Exercise 22, § 4

A regular space $X$ is metrizable if and only if there exists a sequence $(\mathscr{B}_n)$ of locally finite families of open subsets of $X$ such that $\mathscr{B} = \bigcup_n \mathscr{B}_n$ is a base of the topology of $X$. [To show that the condition is necessary, use Lemma 3 of no. 5. To show that it is sufficient, show first that $X$ is paracompact by using Lemmas 4, 5 and 6. For each pair of integers $m, n$ and each $U \in \mathscr{B}_m$, let $U'$ be the union of the sets $V \in \mathscr{B}_n$ such that $\overline{V} \subset U$. Show that $\overline{U'} \subset U$. Let $f_U : X \to [0, 1]$ be a continuous mapping which is equal to 0 on $X - U$ and to 1 on $U'$, and let

show that the pseudometrics $d_{mn}$ define the topology of $X$. ("Theorem of Nagata-Smirnov.") In particular, every regular space with a countable base is metrizable.