See here for definitions.

Kechris, p. 26

(B) Exercise (4.23). Let be metric space with . Show that for nonempty ,

(i) .

Proof.

Suppose .

Thus, for all , there exists , such that implies .

That is, .

Let . Now, choose such that implies .

Now, choose a sequence in , such that , in the following way: For ,

, for some . Pick any such that .

Then , so .

(ii) .

Proof.

Assume . Let . Since

and since is compact, hence closed, .

.

some such that , and .

Therefore, . Therefore .

Therefore , which is a contradiction.

Thus, .

(i) and (ii) show that .

To show that the converse fails, consider the real numbers with the trivial metric.

Let , which is clearly compact. Then

And .

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