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I read the following result in an article.

Let $X$ be a regular space. Let $\mathcal{M}$ be free closed ultrafilter on $X$. Set $\mathcal{U=}\left\{ U:U\text{ is open and there exists a }F\in \mathcal{M}\text{ such that }F\subset U\right\} $. $\mathcal{U% }$ is contained in an open ultrafilter $\mathcal{W}$. By regularity, $\bigcap \overline{\mathcal{W}}=\emptyset $.

Why the regularity implies $\bigcap \overline{\mathcal{W}}=\emptyset $?

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    $\begingroup$ Only to clarify terminology: By closed/open ultrafilter you mean a maximal filter in the set of closed/open subsets of $X$? $\endgroup$
    – Dieter Kadelka
    Sep 1, 2020 at 16:08
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    $\begingroup$ By regularity each point will have an open neighborhood whose closure is disjoint to a set in $\mathcal{M}$. The complement of this closure is in $\mathcal{W}$, and the closure of this complement omits the original open set around the point. $\endgroup$
    – Todd Eisworth
    Sep 1, 2020 at 16:32
  • $\begingroup$ @DieterKadelka, yes, I mean a maximal filter in the set of closed/open subsets of $X$. $\endgroup$
    – Mehmet Onat
    Sep 2, 2020 at 7:51

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Maybe that this answer is essentially the not elaborated answer of Todd Eisworth. Assume that $x \in \bar{W}$ for all $W \in \mathcal{W}$. Since $\mathcal{M}$ is free there is $M \in \mathcal{M}$ with $x \not\in M$. By regularity of $X$ there are open $U,V \subset X$ with $x \in U$, $M \subset V$ and $U \cap V = \emptyset$. Therefore $V \in \mathcal{U}$, but $U \not\in \mathcal{U}$, in particular $V \in \mathcal{W}$ and $x \not\in \bar V$, since $U \cap V = \emptyset$ and $x \in U$ open.

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