Corollary

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

Let d be a metric compatible with the topology of X. We shall show that the space $\overline{A}$ is compact, by applying the criterion of Proposition 15. Let $(x_n)$ be a sequence of points of $\overline{A}$; then for each index n there exists $y_n \in A$ such that $d(x_n, y_n) < 1/n$; the sequence $(y_n)$ has, by hypothesis, a cluster point $a \in X$, and a is also a cluster point of the sequence $x_i$, for if $y_m$ lies in the ball with centre $a$ and radius $1/n$, for some $m > n$, then $x_m$ lies in the ball with centre $a$ and radius $2/n$.

It should be remarked that Proposition 15 is not a consequence of the existence of a countable fundamental system of neighbourhoods at each point of $X$; there are examples of non-metrizable, non-compact spaces in which every point has a countable fundamental system of neighbourhoods and every infinite sequence of points has a cluster point (Exercise 15).