# Is every topological manifold completely metrizable?

Is every (second-countable) topological manifold completely metrizable?

It is known that every smooth manifold possess a complete Riemannian metric, hence in particular it is completely metrizable, however there are non smoothable manifolds.

Yes.

The proof I know is a little roundabout. Let $M$ be your manifold. It is locally compact and Hausdorff, so it has a one-point compactification $M^*$ which is compact Hausdorff. Now $M^*$ is again second countable (see One point compactification is second contable), and (locally) compact Hausdorff spaces are regular, so by the Urysohn metrization theorem, $M^*$ is metrizable with some metric $d^*$. Since $M^*$ is compact, then $(M^*, d^*)$ is of course complete. Now $M$ is an open subset of $M^*$, and every open (or even $G_\delta$) subset of a complete metric space is completely metrizable (with a different metric). See Theorem 1.2 of this note for a proof; it's also in Kechris's Classical Descriptive Set Theory and probably many other standard texts.

In fact, unless I am mistaken, we just showed any locally compact Hausdorff second countable space is completely metrizable.

If there is a more direct proof, I would be interested to see it!

• A more direct proof might show that every manifold has a proper embedding into Euclidean space. For $n \neq 4$-manifolds, every topological manifold has a handlebody structure, so you should be able to use this to construct a proper embedding by hand. Or one could use dimension theory to show that $M^*$ embeds into a Euclidean space, then enbed it into $S^N$ such that $\infty$ maps to $\infty$; then this gives a proper embedding into $\Bbb R^N$. – user98602 Aug 5 '15 at 16:43
• @MikeMiller: This question is relevant - it seems that this is possible but difficult. It also only solves part of the question - not every subset of $\mathbb{R}^n$ is completely metrizable, so we would have to verify that the image of the embedding is $G_\delta$ and then apply the result I mention above. – Nate Eldredge Aug 10 '15 at 18:17
• I was suggesting a closed embedding. If you can embed the one-point compactification into $S^n$ with the point at infinity mapping to $\infty$, then this restricts to a closed embedding of the manifold into $\Bbb R^n$.. Then the restriction of the Euclidean metric provides a complete metric. – user98602 Aug 10 '15 at 18:19
• @MikeMiller: Hmm. Is it a problem that the one-point compactification of a topological manifold need not be a topological manifold? Or is this covered by the "dimension theory" argument (I don't know much about dimension theory)? – Nate Eldredge Aug 10 '15 at 18:26
• I have no idea! I, too, am a dimension theory ignoramus. Probably it will work? – user98602 Aug 10 '15 at 18:29