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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](http://books.google.com/books?id=pPv9KCEkklsC&pg=PA29), p.29 for a more precise result. Moreover, compact metric spaces are separable, hence so are its subspaces, so separability is necessary.2011-09-08
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    @Theo: So it is worse than I thought, eh? :-)2011-09-08
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    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](http://math.stackexchange.com/questions/19344/polish-spaces-and-the-hilbert-cube) (that took too long to find the link...). By the way: I should probably elaborate that into an answer, no?2011-09-08
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    See: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=2234.00012011-09-08
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    @Theo: You may assume the answer is yes. :-)2011-09-08

1 Answers 1

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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 :-)2011-09-08
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    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.2011-09-08
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    @Asaf: Okay, here we go: [Theorem 2, p.35](http://books.google.com/books?id=TAFPsu4eG6wC&pg=PA35).2011-09-08
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    I have Engleking, he uses an internal characterization of Cech completeness, which I prefer to avoid. I guess I have no choice.2011-09-08
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    As for the typo, it is horrible to edit from an iPhone. Feel free to correct. :-)2011-09-08
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    What would be the non-internal one? A space is Čech-complete if and only if it is a $G_{\delta}$ in a compact Hausdorff space? Then Čech himself proves the equivalence in *[On bicompact spaces](http://www.jstor.org/stable/1968839)* Ann. Math. **38**, No. 4 (Oct., 1937), pp. 823-844. on p.838ff.2011-09-08
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    That is the external one. The internal is some countable family of covering, and families of closed sets with FIP. Engelking does that this way.2011-09-08
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    I'm supposed to write a short paper for some proof about Polish spaces, but I am required to do so through a theorem about Cech-complete spaces. I have the needed papers and proofs, but I was hoping to find something which is notably shorter and does not require a vast and extensive net of definitions... I'll just do it the hard way eventually. :-)2011-09-08
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    @Asaf: I still don't know what exactly you're looking for. However, the proof of Kuratowski's theorem I give in the linked answer applies with $X$ an arbitrary topological space. From there it is easy to conclude that if a subset of a Hausdorff topological space is completely metrizable then it must be a $G_{\delta}$. In particular a completely metrizable space must be a $G_{\delta}$ in its Stone-Čech compactification, hence it is Čech-complete in the external definition. This gives you one direction essentially gratis.2011-09-08
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    Interesting! The other direction appears in Engelkind and is quite straightforward. Assuming $X$ a metrizable space is Cech-complete, take the completion and then any compactification of it, then it is a compactification of $X$, thus it is $G_\delta$ there, which in turn means being $G_\delta$ in the metric completion, and therefore $X$ was completely metrizable to begin with.2011-09-08
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    No need to ping me, if you add it I'll see. If not, I'll use the argument given in your link (which I admit that I did not yet see).2011-09-08
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    @Asaf: Real life duties called me away and I don't understand my notes from before anymore. Sorry about that. I then remembered that I saw something like that in Royden's Real Analysis (the section on absolute $G_{\delta}$'s on p.164f of the third edition). However, I can't follow that argument at the moment either...2011-09-08
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    No need to apologize. And as I said, it may very well be the internal characterizations that is needed after all. I will take the time to consider these theorems before that.2011-09-09
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    @Señor Bibliotecario: Do you know a short proof to the Michael theorem about open surjective maps from cech-complete onto paracompact imply range is Cech-complete too?2011-09-14
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    @Asaf: No, I don't. But if Michael's selection theorem is involved, there's a relatively short and clean argument for that in Benyamini-Lindenstrauss, *[Geometric nonlinear functional analysis](http://books.google.com/books?id=lXZ95EKwjYUC&pg=PA21)*, p21ff.2011-09-14
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    I'm afraid that might not be too helpful. I'm just trying to avoid going through sieved and sieve-complete spaces.2011-09-15
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    @Mr. Librarian: One more favor, if I may. I tried finding online J. M. Worrell Jr. paper from Notices of the AMS, vol 13 (1966) p. 858 titled "The paracompact open continuous images of Cech-complete spaces". My local library only have the notices from vol. 16, and I couldn't find it online...2011-09-18
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    @Asaf: As far as I know the Notices are only available online from 1995 on. I just checked in our library and the volume is missing there, too, so I can't help you at the moment. While doing some Googling I found [this](http://projecteuclid.org/euclid.pjm/1102970759) but I doubt it really helps, despite its title being a permutation of the one you're looking for. (It's also strange that two Notices articles by Wicke and Worrell are cited but the one you want isn't...)2011-09-18
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    I found this paper, however it is the opposite result since I need the image to be paracompact and not the domain of the map. MathSciNet gave a similar title using "metacompact", I have never heard this terminology before though. Are you familiar with it?2011-09-18
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    I just found the two papers by Worrell appearing on the 1966 Portuguese journal, they are of no real use (the "metacompact" ones).2011-09-18
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    @Asaf: If I remember correctly metacompact = Hausdorff + every open cover has a point finite refinement (as opposed to a locally finite refinement). That's why *metacompact* is sometimes called *weakly paracompact* (I think Engelking does so).2011-09-18
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    I'll just ask another question, perhaps GEdgar or Brian M. Scott know of a proof (or a sketch of a proof) without resorting to Michael's generalization which requires the introduction of several new definitions.2011-09-18
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    @Asaf You wrote: *I'm supposed to write a short paper for some proof about Polish spaces.* is this write-up/essay (or what should I call it) of yours available online somewhere?2012-06-09
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    @Martin: No and I wouldn't dream of posting it. It is far from being self-contained and relies on several claims proved in class. Instead I should probably finish something else which may be posted online. Also, Stefan says hello.2012-06-09
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    An uncountable discrete topology is locally compact metrisable but it’s not 2nd countable.2018-09-06