Every complete Boolean algebra arises as the lattice of regular open sets in some topological space, namely given a complete Boolean algebra $B$, the corresponding Stone space $S(B)$ will be extremally disconnected, so in particular its regular open sets are precisely its clopen sets.

A semi-common generalization of topological spaces is the class of spaces $(X,\tau)$ such that $\varnothing,X\in\tau$ and such that $\tau$ is closed under arbitrary unions (no assumption about intersections). This is equivalent to a weakening of the Kuratowski closure axioms to the axioms of a general closure operator. (Wikipedia says that these are sometimes called 'closure spaces,' but I've also seen them called 'generalized topological spaces' in some papers. If anyone knows another name for these things I would really like to know it because searching for relevant literature has been difficult.)

A closure operator is equivalently specified by an interior operator, a function $\mathrm{int}:\mathcal{P}(X)\rightarrow \mathcal{P}(X)$ satisfying:

- $\mathrm{int}(A)\subseteq A$
- $A\subseteq B \Rightarrow \mathrm{int}(A)\subseteq \mathrm{int}(B)$
- $\mathrm{int}(\mathrm{int}(A))=\mathrm{int}(A)$

(This definition makes sense in an arbitrary poset.) Under such an operator, the collection of open sets (sets such that $\mathrm{int}(U)=U$) forms a complete lattice with arbitrary meets given by set theoretic union. Unlike in a topological space finite joins are given by $U \wedge V = \mathrm{int}(U \wedge V)$. (This lattice is dual to the lattice of closed sets, but I'm phrasing this question in terms of open sets since I find it a little more familiar and because there's no standard name for the closure of complement operation.)

An interior operator gives rise to a natural exterior operator: $\mathrm{ext}(A)=\mathrm{int}(X \setminus A)$. Just like in a topological space one can show that for an open set $U$, $\mathrm{ext}(\mathrm{ext}(\mathrm{ext}(U)))=\mathrm{ext}(U)$, so call an open set $U$ 'regular' if it satisfies $\mathrm{ext}(\mathrm{ext}(U))=U$. (Note that unlike in topology it seems that the exterior operator is not definable from the structure of the lattice of opens alone.)

The map $U\mapsto \mathrm{ext}(\mathrm{ext}(U))$ gives a closure operator on the lattice of open sets, so we get that the collection of regular opens also forms a complete lattice whose join is the join of open sets (although again unlike in topology this is not set theoretic intersection) but whose meet is different, so in particular neither operation is set theoretic.

One can show that the collection of regular opens together with their meet, join, and the exterior operator form an orthocomplemented lattice, with the exterior playing the role of orthocomplementation, i.e. $ext$ is an order-reversing involution such that $U$ and $\mathrm{ext}(U)$ are always complements. This is a purely algebraic fact about Boolean algebras with interior operators satisfying the above axioms. With a finite counterexample search I found that in general the lattice does not need to be orthomodular.

I convinced myself that every complete lattice arises as the lattice of opens of a closure space, although it's easier to phrase in terms of closed sets and the dual lattice. Given a lattice $(L,0,1,\wedge,\vee)$ we can consider the collection of subsets of $L\setminus \{0\}$ and let a set be closed if it is of the form $(0,a]$ for some $a\in L\setminus \{0\}$ (so in particular the closure operator is $\mathrm{cl}(A)=(0,\bigvee A]$). By completeness of the lattice any intersection of sets of this form is of this form or empty and in particular the intersection's top is the greatest lower bound of the tops of the original sets, so the lattices agree (if we identify $\varnothing$ with $0$, this is why we need to remove $0$, otherwise we get an extra bottom element). To get a lattice of opens we can just start with the dual lattice.

That construction always produces trivial lattices of regular opens (no non-trivial open sets are disjoint, so the exterior of any non-empty set is empty), which shows that the exterior operator really does depend on the underlying set and not just the lattice of opens (since there are closure spaces with non-trivial lattices of regular opens). So finally we can come to the question:

Which complete orthocomplemented lattices arise as the lattice of regular opens in a closure space?