ńņš. 13 |

Notice that we do not demand that ā˜all points are comparableā™, that is,

we do not demand that, if x = y, then either x y or y x. (You should

reread Deļ¬nition D.5 bearing this in mind.) We will use this extra liberty in

Exercises D.14 and D.16.

Exercise D.6. Consider the set N+ of strictly positive integers. Write n

m if m divides n. In Figure D.1, I draw a family tree of the integers 1 to

6 with n connected to m by a descending line if n m. Extend the tree to

cover the integers 1 to 16.

Show that is a direction.

Give an example of two incomparable integers (that is, integers n and m

such that it is not true that n = m or m n or n m).

We can use the notion of direction to deļ¬ne a limit.

Deļ¬nition D.7. If is a direction on a non-empty set X and f is a func-

tion from X to F (where F is an ordered ļ¬eld such as R or Q), we say that

f (x) ā’ a, with respect to the direction , if a ā F and, given any > 0, we

can ļ¬nd an x0 ( ) ā X such that |f (x) ā’ a| < for all x x0 ( ).

The reader should have no diļ¬culty in proving the following version of

Lemma 1.2.2.

2

Interestingly, both [11] and [5] though radical in style are entirely standard in content.

397

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

Exercise D.8. Let be a relation on a non-empty set X, let f , g be func-

tions from X to F and let a, b, c be elements of F. Prove the following

results.

(i) The limit is unique. That is, if f (x) ā’ a and f (x) ā’ b, with respect

to the direction , then a = b.

(ii) Suppose Y is a non-empty subset of X with the property that if when-

ever x, y ā Y , we can ļ¬nd z ā Y such that z x and z y. Then, if Y

is the restriction of the relation to Y , Y is a direction on Y .

Suppose, in addition, that, given any x ā X we can ļ¬nd a y ā Y such that

y x. If f (x) ā’ a, with respect to the direction , and fY is the restriction

of f to Y , then fY (x) ā’ a, with respect to the direction Y .

(iii) If f (x) = c for all x ā X, then f (x) ā’ c, with respect to the direction

.

(iv) If f (x) ā’ a and g(x) ā’ b, with respect to the direction , then,

f (x) + g(x) ā’ a + b.

(v) If f (x) ā’ a and g(x) ā’ b, with respect to the direction , then

f (x)g(x) ā’ ab.

(vi) Suppose that f (x) ā’ a, with respect to the direction . If f (x) = 0

for each x ā X and a = 0, then f (x)ā’1 ā’ aā’1 .

(vii) If f (x) ā¤ A for each x ā X and f (x) ā’ a, with respect to the

direction , then a ā¤ A. If g(x) ā„ B for each x ā X and g(x) ā’ b, with

respect to the direction , then b ā„ B.

As one might expect, we can recover Lemma 1.2.2 from Exercise D.8.

Exercise D.9. (i) If N+ is the set of strictly positive integers, show that >

(with its ordinary meaning) is a direction on N+ . Show further that, if f is

a function from N+ to F (an ordered ļ¬eld) and a ā F, then f (n) ā’ a, with

respect to the direction >, if and only if f (n) ā’ a as n ā’ ā in the sense of

Deļ¬nition 1.2.1.

(ii) Deduce Lemma 1.2.2 from Exercise D.8.

(iii) Show that (i) remains true if we replace > by ā„. Show that (i)

m if n ā„ 10m + 4. Thus

remains true if we replace > by with n

diļ¬erent succession relations can produce the same notion of limit.

The real economy of this approach appears when we extend it.

Exercise D.10. (i) Let a, t and b be real with a < t < b. Show that, if we

on (a, b) \ {t} by x y if |x ā’ t| < |y ā’ t|, then

deļ¬ne the relation

is a direction. Suppose f : (a, b) ā’ R is a function and c ā R. Show that

f (x) ā’ c with respect to the direction , if and only if f (x) ā’ c as x ā’ t,

in the traditional sense of Deļ¬nition D.3.

398 A COMPANION TO ANALYSIS

(ii) Deduce the properties of the traditional limit of Deļ¬nition D.3 from

Exercise D.8.

(iii) Give a treatment of the classical ā˜limit from aboveā™ deļ¬ned in Deļ¬-

nition D.4 along the lines laid out in parts (i) and (ii).

Exercise D.11. Obtain a multidimensional analogue of Deļ¬nition D.7 along

the lines of Deļ¬nition 4.1.8 and prove a multidimensional version of Exer-

cise D.8 along the lines of Lemma 4.1.9.

A little thought shows how to bring Deļ¬nition D.2 into this circle of ideas.

Deļ¬nition D.12. If X is a direction on a non-empty set X, Y is a di-

rection on a non-empty set Y and f is a function from X to Y , we say that

f (x) ā’ ā—Y as x ā’ ā—X if, given y ā Y , we can ļ¬nd x0 (y) ā X such that

f (x) Y y for all x x0 (y).

Exercise D.13. (i) Show that if we take X = N+ , X to be the usual rela-

tion > on X, Y = R and Y to be the usual relation > on Y , then saying that

a function f : X ā’ Y has the property f (x) ā’ ā—Y as x ā’ ā—X is equivalent

to the classical statement f (n) ā’ ā as n ā’ ā.

(ii) Give a similar treatment for the classical statement that a function

f : R ā’ R satisļ¬es f (x) ā’ ā’ā as x ā’ ā.

(iii) Give a similar treatment for the classical statement that a function

f : R ā’ R satisļ¬es f (x) ā’ a as x ā’ ā.

In all the examples so far, we have only used rather simple examples of

direction. Here is a more complicated one.

Exercise D.14. This exercise assumes a knowledge of Section 8.2 where we

deļ¬ned the Riemann integral. It uses the notation of that section. Consider

the set X of ordered pairs (D, E) where D is the dissection

D = {x0 , x1 , . . . , xn } with a = x0 ā¤ x1 ā¤ x2 ā¤ Ā· Ā· Ā· ā¤ xn = b,

and

E = {t0 , t1 , . . . , tn } with xjā’1 ā¤ tj ā¤ xj .

If f : [a, b] ā’ R is a bounded function, we write

n

Ļ(f, D, E) = f (tj )(xj ā’ xjā’1 ).

j=1

399

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

Show that, if we write (D , E ) (D, E) when D ā D, then is a

direction on X. (Note that we place no conditions on E and E .) Show that

not all dissections are comparable.

Using the results of Section 8.2, show that f is Riemann integrable, in

the sense of Section 8.2, with integral I if and only if

Ļ(f, D, E) ā’ I

with respect to the direction .

Here is another example, this time depending on the discussion of metric

spaces in Section 10.3. If we wish to deļ¬ne the notion of a limit for a func-

tion between two metric spaces the natural classical procedure is to produce

something along the lines of Deļ¬nition 10.3.22

Deļ¬nition D.15. Let (X, d) and (Z, Ļ) be metric spaces and f be a map

from X to Z. Suppose that x ā X and z ā Z. We say that f (y) ā’ z as

y ā’ x if, given > 0, we can ļ¬nd a Ī“( , x) > 0 such that, if y ā X and

d(x, y) < Ī“( , x), we have

Ļ(f (y), z) < .

Here is an alternative treatment using direction.

Exercise D.16. Let (X, d) and (Z, Ļ) be metric spaces and x ā X and z ā

Z. We take X to be the collection of open sets in X which contain x and Z

to be the collection of open sets in Z which contain z.

Show that if we deļ¬ne a relation on X by U X V if V ā U , then X is

a direction on X . Is it always true that two elements of X are comparable?

Let Z be deļ¬ned similarly. If f is a map from X to Z, show that

f (y) ā’ z as y ā’ x in the sense of Deļ¬nition D.15 if and only if f (y) ā’ ā—Z

as x ā’ ā—X in the sense of Deļ¬nition D.12.

The advantage of the approach given in Exercise D.16 is that it makes

no reference to the metric and raises the possibility of of doing analysis on

more general objects than metric spaces.

We close with a couple of interesting observations (it will be more conve-

nient to use Deļ¬nition D.7 than our more general Deļ¬nition D.12).

Exercise D.17. Suppose that is a direction on a non-empty set X and f

is a function from X to R.

Suppose that there exists an M ā R such that f (x) ā¤ M for all x ā X and

suppose that f is ā˜increasingā™ in the sense that x y implies f (x) ā¤ f (y).

Show that there exists an a ā F such that that f (x) ā’ a with respect to the

direction .

[Hint. Think about the supremum.]

400 A COMPANION TO ANALYSIS

If the reader thinks about the matter she may recall points in the book

where such a result would have been useful.

Exercise D.18. Prove Lemma 9.2.2 using using Lemma D.17.

Exercise D.19. The result of Lemma D.17 is the generalisation of the state-

ment that every bounded increasing sequence in R has a limit. Find a similar

generalisation of the general principle of convergence. Use it to do Exer-

cise 9.2.10.

If the reader wishes to see more, I refer her to the elegant and eļ¬cient

treatment of analysis in [2].

Appendix E

Traditional partial derivatives

One of the most troublesome culture clashes between pure mathematics and

applied is that to an applied mathematician variables like x and t have mean-

ings such as position and time whereas to a pure mathematician all variables

are ā˜dummy variablesā™ or ā˜place-holdersā™ to be interchanged at will. To a pure

mathematician, v is an arbitrary function deļ¬ned by its eļ¬ect on a variable

so that v(t) = At3 means precisely the same thing as v(x) = Ax3 whereas,

to an applied mathematician who thinks of v as a velocity, the statements

v = At3 and v = Ax3 mean very diļ¬erent (indeed incompatible) things.

dv Ī“v

The applied mathematician thinks of as representing ā˜when every-

dt Ī“t

thing is so small that second order quantities can be neglectedā™. Since

Ī“v Ī“t Ī“v

= ,

Ī“t Ī“x Ī“x

it is obvious to the applied mathematician that

dv dv dx

= , (A)

dt dx dt

but the more rigid notational conventions of the pure mathematicians prevent

them from thinking of v as two diļ¬erent functions (one of t and one of x) in

the same formula. The closest a pure mathematician can get to equation (A)

is the chain rule

d

v(x(t)) = v (x(t))x (t).

dt

Now consider a particle moving along the x-axis so its position is x at

time t. Since

Ī“x Ī“t

= 1,

Ī“t Ī“x

401

402 A COMPANION TO ANALYSIS

it is obvious to the applied mathematician that

dx dt

= 1,

dt dx

and so

dt dx

=1 (B)

dx dt

dx

What does equation (B) mean? In the expression we treat x as a function

dt

dt

of t, which corresponds to common sense, but in the expression we treat t

dx

as a function of x, which seems a little odd (how can the position of a particle

inļ¬‚uence time?). However, if the particle occupies each particular position

at a unique time, we can read oļ¬ time from position and so, in this sense, t

is indeed a function of x. A pure mathematician would say that the function

x : R ā’ R is invertible and replace equation (B) by the inverse function

formula

d ā’1 1

x (t) = .

x (xā’1 (t))

dt

One of the reasons why this formula seems more complicated than equa-

dx dt

tion (B) is that the information on where to evaluate and has been

dt dx

suppressed in equation (B), which ought to read something like

dt dx

(x0 ) = 1 (t0 ) ,

dx dt

or

dt dx

=1 ,

dx dt

x=x0 t=t0

where the particle is at x0 at time t0 .

The clash between the two cultures is still more marked when it comes

to partial derivatives. Consider a gas at temperature T , held at a pressure

P in a container of volume V and isolated from the outside world. The

applied mathematician knows that P depends on T and V and so writes P =

P (T, V ) or P = P (V, T ). (To see the diļ¬erence in conventions between pure

and applied mathematics, observe that, to a pure mathematician, functions

f : R2 ā’ R such that f (x, y) = f (y, x) form a very restricted class!) Suppose

403

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

that, initially, T = T0 , P = P0 , V = V0 . If we change T0 to T0 + Ī“T whilst

keeping V ļ¬xed, then P changes to P0 + Ī“P and the applied mathematician

ā‚P Ī“P

thinks of as representing ā˜when everything is so small that

ā‚T V =V0 ,T =T0 Ī“T

ā‚P

second order quantities can be neglectedā™. In other words, is

ā‚T V =V0 ,T =T0

the rate of change of P when T varies but V is kept ļ¬xed. Often applied

mathematicians write

ā‚P ā‚P

=

ā‚T ā‚T V =V0 ,T =T0

but you should note this condensed notation suppresses the information that

V (rather than, say, V + T ) should be kept constant when T is varied.

There is good reason to suppose that there is a well behaved function

g : R3 ā’ R such that, if a particular gas is at temperature T , held a pressure

P in a container of volume V , then the temperature, pressure, volume triple

(T, P, V ) satisļ¬es

g(T, P, V ) = 0

(at least for a wide range of values of T , P and V ). We may link the pure

mathematicianā™s partial derivatives with those of the applied mathematician

by observing that, if (T0 , P0 , V0 ) and (T0 + Ī“T, P0 + Ī“P, V0 ) are possible

temperature, pressure, volume triples, then, to ļ¬rst order,

0 = g(T0 + Ī“T, P0 + Ī“P, V0 )

= g(T0 , P0 , V0 ) + g,1 (T0 , P0 , V0 )Ī“T + g,2 g(T0 , P0 , V0 )Ī“P

= g,1 (T0 , P0 , V0 )Ī“T + g,2 (T0 , P0 , V0 )Ī“P

and so, to ļ¬rst order,

Ī“P g,1 (T0 , P0 , V0 )

=ā’ .

Ī“T g,2 (T0 , P0 , V0 )

It follows that

ā‚P g,1 (T0 , P0 , V0 )

=ā’ .

ā‚T g,2 (T0 , P0 , V0 )

V =V0 ,T =T0

Essentially identical calculations show that

ā‚T g,3 (T0 , P0 , V0 )

=ā’

ā‚V g,1 (T0 , P0 , V0 )

P =P0 ,V =V0

404 A COMPANION TO ANALYSIS

and

ā‚V g,2 (T0 , P0 , V0 )

=ā’ .

ā‚P g,3 (T0 , P0 , V0 )

T =T0 ,P =P0

Putting the last three equations together, we obtain

ā‚P ā‚T ā‚V

= ā’1.

ā‚T ā‚V ā‚P

V =V0 ,T =T0 P =P0 ,V =V0 T =T0 ,P =P0

This is a very beautiful equation. It can be made much more mysterious by

leaving implicit what we have made explicit and writing

ā‚P ā‚T ā‚V

= ā’1. (C)

ā‚T ā‚V ā‚P

If we further neglect to mention that T , P and V are restricted to the surface

g(T, P, V ) = 0, we get ā˜an amazing result that common sense could not

possibly have predicted . . . It is perhaps the ļ¬rst time in our careers, for

most of us, that we do not understand 3-dimensional spaceā™ ([34], page 65).

Exercise E.1. Obtain the result corresponding to equation (C) for four vari-

ables. Without going into excessive details, indicate the generalisation to n

variables with n ā„ 2. (Check that if n = 2 you get a result corresponding to

equation (B).)

The diļ¬erence between the ā˜pureā™ and ā˜appliedā™ treatment is reļ¬‚ected in

a diļ¬erence in language. The pure mathematician speaks of ā˜diļ¬erentiation

of functions on many dimensional spacesā™ and the applied mathematician of

ā˜diļ¬erentiation of functions of many variablesā™. Which approach is better de-

pends on the problem at hand. Although (T, P, V ) is a triple of real numbers,

it is not a vector, since pressure, volume and temperature are quantities of

diļ¬erent types. Treating (T, P, V ) as a vector is like adding apples to pears.

The ā˜geometricā™ pure approach will yield no insight, since there is no ge-

ometry to consider. On the other hand theories, like electromagnetism and

relativity with a strong geometric component, will beneļ¬t from a treatment

which does not disguise the geometry.

I have tried to show that the two approaches run in parallel but that, al-

though statements in one language can be translated ā˜sentence by sentenceā™

into the other, there is no word for word dictionary between them. A good

mathematician can look at a problem in more than one way. In particular a

good mathematician will ā˜think like a pure mathematician when doing pure

mathematics and like an applied mathematician when doing applied math-

ematicsā™. (Great mathematicians think like themselves when doing mathe-

matics.)

405

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

Exercise E.2. (In this exercise you should assume that everything is well

behaved.) Rewrite the following statement in pure mathematics notation and

prove it using whichever notation (pure, applied or mixed) you prefer.

Suppose that the equations

f (x, y, z, w) = 0

g(x, y, z, w) = 0

can be solved to give z and w as functions of (x, y). Then

ā‚f ā‚g ā‚f ā‚g

ā’

ā‚z ā‚x ā‚w ā‚w ā‚x

=ā’ ā‚f ā‚g ā‚f ā‚g

ā‚x ā’

ā‚z ā‚w ā‚w ā‚z

ā‚f ā‚g ā‚f ā‚g

ā’

ā‚z ā‚y ā‚w ā‚w ā‚y

=ā’ .

ā‚f ā‚g ā‚f ā‚g

ā‚y ā’

ā‚z ā‚w ā‚w ā‚z

Exercise E.3. Each of the four variables p, V , T and S can be regarded as a

well behaved function of any two of the others. In addition we have a variable

U which may be regarded as a function of any two of the four variables above

and satisļ¬es

ā‚U ā‚U

= ā’p.

= T,

ā‚S ā‚V

V S

(i) Show that

ā‚V ā‚T

= .

ā‚S ā‚p

p S

ā‚2U

(ii) By ļ¬nding two expressions for , or otherwise, show that

ā‚pā‚V

ā‚S ā‚T ā‚S ā‚T

ā’ = 1.

ā‚V ā‚p ā‚p ā‚V

p V V p

Exercise E.4. You should only do this exercise if you have met the change

of variable formula for multiple integrals. Recall that, if (u, v) : R 2 ā’ R2 is

a well behaved bijective map, we write

ā‚u ā‚u

ā‚(u, v) ā‚x ā‚y

J= = det ā‚v ā‚v

ā‚(x, y) ā‚x ā‚y

406 A COMPANION TO ANALYSIS

and call J the Jacobian determinant1 . Recall the the useful formula

ā‚(u, v) ā‚(x, y)

= 1.

ā‚(x, y) ā‚(u, v)

Restate it in terms of Df and Df ā’1 evaluated at the correct points. (If you

can not see what is going on, look at the inverse function theorem (Theo-

rem 13.1.13 (ii)) and at our discussion of equation (B) in this appendix.)

Restate and prove the formula

ā‚(u, v) ā‚(s, t) ā‚(u, v)

=

ā‚(s, t) ā‚(x, y) ā‚(x, y)

in the same way.

1

J is often just called the Jacobian but we distinguish between the Jacobian matrix

(see Exercise 6.1.9) and its determinant.

Appendix F

Another approach to the

inverse function theorem

The object of this appendix is to outline a variant on the approach to the

inverse function theorem given in section 13.1.

We begin with a variation on Lemma 13.1.2.

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

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

(f (x) ā’ f (y)) ā’ (x ā’ y) ā¤ Ī· x ā’ y

for all x , y ā¤ Ī“. Then there exists one and only one continuous function

ĀÆ ĀÆ

g : B(0, (1 ā’ Ī·)Ī“) ā’ B(0, Ī“) such that g(0) = 0 and

f (g(y)) = y

ĀÆ

for all y ā B(0, (1 ā’ Ī·)Ī“).

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

ĀÆ

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

with the uniform norm

h = sup h(t) .

ā

tāX

We know that (C, ā) is complete. Let

E = {h : h ā¤ Ī“ and h(0) = 0}.

ā

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

407

408 A COMPANION TO ANALYSIS

Let h ā E. We deļ¬ne T h as a function on X by

T h(y) = h(y) + (y ā’ 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) ā¤ y + h(y) ā’ f (h(y))

= y + (f (h(y)) ā’ f (0)) ā’ (h(y) ā’ 0)

ā¤ y + Ī· h(y) ā’ 0

ā¤ (1 ā’ Ī·)Ī“ + Ī· h(y) < Ī“

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

T h ā X and T is a well deļ¬ned function T : X ā’ X.

If h, k ā 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) ā’ T k(y) = (h(y) ā’ k(y)) ā’ (f (h(y)) ā’ f (k(y))) ā¤ Ī· h(y) ā’ k(y)

for all y ā X. Thus

Th ā’ Tk ā¤Ī· hā’k

ā ā

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

g ā E such that

g = T g,

and so

g(y) = T g(y) = g(y) + (y ā’ f (g(y)),

that is,

f (g(y)) = y

for all y ā 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 ā’ Rm such that f (0) = 0 and

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

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

a Ī“1 with Ī“0 ā„ Ī“1 > 0 and a Ļ > 0 such that there exists one and only one

ĀÆ ĀÆ

continuous function g : B(0, (1 ā’ Ī·)Ī“) ā’ B(0, Ļ) with g(0) = 0 and

f (g(y)) = y

ĀÆ

for all y ā B(0, (1 ā’ Ī·)Ī“).

409

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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 diļ¬erentiable at 0. We deal with this by proving the following

lemma.

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

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

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

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

Proof. Observe that

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

(h) ā’ 0 as h ā’ 0. Thus, if k ā U ,

where

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

and so

k = g(k) + (g(k)) g(k) . (ā )

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

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

(g(k)) < 1/2

whenever k ā V , and so, using (ā ),

k ā„ g(k) ā’ g(k) /2 = g(k) /2

for all k ā V .

Using (ā ) again, we see that

g(k) ā’ k ā¤ (g(k)) g(k) ā¤ 2 (g(k)) k

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

(g(k))

k ā’ 0, it follows that

g(k) = k + Ī·(k) k ,

where Ī·(k) ā’ 0 as k ā’ 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 ā’ Rm such that f (0) = 0 and

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

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

Ī“1 with Ī“0 ā„ Ī“1 > 0 and a Ļ > 0 such that there exists a unique continuous

ĀÆ ĀÆ

function g : B(0, (1 ā’ Ī·)Ī“) ā’ B(0, Ļ) with g(0) = 0 and

f (g(y)) = y

ĀÆ

for all y ā B(0, (1 ā’ Ī·)Ī“). Moreover, g is diļ¬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 ļ¬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

speciļ¬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 speciļ¬c properties of the rationals.

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

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

fact that Īø(j ā’1 ) ā’ Īø(0) is the axiom of Archimedes and required speciļ¬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

[Ī·(j)] > Īø(0) with [Ī·(j)] ā’ Īø(0) as j ā’ ā. The situation is clariļ¬ed by the

following observation.

Lemma G.1. Let (F, +, Ć—, >) be an ordered ļ¬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 ā„ N .

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

as n ā’ ā.

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

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

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

a sequence with the properties required by (B).

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

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

(ii) Show that we can replace the words ā˜at least oneā™ by ā˜exactly oneā™ 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 ļ¬nd [Ī·(j)] ā R such that [Ī·(j)] > Īø(0) and

[Ī·(j)] ā’ Īø(0) as j ā’ ā. By Lemma 14.4.9, it follows that we can ļ¬nd an

xj ā Q such that

[y(j)] + [Ī·(j)] > Īø(xj ) > [y(j)].

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

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

[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 ļ¬eld yield a more general construction.

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

ļ¬eld (H, +, Ć—, >) and an injective map Īø : F ā’ H which preserves addition,

multiplication and order with the following properties.

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

(ii) If a, b ā H and b > a, we can ļ¬nd an x ā F with b > Īø(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 ļ¬eld (F, +, Ć—, >) and and an injec-

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

that there exists an Ī· ā F with Ī· > 0 and Ļ(1/n) > Ī· for all n ā„ 1 [n ā Z].

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

result.

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

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

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

and Ļ(1/n) > Ī· for all n ā„ 1 [n ā 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 ļ¬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 ļ¬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 ļ¬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 ļ¬nd. In our new analysis the objects that we consider must

have a ā˜meaning as a calculationā™. 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 eļ¬ect

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

|xn ā’ xm | ā¤ 10ā’n + 10ā’m .

An unreconstructed classical analyst or unsophisticated computer scientist

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

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

is no such object as the ā˜true answerā™1 .

Here are some examples.

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

(b) We deļ¬ne e by en = n+10 1/j!.

j=0

Exercise H.1. Verify that 1ā— 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

oļ¬ 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 satisļ¬es the consistency condition .

Multiplication is slightly more complicated. Given the programs for x

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

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

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

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

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

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

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

such that |xN ā’ yN | > 2 Ć— 10ā’N .

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

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

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

j=0

e = e.

(iv) Show that e Ć— 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 SchrĀØdingerā™s cat.

o

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

analysis. A function f : R ā’ 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 deļ¬ne a function m corresponding to the classi-

cal ā˜modulusā™ 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. Deļ¬ne a function exp corresponding to the classical exp. [It

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

of the size of xā™.]

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

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

R ā’ R given by H(x) = 0 for x < 0, H(x) = 1 for x ā„ 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 ļ¬xed (otherwise the program might not terminate). Suppose

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

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

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

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

case H(x) = 1ā— 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 speciļ¬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 analystā™s proof (which is a development

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

on such a ā˜proofā™ 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

ā˜zero-ļ¬ndingā™ 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 (0ā— ) = 1ā— and f (1ā— ) = ā’1ā— as a

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

with 0 ā¤ zn ā¤ 1 such that

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

ā—

(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 ā¤ 0, fp,q (1/3) = p, fp,q (2/3) = q and fp,q (t) = ā’1 for t ā„ 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 ļ¬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 ļ¬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

diļ¬culty in proving that

exp(x) = y

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

When English students ļ¬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 oļ¬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. ā˜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.ā™ (See Gibbonā™s Decline and Fall Chapter LI and note the accompanying

commentary.)

419

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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

on its own terms.

Bishop called his version of radical analysis ā˜Constructive Analysisā™ and

the reader will ļ¬nd an excellent account of it in the ļ¬rst few chapters of [7].

The ļ¬rst serious attempt to found a radical analysis was due to Brouwer who

called it ā˜Intuitionismā™.

Appendix I

Miscellany

This appendix consists of short notes on various topics which I feel should

be mentioned but which did not ļ¬t well into the narrative structure of this

book.

Compactness When mathematicians generalised analysis from the study of

metric spaces to the study of more general ā˜topological spacesā™, they needed a

concept to replace the Bolzano-Weierstrass property discussed in this book.

After some experiment, they settled on a property called ā˜compactnessā™.

It can be shown that a metric space has the property of compactness

(more brieļ¬‚y, a metric space is compact) when considered as a topologi-

cal space if and only if it has the Bolzano-Weierstrass property. (See Ex-

ercises K.196 and K.197 if you would like to know more.) We showed in

Theorem 4.2.2 that a subset of Rn with the standard metric has the Bolzano-

Weierstrass property if and only if it is closed and bounded.

Thus, if you read of a ā˜compact subset E of Rn ā™, you may translate

this as ā˜a closed and bounded subset of Rn ā™ and, if you read of a ā˜com-

pact metric spaceā™, you may translate this as ā˜a metric space having the

Bolzano-Weierstrass propertyā™. However, you must remember that a closed

bounded metric space need not have the Bolzano-Weierstrass property (see

Exercise 11.2.4) and that, for topological spaces, the Bolzano-Weierstrass

property is not equivalent to compactness.

Abuse of language The language of mathematics is a product of history as

well as logic and sometimes the forces of history are stronger than those of

logic. Logically, we need to talk about the value sin x that the function sin

takes at the point x, but, traditionally, mathematicians have talked about the

function ā˜sin xā™. To avoid this problem mathematicians tend to use phrases

like ā˜the function f : R ā’ R given by f (x) = x2 ā™ or ā˜the map x ā’ x2 ā™. In

421

422 A COMPANION TO ANALYSIS

Section 13.3, when we wish to talk about the map x ā’ L(Ī», x) = t(x)ā’Ī»f (x)

we write ā˜L(Ī», ) = t ā’ Ī»f ā™. (Many mathematicians dislike leaving a blank

space in a formula and use a place holder ā˜Ā·ā™ instead. They write ā˜L(Ī», Ā·) =

t(Ā·) ā’ Ī»f (Ā·)ā™.)

To a 19th century mathematician and to most of my readers this may

appear unnecessary but the advantage of the extra care appears when we need

to talk about the map x ā’ 1. However, from time to time, mathematicians

revert to their traditional habits.

Bourbaki calls such reversions ā˜abuses of language without which any

mathematical text runs the risk of pedantry, not to say unreadability.ā™

Perhaps the most blatant abuse of language in this book concerns se-

quences. Just as f (x) is not a function f , but the value of f at x, so an is

not the sequence

a1 , a 2 , a 3 , a 4 . . . ,

but the nth term in such a sequence. Wherever I refer to ā˜the sequence

an ā™, I should have used some phrase like ā˜the sequence (an )ā ā™. Perhaps

n=1

future generations will write like this, but it seemed to me that the present

generation would simply ļ¬nd it distracting.

Non-uniform notation When Klein gave his lectures on Elementary Mathe-

matics from An Advanced Standpoint [28] he complained

There are a great many symbols used for each of the vector op-

erations and, so far, it has proved impossible to produce a gener-

ally accepted notation. A commission was set up for this purpose

at a scientiļ¬c meeting at Kassel (1903). However, its members

were not even able to come to a complete agreement among them-

selves. None the less, since their intentions were good, each mem-

ber was willing to meet the others part way and the result was to

bring three new notations into existence1 ! My experience in such

things inclines me to the belief that real agreement is only possi-

ble if there are powerful economic interests behind such a move.

. . . But there are no such interests involved in vector calculus and

so we must agree, for better or worse, to let every mathematician

cling to the notation which he ļ¬nds most convenient or ā“ if he is

dogmatically inclined ā“ the only correct one.

1

In later editions Klein recorded the failure of similar committees set up at the Rome

International Mathematical Congress (1908).

423

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Most mathematics students believe that there should be a unique notation

for each mathematical concept. This is not entirely reasonable. Consider the

derivative of a function f : R ā’ R. If I am interested in the slope of the

dy

curve y = f (x) it is natural to use the suggestive Leibniz notation . If I

dx

am interested in the function which is the derivative of f it is is more natural

to consider f , and if I am interested in the operation of diļ¬erentiation rather

than the functions it operates on I may well write Df . Diļ¬erent branches of

mathematics may have genuine reasons for preferring diļ¬erent notations for

the same thing.

Even if she disagrees, the student must accept that diļ¬erent notations

exist and there is nothing that she can do about it. A quick visit to the

library reveals the following notations for the same thing (here f : R3 ā’ R

is a well behaved function) :-

ā‚3f

, fxxz , fxxz , Dxxz f, f,113 , f113 , f113 , D113 f, D(2,0,1) f, D(2,0,1) f

2 ā‚z

ā‚x

and several others.

Here are three obvious pieces of advice.

(1) If you are writing a piece of mathematics and several notations exist

you must make it clear to the reader which one you are using.

(2) If you are reading a piece of mathematics which uses an unfamiliar

notation try to go along with it rather than translating it back into your

favourite notation. The advantage of a familiar notation is that it carries you

along without too much thought, the advantage of an unfamiliar notation is

that it makes you think again ā” and, who knows, the new notation might

turn out to better than the old.

(3) Do not invent a new notation when a reasonably satisfactory one

already exists.

Left and right derivatives In many ways, the natural subsets of Rm to do

analysis on are the open sets U , since, given a function f : U ā’ Rn and a

point x ā U we can then examine the behaviour of f ā˜in all directions round

xā™, on a ball

B(x, ) = {y : y ā’ x < },

for some, suļ¬ciently small, > 0.

However, as we saw in Theorem 4.3.1 and Theorem 4.5.5, continuous

functions behave particularly well on closed bounded sets. This creates a

certain tension, as the next exercise illustrates.

424 A COMPANION TO ANALYSIS

Exercise I.1. (i) Show that the map Rm ā’ R given by x ā’ x is contin-

uous.

(ii) If E is a closed bounded set, show that there exists an e0 ā E such

that e0 ā„ x for all x ā E. Give an example to show that e0 need not be

unique.

(iii) Show that the only subset of Rm which is both open and closed and

bounded is the empty set.

This tension is often resolved by considering functions deļ¬ned on an open

set U , but working on closed bounded subsets of U . In the special case, when

we work in one dimension and deal with intervals, we can use a diļ¬erent trick.

Deļ¬nition I.2. Let b > a and consider f : [a, b] ā’ R. We say that f has

right derivative f (a+) at a if

f (a + h) ā’ f (a)

ā’ f (a+)

h

as h ā’ 0 through values h > 0.

Exercise I.3. Deļ¬ne the left derivative f (bā’), if it exists.

When mathematicians write that f is diļ¬erentiable on [a, b] with deriva-

tive f they are using the following convention.

Deļ¬nition I.4. If f : [a, b] ā’ R is diļ¬erentiable at each point of (a, b) and

has right derivative f (a+) at a and left derivative f (bā’) at b, we say that

f is diļ¬erentiable on [a, b] and write f (a) = f (a+), f (b) = f (bā’).

Observe that this is not radically diļ¬erent from our other suggested ap-

proach.

Lemma I.5. If f : [a, b] ā’ R is diļ¬erentiable on [a, b] we can ļ¬nd an every-

Ė Ė

where diļ¬erentiable function f : R ā’ R with f (t) = f (t) for t ā [a, b].

Proof. Set

Ė

f (t) = f (a) + f (a)(t ā’ a) for t < a,

Ė for a ā¤ t ā¤ b,

f (t) = f (t)

Ė

f (t) = f (b) + f (b)(t ā’ b) for b < t.

Ė

It is easy to check that f has the required properties.

Exercise I.6. If f : [a, b] ā’ R is diļ¬erentiable on [a, b] with continuous

Ė

derivative, show that we can ļ¬nd an everywhere diļ¬erentiable function f :

Ė

R ā’ R with continuous derivative such that f (t) = f (t) for t ā [a, b].

425

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If f : [a, b] ā’ R is diļ¬erentiable on [a, b] and f : [a, b] ā’ R is diļ¬eren-

tiable on [a, b], it is natural to say that f is twice diļ¬erentiable on [a, b] with

derivative f = (f ) , and so on.

Exercise I.7. If f : [a, b] ā’ R is twice diļ¬erentiable on [a, b] with continuous

second derivative, show that we can ļ¬nd an everywhere twice diļ¬erentiable

Ė Ė

function f : R ā’ R with continuous second derivative such that f (t) = f (t)

for t ā [a, b]. Generalise this result.

Piecewise deļ¬nitions Occasionally mathematicians deļ¬ne functions as piece-

wise continuous, piecewise continuously diļ¬erentiable and so on. The under-

lying idea is that the graph of the function is made up of a ļ¬nite number of

well behaved pieces.

Deļ¬nition I.8. A function f : [a, b] ā’ R is piecewise continuous if we can

ļ¬nd

a = x0 < x1 < Ā· Ā· Ā· < x n = b

and continuous functions gj : [xjā’1 , xj ] ā’ R such that f (x) = gj (x) for all

x ā (xjā’1 , xj ) and all 1 ā¤ j ā¤ n

Deļ¬nition I.9. A function f : [a, b] ā’ R is piecewise linear (respectively

diļ¬erentiable, continuously diļ¬erentiable, inļ¬nitely diļ¬erentiable etc.) if it

is continuous and we can ļ¬nd

a = x0 < x1 < Ā· Ā· Ā· < x n = b

such that f |[xjā’1 ,xj ] : [xjā’1 , xj ] ā’ R is linear (respectively diļ¬erentiable, con-

tinuously diļ¬erentiable, inļ¬nitely diļ¬erentiable etc.) for all 1 ā¤ j ā¤ n.

Notice that Deļ¬nition I.8 does not follow the pattern of Deļ¬nition I.9.

Exercise I.10. (i) Show by means of an example that a piecewise continuous

function need not be continuous.

(ii) Show that a piecewise continuous function is bounded.

Exercise I.11. (This is a commentary on Theorem 8.3.1.) (i) Show that

any piecewise continuous function f : [a, b] ā’ R is Riemann integrable.

(ii) Show that, if f : [a, b] ā’ R is continuous on (a, b) and bounded on

[a, b], then f is Riemann integrable.

The deļ¬nitions just considered are clearly rather ad hoc. Numerical ana-

lysts use a more subtle approach via the notion of a spline (see, for example

Chapters 18 and onwards in [42]).

Appendix J

Executive summary

The summary is mainly intended for experts but may be useful for revision.

It may also be more useful than the index if you want to track down a

particular idea. Material indicated ā™„ . . . ā™„ or ā™„ā™„ . . . ā™„ā™„ is not central

to the main argument. Material indicated [ . . . ] is in appendices or exercises;

this material is either not central to the main argument or is such that most

students will have met it in other courses.

Introduction to the real number system

Need for rigorous treatment of analysis (p. 1). Limits in R, subsequences,

sums and products (p. 3). Continuity of functions from R to R (p. 7). The

real numbers R form an ordered ļ¬eld obeying the fundamental axiom that

every increasing bounded sequence converges (p. 9). Axiom of Archimedes

(p. 10). [Decimal expansion (Exercise 1.5.12, p. 13).] The intermediate value

theorem, proof by lion hunting (p. 14). [Countability (Appendix B, p. 383).

The real numbers are uncountable (Exercise 1.6.7, p. 17). Cantorā™s proof of

the existence of transcendentals (Exercise B.7, p. 385). Explicit construction

of a transcendental number (Exercise K.12, p. 435).] Diļ¬erentiation and the

mean value inequality (one dimensional case) (p. 18). Intermediate value

theorem equivalent to fundamental axiom (p. 22). ā™„ā™„Further informal

discussion of the status of the fundamental axiom (p. 25).ā™„ā™„

Equivalents of the fundamental axiom

Supremum, existence for bounded non-empty sets equivalent to fundamental

axiom, use as proof technique (p. 31). Theorem of Bolzano-Weierstrass,

equivalent to fundamental axiom, use as proof technique (p. 37).

Higher dimensions

Rm as an inner product space, Cauchy-Schwarz and the Euclidean norm

427

428 A COMPANION TO ANALYSIS

(p. 43). Limits in Rm (p. 46). Theorem of Bolzano-Weierstrass in Rm (p. 47).

Open and closed sets (p. 48). Theorem of Bolzano-Weierstrass in the context

of closed bounded subsets of Rm (p. 49). Continuity for many dimensional

spaces (p. 53). The image of a continuous function on a closed bounded

subset of Rm is closed and bounded, a real-valued continuous function on a

closed bounded subset of Rm is bounded and attains its bounds (p. 57). [The

intersection of nested, non-empty, closed, bounded sets is non-empty (Exer-

cise 4.3.8, p. 59).] Rolleā™s theorem and the one dimensional mean value theo-

rem (p. 60). Uniform continuity, a continuous function on a closed bounded

subset of Rm is uniformly continuous (p. 64).

Sums

(This material is treated in Rm .) General principle of convergence (p. 66).

Absolute convergence implies convergence for sums (p. 69). Comparison test

(p. 70). Complex power series and the radius of convergence (p. 71). ā™„Ratio

test and Cauchyā™s condensation test (p. 70). Conditional convergence, al-

ternating series test, Abelā™s test, rearrangement of conditionally convergent

series (p. 78). Informal discussion of the problem of interchanging limits

(p. 81). Dominated convergence theorem for sums, rearrangement of abso-

lutely convergent series, Fubiniā™s theorem for sums (p. 84). The exponential

function mainly for R but with mention of C, multiplication of power series.

(p. 91). The trigonometric functions, notion of angle (p. 98). The logarithm

on (0, ā), problems in trying to deļ¬ne a complex logarithm (p. 102). Powers

(p. 109). Fundamental theorem of algebra (p. 113).ā™„

Diļ¬erentiation from Rn to Rm

Advantages of geometric approach, deļ¬nition of derivative as a linear map,

Jacobian matrix (p. 121). Operator norm, chain rule and other elementary

properties of the derivative (p. 127). Mean value inequality (p. 136). Simple

local and global Taylor theorems in one dimensions, Cauchyā™s example of

a function with no non-trivial Taylor expansion, Taylor theorems depend

on the fundamental axiom (p. 141). Continuous partial derivatives imply

diļ¬erentiability, symmetry of continuous second order derivatives, informal

treatment of higher order local Taylor theorems, informal treatment of higher

order derivatives as symmetric multilinear maps (p. 146). Discussion, partly

informal, of critical points, hill and dale theorem (p. 154).

Riemann integration

Need for precise deļ¬nition of integral and area, Vitaliā™s example (p. 169).

Deļ¬nition of the Riemann integral via upper and lower sums, elementary

properties, integrability of monotonic functions (p. 172). Integrability of

429

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

continuous functions, fundamental theorem of the calculus, Taylorā™s theorem

with integral form of remainder (p. 182). ā™„ Diļ¬erentiation under the inte-

gral for ļ¬nite range, Euler-Lagrange equation in calculus of variations, use

and limitations, Weierstrass type example (p. 190).ā™„ Brief discussion of the

Riemann integral of Rm -valued functions, fā¤ f (p. 202).

ā™„ Class of Riemann integrable functions not closed under pointwise

convergence (p. 205). Informal discussion of improper Riemann integration

(p. 207). Informal and elementary discussion of multiple integrals (no change

of variable formula), Fubini for continuous functions on a rectangle (p. 212),

Riemann-Stieltjes integration (p. 217). Rectiļ¬able curves and line integrals,

Schwarzā™s example showing the problems that arise for surfaces (p. 224).ā™„

Metric spaces

ā™„ Usefulness of generalising notion of distance illustrated by Shannonā™s

theorem on the existence of good codes (p. 233).ā™„ Metric spaces, norms,

limits, continuity, open sets (p. 241). All norms on a ļ¬nite-dimensional

space are Lipschitz equivalent (p. 246). Continuity of functions between

normed spaces (p. 251). ā™„ Informal discussion of geodesics illustrated by

the PoincarĀ“ metric on the upper half plane (p. 254).ā™„

e

Complete metric spaces

Deļ¬nition of completeness, examples of complete and incomplete metric

spaces including among incomplete ones the L1 norm on C([a, b]) (p. 263).

Completeness and total boundedness, equivalence of the conjunction of these

properties with the Bolzano-Weierstrass property (p. 272).

Uniform metric is complete, restatement of result in classical terms (uni-

form limit of continuous functions is continuous, general principle of uni-

form convergence) (p. 275). Uniform convergence, integration and diļ¬erenti-

ation, restatement for inļ¬nite sums, diļ¬erentiation under an inļ¬nite integral

(p. 282). Local uniform convergence of power series, power series diļ¬eren-

tiable term by term, rigorous justiļ¬cation of power series solution of diļ¬eren-

tial equations (p. 288). ā™„ An absolutely convergent Fourier series converges

to the appropriate function (p. 298).ā™„

Contraction mapping theorem

Banachā™s contraction mapping theorem (p. 303). Existence of solutions of

diļ¬erential equations by Picardā™s method (p. 305). ā™„ Informal discussion

of existence and non-existence of global solutions of diļ¬erential equations

(p. 310). Greenā™s function solutions for second order linear diļ¬erential equa-

tions (p. 318).ā™„

The inverse function theorem (p. 329). ā™„ The implicit function theorem

430 A COMPANION TO ANALYSIS

(p. 339). Lagrange multipliers and Lagrangian necessary condition (p. 347).

Lagrangian suļ¬cient condition, problems in applying Lagrange multiplier

methods (p. 353).ā™„

Completion of metric spaces

Density, completion of metric space, inheritance of appropriate structures

such as inner product (p. 355). Proof of existence of completion (p. 362).

ā™„ Informal discussion of construction of Z from N, Q from Z, C from R

(p. 364). Construction of R from Q (p. 369).ā™„ ā™„ā™„ Rapid, optimistic and

informal discussion of foundational issues (p. 375).ā™„ā™„

Appendix K

Exercises

At an elementary level, textbooks consist of a little explanation (usually sup-

plemented by a teacher) and a large number of exercises of a routine nature.

At a higher level the number of of exercises decreases and the exercises be-

come harder and more variable in diļ¬culty. At the highest level there may be

no exercises at all. We may say that such books consist of a single exercise:-

read and understand the contents. Because I would be happy if students

treated my text in this manner I have chosen to put most of the exercises in

an appendix.

I suspect that readers will gain most by tackling those problems which

interest them. To help them make a choice, I have labeled them in the

following manner [2.1, P]. The number 2.1 tells you that Section 2.1 may

be relevant, and the letters have the meanings given below. Like many similar

labeling systems it is not entirely satisfactory.

ā‘ Follows on from the preceding question.

ā‘ā‘ Follows on from an earlier question.

S Rather shorter or easier than the general run of questions.

M Methods type question. Forget theoretical niceties and concentrate on

getting an answer.

M! Just try and get an answer by fair means or foul.

T This question leads you through a standard piece of theory.

T! This question leads you through a standard piece of theory but in a

non-standard way.

P Problem type question.

G Uses general background rather than material in this book.

H The result of this exercise is not standard.

H! The result of this exercise is highly non-standard. Only do this exercise

if you are really interested.

431

432 A COMPANION TO ANALYSIS

Figure K.1: Apostolā™s construction.

ā

Exercise K.1. (irrationality of 2.) [1.1, G, T] The reader presumably

knows the classic proof that the equation n2 = 2m2 has no non-zero integer

solutions (in other words, x2 = 2 has no solution in Q). Here are two others.

(i) Show that, if n2 = 2m2 , then

(2m ā’ n)2 = 2(n ā’ m)2 .

Deduce that, if n and m are strictly positive integers with n2 = 2m2 , we

can ļ¬nd strictly positive integers n and m with n 2 = 2m 2 and n < n.

Conclude that the equation n2 = 2m2 has no non-zero integer solutions.

(ii) Our second argument requires more thought but is also more powerful.

We use it to show that, if N is a positive integer which is not a perfect square,

then the equation x2 = N has no rational solution.

To this end we suppose that x is a positive rational with x2 = N . Explain

why we can ļ¬nd a least positive integer m such that mx is an integer and

why we can ļ¬nd an integer k with k + 1 > x > k. Set m = mx ā’ mk and

show that m is an integer, that m x is an integer and that m > m ā„ 1. The

required result follows by contradiction. (This argument and its extensions

are discussed in [3].)

(iii) Apostol gave the following beautiful geometric version of the argu-

ment of part (i) (see Figure K.1). It will appeal to all fans of Euclidean

geometry and can be ignored by everybody else.

Suppose that n and m are strictly positive integers with n2 = 2m2 , Draw

a triangle ABC with AB and BC of length m units and CA of length n

units. Explain why ABC is a right angled triangle. Let the circle Ī“ with

centre A passing through B cut AC at D and let the tangent to Ī“ at D cut

BC at E. Show that AE, ED and DC all have the same length. Deduce

that ECD is a right angled triangle with sides ED and DC of integer length

m units and side CE of integer length n units. Show that n 2 = 2m 2 and

n > n ā„ 1. Conclude that the equation n2 = 2m2 has no non-zero integer

solutions.

433

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

Exercise K.2. [1.2, P] If dn ā„ 1 for all n and dn + dā’1 tends to a limit as

n

n ā’ ā, show that dn tends to a limit.

Show, by giving an example, that the result is false if we replace the

condition dn ā„ 1 by dn ā„ k for some ļ¬xed k with 0 < k < 1.

Can you ļ¬nd a result similar to that of the ļ¬rst paragraph involving

dn ā C?

Exercise K.3. [1.4, P] Prove that, if

an + b n 2an bn

a1 > b1 > 0 and an+1 = , bn+1 = ,

2 an + b n

then an > an+1 > bn+1 > bn . Prove that, as n ā’ ā, an and bn both tend to

ā

the limit (a1 b1 ).

Use this result to give an example of an increasing sequence of rational

numbers tending to a limit l which is not rational.

Exercise K.4. [1.3, P] Let f, g : R ā’ R be functions such that the com-

position g ā—¦ f is continuous at x.

(i) Show, by means of examples, that the continuity of f at x is neither

a necessary nor a suļ¬cient condition for the continuity of g at f (x).

(ii) Show, by means of examples, that the continuity of g at f (x) is neither

a necessary nor a suļ¬cient condition for the continuity of f at x.

Exercise K.5. [1.3, P] The function f : [0, 1] ā’ [0, 1] is deļ¬ned by

1 ā’ x if x is rational,

f (x) =

x if x is irrational

Consider the following statements, prove those which are true and demon-

strate the falsity of the remainder.

(i) f (f (x)) = x for all x ā [0, 1].

(ii) f (x) + f (1 ā’ x) = 1 for all x ā [0, 1].

(iii) f is bijective.

(iv) f is everywhere discontinuous in [0, 1].

(v) f (x + y) ā’ f (x) ā’ f (y) is rational for all x, y ā [0, 1] such that

x + y ā [0, 1].

Exercise K.6. [1.3, P] Enumerate the rationals in [0, 1] as a sequence q1 ,

q2 , . . . and let H be the usual Heaviside function (so H(t) = 0 for t ā¤ 0,

H(t) = 1 for t > 0). Show that the equation

ā

2ā’n H(x ā’ qn )

f (x) =

n=1

434 A COMPANION TO ANALYSIS

gives a well deļ¬ned function f : [0, 1] ā’ R which is discontinuous at rational

points and continuous at irrational points.

Exercise K.7. [1.4, P] A bounded sequence of real numbers an satisļ¬es

the condition

anā’1 + an+1

an ā¤

2

for all n ā„ 1. By showing that the sequence an+1 ā’ an is decreasing, or

otherwise, show that the sequence an converges.

What, if anything, can we deduce if the sequence can be unbounded?

What, if anything, can we deduce if the sequence an is bounded but we have

the reverse inequality

anā’1 + an+1

an ā„

2

for all n ā„ 1? Prove your answers.

Exercise K.8. [1.4, P] The function f : (0, 1) ā’ R is continuous on the

open interval (0, 1) and satisļ¬es 0 < f (x) < x. The sequence of functions

fn : (0, 1) ā’ R is deļ¬ned by

f1 (x) = f (x), fn (x) = f (fnā’1 (x)) for n ā„ 2.

Prove that fn (x) ā’ 0 as n ā’ ā for all x ā (0, 1).

Show, by means of a counterexample, that the result need not be true if

f is not continuous everywhere in (0, 1).

Enter any number on your calculator. Press the sine button repeatedly.

What appears to happen? Prove your conjecture.

Exercise K.9. [1.4, T] (i) If Ī“ > 0, use the binomial theorem to show that

n2

(1 + Ī“)n ā„ Ī“.

2

Deduce that nā’1 (1 + Ī“)n ā’ ā as n ā’ ā. Show that, if |x| < 1, then

nxn ā’ 0 as n ā’ ā.

(ii) Show, more generally, that, if Ī“ > 0, and k is any integer, nā’k (1 +

Ī“)n ā’ ā as n ā’ ā. Show that, if |x| < 1, then nk xn ā’ 0 as n ā’ ā.

Exercise K.10. [1.4, P] Let x1 , x2 , . . . be a sequence of positive real num-

bers and suppose that

x1 + x 2 + Ā· Ā· Ā· + x n

ā’1

n

435

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

as n ā’ ā. Choose m(n) so that 1 ā¤ m(n) ā¤ n and xm(n) = max(x1 , x2 , . . . , xn ).

Show that xn /n ā’ 0 and xm(n) /n ā’ 0 as n ā’ ā.

By comparing xĪ± with xr xĪ±ā’1 , or otherwise, show that

r m(n)

xĪ± + x Ī± + Ā· Ā· Ā· + x Ī± 0 if Ī± > 1,

1 2 n

ā’

nĪ± ā if 0 ā¤ Ī± < 1,

as n ā’ ā.

Exercise K.11. [1.5, P] Which of the following statements are true and

which are false? Give a proof or counterexample.

(i) The sum of a rational number and an irrational number is irrational.

(ii) The product of a rational number and an irrational number is irra-

tional.

(iii) If x is a real number and > 0, we can ļ¬nd an irrational number y

with |x ā’ y| < .

(iv) If each xn is irrational and xn ā’ x, then x is irrational.

(v) If x and y are irrational then xy is irrational.

[Hint. This is an old chestnut. Consider a = 21/2 and observe that (aa )a = 2.]

Exercise K.12. (Liouvilleā™s theorem.) [1.5, T] A real number x is called

algebraic if

n

ar x r = 0

r=0

for some ar ā Z [0 ā¤ r ā¤ n], an = 0 and n ā„ 1, (in other words, x is a root

of a non-trivial polynomial with integer coeļ¬cients). Liouville proved that

not all real numbers are algebraic. Here is a version of his proof.

(i) If x and y are real and n is a strictly positive integer, show, by ex-

tracting the factor x ā’ y from xn ā’ y n , that

|xn ā’ y n | ā¤ n(max(|x|, |y|))nā’1 |x ā’ y|.

(ii) If P is a polynomial and R > 0 show, using (i), that there exists a K,

depending on P and R, such that

|P (x) ā’ P (y)| ā¤ K|x ā’ y|

whenever |x|, |y| ā¤ R.

(iii) Suppose that

n

ar t r = 0

P (t) =

r=0

436 A COMPANION TO ANALYSIS

for some ar ā Z [0 ā¤ r ā¤ n], an = 0 and n ā„ 1. Show that N n P (M/N )

is an integer whenever M is an integer and N is a strictly positive integer.

Conclude that, if M and N are as stated, either P (M/N ) = 0 or |P (M/N )| ā„

N ā’n .

(iv) Let P , R, K, M and N be as in parts (ii) and (iii). Suppose that Ī·

is a real number with P (Ī·) = 0. If Ī· and M/N lie in the interval [ā’R, R],

show that

N ā’n ā¤ K|Ī· ā’ M/N |.

(v) Use (iv) to prove that, if Ī· is an algebraic number which is not a

rational number, then we can ļ¬nd n and L (depending on Ī·) such that

|Ī· ā’ M/N | ā„ LN ā’n

whenever M is an integer and N is a strictly positive integer.

(vi) Explain why ā 10ā’k! converges. Let us write Ī» = ā

10ā’k! .

k=1 k=1

Show that

J

10ā’k! | ā¤ 2 Ć— 10ā’(J+1)! .

|Ī» ā’

k=1

By considering N = 10J! and M = J 10J!ā’k! , or otherwise, show that (a)

k=1

Ī» is irrational and (b) Ī» is not an algebraic number.

Note that we have shown that Ī» is too well approximable by rationals

to be an algebraic number. We discuss approximation problems again in

Exercises K.13 and K.14.

(vii) In the ļ¬rst part of the question I deliberately avoided using the

mean value inequality (Theorem 1.7). Prove (ii) directly, using the mean

value inequality.

(viii) Show that no real number of the form

ā

mk 10ā’k!

k=1

with mk taking the value 1 or 2 can be algebraic. If you know enough about

countability, show that the collection of numbers of the form just given is

uncountable.

Exercise K.13. (Continued fractions.) [1.5, T] In this exercise we take

a ļ¬rst glance at continued fractions.

437

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

Consider the following process. If x = x0 ā [0, 1), either x0 = 0 and we

stop or we set

1

r1 =

x0

(that is, r1 is the integer part of 1/x0 ). We set

1

ā’ r1 ,

x1 =

x0

so that x1 is the fractional part of 1/x0 and x1 ā [0, 1). We now repeat the

process so that (unless the process terminates) we have xn ā [0, 1),

1 1

ā’ rn+1 .

rn+1 = , xn+1 =

xn xn

Show that, if the process terminates, x must be rational.

For the rest of the question we shall consider the case in which the process

does not terminate unless we explicitly state otherwise. Show, by induction,

or otherwise that

x = (r1 + (r2 + (r3 + (r4 + . . . (rn + (rn+1 + xn+1 )ā’1 )ā’1 . . . )ā’1 )ā’1 )ā’1 )ā’1 .

This formula is usually written out as

1

ńņš. 13 |