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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.

Asaf Shachar
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2 Answers2

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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!

Community
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Nate Eldredge
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    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 $\Bbb R^N$. –  Aug 05 '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
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    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. –  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? –  Aug 10 '15 at 18:29
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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.

Chill2Macht
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Stacey Harris
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    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. – Moishe Kohan Feb 06 '22 at 19:19