A topological space $X$ is totally disconnected if the connected components in $X$ are the one-point sets, and a topological space, $X$ is called completely regular exactly in case points can be separated from closed sets via continuous real-valued 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 $X$?


No; $X$ may have a quasi-component with more than one point, and each quasi-component of $X$ is contained in a connected subset of $\beta X$. It's easy to construct examples in $\mathbb R ^2$.

Even if the quasi-components of $X$ are trivial, $\beta X$ may have a connected subset with more than one point. The Erdös space $\mathfrak E$ has singleton quasi-components, but is not zero-dimensional, therefore $\beta \mathfrak E$ is not zero-dimensional, equivalently, $\beta \mathfrak E$ is not totally disconnected.

Even if $X$ is zero-dimensional, $\beta X$ 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 Hart Dec 19 '18 at 11:42
  • $\begingroup$ @KPHart What is the Dowker example and Dowker's paper you refer to? $\endgroup$ – Forever Mozart Dec 19 '18 at 17:05
  • $\begingroup$ There is a reference in the second answer to that question. $\endgroup$ – KP Hart Dec 19 '18 at 17:07

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