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$\newcommand{\cl}{\operatorname{cl}}$ $\newcommand{\i}{\operatorname{int}}$

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?

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It remains to show that

$$\overset{\circ}{\overline{A}} \cap \overset{\circ}{\overline{B}} \subset \overline{A\cap B}.$$

Since the set on the left is open, it will then also be a subset of the interior of the set on the right.

So let $x \in \overset{\circ}{\overline{A}} \cap \overset{\circ}{\overline{B}}$. Then there are open neighbourhoods $V, W$ of $x$ with $V \subset \overline{A}$ and $W \subset \overline{B}$. Let $U = V\cap W$. Then $U$ is an open neighbourhood of $x$ contained in $\overline{A} \cap \overline{B}$. Let $Z$ be an arbitrary open neighbourhood of $x$, we have to show that $Z\cap A\cap B \neq \varnothing$. By intersecting with $U$, we can assume that $Z \subset U$. Now, $x \in \overline{A}$, so $Y = Z\cap A$ is a nonempty open set. But we also have $Y \subset U \subset \overline{B}$, hence $Y\cap B \neq \varnothing$, and that means $Z\cap A \cap B \neq \varnothing$, i.e., since $Z$ was arbitrary, $x \in \overline{A\cap B}$.

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