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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.

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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.

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  • $\begingroup$ @Gord452 That's what the lemma is for. $\endgroup$
    – Doctor Who
    Jul 29, 2020 at 22:51
  • $\begingroup$ can you clarify why $V\subseteq (\overline{A^c})$? Also can you further justify that if $V\subseteq (\overline{A^c})$ and $V\subseteq \overline{A},$ then $V\subseteq \partial A$? I get why that's true intuitively, but I would like to see a formal justification for that. $\endgroup$
    – user763400
    Jul 29, 2020 at 23:05
  • $\begingroup$ @Gord452 Because for every point $w \in V$, and for every open set $W$ s.t. $w \in W$, the set $W \cap V \subseteq V$ is an open, nonempty set and therefore contains an element of $A^c$; hence, such $w$ is a limit point of $A^c$. Furthermore, we have $\partial A = \bar{A} \cap \bar{A^c}$; thus, if $V$ is a subset of both $\bar{A}$ and $\bar{A^c}$, it is a subset of their intersection. $\endgroup$
    – Doctor Who
    Jul 29, 2020 at 23:11
  • $\begingroup$ Thanks. The first claim may be false, but my point was that if $Vol(A\cap B) = Vol(\overline{A\cap B}) = 0,$ then $Vol(\overline{A}\cap \overline{B}) = 0$. See this post. Your counterexample doesn't work for this claim. My point is that $\overline{A\cap B}$ and $\overline{A}\cap \overline{B}$ get rly close wrt volume. $\endgroup$
    – user763400
    Jul 29, 2020 at 23:20

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