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For $X$ a topological space, let us denote by $\newcommand{\cl}{\operatorname{cl}} A\mapsto \cl(A)$ the closure operation and $\newcommand{\inter}{\operatorname{int}} A\mapsto \inter(A)$ the interior operation. Recall that an open set $U \subseteq X$ is said to be regular open when $U = \inter(\cl(U))$, and that the space $X$ is said to be extremally disconnected when $\cl(U)$ is open for each open $U \subseteq X$ (equivalently, $\inter(F)$ is closed for each closed $F \subseteq X$).

Clearly, when $X$ is extremally disconnected, every regular open set is closed (because by assumption $\inter(\cl(U)) = \cl(U)$ for open $U$).

I am looking for a reference for the converse: a space in which every regular open set is closed is, in fact, extremally disconnected. Hopefully with a proof that avoids any mention of Boolean algebras.

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    $\begingroup$ If $F$ is closed, then $\text{int}(F)$ is regular open, so closed. Hence closure of every open set is open. $\endgroup$
    – Jakobian
    Commented Jul 29 at 16:32
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    $\begingroup$ As a reference you can see Encyclopedia of General Topology by Hart, Nagata, Vaughan, Chapter A, a-2, 3. Preopen sets. The following are equivalent: a) $X$ is extremally disconnected, b) every regular closed subset is preopen, c) every semi-open set is preopen, d) every closure of preopen set is open. The condition b) is easily seen to be the same as b') every regular closed subset is open. Taking complements, its equivalent to b'') every regular open subset is closed. $\endgroup$
    – Jakobian
    Commented Jul 29 at 16:51

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