A topological space $X$ is totally disconnected if the connected components in $X$ are the onepoint sets, and a topological space, $X$ is called completely regular exactly in case points can be separated from closed sets via continuous realvalued functions. Let $X$ be a totally disconnected and completely regular topological space. Can we deducd that $\beta X$ is also totally disconnected, where $\beta X$ is the Stone–Čech compactification of $X$?
1 Answer
No; $X$ may have a quasicomponent with more than one point, and each quasicomponent of $X$ is contained in a connected subset of $\beta X$. It's easy to construct examples in $\mathbb R ^2$.
Even if the quasicomponents of $X$ are trivial, $\beta X$ may have a connected subset with more than one point. The Erdös space $\mathfrak E$ has singleton quasicomponents, but is not zerodimensional, therefore $\beta \mathfrak E$ is not zerodimensional, equivalently, $\beta \mathfrak E$ is not totally disconnected.
Even if $X$ is zerodimensional, $\beta X$ may have nondegenerate connected subsets. There is no separable metrizable $X$ to this effect, but there does exist a Tychonoff (completely regular T$_1$) example.

$\begingroup$ The answer to this question, mathoverflow.net/questions/93719, has a few more examples. $\endgroup$– KP HartDec 19, 2018 at 11:42

$\begingroup$ @KPHart What is the Dowker example and Dowker's paper you refer to? $\endgroup$ Dec 19, 2018 at 17:05

$\begingroup$ There is a reference in the second answer to that question. $\endgroup$– KP HartDec 19, 2018 at 17:07