# interiors and closures of sets where the interior of the boundary is empty

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.

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 $$ \cup [3, 4]$$. So the first claim is false.

• @Gord452 That's what the lemma is for. Jul 29, 2020 at 22:51
• 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.
– user763400
Jul 29, 2020 at 23:05
• @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. Jul 29, 2020 at 23:11
• 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.
– user763400
Jul 29, 2020 at 23:20