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If $X$ is a Tychonoff space and $\beta X$ is it's Stone-Cech compactification, should $X$ be a dense $G_\delta$ subset of $\beta X$?

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This is one of the characterizations of Čech-complete spaces, about which you can read e.g. in Engelking's General Topology. A counterexample would be $\mathbb{Q}$ (Čech-complete spaces are Baire spaces, but $\mathbb{Q}$ isn't a Baire space). – t.b. May 26 '12 at 15:35
I suppose that you mean $X$ being a dense $G_\delta$ subset of $\beta X$. – Asaf Karagila May 26 '12 at 15:36
up vote 3 down vote accepted

We say that a Tychonoff space is Cech complete if and only if it is a dense $G_\delta$ subset of $\beta X$ if and only if $X$ is a $G_\delta$ of any compactification.

It is a nice exercise to see that for a metrizable space being Cech complete is equivalent to being completely metrizable.

From this one can take the counterexample of $\mathbb Q$, as t.b. suggests in the comments. This space is not completely metrizable and therefore not $G_\delta$ in $\beta X$.

Sketch of the proof for the equivalence for metric spaces:

First we prove the following lemmas:

  1. If $X$ is Hausdorff and $A\subseteq X$ is dense and homemorphic to a completely metrizable space then $A$ is a $G_\delta$ subset of $X$.

  2. If $X$ is completely metrizable and $Y\subseteq X$ then $Y$ is completely metrizable if and only if it is $G_\delta$ in $X$.

Now suppose that $X$ is completely metrizable, it is dense in $\beta X$ - which is Hausdorff. Therefore it is $G_\delta$ in $\beta X$ and therefore Cech-complete.

In the other direction, if $X$ is a metrizable Cech-complete space, consider $\widetilde{X}$ to be the metric completion of $X$ and $Y$ some compactification of $\widetilde{X}$. We have that $X$ is dense in $Y$ (dense subset of a dense subset). We have if so that $X$ is a $G_\delta$ subset of $\widetilde{X}$ and therefore by the second lemma - completely metrizable.

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So, now I would like to add a question: If $X$ is a complete metric space, what can you tell about $\beta X$? Would it be metrizable? if so, would it be complete? – user25640 May 26 '12 at 16:32
@user25640: The compactification need not metrizable. If it is then it is a compact metric space and therefore complete and separable. – Asaf Karagila May 26 '12 at 16:41
Also relevant:… – Asaf Karagila May 26 '12 at 16:43
@user25640: Note that $\Bbb N$ with the discrete metric $d(m,n)=1$ if $m\ne n$ is a complete metric space, since the only Cauchy sequences are eventually constant, but $\beta\Bbb N$ fails miserably to be metrizable: it isn’t even first countable. – Brian M. Scott May 27 '12 at 7:09
@user25640: This answer should have been accepted. – daniel Jul 8 '12 at 13:56

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