Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X$ is a metric space. When does it have a metrizable compactification?

Of course it is enough to discuss complete metric spaces, but separability may not be assumed here.

I know that locally compact spaces have one point compactification, however I am not even sure if those are always metrizable. In the separable case I think I can prove it, however these are two extra assumptions.

share|cite|improve this question
If a locally compact space isn't $\sigma$-compact, the point at infinity in the one-point compactification doesn't have a countable base of neighborhoods. For locally compact metric spaces $\sigma$-compactness is equivalent to metrizability of the one-point compactification, see Kechris, Theorem 5.3, p.29 for a more precise result. Moreover, compact metric spaces are separable, hence so are its subspaces, so separability is necessary. – t.b. Sep 8 '11 at 12:40
@Theo: So it is worse than I thought, eh? :-) – Asaf Karagila Sep 8 '11 at 12:49
Yes, sorry about the non-optimal formulation. Now we have reduced the situation to Polish spaces by passing to the completion, and we're happy since every Polish space is a $G_\delta$ in the Hilbert cube (that took too long to find the link...). By the way: I should probably elaborate that into an answer, no? – t.b. Sep 8 '11 at 12:52
See:… – gary Sep 8 '11 at 12:52
@Theo: You may assume the answer is yes. :-) – Asaf Karagila Sep 8 '11 at 13:06
up vote 11 down vote accepted

First of all, a compact metric space is second countable, hence a metrizable space can be homeomorphic to a subspace of a compact metrizable space only if it is second countable.

So let's assume $X$ is a separable metrizable space. Choose a compatible metric $0 \leq d \leq 1$, and complete $X$ to get a Polish space $\overline{X}$ with a homeomorphic copy of $X$ inside. Now, as I argued in my answer here, every Polish space is homeomorphic to a $G_{\delta}$ inside the Hilbert cube $[0,1]^{\mathbb{N}}$.

The embedding itself is easy, simply choose a dense subset $(x_n)_{n \in \mathbb{N}} \subset X$ and map $x$ to $(d(x,x_n))_{n \in \mathbb{N}} \in [0,1]^{\mathbb{N}}$ (recall that we chose a bounded metric $0 \leq d \leq 1$). This is obviously continuous, and it is not hard to show that it's a homeomorphism onto its image. To see that the image of $\overline{X}$ is a $G_{\delta}$ is harder and given in detail in the answer to Apostolos's question I mentioned above.

Upshot: every separable metrizable space is homeomorphic to a subspace of the Hilbert cube (this one of the 100 variants and refinements of the Urysohn theorem).


  • The one-point compactification is not a viable option, as it is Hausdorff only if $\overline{X}$ is locally compact.
  • We can't do better than a $G_{\delta}$ for complete spaces, since open subsets of (locally) compact spaces are locally compact, hence in order to have a compactification in the stricter sense that $\overline{X}$ be open and dense, local compactness of $\overline{X}$ is necessary.

Finally, if a locally compact space is metrizable, then it is necessarily second countable, hence its one-point compactification is second countable as well, and, again by Urysohn metrizable and $\overline{X}$ is an open subset of its one-point compactification.

For more on this, consult Kechris, Classical descriptive set theory, Springer GTM 156, Springer 1994. See in particular Theorem 5.3 on page 29.

share|cite|improve this answer
Oh it is fine. I am in pursue of equivalence of completely metrizible with Cech-complete and metrizible. I guess this is not the direction to march in :-) – Asaf Karagila Sep 8 '11 at 13:36
It's metrizAble :) I thought that Čech-complete spaces are Baire and a metrizable space is Čech-complete space iff it is completely metrizable. I'd have to dig for a reference but it may be in Engelking's topology book. – t.b. Sep 8 '11 at 13:49
@Asaf: Okay, here we go: Theorem 2, p.35. – t.b. Sep 8 '11 at 13:57
I have Engleking, he uses an internal characterization of Cech completeness, which I prefer to avoid. I guess I have no choice. – Asaf Karagila Sep 8 '11 at 14:03
As for the typo, it is horrible to edit from an iPhone. Feel free to correct. :-) – Asaf Karagila Sep 8 '11 at 14:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.