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I take $X=\kappa\Bbb N$ to be $\Bbb N\cup\{p:p \text{ is a free ultrafilter on }\Bbb N\}$. Each singleton in $\Bbb N$ is open and a local base at any free ultrafilter $p$ is given by $\{\{p\}\cup A:A\in p\}$.

Is X zero-dimensional?

In other words, is it true that each point of $X$ has a neighborhood basis of clopen sets?

It is claimed for example here that each basic open set $\{p\}\cup A$ is clopen.

I am not sure about this claim, as it seems that from basic properties of ultrafilters one can prove it is not the case.

Fact: Given an infinite subset $A$ of $\Bbb N$ there is a free ultrafilter containing $A$.

(Take the set of all cofinite subsets of $A$. That forms a filter basis that generates a free filter on $\Bbb N$. Any ultrafilter that extends that filter will be a free ultrafilter containing $A$.)

Now take a free ultrafilter $p\in X$ and an (open) neighborhood $U=\{p\}\cup A$ with $A\in p$. I claim that $U$ is never closed in $X$. $A$ must be infinite because $p$ is free. Partition $A$ into two infinite sets: $A=B\cup C$. By standard ultrafilter properties exactly one of the two subsets, say $B$, must be in $p$. By the Fact above, there is an ultrafilter $q$ containing $C$, and $q$ is necessarily distinct from $p$. This $q$ is in the closure of $U$. Indeed, take any neighborhood $\{q\}\cup D$ with $D\in q$. $D\cap C\in q$ so $D\cap C$ is not empty and $D$ meets $A$, so any neighborhood of $q$ meets $U$. This shows that $U$ is not clopen.

Can you see anything wrong with this argument?

Added: My original question was the one above about $\kappa\Bbb N$, but I also mistakenly thought $\kappa\Bbb N$ was the same as the Cech-Stone compactification $\beta\Bbb N$. Thanks @EricWofsey for setting me straight.

PatrickR
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  • The space you describe is not the Stone-Cech compactification of $\mathbb{N}$. To get the Stone-Cech compactification, the basic open sets would instead be of the form $A\cup{q:A\in q}$. – Eric Wofsey May 21 '20 at 22:35
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    In particular, the first question you linked is not about $\beta\mathbb{N}$ but instead about a different topology on the same set. – Eric Wofsey May 21 '20 at 22:37
  • @EricWofsey Amazing. So what I described is the Katetov extension $\kappa\Bbb N$, which is really different from $\beta\Bbb N$, even though they are defined on the same set. Would you agree that $\kappa\Bbb N$ is not zero-dimensional, contrary to what the link was saying? – PatrickR May 21 '20 at 22:58
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    If $X$ is a $T_4$ space, then $\dim X=\dim \beta X$, where $\dim$ is the Lebesgue covering dimension, see thm 6.4.3 of Pears' "dimension theory of general spaces" (and $\dim X=0$ implies that $X$ has a basis of clopen sets as long as $X$ is $T_1$) – Alessandro Codenotti May 21 '20 at 23:02
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    Yes, this is correct. My statement in that old answer is incorrect; I suspect that I was thinking of $\beta\Bbb N$ when I made it, but after eight years I really don’t know. I’m heading off to correct it now. – Brian M. Scott May 22 '20 at 01:09
  • And thank you for bringing it to my attention so that I could fix it! – Brian M. Scott May 22 '20 at 01:22
  • @BrianM.Scott Thank you! – PatrickR May 22 '20 at 01:41
  • @PatrickR: My pleasure! – Brian M. Scott May 22 '20 at 01:45

2 Answers2

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Your argument is correct, but the space you describe is not the Stone-Čech compactification! Indeed, you can see quite quickly that it is not compact: for each free ultrafilter $p$, the set $\{p\}\cup\mathbb{N}$ is open, and these form an open cover with no finite subcover.

For the Stone-Čech compactification, the basic open sets are those of the form $U_A=A\cup\{p:A\in p\}$ for $A\subseteq\mathbb{N}$. (Or, identifying points of $\mathbb{N}$ with the principal ultrafilters, you just take the set of all ultrafilters that contain $A$.) It is immediate that these are clopen, since the complement of $U_A$ is just $U_{\mathbb{N}\setminus A}$. (Of course, it is not immediate that this space really is the Stone-Čech compactification of $\mathbb{N}$, but that's a longer story.)

Eric Wofsey
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  • Just a small typo: $p\in A$ instead of $A\in p$ – Alessandro Codenotti May 22 '20 at 06:43
  • It's the Katětov extension $\kappa \Bbb N$ that the OP describes. Also interesting but quite different. The general construction for Hausdorff $X$ uses open ultrafilters on $X$, and is $H$-closed Hausdorff (and not compact, typically). – Henno Brandsma May 22 '20 at 07:22
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$\kappa \Bbb N$ cannot be zero-dimensional, it then would be a regular $H$-closed space and thus (by standard results) compact, which your space is not (as $\kappa \Bbb N\setminus \Bbb N$ is infinite, closed and discrete).

Henno Brandsma
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