Boolean algebra of regular opens Let $H$ be a compact Hausdorff space. Let $RO := \{ A\subseteq H \mid \operatorname{Int}(\operatorname{Cl}(A)) = A \}$ be the set of regular opens of $H$. Then $RO$ is a base for the topology of $H$. Moreover, it is known that $(RO, \neg, \wedge, \vee, \emptyset, H)$ forms a Boolean algebra where
$$
A \wedge A' := A \cap A', \quad A \vee A' := \operatorname{Int}(\operatorname{Cl}(A \cup A')) \quad\text{and}\quad \neg A := H - \operatorname{Cl}(A).
$$
Now I am wondering about the converse: Given a Boolean algebra $B$, does there exist a (unique) compact Hausdorff space such that its regular opens form a Boolean algebra that is isomorphic to $B$?
 A: No: the regular open subsets of a topological space $H$ always form a complete Boolean algebra.  Indeed, if $\{U_i\}$ is any collection of regular open sets, then it is straightforward to verify that the interior of the closure of $\bigcup U_i$ is the smallest regular open set containing all of them, and thus their join in $RO(H)$.
So any Boolean algebra which is not complete cannot have the form $RO(H)$.  For an explicit example, consider the Boolean algebra $B$ of subsets of $\mathbb{N}$ that are either finite or cofinite.  This $B$ is not complete: for instance, the collection of all finite sets of even numbers does not have a join in $B$.  So this $B$ is not isomorphic to $RO(H)$ for any space $H$.
However, every complete Boolean algebra $B$ is isomorphic to $RO(H)$ for some compact Hausdoff space $H$.  Namely, let $H$ be the space of Boolean homomorphisms $B\to \{0,1\}$, considered as a subspace of $\{0,1\}^B$ with the product topology (this is known as the Stone space of $B$).  Then there is an isomorphism $B\to RO(H)$ which sends $b\in B$ to the set of homomorphisms $f:H\to\{0,1\}$ such that $f(b)=1$.  (Proving that this is an isomorphism takes some work; it will be an embedding for any Boolean algebra $B$, and completeness of $B$ is needed to show it is surjective.)
This space $H$ is typically not unique, though.  Indeed, the $H$ constructed above always has the property that it is extremally disconnected, meaning the closure of every open set is open (so actually every regular open set is clopen!).  So if you take any compact Hausdorff space $H'$ which is not extremally disconnected and let $B=RO(H')$, the space $H$ constructed above will not be homeomorphic to $H'$ but will also satisfy $B\cong RO(H)$.
However, if you require $H$ to be extremally disconnected, then it is unique.  Indeed, you can show that if $H$ is an extremally disconnected compact Hausdorff space, then $H$ is canonically homeomorphic to the Stone space of $RO(H)$.  More generally, if $H$ is a totally disconnected compact Hausdorff space, then $H$ is canonically homeomorphic to the Stone space of the Boolean algebra of clopen subsets of $H$.
All of this is part of the story of "Stone duality", which is a rich and beautiful theory which says the category of Boolean algebras is dual to the category of totally disconnected compact Hausdorff spaces.
