Question

7. Recall the space m of bounded sequences of real numbers together with the metric d(х, у) — suр |2; — Ук). k 1,2. (a) Give a simple proof to show that m is complete by showing that m = suitable space X. (Recall that C(X) denotes the space of continuous bounded real- valued functions on X together with the supremum norm.) C(X) for some (b) Let A denote the unit ball in m given by А 3 (x€ т: sup |zk < 1}. k-1,2, Show that A is not compact. (Hint: Consider the idea of Problem 7(b).)

Answer 1

7a. Let $X = \mathbb{N}$ the set of natural numbers with the discrete topology. Let be the space of all real valued continuous bounded functions.

m C(X)

Since X has the discrete topology, every function is continuous.

If
FECX
, then it can be thought of as the bounded sequence
{f(n)}n>1
.

If
n fn
is a bounded sequence,
F(n) n
gives a bounded function which is continuous. Hence
FECX
.

m with the sup metric is isometric to the set of bounded sequences and the given metric d. Since is complete, so is the given space.

C(X)

b) A is the closed unit ball. Let $e_n$ be a sequence which has 1 in the nth term and 0 otherwise.

Now $e_n$ is a sequence in A and
d(en, Em) = sup le (k) e(k) n
for
m n
. This means there can never be a convergent subsequence in
fen}
. Therefore A cannot be compact.