To prove that regular open (Boolean) algebra is complete, I tried to show following claim, but I couldn't. I saw this statement in Kunen's 'Set Theory' p.64 but in other books what I checked, describing regular open algebra, there is no mention of the infimum of an arbitrary family of regular open sets. Is the claim really true? or an errata?
Context. $X$ is a topological space.
Definition. $b\subset X$ is a regular open set iff $b=\rm{int}(\rm{cl}(b))$ and $b$ is open. $\rm{ro}(X)$ is the set of all regular open sets in $X$.
Context. $\{{b_i}\}_{i\in I}$ is a family of regular open sets
Claim. $\rm{int}(\underset{i\in I}{\bigcap}b_i)$ is a regular open set. (so that any subset of $\rm{ro}(X)$ has infimum)
It’s correct. Let $U=\operatorname{int}\bigcap_{i\in I}b_i$. Then $U\subseteq b_i$ for each $i\in I$, so $\operatorname{cl}U\subseteq\operatorname{cl}b_i$ for each $i\in I$. Let $V$ be any open subset of $\operatorname{cl}U$; then for each $i\in I$ we have $V\subseteq\operatorname{cl}b_i$ and hence $V\subseteq\operatorname{int}\operatorname{cl}b_i=b_i$. Thus, $V\subseteq\bigcap_{i\in I}b_i$, and it follows that $V\subseteq U$ and hence that $U=\operatorname{int}\operatorname{cl}U\in\operatorname{ro}(X)$.