# Is every compact, homogeneous, totally disconnected space a power of a finite discrete space?

In a comment of this question, t.b. suggests that every compact, homogeneous, totally disconnected, second countable space is either a finite discrete space or the Cantor set; equivalently, it must be a countable power of a finite discrete space. How can this be proven?

Furthermore, $$2^\kappa$$ is compact, homogeneous, and totally disconnected. Is every compact, homogeneous and totally disconnected space (not necessarily second countable) a power of a finite discrete space?

• If $X$ is compact, homogeneous, totally disconnected, second countable but not discrete, then $X$ has no isolated points, so it is a Cantor set by Brouwer's characterization theorem Apr 24, 2023 at 4:45
• It's not clear to me that a compact, homogeneous, totally disconnected space is Hausdorff. Might be that Hausdorff must be added as a prerequisite, even in case of second countable?
– Ulli
Apr 24, 2023 at 8:18
• totally disconnected gets you T1, so proving regular gets you Hausdorff. Apr 24, 2023 at 11:12
• why is it regular?
– Ulli
Apr 24, 2023 at 11:20
• Sorry, I should have said "would get you". I have not been successful in verifying this. Apr 24, 2023 at 12:54

Let $$X$$ be the Double arrow space. It is well-known that $$X$$ is compact, Hausdorff, zero-dimensional, first countable, but not metrizable. By Theorem 2 in this paper of Alan Dow and Elliot Pearl, $$X^\omega$$ is homogeneous. Of course, $$X^\omega$$ is compact, Hausdorff, zero-dimensional (hence totally disconnected) and first countable. Assume $$X^\omega$$ is homeomorphic to $$D^\kappa$$ for a discrete, finite space $$D$$ and a cardinal $$\kappa$$. By first countability, $$\kappa = \omega$$, hence $$X^\omega$$ is metrizable, hence $$X$$ is metrizable. Contradiction.
The endpoints of the Double arrow space are isolated, hence $$Z := X \setminus \{\text{min } X, \text{max } X \}$$ is also compact. It is well-known (and not too difficult to see) that $$Z$$ is homogeneous. Of course, it is also zero-dimensional, Hausdorff, first countable. Hence it is not homeomorphic to $$D^\kappa$$ for a discrete, finite space $$D$$.