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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.

Asaf Karagila
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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 08 '11 at 12:40
  • @Theo: So it is worse than I thought, eh? :-) – Asaf Karagila Sep 08 '11 at 12:49
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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 (that took too long to find the link...). By the way: I should probably elaborate that into an answer, no? – t.b. Sep 08 '11 at 12:52
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    See: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=2234.0001 – gary Sep 08 '11 at 12:52
  • @Theo: You may assume the answer is yes. :-) – Asaf Karagila Sep 08 '11 at 13:06

1 Answers1

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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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t.b.
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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 08 '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 08 '11 at 13:49
  • @Asaf: Okay, here we go: Theorem 2, p.35. – t.b. Sep 08 '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 08 '11 at 14:03
  • As for the typo, it is horrible to edit from an iPhone. Feel free to correct. :-) – Asaf Karagila Sep 08 '11 at 14:04
  • 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 Ann. Math. 38, No. 4 (Oct., 1937), pp. 823-844. on p.838ff. – t.b. Sep 08 '11 at 14:15
  • 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. – Asaf Karagila Sep 08 '11 at 14:27
  • 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. :-) – Asaf Karagila Sep 08 '11 at 15:19
  • @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. – t.b. Sep 08 '11 at 21:39
  • 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. – Asaf Karagila Sep 08 '11 at 21:55
  • 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). – Asaf Karagila Sep 08 '11 at 22:01
  • @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... – t.b. Sep 08 '11 at 23:01
  • 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. – Asaf Karagila Sep 09 '11 at 06:06
  • @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? – Asaf Karagila Sep 14 '11 at 18:38
  • @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, p21ff. – t.b. Sep 14 '11 at 21:28
  • I'm afraid that might not be too helpful. I'm just trying to avoid going through sieved and sieve-complete spaces. – Asaf Karagila Sep 15 '11 at 19:56
  • @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... – Asaf Karagila Sep 18 '11 at 10:34
  • @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 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...) – t.b. Sep 18 '11 at 11:16
  • 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? – Asaf Karagila Sep 18 '11 at 11:30
  • I just found the two papers by Worrell appearing on the 1966 Portuguese journal, they are of no real use (the "metacompact" ones). – Asaf Karagila Sep 18 '11 at 11:38
  • @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). – t.b. Sep 18 '11 at 12:00
  • 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. – Asaf Karagila Sep 18 '11 at 12:21
  • @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? – Martin Sleziak Jun 09 '12 at 05:40
  • @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. – Asaf Karagila Jun 09 '12 at 07:51
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    An uncountable discrete topology is locally compact metrisable but it’s not 2nd countable. – user6 Sep 06 '18 at 17:55
  • Why build $\overline X$? Surely with this embedding, $[0,1]^{\Bbb N}$ is itself a metrizable compactification of $X$? @t.b. – Akiva Weinberger Jun 18 '23 at 21:11