Exercise 36.1

Proof. Let $X$ be our $m$-manifold. We first show $X$ is locally compact. Let $x\in X$ and a neighborhood $U\ni x$ be given. Since $X$ is a manifold, there exists a homeomorphism $f:U\to f(U)\subseteq {ℝ}^m$. Since ${ℝ}^m$ is locally compact by Example 29.2, there exists a neighborhood $V\subseteq f(U)$ of $f(x)$ such that $\overline{V}$ is compact and $\overline{V}\subseteq f(U)$ by Theorem $29.2$. Then, $x\in f^{-1}(V)\subseteq f^{-1}(\overline{V})\subseteq U$. But then, $f^{-1}(\overline{V})$ is compact and therefore closed by Theorems $26.3$ and $26.5$, and so $\overline{f^{-1}(V)}\subseteq f^{-1}(\overline{V})$ since the closure of a set is the intersection of all closed sets containing it, and moreover $\overline{f^{-1}(V)}=f^{-1}(\overline{V})$ by Theorem $18.1$. Finally, we have $x\in f^{-1}(V)\subseteq \overline{f^{-1}(V)}\subseteq U$, with $\overline{f^{-1}(V)}$ compact, and so $X$ is locally compact by Theorem $29.2$.

Now since $X$ is locally compact and Hausdorff, $X$ is regular by Exercise 32.3. Since $X$ is regular and has a countable basis, it is metrizable by the Urysohn metrization theorem (Theorem $34.1$). Note we used that $X$ is Hausdorff in showing $X$ is regular, for the characterization of local compactness, and the assertion that compact implies closed. □