# When Stone–Čech compactification is totally disconnected

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.

• The answer to this question, mathoverflow.net/questions/93719, has a few more examples. Dec 19, 2018 at 11:42
• @KPHart What is the Dowker example and Dowker's paper you refer to? Dec 19, 2018 at 17:05
• There is a reference in the second answer to that question. Dec 19, 2018 at 17:07