Proposition 15

A metrizable topological space X is compact if and only if every infinite sequence of points of X has a cluster point in X.

Axiom (C) of Chapter I, § 9, no. 1 shows that the condition is necessary. To show sufficiency, let d be a metric compatible with the topology of X. We show first that the metric space X so defined is complete: every Cauchy sequence in X has a cluster point and is therefore convergent (Chapter II, § 3, no. 2, Proposition 5, Corollary 2); hence X is complete, by Proposition 9. Next we shall show that X is precompact; if this were not so, then by Proposition 14 there would exist a real number $\alpha > 0$ such that X could not be covered by any finite number of subsets of X of diameter $\leq \alpha$. We could then define by induction on n an infinite sequence $(x_n)$ of points of X by the condition $d(x_p, x_n) > \frac{1}{2} \alpha$ for all $p < n$; and such a sequence can have no cluster point, since every ball of radius $< \frac{1}{2} \alpha$ contains at most one point of the sequence.