Exercise 23, § 2

A topological space is said to be submetrizable if its topology is finer than the topology of a metrizable space. A submetrizable space is Hausdorff but not necessarily regular (Chapter I, § 8, Exercise 20).

a) A completely regular space $X$ is submetrizable if and only if there exists a uniformity on $X$, compatible with the topology of $X$ and defined by a family $\Phi$ of pseudometrics which contains at least one metric $d$ (cf. § 1, Exercise 3). Let $\hat{X}$ be the completion of $X$ with respect to this uniformity; the metric $d$ extends to a pseudometric $\bar{d}$ on $\hat{X}$; show that for each $x \in \hat{X}$ there is at most one point $y \in X$ such that $\bar{d}(x, y) = 0$.

b) Show that if $X$ is a completely regular submetrizable space, there exists a uniformity, compatible with the topology of $X$, and with respect to which $X$ is complete [use $a$ and $§ 1$, Exercise 16 c)]. Hence show that if in addition $X$ is pseudo-compact ($§ 1$, Exercise 21), then $X$ is compact.

c) Show that a locally compact, paracompact, submetrizable space $X$ is metrizable (reduce to the case where $X$ is $\sigma$-compact by using Chapter I, $§ 9$, no. 10, Theorem 5; then use the Corollary to Proposition 16) [cf. $§ 5$, Exercise 15].