Let $A, B\subseteq \mathbb{R}^n, int(\partial A) = int(\partial B) = \emptyset.$If $A\cap B\neq \emptyset,$ is it necessarily true that $\overline{A\cap B} = \overline{A}\cap \overline{B}$? Is it true that if $A\cap B = \emptyset,$ then $int(\overline{A}\cap \overline{B}) = \emptyset$? Prove or disprove.
The proposition is not true if both of $int(\partial A), int(\partial B)\neq \emptyset$; it might be true if only one of $int(\partial A), int(\partial B) = \emptyset.$ By definition, the interior of a set is the set of interior points and the boundary of a set is the set difference between the closure (the set of limits points) and the interior. I already know how to prove that $\overline{A\cap B} \subseteq \overline{A}\cap \overline{B}.$ So I just need to find a way to show $\supseteq$.
For the second problem, I know that $int(\overline{A}\cap \overline{B}) = int(\overline{A})\cap int(\overline{B}),$ and I think that $int(\overline{A}) = int(\overline{A})$ and similarly for $B$, which I might be able to show using the fact that the interiors of the boundaries are zero.
Best Answer
Lemma: suppose $int(\partial A) = \emptyset$ and open, nonempty $V \subseteq \bar{A}$. Then there is an open, nonempty $U \subseteq V$ such that $U \subseteq A$. Proof: suppose on the other hand that for every open, nonempty $U \subseteq V$, we have some $x \in U \cap A^c$. Then $V \subseteq \bar{A^c}$. Then $V \subseteq \partial A$. Then $int(\partial A)$ is nonempty. Contradiction.
The second claim is true. For suppose $x \in int(\bar{A} \cap \bar{B})$. Take open $V$ s.t. $x \in V \subseteq \bar{A} \cap \bar{B}$. Then take open, nonempty $U \subseteq V \cap A$. Then take open, nonempty $W \subseteq U \cap B \subseteq V \cap A \cap B \subseteq A \cap B$. Then $A \cap B$ is nonempty; contradiction.
I haven't found a proof one way or the other for the first problem yet.
Edit: take $A = (0, 1) \cup (3, 4)$, $B = (1, 2) \cup (3, 4)$. The closure of the intersection is $[3, 4]$; the intersection of the closures is $[1] \cup [3, 4]$. So the first claim is false.