Lemma F.1. Consider a function f : Rm ’ Rm such that f (0) = 0. Sup-

pose that there exists a δ > 0 and an · with 1 > · > 0 such that

(f (x) ’ f (y)) ’ (x ’ y) ¤ · x ’ y

for all x , y ¤ δ. Then there exists one and only one continuous function

¯ ¯

g : B(0, (1 ’ ·)δ) ’ B(0, δ) such that g(0) = 0 and

f (g(y)) = y

¯

for all y ∈ B(0, (1 ’ ·)δ).

Proof. This is a clever modi¬cation of the proof of Lemma 13.1.2. Let X =

¯

B(0, (1 ’ ·)δ). Consider the space C of continuous functions h : X ’ Rm

with the uniform norm

h = sup h(t) .

∞

t∈X

We know that (C, ∞) is complete. Let

E = {h : h ¤ δ and h(0) = 0}.

∞

Since E is closed in C, we know that E is complete under the uniform norm.

407

408 A COMPANION TO ANALYSIS

Let h ∈ E. We de¬ne T h as a function on X by

T h(y) = h(y) + (y ’ f (h(y)))

so (following exactly the same calculations as in Lemma 13.1.2, but with

h(y) in place of x)

T h(y) ¤ y + h(y) ’ f (h(y))

= y + (f (h(y)) ’ f (0)) ’ (h(y) ’ 0)

¤ y + · h(y) ’ 0

¤ (1 ’ ·)δ + · h(y) < δ

whenever y ∈ X. It is easy to check that, in addition, T h(0) = 0. Thus

T h ∈ X and T is a well de¬ned function T : X ’ X.

If h, k ∈ X then (following exactly the same calculations as in Lemma 13.1.2,

but with h(y) and k(y) in place of x and x ) we have

T h(y) ’ T k(y) = (h(y) ’ k(y)) ’ (f (h(y)) ’ f (k(y))) ¤ · h(y) ’ k(y)

for all y ∈ X. Thus

Th ’ Tk ¤· h’k

∞ ∞

and T is a contraction mapping. If follows that T has a unique ¬xed point

g ∈ E such that

g = T g,

and so

g(y) = T g(y) = g(y) + (y ’ f (g(y)),

that is,

f (g(y)) = y

for all y ∈ X, as required.

The new version of Lemma 13.1.2 gives rise to a new version of Lemma 13.1.4.

Lemma F.2. Consider a function f : Rm ’ Rm such that f (0) = 0 and

there exists a δ0 > 0 such that f is di¬erentiable in the open ball B(0, δ0 ). If

Df is continuous at 0 and Df (0) = I (the identity map), then we can ¬nd

a δ1 with δ0 ≥ δ1 > 0 and a ρ > 0 such that there exists one and only one

¯ ¯

continuous function g : B(0, (1 ’ ·)δ) ’ B(0, ρ) with g(0) = 0 and

f (g(y)) = y

¯

for all y ∈ B(0, (1 ’ ·)δ).

409

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

We leave the proof to the reader.

The reader will note that, though Lemma F.2 is stronger in some ways

than Lemma 13.1.4, it is weaker in others. In particular, we have not shown

that g is di¬erentiable at 0. We deal with this by proving the following

lemma.

Lemma F.3. Consider a function f : Rm ’ Rm such that f (0) = 0 and f is

di¬erentiable at 0 with Df (0) = I. Suppose that U is an open set containing

0 and g : U ’ Rm is function such that g(0) = 0, f (g(y)) = y for all y ∈ U

and g is continuous at 0. Then f is di¬erentiable at 0 with Df (0) = I.

Proof. Observe that

f (h) = h + (h) h ,

(h) ’ 0 as h ’ 0. Thus, if k ∈ U ,

where

f (g(k)) = g(k) + (g(k)) g(k)

and so

k = g(k) + (g(k)) g(k) . (†)

Since g is continuous at 0, we know that (g(k)) ’ 0 as k ’ 0. In

particular, we can ¬nd an open set V ‚ U with 0 ∈ V such that

(g(k)) < 1/2

whenever k ∈ V , and so, using (†),

k ≥ g(k) ’ g(k) /2 = g(k) /2

for all k ∈ V .

Using (†) again, we see that

g(k) ’ k ¤ (g(k)) g(k) ¤ 2 (g(k)) k

for all k ∈ V . Thus, since, as already noted above, ’ 0 as

(g(k))

k ’ 0, it follows that

g(k) = k + ·(k) k ,

where ·(k) ’ 0 as k ’ 0. The result is proved.

Combining this result with Lemma F.2 we obtain the following version of

Lemma 13.1.4

410 A COMPANION TO ANALYSIS

Lemma F.4. Consider a function f : Rm ’ Rm such that f (0) = 0 and

there exists a δ0 > 0 such that f is di¬erentiable in the open ball B(0, δ0 ). If

Df is continuous at 0 and Df (0) = I (the identity map), then we can ¬nd a

δ1 with δ0 ≥ δ1 > 0 and a ρ > 0 such that there exists a unique continuous

¯ ¯

function g : B(0, (1 ’ ·)δ) ’ B(0, ρ) with g(0) = 0 and

f (g(y)) = y

¯

for all y ∈ B(0, (1 ’ ·)δ). Moreover, g is di¬erentiable at 0 with Dg(0) = I.

We can now rejoin the path taken in Section 13.1 to obtain the inverse

function theorem (Theorem 13.1.13).

Appendix G

Completing ordered ¬elds

In section 14.4 we constructed the reals from the rationals. However, our

arguments, up to and including the proof Lemma 14.4.9, made no use of

speci¬c properties of the rationals. In fact, we can push our arguments a

little further and prove the general principle of convergence (Lemma 14.4.12)

without using speci¬c properties of the rationals.

If we examine our proof of Lemma 14.4.12, we see that it makes use of

the sequence θ(j ’1 ) and the fact that θ(j ’1 ) > θ(0) and θ(j ’1 ) ’ θ(0). The

fact that θ(j ’1 ) ’ θ(0) is the axiom of Archimedes and required speci¬c

properties of the rationals in its proof. To provide a proof of Lemma 14.4.12

independent of the properties of the rationals, all we need is a sequence

[·(j)] > θ(0) with [·(j)] ’ θ(0) as j ’ ∞. The situation is clari¬ed by the

following observation.

Lemma G.1. Let (F, +, —, >) be an ordered ¬eld. Then at least one of

the two following statements is true.

(A) If xn is a Cauchy sequence in F, then there exists an N such that

xn = xN for all n ≥ N .

(B) We can ¬nd a sequence ·n in F with ·n = 0 for all n but with ·n ’ 0

as n ’ ∞.

Proof. If (A) is false, we can ¬nd a Cauchy sequence xn such that, given any

N we can ¬nd an M > N , with xM = xN . We can thus ¬nd n(1) < n(2) <

n(3) < . . . such that xn(j+1) = xn(j) . Setting ·j = |xn(j+1) ’ xn(j) |, we obtain

a sequence with the properties required by (B).

Exercise G.2. (i) Explain why any (F, +, —, >) satisfying alternative (A)

in Lemma G.1 automatically satis¬es the general principle of convergence.

(ii) Show that we can replace the words ˜at least one™ by ˜exactly one™ in

the statement of Lemma G.1.

411

412 A COMPANION TO ANALYSIS

Proof of Lemma 14.4.12 without using the axiom of Archimedes. If alterna-

tive (A) of Lemma G.1 applies, the result is automatic. If not, then alter-

native (B) applies and we can ¬nd [·(j)] ∈ R such that [·(j)] > θ(0) and

[·(j)] ’ θ(0) as j ’ ∞. By Lemma 14.4.9, it follows that we can ¬nd an

xj ∈ Q such that

[y(j)] + [·(j)] > θ(xj ) > [y(j)].

From now on the proof follows the same path as our original proof. More

speci¬cally, we ¬rst show that xj is Cauchy in Q and then that [y(j)] ’

[x].

Exercise G.3. Fill in the details of the proof just given.

Taken together, the parts of the construction of R which only used the

fact that Q is an ordered ¬eld yield a more general construction.

Lemma G.4. Given any ordered ¬eld (F, +, —, >) we can ¬nd an ordered

¬eld (H, +, —, >) and an injective map θ : F ’ H which preserves addition,

multiplication and order with the following properties.

(i) The general principle of convergence holds for H.

(ii) If a, b ∈ H and b > a, we can ¬nd an x ∈ F with b > θ(x) > a.

In Exercise 1.5.11 we constructed an object whose properties we restate

in the next lemma.

Lemma G.5. There exists an ordered ¬eld (F, +, —, >) and and an injec-

tive map φ : Q ’ F which preserves addition, multiplication and order such

that there exists an · ∈ F with · > 0 and φ(1/n) > · for all n ≥ 1 [n ∈ Z].

Applying Lemma G.4 to the ¬eld of Lemma G.5, we obtain the following

result.

Lemma G.6. There exists an ordered ¬eld (H, +, —, >) obeying the gen-

eral principle of convergence and an injective map ψ : Q ’ H which preserves

addition, multiplication and order such that there exists an · ∈ F with · > 0

and ψ(1/n) > · for all n ≥ 1 [n ∈ Z].

Thus we cannot deduce the axiom of Archimedes from the general prin-

ciple of convergence and so we cannot deduce the fundamental axiom of

analysis from the general principle of convergence.

The reader may ask whether there actually exist ordered ¬elds satisfy-

ing alternative (A) of Lemma G.1. She may also remark that, since any

413

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

convergent sequence is a Cauchy sequence, it follows that the only conver-

gent sequences for such a ¬eld are constant from some point on. Since the

treatment of analysis in this book is based on sequences, she may therefore

enquire if it is possible to do analysis on a space where there are no inter-

esting convergent sequences. The answers to her questions are that there do

exist ordered ¬elds satisfying alternative (A) and that it is possible to do

analysis without using sequences. Analysis without sequences and without

metrics is the subject of general topology. But that, as they say, is another

story.

Appendix H

Constructive analysis

The purpose of this appendix is to try and give the reader a vague idea of how

a radical reconstruction of analysis might work. It is based on the version

proposed by Bishop. (A follower of Bishop might well call it a caricature. If

so, it is not intended as a hostile caricature.)

In classical analysis we often talk about objects that we cannot program

a computer to ¬nd. In our new analysis the objects that we consider must

have a ˜meaning as a calculation™. The old notion of a real number x has no

such meaning and is replaced by the following version. A real number x is

a computer program, which, given a positive integer n, produces a rational

number xn subject to the consistency condition, that if we consider the e¬ect

of trying our computer program on two integers n and m we have

|xn ’ xm | ¤ 10’n + 10’m .

An unreconstructed classical analyst or unsophisticated computer scientist

would say that ˜given a positive integer n™ our program provides a rational

xn correct to within 10’n of the true answer™ but to the radical analyst there

is no such object as the ˜true answer™1 .

Here are some examples.

(a) If p is a rational number, we de¬ne de¬ne p— = x by xn = p.

(b) We de¬ne e by en = n+10 1/j!.

j=0

Exercise H.1. Verify that 1— and e satisfy the consistency condition .

We can add two real numbers as follows. Given the programs for x and

y, we construct a program for x + y as follows. Given a positive number n,

1

Of course we can imagine the true answer but we can also imagine unicorns. Or we

can say that the true answer exists but is unknowable, in which case we might be better

o¬ doing theology rather than mathematics.

415

416 A COMPANION TO ANALYSIS

we use the old programs to give us xn+2 and yn+2 . We set zn = xn+2 + yn+2

and write z = x + y.

Exercise H.2. Verify that x + y satis¬es the consistency condition .

Multiplication is slightly more complicated. Given the programs for x

and y, we construct a program for x — y as follows. Given a positive number

n, we ¬rst use the old programs to give us x0 and y0 . We ¬nd the smallest

positive integers m1 and m2 such that 10m1 ≥ |x0 | and 10m2 ≥ |y0 | and take

N = m1 + m2 + n + 8. We now set wn = xN — yN and write w = x — y.

Exercise H.3. Verify that x — y satis¬es the consistency condition .

We say that x = y if we can prove, by looking at the programs involved,

that |xn ’ yn | ¤ 2 — 10’n for all n. We say that x = y if we can ¬nd an N

such that |xN ’ yN | > 2 — 10’N .

Exercise H.4. (i) Show that 1— + 1— = 2— .

(ii) Show that if x and y are real numbers, then x + y = y + x.

(iii) We de¬ne e by en = n+9 1/j!. Show that e is a real number and

j=0

e = e.

(iv) Show that e — e = e.

Notice that, given two real numbers, the radical analyst will say that

there are three possibilities

(a) x = y,

(b) x = y,

(c) neither (a) nor (b) is true.

Notice that for a radical analyst the nature of equality changes with time

since new knowledge may enable us to move a pair of real numbers from

class (c) to one of class (a) or class (b). The conservative analyst, in contrast,

believes that, given real numbers x and y, it is true that x = y or x = y

even though it may be true that nobody will ever know which 2 . In general the

radical analyst replaces the dichotomy

true/false

of the conservative analyst by the much more careful trichotomy

proved true/proved false/not proved true or false.

2

Compare someone who believes that he is being spied on by invisible aliens using

undetectable rays.

417

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

The physicist who feels that this is too cautious might care to ask the opinion

of Schr¨dinger™s cat.

o

It is, I hope, clear what we should consider a function to be in our new

analysis. A function f : R ’ R is a program such that, when we give it a

real x as a callable sub-routine, we obtain a real number f (x).

Thus, for example we can de¬ne a function m corresponding to the classi-

cal ˜modulus™ function m with m(x) = |x| as follows. Given x and n compute

xn+2 . Set zn = |xn+2 |. We take m(x) = z.

Exercise H.5. De¬ne a function exp corresponding to the classical exp. [It

would be a good idea for your program to look ¬rst at x0 ˜to gain some idea

of the size of x™.]

Suppose now that a conservative analyst, anxious to be helpful, tries to

de¬ne a function H corresponding to the classical Heaviside function H :

R ’ R given by H(x) = 0 for x < 0, H(x) = 1 for x ≥ 0. We need a

program such that, when we give it a real x as a callable sub-routine, we

obtain a real number z = H(x). Suppose we ask our program for the value

of z2 . Our program cannot ask for the value of xn for all n, so it must ask for

some subset of x0 , x1 , . . . , xN where N may depend on x0 , x1 , . . . , xM but M

itself must be ¬xed (otherwise the program might not terminate). Suppose

x0 = x1 = · · · = xN = 0. We know that H(x) = 0— or H(x) = 1— , so, by

the de¬nition of equality, either z2 > 3/4 and H(x) = 1— or z2 < 1/4 and

H(x) = 0— . But the next term in our sequence might be xN +1 = ’10’(N +1) /2

(in which case H(x) = 0— and z2 < 1/4 ) or xN +1 = 10’(N +1) /2 (in which

case H(x) = 1— and z2 > 3/4) so our program does not produce the required

answer3 .

The reader will now remember that our basic counterexample, given in

Example 1.1.3, to which we returned over and over again, depended on being

able to construct functions of Heaviside type. Someone who believes that

mathematical intuition comes only from the physical world might rush to

the conclusion that every function in our new analysis must satisfy the inter-

mediate value theorem. However the surest source of intuition for our new

analysis is the world of computing (or, more speci¬cally, numerical analysis)

3

The classical analyst now tells the radical analyst that all functions in the new analysis

must be continuous. (This is what the present author, a typical sloppy conservative

analyst, said on page 375.) However, the classical analyst™s proof (which is a development

of the argument just given) depends on classical analysis! The radical analyst™s views

on such a ˜proof™ are those of an anti-smoking campaigner on a statement by a tobacco

company executive. In some radical reconstructions of analysis it is possible to prove that

all functions are continuous, in others (such as the one given here) there is, at present, no

proof that all functions are continuous and no example of a function which is not.

418 A COMPANION TO ANALYSIS

and anyone who has done numerical analysis knows that there is no universal

˜zero-¬nding™ program.

What would be the statement of the intermediate value theorem in our

new analysis? Surely, it would run something like this. We have a program

P which, when given a function f with f (0— ) = 1— and f (1— ) = ’1— as a

callable subroutine and a positive integer n, produces a rational number zn

with 0 ¤ zn ¤ 1 such that

(i) the sequence zn satis¬es our consistency condition and

—

(ii) f (z) = 0 .

Exercise H.6. (i) Let p and q be rational numbers. Show that our new

analysis contains functions fp,q corresponding to the classical function fp,q

which is the simplest continuous piecewise linear function with fp,q (t) = 1 for

t ¤ 0, fp,q (1/3) = p, fp,q (2/3) = q and fp,q (t) = ’1 for t ≥ 1.

(ii) Sketch fp,q when p and q are very small (note that p and q may be

positive or negative). Explain why your favourite root ¬nding program will

have problems with this function.

(iii) Using the same kind of argument as we used to show the non-

existence of H, convince yourself that a program P of the type described

in the paragraph before this exercise does not exist.

Of course, just as numerical analysis contains zero ¬nding algorithms

which will work on functions satisfying reasonable conditions, so our new

analysis contains versions of the intermediate value theorem which will work

on functions satisfying reasonable conditions. The radical analyst has no

di¬culty in proving that

exp(x) = y

has one and only one solution (for y = 0— ).

When English students ¬rst learn that nouns in other languages have

gender they burst into incredulous laughter and few English people entirely

lose the view that the French speak French in public and English in private.

Because I have tried to explain a version of radical analysis in terms of conser-

vative analysis, it looks as though radical analysis is simply a curious version

of conservative analysis4 . However someone trained in the radical tradition

would see conservative analysis as a curious o¬shoot of radical analysis in

4

One problem faced by anyone trying to reconstruct analysis is given in the traditional

reply of the book burner. ˜If these writings of the Greeks agree with the book of God,

they are useless and need not be preserved: if they disagree they are pernicious and ought

to be destroyed.™ (See Gibbon™s Decline and Fall Chapter LI and note the accompanying

commentary.)

419

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

which distorted versions of known objects obeyed twisted versions of familiar

theorems. There is no primal language, or world view, or real line, but there

are many languages, world views and real lines each of which can be under-

stood to a large extent in terms of another but can only be fully understood