Spaces with every compactification $0$-dimensional which aren't locally compact

Question

Recently I've proven the following theorem

Theorem. Let $X$ be a zero-dimensional locally compact Hausdorff space. Then the following are equivalent:

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

2. There exists no continuous surjection $\beta X\setminus X\to [0, 1]$.

The theorem is useful because it allowed me to give various examples of spaces which would have only zero-dimensional compactifications. I haven't been able to prove that 1 implies 2 or 2 implies 1 without the assumption of local compactness, although I do have a very weak result analogous to this one.

Proposition. Let $X$ be a Tychonoff space. If there exists no continuous surjection $\beta X\to [0, 1]$, then every compactification of $X$ is zero-dimensional.

I say this result is weak because its clearly not an equivalence - if we take $X = \{0, 1\}^\omega$ then every compactification of $X$ is zero-dimensional, being $X$ itself, yet a surjection $\beta X = \{0, 1\}^\omega\to [0, 1]$ clearly exists.

I request examples of spaces which would be Tychonoff, not locally compact, and every compactification of them is $0$-dimensional. This would help me better understand the problem.

Answer 1

Let $\omega$ denote the natural numbers and let $N$ be a countably infinite discrete subset of $\beta \omega \setminus \omega$. If $X = \beta \omega \setminus N$, then $X$ is not locally compact, and every compactification of $X$ is zero-dimensional. That $X$ is not locally compact at any point of $(Cl_{\beta \omega}N) \setminus N$ is clear. Suppose $K$ is a compactification of $X$. Then the identity map from $X$ into $K$ extends to a continuous function $f \colon \beta \omega \to K$. Suppose first that $p \in K$ and $U$ is a neighborhood of $p$ in $K$. It is enough to show that $U$ contains a neighborhood $V$ of $p$ which is clopen (= closed and open) in $X$. Then $f^{-1}(U)$ is a neighborhood of the set $f^{-1}\{p\}$ which is finite because it is a compact subset of the discrete set $A$. Since $\beta \omega$ is zero-dimensional, there is a clopen subset $\widehat{V}$ of $\beta \omega$ such that $f^{-1}\{p\} \subseteq \widehat{V} \subseteq f^{-1}(U)$. Letting $V = f(\widehat{V})$ gives the required clopen neighborhood of $p$.

What is left is to show that if $p \in Cl_Kf(N) \setminus f(N)$ and $U$ is a neighborhood of $p$, there is a clopen neighborhood $V$ of $p$ such that $V \subseteq U$. By complete regularity, we may assume that $U$ is a zero-set neighborhood of $p$. Since $\beta \omega$ is an F-space, the $\sigma$-compact set $\{n :f(n) \in U \}$ is completely separated from the cozero-set $\beta \omega \setminus f^{-1}(U)$, that is, they are contained in disjoint zero-sets $Z_1$ and $Z_2$ respectively. Since $\beta \omega$ is strongly zero dimensional, these zero sets are contained in complementary clopen sets $V_1$ and $V_2$ respectively. Then $V = f(V_1)$ has the required property.

Answer 2

This is based on Anonymous answer.

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ a locally compact scattered space. Then $X = \beta X_0\setminus Y$ is such that $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold. Clearly $Y$ is compact iff $X$ is locally compact.

Proof:

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\beta X_0\setminus \text{cl}_{\beta X_0} Y \subseteq K$ is open, given $p\in S \cap (\beta X_0\setminus \text{cl}_{\beta X_0} Y)$ and $q\in S, q\neq p$, there exists a clopen compact subset $U$ of $\beta X_0\setminus \text{cl}_{\beta X_0} Y$ with $p\in U, q\notin U$, which is impossible. So that we need to have $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$. Let $L = \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$, $L$ compact.

Since $Y$ is locally compact, $\text{cl}_{\beta X_0} Y \setminus Y$ is closed in $L$, so $K\setminus X$ is open in $L$. The map $Y\to K\setminus X$ is perfect and so by this answer by K. P. Hart, $K\setminus X$ is open and zero-dimensional. If now $p\in S\cap (K\setminus X)$ and $q\in S, p\neq q$, then we can take a clopen compact subset $U$ of $L$ with $p\in U, q\notin U$, so as previously we can assume $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$.

Since $\text{cl}_{\beta X_0} Y \setminus Y$ is subspace of zero-dimensional space, it's itself zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

It follows that $K$ is totally disconnected and compact, and so zero-dimensional. $\square$

Example. If $Y$ is a strongly zero-dimensional, locally compact, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$ is zero-dimensional. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ has only zero-dimensional compactifications and is such that $\beta X\setminus X \cong Y$. $\square$

Corollary. Let $X$ be a strongly zero-dimensional Tychonoff space with $\beta X\setminus X$ locally compact. Then the following are equivalent:

1. $X$ has only zero-dimensional compactifications.

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

Proof: 2 implies 1 from the theorem.

Suppose $X$ has only zero-dimensional compactifications. Then $\beta X\setminus X$ is zero-dimensional as a subspace of $\beta X$. If $\beta X\setminus X$ weren't scattered, then we would find a compact perfect set $A\subseteq \beta X\setminus X$ as in K. P. Hart answer and so there would exist a surjection $A\to [0, 1]$. Since $\beta X\setminus A$ is locally compact, from Engelking theorem 3.5.13 there exists a compactification $Z$ of $\beta X\setminus A$ with remainder homeomorphic to $[0, 1]$. Its also a compactification of $X$, and $Z$ is not zero-dimensional. The contradiction shows that such set $A$ cannot exist, and so $\beta X\setminus X$ is scattered. $\square$