Regular open algebra

Example

Let $X$ be a topological space, and let $\operatorname{RO}(X)$ be the set of its regular open sets, the open sets $U$ with $U = \operatorname{int} \overline{U}$. Order $\operatorname{RO}(X)$ by inclusion and define:

1. $U \wedge V = U \cap V$;
2. $U \vee V = \operatorname{int} \overline{U \cup V}$;
3. $\neg U = \operatorname{int}(X \setminus U)$;
4. $0 = \emptyset$ and $1 = X$.

Then $\operatorname{RO}(X)$ is a complete Boolean algebra, called the regular open algebra of $X$. The supremum of a family $(U_i)$ is $\operatorname{int} \overline{\bigcup_i U_i}$, and the infimum is $\operatorname{int} \bigcap_i U_i$.

Check of the key points: the intersection of two regular open sets is regular open. Also $\operatorname{int} \overline{W}$ is regular open for every open $W$, so the operations stay in $\operatorname{RO}(X)$. For $U$ regular open, $\neg U$ is regular open, $U \cap \neg U = \emptyset$, and $\operatorname{int} \overline{U \cup \neg U} = X$, because $U \cup \neg U$ is dense in $X$. For any family, $\operatorname{int} \overline{\bigcup_i U_i}$ is regular open, contains every $U_i$, and lies in every regular open $V$ that contains all $U_i$, since then $\overline{\bigcup_i U_i} \subseteq \overline{V}$.