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this has in¬nite area). Equally ˜clearly™, since E has ¬nite volume, we can
¬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
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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

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