more prim-

itive and more natural than the other. (Chapter 1, [37])

We now turn towards analysis.

De¬nition 4.1.8. We work in Rm with the Euclidean norm. 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 > 0, we can ¬nd an n0 ( ) such that

an ’ a < for all n ≥ n0 ( ).

Notice that this shows that De¬nition 1.2.1 was about the distance be-

tween two points and not the absolute value of the di¬erence of two numbers.

We can prove the following results on sequences in Rm in exactly the same

way as we proved the corresponding results for R (and more general ordered

¬elds F) in Lemma 1.2.2.

47

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

Lemma 4.1.9. We work in Rm with the Euclidean norm.

(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 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) Suppose an ∈ Rm , a ∈ Rm , »n ∈ R and » ∈ R. If an ’ a and

»n ’ », then »n an ’ »a.

Proof. Left to the reader.

Exercise 4.1.10. The parts of Lemma 1.2.2 do not all have corresponding

parts in Lemma 4.1.9. Explain brie¬‚y why these di¬erences occur.

Exercise 4.1.11. (i) If an ’ a and bn ’ b as n ’ ∞, show that an · bn ’

a.b. (You may ¬nd the Cauchy-Schwarz inequality useful.)

(ii) If x, y ∈ Rm show that

2

+ x ’ y 2 )/4.

x.y = ( x + y

Prove part (i) following the method of Exercise 1.2.6.

Lemma 4.1.9 is, of course, merely algebra and applies to Qm as much as

to Rm . In order to do analysis we need a more powerful tool and, in keeping

with the spirit of our general programme, we extend the Bolzano-Weierstrass

theorem to Rm .

Theorem 4.1.12. (Bolzano-Weierstrass.) If xn ∈ Rm and there exists a

K such that xn ¤ K for all n, then we can ¬nd n(1) < n(2) < . . . and

x ∈ Rm such that xn(j) ’ x as j ’ ∞.

Once again ˜any bounded sequence has a convergent subsequence™.

Proof. We prove the result for m = 2, leaving it to the reader to prove the

general result. Let us write xn = (xn , yn ). Observe that, since xn ¤ K,

it follows that |xn | ¤ K. By the Bolzano-Weierstrass theorem for the reals

(Theorem 3.2.1), it follows that there exist a real number x and a sequence

m(1) < m(2) < . . . such that xm(k) ’ x as k ’ ∞.

Since xm(k) ¤ K, it follows that |ym(k) | ¤ K so, again by the Bolzano-

Weierstrass theorem for the reals, if follows that there exist a real number y

and a sequence k(1) < k(2) < . . . such that ym(k(j)) ’ y as j ’ ∞.

Setting n(j) = m(k(j)) and x = (x, y) we have

xn(j) ’ x ¤ |xn(j) ’ x| + |yn(j) ’ y| = |xm(k(j)) ’ x| + |ym(k(j)) ’ y| ’ 0 + 0 = 0,

and so xn(j) ’ x as j ’ ∞.

48 A COMPANION TO ANALYSIS

Exercise 4.1.13. Prove Theorem 4.1.12 for general m.

Exercise 4.1.14. We can also prove Theorem 4.1.12 by ˜multidimensional

lion hunting™. In this exercise we again consider the case m = 2, leaving

the general case to the reader. She should compare this exercise with Exer-

cise 3.2.5.

We assume the hypotheses of Theorem 4.1.12. Consider the square

S0 = [a0 , b0 ] — [a0 , b0 ] = [’K, K] — [’K, K].

Explain why xn ∈ S0 for all n.

Set c0 = (a0 + b0 )/2, c0 = (a0 + b0 )/2 and consider the four squares

T0,1 = [a0 , c0 ] — [a0 , c0 ], T0,2 = [c0 , b0 ] — [a0 , c0 ],

T0,3 = [a0 , c0 ] — [c0 , b0 ], T0,4 = [c0 , b0 ] — [c0 , b0 ].

Explain why at least one of the squares T0,p , say, must be such that xr ∈ T0,p

for in¬nitely many values of r. We set [a1 , b1 ] — [a1 , b1 ] = T0,p .

Show that we can ¬nd a sequence of pairs of intervals [an , bn ] and [an , bn ]

such that

xr ∈ [an , bn ] — [an , bn ] for in¬nitely many values of r,

an’1 ¤ an ¤ bn ¤ bn’1 , an’1 ¤ an ¤ bn ¤ bn’1 ,

and bn ’ an = (bn’1 ’ an’1 )/2, bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.

Show that an ’ c as n ’ ∞ for some c ∈ [a0 , b0 ] and an ’ c as

n ’ ∞ for some c ∈ [a0 , b0 ]. Show further that we can ¬nd m(j) with

m(j + 1) > m(j) and xm(j) ∈ [aj , bj ] — [aj , bj ] for each j ≥ 1. Deduce the

Bolzano-Weierstrass theorem.

The proof of Theorem 4.1.12 involves extending a one dimensional result

to several dimensions. This is more or less inevitable because we stated

the fundamental axiom of analysis in a one dimensional form. However the

Bolzano-Weierstrass theorem itself contains no reference as to whether we are

working in R or Rm . It is thus a excellent tool for multidimensional analysis.

4.2 Open and closed sets

When we work in R the intervals are, in some sense, the ˜natural™ sets to

consider. One of the problems that we face when we try to do analysis in

many dimensions is that the types of sets with which we have to deal are

49

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

much more diverse. It turns out that the so called closed and open sets

are both su¬ciently diverse and su¬ciently well behaved to be useful. This

short section is devoted to deriving some of their simpler properties. Novices

frequently ¬nd the topic hard but eventually the reader will appreciate that

this section is a rather trivial interlude in a deeper discussion.

The de¬nition of a closed set is a natural one.

De¬nition 4.2.1. A set F ⊆ Rm is closed if whenever xn ∈ F for each n

and xn ’ x as n ’ ∞ then x ∈ F .

Thus a set is closed in the sense of analysis if it is ˜closed under the

operation of taking limits™. An indication of why this is good de¬nition is

given by the following version of the Bolzano-Weierstrass theorem.

Theorem 4.2.2. (i) If K is a closed bounded set in Rm then every sequence

in K has a subsequence converging to a point of K.

(ii) Conversely, if K is a subset of Rm such that every sequence in K has

a subsequence converging to a point of K, then K is a closed bounded set.

Proof. Both parts of the proof are easy.

(i) If xn is a sequence in K, then it is a bounded sequence and so, by

Theorem 4.1.12, has a convergent subsequence xn(j) ’ x, say. Since K is

closed, x ∈ K and we are done.

(ii) If K is not closed, we can ¬nd xn ∈ K and x ∈ K such that xn ’ x

/

as n ’ ∞. Since any subsequence of a convergent subsequence converges to

the same limit, no subsequence of the xn can converge to a point of K.

If K is not bounded, we can ¬nd xn ∈ K such that xn > n. If x is any

point of Rm , then the inequality

xn ’ x ≥ x n ’ x > n ’ x

shows that no subsequence of the xn can converge.

When working in Rm , the words ˜closed and bounded™ should always elicit

the response ˜Bolzano-Weierstrass™. We shall see important examples of this

slogan in action in the next section (Theorem 4.3.1 and Theorem 4.5.5).

The following remark is sometimes useful.

Exercise 4.2.3. (i) If A is a non-empty closed subset of R with supremum

±, then we can ¬nd an ∈ A with an ’ ± as n ’ ∞.

(ii) If A is a non-empty closed subset of R, then, if supa∈A a exists,

supa∈A a ∈ A.

We turn now to the de¬nition of an open set.

50 A COMPANION TO ANALYSIS

De¬nition 4.2.4. A set U ⊆ Rm is open if, whenever x ∈ U , there exists

an > 0 such that, whenever x ’ y < , we have y ∈ U .

Thus every point of an open set lies ˜well inside the set™. The ability

to ˜look in all directions™ plays an important role in many proofs. The ¬rst

example we shall see occurs in the proof of Rolle™s theorem (Theorem 4.4.4)

and the idea will play a key part in the study of complex analysis.

Exercise 4.2.5. Consider sets in R. Prove the following results. The inter-

val [a, b] = {x ∈ R : a ¤ x ¤ b} is closed, the interval (a, b) = {x ∈ R : a <

x < b} is open, the interval [a, b) = {x ∈ R : a ¤ x < b} is neither open nor

closed, R is both open and closed.

Lemma 4.2.6. Consider sets in Rm . Let x ∈ Rm and r > 0.

(i) The set B(x, r) = {y ∈ Rm : x ’ y < r} is open.

¯

(ii) The set B(x, r) = {y ∈ Rm : x ’ y ¤ r} is closed.

Proof. This is routine3 . There is no loss of generality in taking x = 0.

(i) If z ∈ B(0, r), then z < r so, if we set δ = r ’ z , it follows that

δ > 0. If z ’ y < δ, the triangle inequality gives

y ¤ z + z ’ y < z + δ = r,

so that y ∈ B(0, r). Thus B(0, r) is open.

¯

(ii) If yn ∈ B(0, r) and yn ’ y as n ’ ∞, then, by the triangle inequal-

ity,

y ¤ yn + y ’ yn ¤ r + y ’ yn ’ r + 0 = r

¯

as n ’ ∞. Thus y ¤ r and we have shown that B(0, r) is closed.

¯

We call B(x, r) the open ball of radius r and centre x. We call B(x, r)

the closed ball of radius r and centre x. Observe that the closed and open

balls of R are precisely the closed and open intervals.

The following restatement of the de¬nition helps us picture an open set.

Lemma 4.2.7. A subset U of Rm is open if and only if each point of U is

the centre of an open ball lying entirely within U .

Thus every point of an open set is surrounded by a ball consisting only

of points of the set.

The topics of this section are often treated using the idea of neighbour-

hoods. We shall not use neighbourhoods very much but they come in useful

from time to time.

3

Mathspeak for ˜It may be hard the ¬rst time you see it but when you look at it later

you will consider it to be routine.™

51

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

De¬nition 4.2.8. The set N is a neighbourhood of the point x if we can

¬nd an r > 0 (depending on both x and N ) such that B(x, r) ⊆ N .

Thus a set is open if and only if it is a neighbourhood of every point that

it contains.

Returning to the main theme we note the following remarkable fact.

Lemma 4.2.9. A subset E of Rm is open if and only if its complement

Rm \ E is closed.

Proof. Again, this is only a question of writing things down clearly. We split

the proof into two parts.

Necessity Suppose that E is open. If xn ∈ Rm \ E for all n and xn ’ x

as n ’ ∞, then we claim that x ∈ Rm \ E. For, if not, we must have

x ∈ E and so, since E is open, we can ¬nd an > 0 such that, whenever

a ’ x < , it follows that a ∈ E. Since xn ’ x, we can ¬nd an N such that

xN ’ x < and so xN ∈ E, contradicting the statement that xn ∈ Rm \ E.

Thus x ∈ Rm \ E and we have shown that Rm \ E is closed.

Su¬ciency Suppose that Rm \ E is closed. We show that E is open. For,

if not, there must exist an a ∈ E such that, given any > 0, there exists

a y ∈ E with y ’ a < . In particular, we can ¬nd xn ∈ Rm \ E such

/

that xn ’ a < 1/n. By the axiom of Archimedes, this means that xn ’ a

as n ’ ∞ and so, since Rm \ E is closed, a ∈ Rm \ E, contradicting our

assumption that a ∈ E. Thus E is open.

We observe the following basic results on open and closed sets.

Lemma 4.2.10. Consider the collection „ of open sets in Rm .

(i) … ∈ „ , Rm ∈ „ .

(ii) If U± ∈ „ for all ± ∈ A, then ±∈A U± ∈ „ .

(iii) If U1 , U2 , . . . , Un ∈ „ , then n Uj ∈ „ .

j=1

Proof. This is routine.

(i) Since the empty set contains no points, every point in the empty set

has any property we desire (in this case, that of being the centre of an open

ball lying within the empty set). Thus the empty set is open. If x ∈ Rm

then B(x, 1) ⊆ Rm . Thus Rm is open.

(ii) If x ∈ ±∈A U± , then we can ¬nd a particular ±(0) ∈ A such that

x ∈ U±(0) . Since U±(0) is open, we can ¬nd a δ > 0 such that B(x, δ) ⊆ U±(0) .

Automatically, B(x, δ) ⊆ ±∈A U± . We have shown that ±∈A U± is open.

(iii) If x ∈ n Uj , then x ∈ Uj for each 1 ¤ j ¤ n. Since each Uj is

j=1

open we can ¬nd a δj > 0, such that B(x, δj ) ⊆ Uj for each 1 ¤ j ¤ n.

Setting δ = min1¤j¤n δj , we have δ > 0 (note that this part of the argument

52 A COMPANION TO ANALYSIS

requires that we are only dealing with a ¬nite number of open sets Uj ) and

B(x, δ) ⊆ Uj for each 1 ¤ j ¤ n. Thus B(x, δ) ⊆ n Uj and we have

j=1

shown that n Uj is open.

j=1

Lemma 4.2.11. Consider the collection F of closed sets in Rm .

(i) … ∈ F, Rm ∈ F.

(ii) If F± ∈ F for all ± ∈ A, then ±∈A F± ∈ F.

(iii) If F1 , F2 , . . . , Fn ∈ F, then n Fj ∈ F.

j=1

Proof. This follows from Lemma 4.2.10 by repeated use of Lemma 4.2.9.

(i) Observe that … = Rm \Rm and Rm = Rm \…. Now use Lemma 4.2.10 (i).

(ii) Observe that

F± = R m \ (Rm \ F± )

±∈A ±∈A

and use Lemma 4.2.10 (ii).

(iii) Observe that

n n

Fj = R m \ (Rm \ Fj )

j=1 j=1

and use Lemma 4.2.10 (iii).

Exercise 4.2.12. We proved Lemma 4.2.10 directly and obtained Lemma 4.2.11

by complementation. Prove Lemma 4.2.11 and obtain Lemma 4.2.10 by com-

plementation.

Exercise 4.2.13. (i) We work in R and use the usual notation for intervals

(see Exercise 4.2.5 if necessary). Show that

∞

(’1 ’ j ’1 , 1) = [’1, 1)

j=1

and conclude that the intersection of open sets need not be open. Why does

this not contradict Lemma 4.2.10?

(ii) Let U1 , U2 , . . . be open sets in R such that U1 ⊇ U2 ⊇ U3 ⊇ . . . .

Show, by means of examples, that ∞ Uj may be (a) open but not closed,

j=1

(b) closed but not open, (c) open and closed or (d) neither open nor closed.

(iii) What result do we get from (iii) by complementation?

(iv) Let Fj = [aj , bj ] and F1 ⊆ F2 ⊆ F3 ⊆ . . . Show, by means of exam-

ples, that ∞ Fj may be (a) open but not closed, (b) closed but not open,

j=1

(c) open and closed or (d) neither open nor closed.

53

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

(v) Let a < b and c < d. Show that, if we work in R2 , [a, b] — [c, d] is

closed, (a, b) — (c, d) is open and (a, b) — [c, d] is neither open nor closed.

(vi) Do part (ii) with R replaced by R2 .

(vii) If A is open in Rm and B is open in Rn , show that A — B is open

in Rm+n . State and prove the corresponding result for closed sets.

Of course, analysis deals with (reasonably well behaved) functions as

well as sets. The notion of continuity gives us a natural class of reasonably

well behaved functions. The de¬nition carries over unchanged from the one

dimensional case.

De¬nition 4.2.14. Let E ⊆ Rm . We say that a function f : E ’ Rp is

continuous at some point x ∈ E if, given > 0, we can ¬nd a δ( , x) > 0

such that, whenever y ∈ E and x ’ y < δ( , x), we have

f (x) ’ f (y) < .

If f is continuous at every point x ∈ E, we say that f is a continuous function

on E.

This may be the place to make a comment on vector notation. It is

conventional in elementary analysis to distinguish elements of Rm from those

in R by writing points of Rm in boldface when printing and underlining

them when handwriting. Eventually this convention becomes tedious and,

in practice, mathematicians only use boldface when they wish to emphasise

that vectors are involved.

Exercise 4.2.15. After looking at Lemma 1.3.2 and parts (iii) to (v) of

Lemma 4.1.9, state the corresponding results for continuous functions. (Thus

part (v) of Lemma 4.1.9 corresponds to the statement that, if » : E ’ R and

f : E ’ Rp are continuous at x ∈ E, then so is »f .) Prove your statements

directly from De¬nition 4.2.14.

Suppose that E ⊆ Rm and f : E ’ R is continuous at x. Show that, if

f (t) = 0 for all t ∈ E, then 1/f is continuous at x.

Once again we have the following useful observation.

Lemma 4.2.16. Let E be a subset of Rm and f : E ’ Rp a function.

Suppose 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 the reader.

Another way of looking at continuity, which will become progressively

more important as we proceed, is given by the following lemma.

54 A COMPANION TO ANALYSIS

Lemma 4.2.17. The function f : Rm ’ Rp is continuous if and only if

f ’1 (U ) is open whenever U is an open set in Rp .

The reader may need to be reminded of the de¬nition

f ’1 (U ) = {x ∈ Rm : f (x) ∈ U }.

Proof. As with most of the proofs in this section, this is just a matter of

writing things down correctly. We split the proof into two parts.

Necessity Suppose f is continuous and U is an open set in Rp . If x ∈ f ’1 (U ),

then f (x) ∈ U . But U is open, so there exists an > 0 such that B(f (x), ) ⊆

U . Since f is continuous at x, we can ¬nd a δ > 0 such that

f (x) ’ f (y) < whenever x ’ y < δ.

We thus have B(x, δ) ⊆ f ’1 (U ). It follows that f ’1 (U ) is open.

Su¬ciency Suppose that f ’1 (U ) is open whenever U is an open subset of Rp .

Let x ∈ Rm and > 0. Since B(f (x), ) is open, it follows that f ’1 (B(f (x), ))

is open. But x ∈ f ’1 (B(f (x), )), so there exists a δ > 0 such that B(x, δ) ⊆

f ’1 (B(f (x), )). We thus have

f (x) ’ f (y) < whenever x ’ y < δ.

It follows that f is continuous.

Exercise 4.2.18. Show that sin((’5π, 5π)) = [’1, 1]. Give examples of

bounded open sets A in R such that (a) sin A is closed and not open, (b)

sin A is open and not closed, (c) sin A is neither open nor closed, (d) sin A

is open and closed. (Observe that … is automatically bounded.)

The reader may object that we have not yet derived the properties of sin.

In my view this does not matter if we are merely commenting on or illustrat-

ing our main argument. (I say a little more on this topic in Appendix C.)

However, if the reader is interested, she should be able to construct a poly-

nomial P such that (a), (b), (c) and (d) hold for suitable A when sin A is

replaced by P (A).

The next exercise gives a simple example of how Lemma 4.2.17 can be

used and asks you to contrast the new ˜open set™ method with the old ˜ , δ

method

Exercise 4.2.19. Prove the following result, ¬rst directly from De¬nition 4.2.14

and then by using Lemma 4.2.17 instead.

If f : Rm ’ Rp and g : Rp ’ Rq are continuous, then so is their

composition g —¦ f .

55

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

(Recall that we write g —¦ f (x) = g(f (x)).)

The reader who has been following carefully may have observed that we

have only de¬ned limits of sequences. Here is another notion of limit which

is probably familiar to the reader.

De¬nition 4.2.20. Let E ⊆ Rm , x ∈ E and a ∈ Rp . Consider a function

f : E \ {x} ’ Rp (or4 a function f : E ’ Rp ). We say that f (y) ’ a as

y ’ x through values of y ∈ E if, given > 0, we can ¬nd a δ( , x) > 0

such that, whenever y ∈ E and 0 < x ’ y < δ( , x), we have

f (x) ’ a < .

(We give a slightly more general de¬nition in Exercise K.23.)

Exercise 4.2.21. Let E ⊆ Rm , x ∈ E and a ∈ Rp . Consider a function

f : E \ {x} ’ Rp . De¬ne ˜ : E ’ Rp by ˜(y) = f (y) if y ∈ E \ {x} and

f f

˜(x) = a. Show that f (y) ’ a as y ’ x through values of y ∈ E if and only

f

if ˜ is continuous at x.

f

Exercise 4.2.22. After looking at your solution of Lemma 4.2.15, state and

prove the corresponding results for limits.

Exercise 4.2.23. [In this exercise you should start from De¬nition 1.7.2]

Let U be an open set in R. Show that a function f : U ’ R is di¬erentiable

at t ∈ U with derivative f (t) if and only if

f (t + h) ’ f (t)

’ f (t)

h

as h ’ 0 (through values of h with t + h ∈ U ).

Exercise 4.2.24. In Chapter 6 we approach the properties of di¬erentia-

tion in a more general manner. However the reader will probably already

have met results like the following which can be proved using Exercises 4.2.22

and 4.2.23.

(i) If f, g : (a, b) ’ R are di¬erentiable at x ∈ (a, b), then so is the sum

f + g and we have (f + g) (x) = f (x) + g (x).

(ii) If f, g : (a, b) ’ R are di¬erentiable at x ∈ (a, b), then so is the

product f — g and we have (f — g) (x) = f (x)g(x) + f (x)g (x). [Hint: f (x +

h)g(x + h) ’ f (x)g(x) = (f (x + h) ’ f (x))g(x + h) + f (x)(g(x + h) ’ g(x)).]

(iii) If f : (a, b) ’ R is nowhere zero and f is di¬erentiable at x ∈ (a, b),

then so is 1/f and we have (1/f ) (x) = ’f (x)/f (x)2 .

4

Thus it does not matter whether f is de¬ned at x or not (and, if it is de¬ned, it does

not matter what the value of f (x) is).

56 A COMPANION TO ANALYSIS

(iv) State accurately and prove a result along the lines of (ii) and (iii)

dealing with the derivative of f /g.

(v) If c ∈ R, c = 0 and f : R ’ R is di¬erentiable at x, show that the

function fc de¬ned by fc (t) = f (ct) [t ∈ R] is di¬erentiable at c’1 x and we

have fc (c’1 x) = cf (x). What happens if c = 0?

(vi) Use part (ii) and induction on n to show that if rn (x) = xn , then rn

is everywhere di¬erentiable with rn (x) = nrn’1 (x) [n ≥ 1]. Hence show that

every polynomial is everywhere di¬erentiable. If P and Q are polynomials

and Q(t) = 0 for all t ∈ (a, b) show that P/Q is everywhere di¬erentiable on

(a, b).

Exercise 4.2.25. (i) Use part (ii) of Exercise 4.2.24 to show that, if f, g :

(a, b) ’ R satisfy the equation f (t)g(t) = 1 for all t ∈ (a, b) and are di¬er-

entiable at x ∈ (a, b) then g (x) = ’f (x)/f (x)2 .

(ii) Explain why we can not deduce part (iii) of Exercise 4.2.24 directly

from part (i) of this exercise. Can we deduce the result of part (i) of this

exercise directly from part (iii) of Exercise 4.2.24?

(iii) Is the following statement true or false? If f, g : (a, b) ’ R are

di¬erentiable at x ∈ (a, b) and f (x)g(x) = 1 then g (x) = ’f (x)/f (x)2 .

Give a proof or counterexample.

Exercise 4.2.26. From time to time the eagle eyed reader will observe state-

ments like

˜f (x) ’ ∞ as x ’ ’∞™

which have not been formally de¬ned. If this really bothers her, she is probably

reading the wrong book (or the right book but too early). It can be considered

a standing exercise to ¬ll in the required details.

In Appendix D, I sketch a method used in Beardon™s elegant treatment [2]

which avoids the need for such repeated de¬nitions.

4.3 A central theorem of analysis

In this section we prove Theorem 4.3.4 which says that a real-valued con-

tinuous function on a closed bounded set in Rm is bounded and attains its

bounds. This result together with the intermediate value theorem (proved as

Theorem 1.6.1) and the mean value inequality (proved as Theorem 1.7.1 and

later in a more general context as Theorem 6.3.1) are generally considered

to be the central theorems of elementary analysis.

Our next result looks a little abstract at ¬rst.

57

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

Theorem 4.3.1. Let K be a closed bounded subset of Rm and f : K ’ Rp

a continuous function. Then f (K) is closed and bounded.

Thus the continuous image of a closed bounded set is closed and bounded.

Proof. By Theorem 4.2.2 (ii), we need only prove that any sequence in f (K)

has a subsequence converging to a limit in f (K).

To this end, suppose that yn is a sequence in f (K). By de¬nition, we can

¬nd xn ∈ K such that f (xn ) = yn . Since K is closed and bounded subset,

Theorem 4.2.2 (i) tells us that there exist n(j) ’ ∞ and x ∈ K such that

xn(j) ’ x as j ’ ∞. Since f is continuous,

yn(j) = f (xn(j) ) ’ f (x) ∈ f (K)

and we are done.

Exercise 4.3.2. Let N : Rm ’ R be given by N (x) = x . Show that

N is continuous. Deduce in particular that if xn ’ x as n ’ ∞, then

xn ’ x .

Exercise 4.3.3. (i) Let A be the open interval (0, 1). Show that the map

f : A ’ R given by f (x) = 1/x is continuous but that f (A) is unbounded.

Thus the continuous image of a bounded set need not be bounded.

(ii) Let A = [1, ∞) = {x ∈ R : x ≥ 1} and f : A ’ R be given by

f (x) = 1/x. Show that A is closed and f is continuous but f (A) is not

closed. Thus the continuous image of a closed set need not be closed.

(iii) Show that the function π : R2 ’ R given by π(x, y) = x is continu-

ous. (The function π is called a projection.) Show that the set

A = {(x, 1/x) : x > 0}

is closed in R2 but that π(A) is not.

We derive a much more concrete corollary.

Theorem 4.3.4. Let K be a closed bounded subset of Rm and f : K ’ R a

continuous function. Then we can ¬nd k1 and k2 in K such that

f (k2 ) ¤ f (k) ¤ f (k1 )

for all k ∈ K.

Proof. Since f (K) is a non-empty bounded set, it has a supremum M say.

Since f (K) is closed, M ∈ f (K), that is M = f (k1 ) for some k1 ∈ K. We

obtain k2 similarly.

58 A COMPANION TO ANALYSIS

In other words, a real-valued continuous function on a closed bounded

set is bounded and attains its bounds. Less usefully we may say that, in this

case, f actually has a maximum and a minimum. Notice that there is no

analogous result for vector-valued functions. Much popular economic writing

consists of attempts to disguise this fact (there is unlikely to be a state of

the economy in which everybody is best o¬).

Exercise 4.3.5. When I was an undergraduate, we used another proof of

Theorem 4.3.4 which used lion hunting to establish that f was bounded and

then a clever trick to establish that it attains its bounds.

(i) We begin with some lion hunting in the style of Exercise 4.1.14. As

in that exercise, we shall only consider the case m = 2, leaving the general

case to the reader. Suppose, if possible, that f (K) is not bounded above (that

is, given any κ > 0, we can ¬nd a x ∈ K such that f (x) > κ).

Since K is closed and bounded, we can ¬nd a rectangle S0 = [a0 , b0 ] —

[a0 , b0 ] ⊇ K. Show that we can ¬nd a sequence of pairs of intervals [an , bn ]

and [an , bn ] such that

f (K © [an , bn ] — [an , bn ]) is not bounded,

an’1 ¤ an ¤ bn ¤ bn’1 , an’1 ¤ an ¤ bn ¤ bn’1 ,

and bn ’ an = (bn’1 ’ an’1 )/2, bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.

Show that an ’ c as n ’ ∞ for some c ∈ [a0 , b0 ] and an ’ c as n ’ ∞

for some c ∈ [a0 , b0 ]. Show that c = (c, c ) ∈ K. Use the fact that f is

continuous at c to show that there exists an > 0 such that, if x ∈ K and

x ’ c < , then f (x) < f (c) + 1. Show that there exists an N such that

[an , bn ] — [an , bn ] ⊆ B(c, )

for all n ≥ N and derive a contradiction.

Hence deduce that f (K) is bounded above. Show also that f (K) is bounded

below.

(ii) Since any non-empty bounded subset of R has a supremum, we know

that M = sup f (K) and m = inf f (K) exist. We now produce our clever

trick. Suppose, if possible, that f (x) = M for all x ∈ K. Explain why,

if we set g(x) = 1/(M ’ f (x)), g : K ’ R will be a well de¬ned strictly

positive continuous function. Deduce that there exists a real number M > 0

such that g(x) ¤ M for all x ∈ K and show that f (x) ¤ M ’ 1/M for all

x ∈ K. Explain why this contradicts the de¬nition of M and conclude that

there must exist some k1 ∈ K such that f (k1 ) = M . We obtain k2 similarly.

(The author repeats the remark he made on page 38 that amusing as proofs

59

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

like these are, clever proofs should only be used when routine proofs do not

work.)

Our next theorem is just a particular but useful case of Theorem 4.3.4.

Theorem 4.3.6. Let f : [a, b] ’ R be a continuous function. Then we can

¬nd k1 , k2 ∈ [a, b] such that

f (k2 ) ¤ f (x) ¤ f (k1 )

for all x ∈ [a, b].

Later we will use this result to prove Rolle™s theorem (Theorem 4.4.4)

from which in turn we shall obtain the mean value theorem (Theorem 4.4.1).

Theorem 4.3.4 can also be used to prove the fundamental theorem of

algebra which states that every complex polynomial has a root. If the reader

cannot wait to see how this is done then she can look ahead to section 5.8.

Exercise 4.3.7. (i) Prove Theorem 4.3.6 directly from the one-dimensional

version of the Bolzano-Weierstrass theorem. (Essentially just repeat the ar-

guments of Theorem 4.3.1.)

(ii) Give an example of a continuous f : (a, b) ’ R which is unbounded.

(iii) Give an example of a continuous f : (a, b) ’ R which is bounded but

does not attain its bounds.

(iv) How does your argument in (i) fail in (ii) and (iii)?

(v) Suppose now we work over Q and write [a, b] = {x ∈ Q : a ¤ x ¤ b}.

Show that f (x) = (1+(x2 ’2)2 )’1 de¬nes a continuous function f : [0, 2] ’ Q

which is continuous and bounded but does not attain its upper bound. How

does your argument in (i) fail?

De¬ne a continuous function g : [0, 2] ’ Q which is continuous and

bounded but does not attain either its upper bound or its lower bound. De¬ne

a continuous function h : [0, 2] ’ Q which is continuous but unbounded.

We conclude this section with an exercise which emphasises once again the

power of the hypotheses ˜closed and bounded™ combined with the Bolzano-

Weierstrass method. The result is important but we shall not make much

use of it.

Exercise 4.3.8. (i) By picking xj ∈ Kj and applying the Bolzano-Weierstrass

theorem, prove the following result.

Suppose that K1 , K2 , . . . are non-empty bounded closed sets in Rm such

that K1 ⊇ K2 ⊇ . . . . Then ∞ Kj = …. (That is, the intersection of a

j=1

nested sequence of bounded, closed, non-empty sets is itself non-empty.)

60 A COMPANION TO ANALYSIS

(ii) By considering Kj = [j, ∞), show that boundedness cannot be dropped

from the hypothesis.

(iii) By considering Kj = (0, j ’1 ), show that closedness cannot be dropped

from the hypothesis.

Exercises K.29 to K.36 discuss a substantial generalisation of Exercise 4.3.8

called the Heine-Borel theorem.

4.4 The mean value theorem

Traditionally one of the ¬rst uses of the theorem that every continuous func-

tion on a closed interval is bounded and attains its bounds has been to prove

a slightly stronger version of the mean value inequality.

In common with Dieudonn´ ([13], page 142) and Boas ([8], page 118), I

e

think that the mean value inequality is su¬cient for almost all needs and

that the work required to understand the subtleties in the statement and

proof of Theorem 4.4.1 far outweigh any gain.

However, Theorem 4.4.1 is likely to remain part of the standard analysis

course for many years, so I include it here.

Theorem 4.4.1. (The mean value theorem.) If f : [a, b] ’ R is a

continuous function with f di¬erentiable on (a, b), then we can ¬nd a c ∈

(a, b) such that

f (b) ’ f (a) = (b ’ a)f (c).

Here are some immediate consequences.

Lemma 4.4.2. If f : [a, b] ’ R is a continuous function with f di¬eren-

tiable on (a, b), then the following results hold.

(i) If f (t) > 0 for all t ∈ (a, b) then f is strictly increasing on [a, b].

(That is, f (y) > f (x) whenever b ≥ y > x ≥ a.)

(ii) If f (t) ≥ 0 for all t ∈ (a, b) then f is increasing on [a, b]. (That is,

f (y) ≥ f (x) whenever b ≥ y > x ≥ a.)

(iii) If f (t) = 0 for all t ∈ (a, b) then f is constant on [a, b]. (That is,

f (y) = f (x) whenever b ≥ y > x ≥ a.)

Proof. We prove part (i), leaving the remaining parts to the reader. If b ≥

y > x ≥ a, then the mean value theorem (Theorem 4.4.1) tells us that

f (y) ’ f (x) = (y ’ x)f (c)

for some c with y > c > x. By hypothesis f (c) > 0, so f (y) ’ f (x) > 0.

61

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

Exercise 4.4.3. Prove Theorem 1.7.1 from Theorem 4.4.1.

The key step in proving Theorem 4.4.1 is the proof of the special case

when f (a) = f (b).

Theorem 4.4.4. (Rolle™s theorem.) If g : [a, b] ’ R is a continuous

function with g di¬erentiable on (a, b) and g(a) = g(b), then we can ¬nd a

c ∈ (a, b) such that g (c) = 0.

The next exercise asks you to show that the mean value theorem follows

from Rolle™s theorem.

Exercise 4.4.5. (i) If f is as in Theorem 4.4.1, show that we can ¬nd a real

number A such that, setting

g(t) = f (t) ’ At,

the function g satis¬es the conditions of Theorem 4.4.4.

(ii) By applying Rolle™s theorem (Theorem 4.4.4) to the function g in (i),

obtain the mean value theorem (Theorem 4.4.1). (Thus the mean value the-

orem is just a tilted version of Rolle™s theorem.)

Cauchy produced an interesting variant on the argument of Exercise 4.4.5.

which we give as Exercise K.51.

Exercise 4.4.6. The following very easy consequence of De¬nition 1.7.2 will

be used in the proof of Rolle™s theorem. Let U be an open set in R and let

f : U ’ R be di¬erentiable at t ∈ U with derivative f (t). Show that if

tn ∈ U , tn = t and tn ’ t as n ’ ∞, then

f (tn ) ’ f (t)

’ f (t)

tn ’ t

as n ’ ∞.

We now turn to the proof of Rolle™s theorem.

Proof of Theorem 4.4.4. Since the function g is continuous on the closed in-

terval [a, b], Theorem 4.3.6 tells us that it is bounded and attains its bounds.

More speci¬cally, we can ¬nd k1 , k2 ∈ [a, b] such that

g(k2 ) ¤ g(x) ¤ g(k1 )

for all x ∈ [a, b]. If both k1 and k2 are end points of [a, b] (that is k1 , k2 ∈

{a, b}) then

g(a) = g(b) = g(k1 ) = g(k2 )

62 A COMPANION TO ANALYSIS

and g(x) = g(a) for all x ∈ [a, b]. Taking c = (a + b)/2, we have g (c) = 0

(the derivative of a constant function is zero) and we are done.

If at least one of k1 and k2 is not an end point there is no loss in generality

in assuming that k1 is not an end point (otherwise, consider ’g). Write

c = k1 . Since c is not an end point, a < c < b and we can ¬nd a δ > 0 such

that a < c ’ δ < c + δ < b. Set xn = c ’ δ/n. Since c is a maximum for g,

we have g(c) ≥ g(xn ) and so

g(xn ) ’ g(c)

≥0

xn ’ c

for all n. Since

g(xn ) ’ g(c)

’ g (c),

xn ’ c

it follows that g (c) ≥ 0. However, if we set yn = c + δ/n, a similar argument

shows that

g(yn ) ’ g(c)

¤0

yn ’ c

for all n and so g (c) ¤ 0. Since 0 ¤ g (c) ¤ 0, it follows that g (c) = 0 and

we are done.

(We look more closely at the structure of the preceeding proof in Exer-

cise K.45.)

In his interesting text [11], R. P. Burn writes

Both Rolle™s theorem and the mean value theorem are geomet-

rically transparent. Each claims, with slightly more generality

in the case of the mean value theorem, that for a graph of a

di¬erentiable function, there is always a tangent parallel to the

chord.

My view is that the apparent geometrical transparency is due to our strong

intuitive feeling a function with positive derivative ought to increase ” which

is precisely what we are ultimately trying to prove5 . It is because of this strug-

gle between intuition and rigour that the argument of the second paragraph

of the proof always brings to my mind someone crossing a tightrope above

5

This should not be interpreted as a criticism of Burn™s excellent book. He is writing

a ¬rst course in analysis and is trying to persuade the unwilling reader that what looks

complicated is actually simple. I am writing a second course in analysis and trying to

persuade the unwilling reader that what looks simple is actually complicated.

63

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

a pool full of crocodiles. Let me repeat to any reader tempted to modify

that argument, we wish to use Theorem 4.4.1 to prove that a function with

positive derivative is increasing and so we cannot use that result to prove

Theorem 4.4.1. If you believe that you have a substantially simpler proof of

Rolle™s theorem than the one given above, ¬rst check it against Exercise K.46

and then check it with a professional analyst. Exercise K.43 gives another

use of the kind of argument used to prove Rolle™s theorem.

If the reader uses Theorem 4.4.1, it is important to note that we know

nothing about c apart from the fact that c ∈ (a, b).

Exercise 4.4.7. Suppose that k2 is as in the proof of Theorem 4.4.4. Show

explicitly that, if k2 is not an end point, g (k2 ) = 0.

Exercise 4.4.8. Suppose that g : R ’ R is di¬erentiable, that a < b and

that g(a) = g(b). Suppose k1 and k2 are as in the proof of Theorem 4.4.4.

Show that, if k1 = a, then g (a) ¤ 0 and show by example that we may have

g (a) < 0. State similar results for the cases b = k1 and a = k2 .

Exercise 4.4.9. (This exercise should be compared with Lemma 4.4.2.)

(i) Suppose that f : (a, b) ’ R is di¬erentiable and increasing on (a, b).

Show that f (t) ≥ 0 for all t ∈ (a, b).

(ii) If f : R ’ R is de¬ned by f (t) = t3 , show that f is di¬erentiable and

everywhere strictly increasing yet f (0) = 0.

Exercise 4.4.10. I said above that the mean value inequality is su¬cient

for most purposes. For the sake of fairness here is an example where the

extra information provided by Rolle™s theorem does seem to make a di¬erence.

Here, as elsewhere in the exercises, we assume that reader knows notations

like F (r) for the rth derivative of F and can do things like di¬erentiating a

polynomial which have not been explicitly treated in the main text.

Suppose that f : R ’ R is n times di¬erentiable and that

a < x1 < x2 < · · · < xn < b.

Suppose that P is a polynomial of degree n ’ 1 with P (xj ) = f (xj ). (We say

that P is an interpolating polynomial for f .) We are interested in the error

E(t) = f (t) ’ P (t)

at some point t ∈ [a, b]. Since we already know that E(xj ) = 0, we may also

assume that t, x1 , x2 , . . . , xn are distinct.

We consider the function

n

x ’ xj

F (x) = f (x) ’ P (x) ’ E(t) .

t ’ xj

j=1

64 A COMPANION TO ANALYSIS

(i) Show that F vanishes at t, x1 , x2 , . . . , xn and so vanishes at n + 1

distinct points in (a, b).

(ii) By using Rolle™s theorem (Theorem 4.4.4), show that F vanishes at

n distinct points in (a, b).

(iii) By repeated use of Rolle™s theorem show that F (n) vanishes at some

point c ∈ (a, b).

(iv) By computing F (n) explicitly, deduce that

n

1

(n)

(c) ’ n!E(t)

0=f ,

t ’ xj

j=1

and so

n

f (n) (c)

(t ’ xj ).

E(t) =

n! j=1

Of course, we know nothing about c, but, if we know that |f (n) (x)| ¤ A for

all x ∈ [a, b], we can deduce that

n

A

|f (t) ’ P (t)| ¤ (t ’ xj )

n! j=1

for all t ∈ [a, b]. (We discuss this matter further in Exercise K.48.)

(v) Deduce the weaker inequality

(b ’ a)n

|f (t) ’ P (t)| ¤ A

n!

for all t ∈ [a, b].

A similar argument to the one just given is used in Exercise K.49 to prove

a version of Taylor™s theorem.

A very sharp-eyed reader may observe that we cannot prove Lemma 4.4.2 (i)

from the mean value inequality6 .

4.5 Uniform continuity

The mean value inequality (Theorem 1.7.1) is an excellent example of the

way that many theorems of analysis convert local information into global

6

Having gone to all the bother of proving Theorem 4.4.1 from which Lemma 4.4.2 (i)

follows, we might as well use it. However, Exercise K.27 provides an alternative proof.

65

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

information. We know that f (x) ¤ K so that locally the rate of increase of

f is no greater than K. We deduce that f (u) ’ f (v) ¤ K(u ’ v) so that

globally the rate of increase of f is no greater than K. I remind the reader,

once again, that this conversion of local to global fails for Q and depends on

the fundamental axiom of analysis.

The main theorem of this section (Theorem 4.5.5 on uniform continuity) is

another excellent example of the conversion of local information to global. We

need a couple of de¬nitions and examples. Recall ¬rst from De¬nition 4.2.14

what it means for a function to be continuous on a set

De¬nition 4.5.1. Let E ⊆ Rm . We say that a function f : E ’ Rp is

continuous on E if, given any point x ∈ E and any > 0, we can ¬nd a

δ( , x) > 0 such that, whenever y ∈ E and x ’ y < δ( , x), we have

f (x) ’ f (y) < .

Now compare De¬nition 4.5.1 with our de¬nition of uniform continuity.

De¬nition 4.5.2. Let E ⊆ Rm . We say that a function f : E ’ Rp is

uniformly continuous on E if, given any > 0, we can ¬nd a δ( ) > 0 such

that, whenever x, y ∈ E and x ’ y < δ( ), we have

f (x) ’ f (y) < .

Example 4.5.4 and Theorem 4.5.5 depend on understanding what it means

for a function not to be uniformly continuous.

Exercise 4.5.3. Let E ⊆ Rm . Write down the de¬nition of what it means

for a function f : E ’ Rp not to be uniformly continuous7 .

Example 4.5.4. The following three functions are continuous but not uni-

formly continuous.

(i) f1 : R ’ R given by f1 (x) = x2 .

(ii) f2 : (0, 1) ’ R given by f2 (x) = x’1 .

(iii) f3 : (0, 1) ’ R given by f3 (x) = sin(x’1 ).

Proof. (i) Suppose that δ > 0. If we take x > δ ’1 and y = x + δ/2, we have

|x ’ y| < δ, yet

|x2 ’ y 2 | = y 2 ’ x2 = (y + x)(y ’ x) > 2δ ’1 δ/2 = 1.

Thus f1 is not uniformly continuous.

(ii) and (iii) are left as exercises for the reader.

7

Mathematical educationists call this sort of thing ˜¬nding the contrapositive™.

66 A COMPANION TO ANALYSIS

Theorem 4.5.5. Let K be a closed bounded subset of Rm . If f : K ’ Rp is

continuous on K, then f is uniformly continuous on K.

Proof. Earlier I said that the words ˜closed and bounded™ should elicit the

response ˜Bolzano-Weierstrass™. Let us see how this slogan works here.

If f is not uniformly continuous, then there must exist an > 0 such that,

given any δ > 0, we can ¬nd x, y ∈ K such that

x ’ y < δ but f (x) ’ f (y) > .

Thus we can ¬nd a sequence of pairs of points xn , yn ∈ K such that

xn ’ yn < 1/n but f (xn ) ’ f (yn ) > .

By the Bolzano-Weierstrass theorem, we can ¬nd a k ∈ K and a sequence

n(1) < n(2) < n(3) < . . . such that xn(j) ’ k as j ’ ∞. Since

yn(j) ’ k ¤ yn(j) ’ xn(j) + xn(j) ’ k ’ 0 + 0 = 0,

it follows that yn(j) ’ k as j ’ ∞. Since f is continuous, it follows that

< f (xn(j) ) ’ f (yn(j) ) ¤ f (xn(j) ) ’ f (k) + f (yn(j) ) ’ f (k) ’ 0 + 0 = 0