¬ll the trumpet up with a ¬nite amount of paint and empty it again leaving

the inside nicely painted!

Note that by ˜winding our trumpet like a ball of knitting” we can keep it

within a bounded set (compare parts (i) and (ii)).

(v) (Thor™s drinking horn) As a variation on these themes, consider the

volumes of revolution K obtained by revolving the curve y = x’1/2 for x ≥ 1

around the x axis and K obtained by revolving the curve y = x’1/2 + x’1

for x ≥ 1 around the x axis. Thus

K = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ x’1 , x ≥ 1} and

K = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ (x’1/2 + x’1 )2 , x ≥ 1}.

Show that K has in¬nite volume but that K \ K has ¬nite volume. Sluse

who was the ¬rst to discover such an object wrote to Huygen™s that he could

design a ˜glass of small weight which the hardest drinker could not empty™.

Exercise K.172. [9.5, T] In Section 9.5 we discussed how to de¬ne the line

integral

f (x) ds.

γ

In this exercise we discuss the vector line integral

f (x) ds.

γ

(i) In some sense, we should have

N

f (x) ds ≈ f (xj )(xj ’ xj’1 ).

γ j=1

Discuss informally how this leads to the following variation of the de¬nition

for γ f (x) ds found on page 229.

Let x, y : [a, b] ’ R2 have continuous derivatives (with the usual conven-

tions about left and right derivatives at end points) and consider the curve

γ : [a, b] ’ R2 given by γ(t) = (x(t), y(t)). If f : R2 ’ R is continuous, then

we de¬ne

b

f (x) ds = f (x(t), y(t))γ (t) dt.

a

γ

523

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

(Here, as usual, γ (t) = (x (t), y (t)).)

(ii) Compute γ fj (x) ds for each of the six curves γ k on page 224 and

k

each of the two functions fj given by f1 (x, y) = (x, y) and f2 (x, y) = (y, x).

(iii) (What follows requires Exercise K.168.) We can extend our de¬ni-

tions to general recti¬able curves as follows. Suppose that γ : [a, b] ’ R2

is recti¬able. Show that, if we set γ(t) = (X(t), Y (t)), then the functions

X, Y : [a, b] ’ R are of bounded variation and so we may de¬ne

b b

f (x) ds = f (X(t), Y (t)) dX(t), f (X(t), Y (t)) dY (t) .

a a

γ

Show that, if γ is su¬ciently well behaved that the de¬nition given in part (i)

applies, then the de¬nitions in (i) and (iii) give the same answer.

Exercise K.173. [10.2, H, G] Consider the system described in the ¬rst

few paragraphs of Section 10.2 where n bits are transmitted and each bit has

a probability p of being transmitted wrongly independently of what happens

to any other bit.

(i) Explain why the probability that there are k errors in all is given by

nk

p (1 ’ p)n’k .

un (k) =

k

By looking at un (k + 1)/un (k), or otherwise, show that there is an integer

k0 with un (k + 1) ≥ un (k) for k ¤ k0 ’ 1 and un (k + 1) ≥ un (k) for k0 ¤ k.

Show that np ’ 1 < k0 < np + 1.

(ii) Let > 0. Use the ideas of Lemma 10.2.4 to show that

un (k)

|k’np|<n

’0

un (k)

|k’np|≥n

as n ’ ∞. Deduce that, if we write Nn for the number of errors in a message

of length n,

Pr{|Nn ’ np| ≥ n} ’ 0, †

and so, in particular,

Pr{Nn ≥ (p + )n} ’ 0,

as n ’ ∞.

(iii) If we are prepared to use a little more probability theory then, as the

reader almost certainly knows, there is a general method for obtaining the

524 A COMPANION TO ANALYSIS

result of (ii). (If the notation is unfamiliar do not proceed further.) Write

Xj = 0 if the jth bit is transmitted correctly and Xj = 1 if not. Show that

EXj = p and var Xj = p(1 ’ p). Explain why Nn = n Xj and deduce

j=1

carefully that ENn = pn and var Nn = np(1 ’ p). Now use Tchebychev™s

inequality (Lemma 9.4.14 with X = Nn ’ np) to obtain equation † in (ii).

Exercise K.174. [10.3, P, S] This exercise and the next deal with modi¬-

cations to the axioms for a metric space. Recall that they are

(A) d(x, y) = 0 if and only if x = y.

(B) d(x, y) = d(y, x) for all x, y ∈ X.

(C) d(x, y) + d(y, z) ≥ d(x, z) for all x, y, z ∈ X.

(i) Suppose that d : X 2 ’ R satis¬es axiom (A) and

(C) d(x, y) + d(z, y) ≥ d(x, z) for all x, y, z ∈ X.

Show that (X, d) is a metric space.

(ii) Say everything that there is to say about ˜anti-triangle™ spaces (X, d)

satisfying axioms (A) and (B) together with

(D) d(x, y) + d(z, y) ¤ d(x, z) for all x, y, z ∈ X.

(iii) Suppose (X, d) satis¬es axioms (A) and (B). Show that, if we set

ρ(x, y) = d(x, y) + d(y, x), then (X, ρ) is a metric space.

Exercise K.175. [10.3, T, S] This exercise is more interesting than the

previous one because it deals with a fairly frequent situation. It requires

knowledge of the notions of equivalence relation and equivalence class.

Suppose (X, d) satis¬es axioms (B) and (C) together with

(A) d(x, x) = 0 for all x ∈ X.

Show that the relation x ∼ y if d(x, y) = 0 is an equivalence relation. If

˜

we write [x] for the equivalence class of x and X = X/ ∼ for the set of

equivalence classes, show that

˜

d([x], [y]) = d(x, y)

˜˜

˜

gives a well de¬ned function on X 2 and that (X, d) is a metric space.

Exercise K.176. [10.3, T, G] This is easy but requires enough group the-

ory to understand the statement of the question.

If G is a ¬nite group with identity e generated by a subset X, de¬ne

ρ(e) = 1 and

ρ(g) = min{N : we can ¬nd x1 , x2 , . . . xN ∈ X ∪ X ’1 such that g = x1 x2 . . . xN }

whenever g = e. (Here X ’1 = {x’1 : x ∈ X}.) Show that d(g, h) = ρ(gh’1 )

de¬nes a metric on G.

525

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

Show also that d is left invariant, that is to say d(gk, hk) = d(g, h) for all

g, h, k ∈ G. Is it necessarily true that d is right invariant (i.e. d(kg, kh) =

d(g, h) for all g, h, k ∈ G)? [Hint, if required. Think about the dihedral

group.]

We say that G has diameter maxg∈G (d(e, g)) (with respect to the gener-

ating set X).

(i) Show that Sn , the permutation group on n elements is generated by

the set X1 of elements of the form (ij). Show that Sn has diameter less that

A1 n (for some constant A1 ) with respect to X1 . Interpret this result in terms

of card shu¬„ing.

(ii) Show that Sn , is generated by the set X2 of elements of the form (1j).

Show that Sn has diameter less that A2 n (for some constant A2 ) with respect

to X2 .

(iii) Show that Sn is generated by the set X3 consisting of the two elements

(12) and (123 . . . n). Show that Sn has diameter less that A3 n2 (for some

constant A3 ) with respect to X3 .

(iv) Let n ≥ 2. Consider the dihedral group Dn generated by a and b

subject only to the relations an = e, b2 = e and bab = a’1 . Show that there

exists an A4 > 0 such that Dn has diameter at least A4 n with respect to

X4 = {a, b}.

(v) How many elements does Sn have? How many does Dn have? Com-

ment very brie¬‚y indeed on the relationship between size and diameter in

parts (i) to (iv).

Exercise K.177. [10.3, P] Let (X, d) and (Y, ρ) be a metric spaces and let

X1 and X2 be subsets of X such that X1 ∪ X2 = X. Suppose f : X ’ Y is

such that f |Xj is continuous as a function from Xj to Y . Which, if any, of

the following statements are true and which may be false? (Give proofs or

counterexamples as appropriate.)

(i) The function f is automatically continuous.

(ii) If X1 and X2 are closed, then f is continuous.

(iii) If X1 and X2 are open, then f is continuous.

(iv) If X1 is closed and X2 is open, then f is continuous.

Give Rn its usual Euclidean norm. Suppose g : X ’ Rn is continuous.

Show that there is a continuous function f : X ’ Rn such that f (x) = g(x)

when g(x) < 1 and f (x) = 1 when g(x) ≥ 1.

Exercise K.178. [10.3, P] (i) Give an example of a metric space (X, d)

which contains a set A which is both open and closed but A = X and A = ….

(ii) Suppose (X, d) is a metric space, that A and B are open subsets of

X with A ∪ B = X and that we are given a function f : X ’ R. Show that,

526 A COMPANION TO ANALYSIS

if the restrictions f |A : A ’ R and f |B : B ’ R are continuous, then f is

continuous.

(iii) Consider the interval [0, 1] with the usual metric. Suppose that A

and B are open subsets of [0, 1] with A ∪ B = [0, 1] and A © B = …. De¬ne

f (x) = 1 if x ∈ A and f (x) = 0 if x ∈ B. By considering the result of (ii),

show that A = … or B = …. Deduce that the only subsets of [0, 1] which are

both open and closed are [0, 1] and ….

(iv) Suppose that A is an open subset of Rn and a ∈ A, b ∈ A. By /

considering f : [0, 1] ’ R given by f (t) = 1 if (1 ’ t)a + tb ∈ A and f (t) = 0,

otherwise, show that A is not closed. Thus the only subsets of Rn which are

both open and closed are Rn and ….

Exercise K.179. [10.3, T] Let (X, d) be a metric space. If A is a subset

of X show that the following de¬nitions are equivalent.

¯

(i) The point x ∈ A if and only if we can ¬nd xn ∈ A such that d(xn , x) ’

0 as n ’ ∞.

(ii) Let F be the collection of closed sets F such that F ⊇ A. Then

¯

A = F ∈F F .

¯ ¯

(iii) A is a closed set with A ⊇ A such that, if F is closed and F ⊇ A,

¯

then F ⊇ A.

(It is worth noting that condition (iii) has the disadvantage that it is not

¯

clear that such a set always exists.) We call A the closure of A. We also

¯

write A = Cl A.

If B is a subset of X show that the following de¬nitions are equivalent.

(i) The point x ∈ B —¦ if and only if we can ¬nd > 0 such that y ∈ B

whenever d(x, y) < .

(ii) Let U be the collection of open sets U such that U ⊆ B. Then

B —¦ = U ∈U U .

State a condition (iii) along the lines of (iii) and show that it is equivalent

to (i) and (ii) .

We call B —¦ the interior of B. We also write B —¦ = Int B.

Show that Cl A = X \ Int(X \ A) and Int A = X \ Cl(X \ A). Show that

A = Cl A if and only if A is closed and A = Int A if and only if A is open.

Suppose we work in R with the usual distance. Find the interior and

closure of the following sets Q, Z, {1/n : n ∈ Z, n ≥ 1}, [a, b], (a, b) and

[a, b) where b > a.

Let (X, d) and (Y, ρ) be metric spaces. Show that f : X ’ Y is continu-

¯

ous if and only if f (A) ⊆ f (A) for all subsets A of X.

(An unimportant warning.) Find a metric space (X, d) and an x ∈ X

such that

Cl{y : d(x, y) < 1} = {y : d(x, y) ¤ 1} and Int{y : d(x, y) ¤ 1} = {y : d(x, y) < 1}.

527

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

Exercise K.180. [10.3, P, ‘ ] We use the notation of Exercise K.179. Show

that, if U is open,

Cl(Int(Cl U )) = Cl U.

Show that, starting from a given A, it is not possible to produce more

than 7 distinct sets by repeated application of the operators Cl and Int.

Give an example of a set A in R with the usual metric which gives rise

to 7 distinct sets in this way.

Exercise K.181. [10.4, T] The results of this question are simple but give

some insight into the nature of the norm.

(i) Let be a norm on a real or complex vector space. Show that the

closed unit ball

¯

“ = B(0, 1) = {x ∈ V : x ¤ 1}

has the following properties.

(A) “ is convex, that is to say, if x, y ∈ “ and 1 ≥ » ≥ 0, then »x + (1 ’

»)y ∈ “.

(B) “ is centrally symmetric, that is to say, if x ∈ “, then ’x ∈ “.

(C) “ is absorbing, that is to say, if x ∈ V , then we can ¬nd a strictly

positive real number » with »x ∈ “.

(ii) Show, conversely, that if “ is a closed, convex, centrally symmetric,

absorbing set then “ is the closed unit ball of a norm.

(iii) If A and B are norms on V and K > 0 state and prove a

necessary and su¬cient condition in terms of the closed unit balls of the two

norms for the inequality

¤K x

x A B

to hold for all x ∈ V .

Exercise K.182. [10.4, P] Consider the real vector space l ∞ of real se-

quences a = (a1 , a2 , . . . ) with norm a = supn≥1 |an |. Show that the set

E = {a : there exists an N with an = 0 for all n ≥ N }

is a subspace of l∞ but not a closed subset.

Show that any ¬nite-dimensional subspace of any normed vector space is

closed.

528 A COMPANION TO ANALYSIS

) be a real normed space and let R

Exercise K.183. [10.4, T] Let (E,

have its usual norm. Let T : E ’ R be a linear map.

(i) Show that, if T is not the zero map, the null space

N = T ’1 (0) = {e ∈ E : T e = 0}

is a subspace of E such that, if y ∈ E then every x ∈ E can be written in

/

exactly one way as x = »y + e with » ∈ R and e ∈ N . Comment very brie¬‚y

on what happens when T is the zero map.

(ii) Suppose that T is not the zero map and N is as in (i). Show that N

is closed if and only if there exists a δ > 0 such that »y + e > |»|δ for all

» ∈ R and e ∈ N .

(iii) Show that T is continuous if and only if its null space T ’1 (0) is closed.

(iv) Find T ’1 (0) and check directly that the conclusions of (iii) are satis-

¬ed if we take E = s00 and take T as in part (ii) of Exercise 10.4.14. (There

are ¬ve norms to check.)

Exercise K.184. [10.4, H] This question and the one that follows are in-

cluded more for the bene¬t of the author than the reader. I was worried

whether the proof of Theorem 10.4.6 actually required the use of results

from analysis. In this question I outline a simple example supplied by Imre

Leader which should convince all but the most stubborn that the proof should

indeed use analysis.

Consider the map N : Q2 ’ R given by N (x, y) = |x + y21/2 |. Show that

(1) N (x) ≥ 0 for all x ∈ Q2 .

(2) If N (x) = 0, then x = 0.

(3) If » ∈ Q and x ∈ Q2 , then N (»x) = |»|N (x).

(4) (The triangle inequality) If x, y ∈ Q2 , then N (x + y) ¤ N (x) + N (y).

Let (x, y) = max(|x|, |y|). Show that given any δ > 0 we can ¬nd an

x ∈ Q2 such that

N (x) < δ x .

Explain why, if N : Q2 ’ R is any function satisfying conditions (1)

to (4), there exists a K > 0 such that

N (x) < K x

for all x ∈ Q2 .

Exercise K.185. [10.4, H!] In this question we give another example of

Imre Leader which proves conclusively that Theorem 10.4.6 is indeed a result

of analysis. It is quite complicated and I include it to provide pleasure for

529

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

the expert12 rather than enlightenment for the beginner. Our object is to

prove the following result.

There is a map N : Q3 ’ Q such that

(1) N (x) ≥ 0 for all x ∈ Q3 .

(2) If N (x) = 0, then x = 0.

(3) If » ∈ Q and x ∈ Q3 , then N (»x) = |»|N (x).

(4) (The triangle inequality) If x, y ∈ Q3 , then N (x + y) ¤ N (x) + N (y).

(5) If we write x ∞ = max1¤j¤3 |xj |, then, given any δ > 0, we can ¬nd

a x ∈ Q3 such that

N (x) < δ x ∞.

(i) Why does this result show that Theorem 10.4.6 is a result of analysis?

(ii) (The proof begins here. It will be helpful to recall the ideas and

notation of Question K.181.) We work in R2 with the usual Euclidean norm

. Suppose u1 , u2 , . . . are elements of R2 none of which are scalar multiples

of any other and such that 1/2 ¤ uj ¤ 2 for all j ≥ 1. Show, by using

induction, or otherwise, that we can ¬nd an an ∈ Q such that, writing

vn = an un , we have

(a) 1/2 < vn < 2, and

(b) vn ∈ “n’1

/

¯

where “0 is the closed disc B(0, 1/2) of radius 1/2 and centre 0 and, if n ≥ 1

“n = {»x + µvn : x ∈ “n’1 , |»| + |µ| = 1, », µ ∈ R}.

(For those who know the jargon, “n is the convex hull of “n’1 ∪ {vn , ’vn }.)

(iii) Continuing with the ideas of (ii), write “ for the closure of ∞ “n .

n=0

Show that “ is a closed centrally symmetric convex set with

¯ ¯

B(0, 1/2) ⊆ “ ⊆ B(0, 2).

Show further that, if µ ∈ R, then µvn ∈ “ if and only if |µ| ¤ 1. Use the

— on R such

2

results of Question K.181 to deduce the existence of a norm

that un — ∈ Q for all n and

2x ≥ x ≥ x /2

—

for all x ∈ R2 .

(iv) Let

E = {(a + b21/2 , c + b31/2 ) : a, b, c ∈ Q}.

Who can start by considering the question of whether we can replace Q3 by Q2 in the

12

next paragraph.

530 A COMPANION TO ANALYSIS

By considering a suitable sequence un ∈ E show that there exists a norm

— on R such that e — ∈ Q for all e ∈ E and

2

2x ≥ x ≥ x — /2

—

for all x ∈ R2 .

(v) De¬ne N : Q3 ’ Q by N (x, y, z) = (x + y21/2 , z + y31/2 ) — . Show

that N has properties (1) to (5) as required.

Exercise K.186. [11.1, P] Suppose that (X, d) is a complete metric space.

We say that a subset F of X has bounded diameter if there exists a K such

that d(x, y) < K for all x, y ∈ F and we then say that F has diameter

sup{d(x, y) : x, y ∈ F }.

Suppose that we have sequence of closed subsets Fj of bounded diameter

with F1 ⊇ F2 ⊇ . . . . Show that, if the diameter of Fn tends to 0, then

∞

j=1 Fj contains exactly one point.

Show that ∞ Fj may be empty if

j=1

(a) (X, d) is not complete, or

(b) the Fj do not have bounded diameter, or