This is an exercise in Willard's General Topology.
A subset $B$ of a topological space is called regularly open iff $Int(Cl(B))=B$.
I need to show that if $U$ and $V$ are regularly open then $Int(Cl(U\cap V))=U\cap V$.
I've been using the facts that $Int(Y)=X\setminus Cl(X\setminus Y)$, $Cl(A\cup B)=Cl(A)\cup Cl(B)$ and $Int(A\cap B) =Int(A)\cap Int(B)$ but I always wind up getting back to where I started.
I appreciate your help.