Spaces $X$ with every compactification $0$-dimensional with $\beta X\setminus X$ not locally compact

Question

Previously, in this post I've shown the following characterization of spaces with only zero-dimensional compactifications:

Theorem. Let $X$ be strongly zero-dimensional and $\beta X\setminus X$ locally compact. The following are equivalent:

1. Every compactification of $X$ is zero-dimensional

2. $\beta X\setminus X$ is scattered

So the case of when $\beta X\setminus X$ is locally compact is solved.

Although I don't think this problem can be worked on much further, I am looking for an example of a Tychonoff space $X$ such that every compactification of $X$ is zero-dimensional, but $\beta X\setminus X$ is not locally compact.

Answer 1

Let $X = \beta X_0 \setminus A$ where $A\subseteq \beta X_0\setminus X_0$ and $\overline{A}\setminus A$ is locally compact, $X_0$ is strongly zero-dimensional, $A$ is scattered or countable, $A\to B$ is a perfect map with $B$ Tychonoff then $B$ is zero-dimensional.

Remark. The last condition is satisfied when $A$ is countable or scattered and locally compact.

Then $X$ has only zero-dimensional compactifications:

Suppose that $K$ is a compactification of $X$, let $\beta X_0\to K$ be the induced map, where $\beta X = \beta X_0$.

If $S\subseteq K$ is connected with at least two points then $S\subseteq K\setminus (\beta X_0 \setminus \overline{A}) = \overline{A}\setminus A\cup K\setminus X$.

The map $A\to K\setminus X$ is perfect so $K\setminus X$ is zero-dimensional. By the same argument as previosuly, $S\subseteq \overline{\overline{A}\setminus A}$. The space $\overline{A}\setminus A$ is open in its closure and zero-dimensional so that $S\subseteq \overline{\overline{A}\setminus A}\setminus (\overline{A}\setminus A)\subseteq K\setminus X$.

It follows that $\overline{S}\subseteq K\setminus X$ is a compact connected set and so because $A\to K\setminus X$ is perfect, $A$ contains a compact set surjecting continuously onto $[0, 1]$, and hence not scattered. This is impossible since $A$ is scattered or countable.

Answer 2

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

First note that, since every infinite compact subset of $\beta \omega$ is uncountable, every fiber $f^{-1}(y)$ is finite. The claim is that only finitely many fibers have size greater than one. For suppose not. Let $D$ be an infinite discrete subset of $K$ such that $f^{-1}(d)$ has at least two elements for each $d \in D$. Let $w_d$ and $z_d$ be distinct elements of $f^{-1}(d)$. Then $N = \cup_{d \in D}\{w_d,z_d\}$ is a copy of $\omega$ in $\beta \omega$ and the restriction of $f$ to $N$ is two-to-one. Therefore, by continuity, the restriction of $f$ to $Cl_{\beta \omega}N$ is also (at least) two-to-one, and, in particular, there is an element $y$ of $Cl_KA \setminus A$ such that $f^{-1}(y)$ has more than one element, a contradiction. This proves the claim.

The result now follows from the easy fact that if $X$ is any zero-dimensional space and $Y$ is obtained from $X$ by collapsing each of a finite number of finite sets to a point, then $Y$ is zero-dimensional.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.