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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
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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
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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.)
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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

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