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 $X$?


1 Answer 1


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, 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 Hart
    Dec 19, 2018 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.