Are all Hausdorff extremally disconnected P-spaces also regular?

Question

A space is extremally disconnected if closure of any open set is open. It's a P-space when intersection of any countable amount of open sets is open.

There exist Hausdorff extremally disconnected spaces that are not regular e.g. strong ultrafilter topology (this is not a P-space since if $\mathcal{U}$ is a free ultrafilter on $\mathbb{N}$, then there exist $A_i\in \mathcal{U}$ such that $\bigcap_i A_i = \emptyset$, so that $\bigcap_i (A_i\cup \{\mathcal{U}\}) = \{\mathcal{U}\}$ is not open).

Are all Hausdorff extremally disconnected P-spaces also regular?

Answer 1

If $X$ is a Hausdorff extremally disconnected $P$-space with $|X|$ smaller than first measurable cardinal, then $X$ is discrete.

To see this, note that the proof of point $1.$ given in this answer of mine only uses Tychonoff assumption to obtain a clopen set $U$ such that $x\in U$ and $p\notin U$ in the first part of the answer. But since Hausdorff extremally disconnected spaces are totally separated, this proof still holds if we replace Tychonoff by the weaker separation axiom.

So any example of a Hausdorff extremally disconnected $P$-space that isn't regular, cannot be constructed in ZFC.

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By the same procedure as in David Gao's proof, given $\kappa$ is a measurable cardinal, $\lambda < \kappa$, and $G$ a $\kappa$-complete free ultrafilter, we can obtain $\lambda$ distinct $\kappa$-complete free ultrafilters by decomposing $\kappa = \bigcup_{\alpha < \lambda} X_\alpha$ where $|X_\alpha| = \kappa$ and considering bijections that swap $X_\alpha$ with each other. Because $G$ is $\kappa$-complete and $\kappa\in G$, there must exist $\alpha$ such that $X_\alpha\in G$. Since $\kappa$ is measurable, it's inaccessible, and so $\kappa = \sup\{\lambda : \lambda < \kappa\}$ and so this gives us $\kappa$ distinct $\kappa$-complete free ultrafilters.

In particular, $|\nu(\kappa)\setminus \kappa|\geq \kappa$ where $\kappa$ has discrete topology, and if $\kappa\leq \kappa_0$, then $\kappa$ is $C$-embedded in $\text{cl}_{\nu(\kappa_0)} \kappa$, and so $\nu(\kappa)\setminus \kappa \subseteq \nu(\kappa_0)\setminus \kappa_0$ so that $|\nu(\kappa_0)\setminus \kappa_0|\geq \kappa$.

As a corollary, if $\mathfrak{m}$ is the smallest measurable cardinal and $\kappa_0\geq \mathfrak{m}$, then $\nu(\kappa_0)\setminus \kappa_0$ is not metrizable, since being a closed subset of realcompact space $\nu(\kappa_0)$ it's realcompact, and a realcompact metrizable space has size $< \mathfrak{m}$ (see discussion here). This is, I feel like, a good insight into topology of $\nu(\kappa_0)$, something that I haven't seen anyone studying extensively (I could only find that $\nu(\kappa_0)$ is not a $k$-space).

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Now let $\kappa\geq \mathfrak{m}$ have discrete topology where $\mathfrak{m}$ is the smallest measurable cardinal. Note that $\nu(\kappa)$ is an extremally disconnected $P$-space. Pick non-closed set $C\subseteq \nu(\kappa)\setminus\kappa$, this is possible since $\nu(\kappa)\setminus\kappa$ is a closed non-discrete set. Let the open sets on $X = \nu(\kappa)$ be of the form $V_1\cup V_2\setminus C$ where $V_1, V_2\subseteq \nu(\kappa)$ are open. As in the proof for the K-topology, because $C$ is nowhere dense, $X$ can easily shown to be Hausdorff and not regular. It's also easy to see that $X$ is a $P$-space.

Moreover $\text{cl}_X (V_1\cup V_2\setminus C) = \overline{V_1}\cup \overline{V_2}$ as one can show, so $X$ is extremally disconnected (this construction, in general, preserves extremally disconnected property).

So this is another way to obtain a Hausdorff extremally disconnected $P$-space that is not regular.

Answer 2

Here is an example showing that, if a measurable cardinal is assumed to exist, then there exists a Hausdorff extremally disconnected $P$-space that is not regular. Indeed, this is essentially an adaptation of the strong ultrafilter topology: Let $\kappa$ be the first measurable cardinal and $M$ be the set of $\kappa$-complete free ultrafilters on $\kappa$. Let $X = \kappa \cup M$. We topologize $X$ by declaring all points of $\kappa$ isolated and for each $F \in M$, we choose as a local basis sets of the form $A \cup \{F\}$ where $A \in F$. It is not hard to check $X$ is Hausdorff. To show that it is an extremally disconnected $P$-space, we note that $A \cup N \subset X$, where $A \subset \kappa$ and $N \subset M$, is open iff, for every $F \in N$, $A \in F$. Thus, if we have a sequence of open sets $A_n \cup N_n$, then,

$$\bigcap_n (A_n \cup N_n) = \left(\bigcap_n A_n\right) \cup \left(\bigcap_n N_n\right)$$

is open. Indeed, for any $F \in \bigcap_n N_n$, we have $F \in N_n$ so $A_n \in F$ for all $n$. Thus, as $F$ is $\kappa$-complete, $\bigcap_n A_n \in F$. This shows $X$ is a $P$-space.

Now, let $A \cup N \subset X$ be open. Then one can easily check that $\overline{A \cup N} = A \cup N’$, where $N’ = \{F \in M: A \in F\}$. From this description, it follows that $\overline{A \cup N}$ is open, so $X$ is extremally disconnected.

Finally, we show that $X$ is not regular. Indeed, let $G \in M$. Then $\kappa \cup \{G\}$ is an open neighborhood of $G$. If $A \cup \{G\}$ is an open neighborhood of $G$ contained in $\kappa \cup \{G\}$, then, as we have seen before, $\overline{A \cup \{G\}} = A \cup N$, where $N = \{F \in M: A \in F\}$. We note that $\{G\} \subsetneq N$. Indeed, as $A \in G$ and $G$ is a $\kappa$-complete free ultrafilter, $|A| = \kappa$. Write $A = B \cup C$ where $|B| = |C| = \kappa$ and $B, C$ are disjoint. Then one of $B, C$ must be in $G$, so we may WLOG assume $B \in G$. Hence, if we let $f: \kappa \to \kappa$ be a bijection s.t. $f(B) = C$ and $f(C) = B$, then $f(G)$ is a $\kappa$-complete free ultrafilter s.t. $A \in f(G)$ but, as $C \in f(G)$, $G \neq f(G)$. Thus, $A \cup N \not\subset \kappa \cup \{G\}$, so $X$ is not regular.