Elementary question about the Stone-Cech compactification

Question

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

Answer 1

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 $\beta X$.

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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.