a

gives a norm on C([a, b]).

Show, however, that (C([a, b]), 2) is not a complete normed space.

If we wish to show that a metric space (X, d) is complete, the natural way

to proceed is to take a Cauchy sequence xn and show that it must converge.

However, we must show that xn actually converges to a point in X and not

that ˜the sequence xn looks as though it ought to converge™. In many cases

the proof proceeds through the following steps.

(A) The sequence xn converges in some sense (but not the sense we want)

to an object x.

(B) The object x actually lies in X.

(C) The sequence xn actually converges to x in the sense we want.

Here is an example.

Example 11.1.10. The set l 1 of real sequences a with ∞ |aj | convergent

j=1

forms a vector space if we use the natural de¬nitions of addition and scalar

multiplication

(an ) + (bn ) = (an + bn ), »(an ) = (»an ).

If we set

∞

|aj |,

a =

1

j=1

then (l1 , 1) is a complete normed space.

268 A COMPANION TO ANALYSIS

Proof. We know that the space of all sequences forms a vector space, so

we only have to show that l 1 is a subspace. Clearly 0 ∈ l 1 , and, since

N

|»aj | = |»| N |aj |, we have »a ∈ l1 whenever a ∈ l1 and » ∈ R.

j=1 j=1

1

Suppose a, b ∈ l . We have

∞ ∞

N N N

|aj + bj | ¤ |aj | + |bj | ¤ |aj | + |bj |,

j=1 j=1 j=1 j=1 j=1

so, since an increasing sequence bounded above tends to a limit, ∞ |aj +bj |

j=1

1 1

converges and a + b ∈ l . Hence l is, indeed, a subspace of the space of all

sequences. It is easy to check that 1 is a norm.

I shall label the next three paragraphs in accordance with the discussion

just before this example.

Step A Suppose a(n) is a Cauchy sequence in (l 1 , 1 ). For each ¬xed j,

|aj (n) ’ aj (m)| ¤ a(n) ’ a(m) 1 ,

so aj (n) is a Cauchy sequence in R. The general principle of convergence

tells us that aj (n) tends to a limit aj as n ’ ∞.

Step B Since any Cauchy sequence is bounded, we can ¬nd a K such that

a(n) 1 ¤ K for all n. We observe that

N N N N

|aj | ¤ |aj ’ aj (n)| + |aj (n)| ¤ |aj ’ aj (n)| + a(n) 1

j=1 j=1 j=1 j=1

N

¤ |aj ’ aj (n)| + K ’ K

j=1

∞

as n ’ ∞. Thus N |aj | ¤ K for all N , and so |aj | converges. We

j=1 j=1

have shown that a ∈ l1 .

Step C We now observe that, if n, m ≥ M ,

N N N

|aj ’ aj (n)| ¤ |aj ’ aj (m)| + |aj (m) ’ aj (n)|

j=1 j=1 j=1

N

¤ |aj ’ aj (m)| + a(m) ’ a(n) 1

j=1

N

¤ |aj ’ aj (m)| + sup a(p) ’ a(q) 1

p,q≥M

j=1

’ sup a(p) ’ a(q) 1

p,q≥M

269

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

N

as m ’ ∞. Thus |aj ’ aj (n)| ¤ supp,q≥M a(p) ’ a(q) for all N , and

1

j=1

so

a ’ a(n) ¤ sup a(p) ’ a(q)

1 1

p,q≥M

for all n ≥ M . Recalling that the sequence a(m) is Cauchy, we see that

a ’ a(n) 1 ’ 0 as n ’ ∞ and we are done.

The method of proof of Step C, and, in particular, the introduction of the

˜irrelevant m™ in the ¬rst set of inequalities is very useful but requires some

thought to master.

Exercise 11.1.11. In the proof above we said ˜any Cauchy sequence is bounded™.

Give the one line proofs of the more precise statements that follow.

(i) If (X, d) is a metric space, xn is a Cauchy sequence in X and a ∈ X,

then we can ¬nd a K such that d(a, xn ) < K for all n ≥ 1.

(ii) If (V, ) is a normed vector space and xn is a Cauchy sequence in

V, then we can ¬nd a K such that xn < K for all n ≥ 1.

Alternative proof of Example 11.1.6. We wish to show that s00 , with norm

∞

|aj |,

a =

1

j=1

is not complete. Observe that we can consider s00 as a subspace of l1 and

that the norm on l1 agrees with our norm on s00 . Now set

a(n) = (1, 2’1 , 2’2 , . . . , 2’n , 0, 0, . . . ),

and

a = (1, 2’1 , 2’2 , . . . , 2’n , 2’n’1 , 2’n’2 , . . . ).

Since a(n) ∈ s00 for all n and a(n) ’ a 1 ’ 0 as n ’ ∞, we see that s00 is

not closed in l1 and so, by Exercise 11.1.5, (s00 , 1 ) is not complete.

Here are two exercises using the method of proof of Example 11.1.10

Exercise 11.1.12. Let U be a complete vector space with norm . Show

that the set l1 (U ) of sequences

u = (u1 , u2 , u3 , . . . )

270 A COMPANION TO ANALYSIS

with ∞ uj convergent forms a vector space if we use the natural de¬ni-

j=1

tions of addition and scalar multiplication

(un ) + (vn ) = (un + vn ), »(un ) = (»un ).

Show that, if we set

∞

u = uj ,

1

j=1

then (l1 , 1) is a complete normed space.

Exercise 11.1.13. Show that the set l ∞ of bounded real sequences forms a

vector space if we use the usual de¬nitions of addition and scalar multiplica-

tion.

Show further that, if we set

= sup |aj |,

a ∞

j≥1

then (l∞ , ∞ ) is a complete normed space.

[We shall prove a slight generalisation as theorem 11.3.3, so the reader

may wish to work through this exercise before she meets the extension.]

Exercise 11.1.14. Consider the set c0 of real sequences a = (a1 , a2 , a3 , . . . )

such that an ’ 0. Show that c0 is a subspace of l∞ and a closed subset of

(l∞ , ∞ ). Deduce that (c0 , ∞ ) is a complete normed space.

Exercise K.188 contains another example of an interesting and important

in¬nite dimensional complete normed space.

The ¬nal result of this section helps show why the operator norm is so

useful in more advanced work.

Lemma 11.1.15. Let U and V be vector spaces with norms U and V.

Suppose further that V is complete. Then the operator norm is a complete

norm on the space L(U, V ) of continuous linear maps from U to V .

Proof. Once again our argument falls into three steps.

Step A Suppose that Tn is a Cauchy sequence in (L(U, V ), ). For each

¬xed u ∈ U ,

Tn u ’ T m u = (Tn ’ Tm )u ¤ T n ’ Tm u ,

V V V

so Tn u is a Cauchy sequence in V . Since V is complete, it follows that Tn u

tends to a limit T u, say.

271

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

Step B In Step A we produced a map T : U ’ V . We want to show that, in

fact, T ∈ L(U, V ). Observe ¬rst that, since Tn is linear

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 V

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 ’ Tn (»1 u1 + »2 u2 ) ’ »1 Tn u1 ’ »2 Tn u2

= V

= T (»1 u1 + »2 u2 ) ’ Tn (»1 u1 + »2 u2 ) ’ »1 (T u1 ’ Tn u1 ) ’ »2 (T u2 ’ Tn u2 ) V

¤ T (»1 u1 + »2 u2 ) ’ Tn (»1 u1 + »2 u2 ) V + |»1 | T u1 ’ Tn u1 V + |»2 | T u2 ’ Tn u2 V

’0

as n ’ ∞. Thus

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 = 0,

V

so

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 = 0,

and T is linear.

Next, observe that, since every Cauchy sequence is bounded, we can ¬nd

a K such that Tn ¤ K for all n. It follows that Tn u V ¤ K u U for each

n. Thus

¤ T u ’ Tn u ¤ T u ’ Tn u ’K u

Tu + Tn u +K u

V V V V U U

as n ’ ∞, and so T u V ¤ K u U for all u ∈ U . Thus T is continuous.

Step C Finally we need to show that T ’ Tn ’ 0 as n ’ ∞. To do this we

use the trick of ˜the irrelevant m™ introduced in the proof of Example 11.1.10.

T u ’ Tn u ¤ T u ’ Tm u Tm u ’ Tn u V

+

V V

¤ T u ’ Tm u (Tm ’ Tn )u V

+

V

¤ T u ’ Tm u T m ’ Tn u U

+

V

¤ T u ’ Tm u sup Tp ’ Tq u

+

V U

p,q≥M

’ sup Tp ’ Tq u U

p,q≥M

as m ’ ∞. Thus T u ’ Tn u ¤ supp,q≥M Tp ’ Tq for all u ∈ U ,

u

V U

and so

T ’ Tn ¤ sup Tp ’ Tq

p,q≥M

for all n ≥ M . Recalling that the sequence Tm is Cauchy, we see that

T ’ Tn ’ 0 as n ’ ∞ and we are done.

272 A COMPANION TO ANALYSIS

Remark: In this book we are mainly concerned with the case when U and V

are ¬nite dimensional. In this special case, L(U, V ) is ¬nite dimensional and,

since all norms on a ¬nite dimensional space are equivalent (Theorem 10.4.6),

the operator norm is automatically complete.

Exercise 11.1.16. In Exercise 10.4.16, U and V are not complete. Give an

example along the same lines involving complete normed spaces.

11.2 The Bolzano-Weierstrass property

In the previous section we introduced the notion of a complete metric space

as a generalisation of the general principle of convergence. The reader may

ask why we did not choose to try for some generalisation of the Bolzano-

Weierstrass theorem instead. One answer is that it is generally agreed that

the correct generalisation of the Bolzano-Weierstrass property is via the no-

tion of compactness and that compactness is best studied in the context of

topological spaces (a concept more general than metric spaces). A second an-

swer, which the reader may ¬nd more satisfactory, is given in the discussion

below which concludes in Theorem 11.2.7.

We make the following de¬nition.

De¬nition 11.2.1. A metric space (X, d) has the Bolzano-Weierstrass prop-

erty if every sequence xn ∈ X has a convergent subsequence.

Lemma 11.2.2. A metric space (X, d) with the Bolzano-Weierstrass prop-

erty is complete.

Proof. Suppose that xn is a Cauchy sequence. By de¬nition, given any > 0,

we can ¬nd n0 ( ) such that d(xp , xq ) < for all p, q ≥ n0 ( ). By the Bolzano-

Weierstrass property, we can ¬nd n(j) ’ ∞ and x ∈ X such that n(j) ’ ∞

and xn(j) ’ x as j ’ ∞.

Thus, given any > 0, we can ¬nd a J such that n(J) ≥ n0 ( /2) and

d(x, xn(J) ) < /2. Since n(J) ≥ n0 ( /2), we know that, whenever m ≥

n0 ( /2), we have d(xn(J) , xm ) < /2, and so

d(x, xm ) ¤ d(x, xn(J) ) + d(xn(J) , xm ) < /2 + /2 = .

Thus xn ’ x as n ’ ∞.

Exercise 11.2.3. We work in a metric space (X, d).

(i) Show that if xn ’ x as n ’ ∞ then the sequence xn is Cauchy. (Any

convergent sequence is Cauchy.)

273

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

(ii) If xn is a Cauchy sequence and we can ¬nd n(j) ’ ∞ and x ∈ X

such that n(j) ’ ∞ and xn(j) ’ x as j ’ ∞, then xn ’ x. (Any Cauchy

sequence with a convergent subsequence is convergent.)

To show that the converse of Lemma 11.2.2 is false it su¬ces to consider

R with the usual topology. To give another counter example (in which,

additionally, the metric is bounded) we introduce a dull but useful metric

space.

Exercise 11.2.4. Let X be any set. We de¬ne d : X 2 ’ R by

d(x, y) =1 if x = y,

d(x, x) =0.

(i) Show that d is a metric.

(ii) Show that d(xn , x) ’ 0 as n ’ ∞ if and only if there exists an N

such that xn = x for all n ≥ N .

(iii) Show that (X, d) is complete.

¯

(iv) If y ∈ X, show that the closed unit ball B(y, 1) = {x : d(x, y) ¤

1} = X.

(v) If X is in¬nite, show, using (ii) or otherwise, that X does not have

the Bolzano-Weierstrass property.

(vi) Show also that every subset of X is both open and closed.

We call the metric d of the previous lemma the discrete metric.

In order to characterise metric spaces having the Bolzano-Weierstrass

property, we must introduce a further de¬nition.

De¬nition 11.2.5. We say that (X, d) is totally bounded if, given any > 0,

we can ¬nd y1 , y2 , . . . , yN ∈ X such that N B(yj , ) = X.

j=1

In other words, (X, d) is totally bounded if, given any > 0, we can ¬nd

a ¬nite set of open balls of radius covering X.

Lemma 11.2.6. If (X, d) is a metric space with the Bolzano-Weierstrass

property, then it is totally bounded.

Proof. If (X, d) is not totally bounded, then we can ¬nd an > 0 such that

no ¬nite set of open balls of radius covers X. Choose any x1 ∈ X. We

obtain x2 , x3 , . . . inductively as follows. Once x1 , x2 , . . . , xn have been ¬xed,

we observe that n B(xj , ) = X so we can choose xn+1 ∈ n B(xj , ). / j=1

j=1

Now consider the sequence xj . By construction, d(xi , xj ) ≥ for all i = j

and so, if x ∈ X, we have max(d(x, xi ), d(x, xj )) ≥ /2 for all i = j. Thus

the sequence xj has no convergent subsequence and (X, d) does not have the

Bolzano-Weierstrass property.

274 A COMPANION TO ANALYSIS

Theorem 11.2.7. A metric space (X, d) has the Bolzano-Weierstrass prop-

erty if and only if it is complete and totally bounded.

Proof. Necessity follows from Lemmas 11.2.2 and 11.2.6.

To prove su¬ciency, suppose that (X, d) is complete and totally bounded.

Let xn be a sequence in X. We wish to show that it has a convergent

subsequence.

The key observation is contained in this paragraph. Suppose that A is a

subset of X such that xn ∈ A for in¬nitely many values of n and suppose

> 0. Since X is totally bounded we can ¬nd a ¬nite set of open balls B1 ,

B2 , . . . , BM , each of radius , such that M Bm = X. It follows that

m=1

M

m=1 A © Bm = A, and, for at least one of the balls Bm , it must be true that

xn ∈ A © Bm for in¬nitely many values of n. Thus we have shown that that

there is an open ball of radius such that xn ∈ A © B for in¬nitely many

values of n.

It follows that we can construct inductively a sequence of open balls B1 ,

B2 , . . . such that Br has radius 2’r and xn ∈ r Bs for in¬nitely many

s=1

values of n [r = 1, 2, . . . ]. Pick n(1) < n(2) < n(3) < . . . such that xn(r) ∈

r ’r+1

s=1 Bs . If p, q > r, then xn(p) , xn(q) ∈ Br , and so d(xn(p) , xn(q) ) < 2 .

Thus the sequence xn(r) is Cauchy and, since X is complete, it converges.

Exercise 11.2.8. Show that the open interval (0, 1) with the usual Euclidean

metric is totally bounded but does not have the Bolzano-Weierstrass property.

Exercise 11.2.9. Use the completeness of the Euclidean norm on Rm and

Theorem 11.2.7 to show that a closed bounded subset of Rm with the usual

Euclidean norm has the Bolzano-Weierstrass property. (Thus we have an al-

ternative proof of Theorem 4.2.2, provided we do not use Bolzano-Weierstrass

to prove the completeness of Rm .)

Exercise 11.2.10. Show that a metric space (X, d) is totally bounded if and

only if every sequence in X has a Cauchy subsequence.

We shall not make a great deal of use of the concept of the Bolzano-Weierstrass

property in the remainder of this book. Thus, although the results that fol-

low are quite important, the reader should treat them merely as a revision

exercise for some of the material of Section 4.3.

Our ¬rst result is a generalisation of Theorem 4.3.1.

De¬nition 11.2.11. If (X, d) is a metric space, we say that a subset A has

the Bolzano-Weierstrass property2 if the metric subspace (A, dA ) (where dA

is the restriction of the metric d to A) has the Bolzano-Weierstrass property.

2

It is more usual to say that A is compact. However, although the statement ˜A has

the Bolzano-Weierstrass property™ turns out to be equivalent to ˜A is compact™ for metric

spaces, this is not true in more general contexts.

275

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Exercise 11.2.12. (i) Let (X, d) and (Z, ρ) be metric spaces. Show that, if

K is a subset of X with the Bolzano-Weierstrass property and f : X ’ Z is

continuous, then f (K) has the Bolzano-Weierstrass property.

(ii) Let (X, d) be a metric space with the Bolzano-Weierstrass property

and let Rp have the usual Euclidean norm. Show, by using part (i), or oth-

erwise, that, if f : X ’ Rp is a continuous function, then f (K) is closed and

bounded.

Exercise 11.2.13. State and prove the appropriate generalisation of Theo-

rem 4.3.4.

Exercise 11.2.14. (This generalises Exercise 4.3.8.) Let (X, d) be a metric

space with the Bolzano-Weierstrass property. Show that if K1 , K2 , . . . are

closed sets such that K1 ⊇ K2 ⊇ . . . , then ∞ Kj = ….

j=1

The following is a natural generalisation of De¬nition 4.5.2.

De¬nition 11.2.15. Let (X, d) and (Z, ρ) be metric spaces. We say that a

function f : X ’ Z is uniformly continuous if, given > 0, we can ¬nd a

δ( ) > 0 such that, if x, y ∈ X and d(x, y) < δ( ), we have

ρ(f (x), f (y)) < .

The next exercise generalises Theorem 4.5.5.