In my textbook, I can find the following proposition:
The family ${\rm R}(X)$ of regular closed subsets of a space $X$ is a complete Boolean algebra with the following operations:
(1) $A\leq B$ if and only if $A\subseteq B$.
(2) $\bigvee_\alpha A_\alpha={\rm cl}\ \big(\bigcup_\alpha {\rm int}\ A_\alpha\big)$.
(3) $\bigwedge_\alpha A_\alpha={\rm cl}\ {\rm int}\ \big(\bigcap_\alpha A_\alpha\big)$.
(4) $A'={\rm cl}\ (X\setminus A)$.
Is statement (2) correct? It seems wrong to me, for I would have rather expected:
(2') $\bigvee_\alpha A_\alpha={\rm cl}\ {\rm int}\ \big(\bigcup_\alpha A_\alpha\big)$.
(2) and (2') are the same set. For any family $F$ of subsets of a space $X$ we have $$cl (\cup F)=cl (\;\cup \{\;cl( f): f\in F\}).$$ So for any family $\{A_a:a\in B\}$ of regular closed sets we have $$cl (\;\cup_{a\in B}\; int \;A_a)=$$ $$=cl (\;\cup_{a\in B}\;cl (\;int A_a))=$$ $$=cl (\;\cup_{a\in B}\;A_a)\; \supset$$ $$ \supset cl (\;int (\;\cup_{a\in B}\;A_a)\supset$$ $$\supset cl (\;\cup_{a\in B}\; int \;A_a).$$
BTW. Each $int (A_a)$ is an open subset of $\cup_{a\in B}\;A_a.$ So $int (\;\cup_{a\in B}\;A_a\;)\supset \cup_{a\in B} \; int \;A_a,$ which justifies the last line above.