A set $U$ in a topological space $X$ is called regular open if $U=\text{Int}\left(\overline{U}\right)$. Similarly, a set $F$ is regular closed if $X\setminus F$ is regular open or equivalently $F=\overline{\text{Int}(U)}$.
If $A\subseteq X$. Shows $U(A)$ is regular open. Where $U(A)=\bigcup\{U \space \text{open} :U\Vdash A\}$
A suggestion for this exercise, I have had many problems with this fact. Thanks
HINT: First prove that $U(A)\Vdash A$. Then show that $\operatorname{int}\operatorname{cl}U(A)\Vdash U(A)$, and conclude that $\operatorname{int}\operatorname{cl}U(A)\Vdash A$. It may be useful to recall that a set is co-meagre in $V$ if and only if it contains the intersection of countably many dense open subsets of $V$.