# 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, 2015 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. Aug 10, 2015 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, 2015 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)? Aug 10, 2015 at 18:26
• I have no idea! I, too, am a dimension theory ignoramus. Probably it will work?
– user98602
Aug 10, 2015 at 18:29

Express $$M$$ as a union of open sets $$\{U_n\}$$ with $$clos(U_n)$$ compact and contained within $$U_{n+1}$$ (always possible if $$M$$ is not itself compact).

Put a Riemannian metric $$g_0$$ on $$M$$. Revise $$g_0$$ to $$g_1$$ as follows: $$g_1 = f_1 g_0$$ for $$f_1$$ a positive scalar function obeying:
(a) $$f_1 = 1$$ on $$U_1$$ and
(b) on $$U_3 - U_2$$, $$f_1 < 1/\sqrt{d_0(U_2, M - U_3)}$$ (where $$d_0$$ = distance function generated by $$g_0$$).

Note that for $$d_1$$ the distance function generated by $$g_1$$, $$d_1(U_2, M - U_3)$$ is at least $$1$$; in particular, any curve starting in $$U_1$$ and escaping to infinity must have at least length $$1$$, because it crosses from the boundary of $$U_2$$ to the boundary of $$U_3$$, a distance of at least $$1$$.

Now continue inductively: treating $$U_3$$, $$U_4$$, and $$U_5$$ as we just did $$U_1$$, $$U_2$$, and $$U_3$$ and creating $$g_2 = f_2 g_1$$, with any curve from $$U_1$$ escaping to infinity having length at least $$2$$, as it crosses both from boundary of $$U_2$$ to $$U_3$$ (same metric as $$g_1$$) as well as from boundary of $$U_4$$ to boundary of $$U_5$$ (again with distance at least $$1$$); and so on.

We end with a well-defined metric $$g_{\infty} = \lim g_n$$. Any curve escaping to infinity must cross an infinite number of bands of width at least $$1$$ in $$g_{\infty}$$, so it must have infinite length in $$g_{\infty}$$. That means $$(M, g_{\infty})$$ is complete.

• Note: $M$ here is only assumed to be a topological manifold. Thus, you cannot put a Riemannian metric on it (is n general). Also, OP explicitly stated that he knows a proof in the smooth setting. Feb 6 at 19:19