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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.

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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
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See: at.yorku.ca/cgi-bin/… –  gary Sep 8 '11 at 12:52
    
@Theo: You may assume the answer is yes. :-) –  Asaf Karagila Sep 8 '11 at 13:06
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1 Answer

up vote 5 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).

Note:

  • 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.

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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
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