# Regular opens of compact Hausdorff space

Setting: Let $H$ be a compact Hausdorff space and suppose the collection $\{ a_i \mid i \in I \}$ is a subbasis for the topology on $H$. Moreover, suppose each of the $a_i$ is regular open.

Let $\mathbf{I}u$ and $\mathbf{C}u$ denote the interior and closure of a set $u$. It is known that the collection of all regular open sets on a compact Hausdorff space forms a Boolean algebra with join $\bigvee u_j = \mathbf{IC}(\bigcup u_j)$, meet $\bigwedge u_j = \mathbf{I}(\bigcap u_j)$ and negation $\neg u = \mathbf{I}(H \setminus u)$. Call this Boolean algebra $\operatorname{RO}(H)$.

Let $A$ be the collection of subsets of $H$ generated by the $a_i$ and closed under the intersection, union and negation ($\bigwedge, \bigvee, \neg$) as described in the previous alinea. (That is, the ones used to make the collection of regular opens a Boolean algebra.) Then $A$ carries a Boolean algebra structure.

Question: Is $A = \operatorname{RO}(H)$?

Ideas: It is obvious that $A$ is a sub-Boolean algebra of $\operatorname{RO}(H)$. I suspect the converse might be true because the $a_i$ generate the topology on $H$, but I have not found a proof yet.

• No compactness is needed to see that $\textrm{RO}(X)$ is a Boolean algebra. This holds for all topological spaces. – Henno Brandsma Jan 21 '18 at 14:03
• Do you mean the statement in my question holds for all topological spaces? Or the fact that RO is a Boolean algebra? I am working specifically with compact Hausdorff spaces that is why I mentioned compactness. – Math Student 020 Jan 21 '18 at 14:44
• The finite intersections are the same as finite meets, so the generated set contains a base. As $X$ is regular we can write each regular open set as the join of base elements and so the subbase generates all regular open sets. – Henno Brandsma Jan 21 '18 at 14:50
• thanks, now I see it! – Math Student 020 Jan 21 '18 at 14:54