Is every Extremally Disconnected Hausdorff Space Regular? I was wondering, is it true that a Extremally Disconnected Hausdorff Space is Regular? Let $A$ be an open set. I must find a open set $V$ such that $\bar V \subset A$. Since $\bar V$ will be open I think we can do it somehow but I'm not managing to construct it.
Recall that $X$ is extremally disconnected iff the closure of every open set is open.
 A: Vinicius,
see if this link answers your question:
http://www.ams.org/journals/proc/1967-018-02/S0002-9939-1967-0210066-0/S0002-9939-1967-0210066-0.pdf
A: Not necessarily.

$p \subseteq \mathcal{P}(\mathbb{N})$ is an ultrafilter on $\mathbb{N}$ if 
  
  
*
  
*for all $A \subseteq \mathbb{N}$ either $A \in p$ or $\mathbb{N} \setminus A \in p$ (but not both);
  
*whenever $A \subseteq B \subseteq \mathbb{N}$ are such that $A \in p$, then $B \in p$;
  
*whenever $A , B \in p$, then $A \cap B \in p$.
  
  
  We say that $p$ is nonprincipal if no finite set belongs to $p$.
A basic fact about ultrafilters is that if $\mathcal{A} \subseteq \mathcal{P}(\mathbb{N})$ is nonempty and the intersection of any finite number of elements of $\mathcal{A}$ is nonempty, then there is an ultrafilter $p$ on $\mathbb{N}$ including $\mathcal{A}$ as a subset.  It follows that given any infinite $A \subseteq \mathbb{N}$ there is a nonprincipal ultrafilter on $\mathbb{N}$ containing $A$.

Let $\mathcal{U}$ denote the collection of all nonprincipal ultrafilters on $\mathbb{N}$, and define $X = \mathbb{N} \cup \mathcal{U}$. We topologise $X$ as follows:


*

*each $n \in \mathbb{N}$ is isolated;

*given $p \in \mathcal{U}$, the basic open neighbourhoods of $p$ are of the form $A \cup \{ p \}$ for $A \in p$.


Claim. This space is Hausdorff but not regular.
sketch. Hausdorffness is fairly straightforward, you just have to handle the cases.  For example, if $p , q \in \mathcal{U}$ are distinct, then there is some $A \subseteq \mathbb{N}$ which is in $p$ but not $q$.  But then $\mathbb{N} \setminus A \in q$ (but not in $p$), and so $A \cup \{ p \}$, $( \mathbb{N} \setminus A ) \cup \{ q \}$ are disjoint open neighbourhoods of $p$ , $q$, respectively.
To show that it is not regular, it suffices to show that if $A \cup \{ p \}$ is a basic open neighbourhood of $p \in \mathcal{U}$, then there is a $q \in \overline{ A \cup \{ p \}} \cap \mathcal{U}$ which is distinct from $p$.  For this, partition $A$ into two infinite subsets as $B \cup C$. Then one of these, say $B$, does not belong to $p$. Letting $q \in \mathcal{U}$ contain $B$, clearly $q \neq p$.  Note that if $D \in q$, then $D \cap B \neq \varnothing$, and so $(D \cup \{ q \} ) \cap ( A \cup \{ p \} ) \neq \varnothing$.  It follows that $q \in \overline{ A \cup \{ p \} }$.
Claim. This space is extremally disconnected.
sketch. Let $U \subseteq X$ be open.  Note that $\overline{U} \setminus U \subseteq \mathcal{U}$ (since the points in $\mathbb{N}$ are isolated).  Suppose that $p \in \overline{U} \setminus U$. Letting $A = U \cap \mathbb{N}$, it follows that $B \cap A \neq \varnothing$ for all $B \in p$. But this implies that $A \in p$ (since otherwise $\mathbb{N} \setminus A \in p$, but this is disjoint from $A$). And so $A \cup \{ p \}$ is an open neighbourhood of $p$ included in $\overline{U}$.
