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In the Wikipedia article on the interior operator it is stated that on a complete metric space $X$ one has for every sequence of open sets $(A_i)_{i\in \mathbb N}\subset X$ the relation $$ \mathrm{Int}\Big(\bigcap_{i}\overline A_i\Big)=\mathrm{Int}\Big(\overline {\bigcap_{i}A_i}\Big). $$ Now I wonder whether the finite version of this relation, i.e., for any two open sets $A,B\subset X$ to have $$ \mathrm{Int}\big(\overline A\cap \overline B\big)=\mathrm{Int}\big(\overline {A\cap B}\big), $$ does actually hold in any topological space $X$. If so, it should be found in some elementary textbooks, perhaps as an exercise. I searched for a while but it turned out that searching for relations like this is extremely inconvenient - one obtains hundreds of results which are similar but different from the wanted relation. On the other hand, I guess that (if the relation holds in general) there should be an easy direct proof based on playing around with the operations interior, closure, intersection, union, and complement. Since I don't have enough time for this game I'm asking directly for a reference.

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Yes, it's a standard fact used in the theory of so-called regular-open sets (sets such that $\operatorname{int}(\overline{U})=U$; your sets of the form "interior of closure" are regular open..) and forms part of the proof that those sets form a Boolean algebra under the operations $U \land V = U \cap V$, $U \lor V = \operatorname{int}(\overline{U \cup V})$ and $\lnot U = \operatorname{int}U^\complement$. I think Halmos proves it in his Boolean algebra book, and it's also in Engelking somewhere, I'm sure.

As $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$, because $A \subseteq \overline{A}$ and likewise for $B$, so $\overline{A} \cap \overline{B}$ is a closed superset of $A \cap B$, so it contains $\overline{A \cap B}$ too. Taking interior on both sides gives the easy inclusion (we don't need anything on $A$,$B$):

$$\operatorname{int}(\overline{A \cap B})\subseteq \operatorname{int}(\overline{A} \cap \overline{B})$$

For the reverse inclusion we can use a small lemma that for an open $U$ and any $A$ we have $\overline{U \cap A} = \overline{U \cap \overline{A}}$. Think about it.

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