What is an example of a 0-dimensional locally compact Hausdorff space $X$ for which the Stone-Čech compactification $\beta(X)$ is *not* 0-dimensional?


It is known that if $X$ is a 0-dimensional locally compact Hausdorff space which is also paracompact, then $\beta(X)$ is 0-dimensional. (Engelking 1989, Th. 6.2.9). I would expect a counterexample in the non-paracompact case.


Another way of asking the question is to look at a Boolean ring L (without assuming a unit). If X is the Stone space of L then X is a 0-dimensional locally compact Hausdorff space and L is isomorphic with the ring of compact open sets in X. The Boolean algebra $\mathrm{Cl}(X)$ of closed-open subsets is clearly a Boolean algebra extension of L, and the Stone space of $\mathrm{Cl}(X)$ is a certain compactification of $X$. I believe it is easy to see that this compactification of $X$ is just $\beta(X)$ if and only if $\beta(X)$ is 0-dimensional.


Thus the question can be expressed in the algebraic side of the duality. Find conditions on the Boolean ring $L$ so that $\beta(X)$ is 0-dimensional, where $X$ is the Stone space of $L$.


I think this may be true if, for example, $L$ is a Boolean $\sigma$-ring. My original question asks for an example showing this is not true without some additional conditions on $L$.



Is Stone-Čech compactification of 0-dimensional space also 0-dimensional?
In [this paper](http://www.sciencedirect.com/science/article/pii/0166864180900206) (*Spaces $N\cup\mathscr{R}$ and their dimensions, Topol. Appl.
11(1) 1980 93-102*) — I hope the PDF is freely available) Jun Terasawa constructs maximal almost disjoint families on $\mathbb{N}$ whose associated spaces can have any dimension you want. Given an almost disjoint family $\mathcal{A}$ on $\mathbb{N}$ one defines a topology on the union $\mathbb{N}\cup\mathcal{A}$ by declaring each natural number to be isolated and giving each $A\in\mathcal{A}$ a countable local base by putting $$U(A,n)=\lbrace A\rbrace \cup \lbrace i\in A:i\ge n\rbrace$$ for each $n$. This space is locally compact and zero-dimensional but Terasawa could arrange it so that its Čech-Stone Compactification would contain the Hilbert cube, for example.


Dowker's example does have a locally compact version: 
Let us retain the notation of Dowker's paper and use $T$ to denote the
set of countable ordinals and $Q\_\alpha$ to denote the $\alpha$th
congruence class as chosen by Dowker.
In addition let $\mathbb{A}$ denote Alexandroff's double arrow space.
This is the product $[0,1]\times\lbrace0,1\rbrace$, ordered lexicographically
and endowed with its order topology and with the two isolated points
$\langle 0,0\rangle$ and $\langle 1,1\rangle$ deleted.


Now consider the product $T\times \mathbb{A}$ and define a quotient
space $X$ by identifying the points 
$\langle \alpha,x,0\rangle$ and $\langle \alpha,x,1\rangle$ 
whenever $x\notin\bigcup\_{\beta\ge\alpha}Q\_\beta$.


It is elementary to verify that this is an upper semicontinuous decomposition
and that the resulting space is locally compact and zero-dimensional.
The key observation is that for every $\alpha$ the product 
$T\_\alpha\times\mathbb{A}$ is compact and open in the domain and its
image is compact and open in $X$.


Furthermore, arguments similar to those given by Dowker will show that
each finite open cover of $T\times\mathbb{A}$ has a refinement of the
form $\mathcal{U}\cup\mathcal{V}$, where $\mathcal{U}$ is a disjoint
open cover of $T\_\alpha\times\mathbb{A}$ for some $\alpha$
and $\mathcal{V}$ consists of finitely many sets of the 
form $T\times C$, where $C$ is a clopen interval in $\mathbb{A}$.


The latter can then be used, just as for Dowker's $M$, to show that the bottom
and top lines in $X$ cannot be separated by clopen sets and hence
that $\beta X$ is not zero-dimensional. 


