Prove that if $A$ and $B$ are open, then $\i(\cl(A\cap B))=\i(\cl A) \cap \i(\cl B)$.
One way implication is easy as we have $$\cl(A\cap B)\subseteq \cl A \cap \cl B \Rightarrow \i(\cl(A\cap B))\subseteq \i(\cl(A)\cap \cl(B))= \i \cl A \cap \i\cl B. $$
I had difficulties in proving the other way. Let $x \in \i\cl(A) \cap \i\cl(B).$ Then there are open neighborhoods $U$, $V$ of $x$ such that $x\in U\subseteq \cl A$ and $x\in V\subseteq \cl B.$ I want to show that$ U\cap V \subseteq \cl(A \cap B)$ hence implies that this way implication is true. let $y\in U\cap V$ and $R$ to be any open neighborhood of $y$. Then $R\cap A$ and$ R\cap B$ are non-empty. But I am stuck here when I try to prove that $R \cap A \cap B$ is non-empty. (Because if $R \cap A\cap B$ is non-empty were to be true, then I am done). Can anyone give me some hints to proceed in this proof?Or actually I am heading in the wrong direction?