6
$\begingroup$

Let $T$ be a compact Hausdorff extremally disconnected set (so $T$ is a compact Hausdorff space, such that the closure of each open subset is again open). Let $S \subseteq T$ be a closed subset.

Question: Is $S$ extremally disconnected?

For me, this looks like a very natural question about extremally disconnected sets. However, on the spot, I could neither prove this, nor find a counterexample. Also, I was not able to find anything on this in the literature.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ The answer is there, let me just mention that since extremally disconnected compact Hausdorff spaces are in particular Stone spaces, you may invoke Stone duality and ask the equivalent question about boolean algebras. Which is whether any homomorphic image of a complete boolean algebra is complete. And this is not true: you may destroy completeness if you identify some elements not carefully enough. $\endgroup$
    – მამუკა ჯიბლაძე
    Nov 27, 2021 at 6:49
  • $\begingroup$ What is true is that a clopen subset of an extremally disconnected set is extremally disconnected (straight from the definition). Likewise the quotient of a complete Boolean algebra by a principal ideal is complete, and the homomorphism is a complete homomorphism. $\endgroup$
    – Robert Furber
    Nov 28, 2021 at 12:37

1 Answer 1

Reset to default
10
$\begingroup$

No, the Stone-Cech compactification $\beta\mathbf{N}$ of $\mathbf{N}$ is extremally disconnected, but not the Stone-Cech boundary $\beta\mathbf{N}\smallsetminus\mathbf{N}$.

To see this, it is enough to find an increasing sequence $(F_n)$ of clopen subsets with no supremum (=least upper bound) in the Boolean algebra of clopen subsets, or equivalently of the Boolean algebra of subsets modulo symmetric difference finite subsets.

For this, it is more convenient to work with $\mathbf{N}^2$: then $F_n$ is just the boundary of $\mathbf{N}\times\{0,\dots,n\}$. An upper bound for this sequence is just a subset $Y$ intersecting each horizontal line in a cofinite subset. By given any such $Y$ one can remove one point in each horizontal layer and get another upper bound $Y'\subset Y$ with $Y\smallsetminus Y'$ infinite. So there is no least upper bound.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see. Thank you! As far as I understand, in the answer to this question you prove why this is the case. (Oh, I just wrote the previous version of this comment, before you added more details to the answer...) $\endgroup$
    – AlexIvanov
    Nov 18, 2021 at 13:23
  • $\begingroup$ @AlexIvanov ah indeed, I forgot this one (I was adding the proof while you wrote your comment). In terms of Boolean algebra, it indeed states that the power set of $\mathbf{N}$ is complete, while modding out by finite subsets yields a non-complete Boolean algebra. $\endgroup$
    – YCor
    Nov 18, 2021 at 13:24

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