<< . .

. 31
( : 70)



. . >>

2
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
Please send corrections however trivial to twk@dpmms.cam.ac.uk

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.

<< . .

. 31
( : 70)



. . >>