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Let $A$ and $B$ be subsets of a closed set $F \subseteq \mathbb{R}^n$ such that $\overline{A} = \overline{B} = F$. Show that if $A$ and $B$ are open, $\ \overline{A \cap B} = F$.

I've proved $\overline{A \cap B} \subseteq F$ so far:

$$p \in \overline{A \cap B} \iff \forall \epsilon > 0, \ B(p,\epsilon) \cap (A \cap B) \neq \emptyset$$ so that implies $$ \forall \epsilon > 0, \ B(p,\epsilon) \cap A \neq \emptyset$$ and therefore $p \in \overline{A} = F$.

I'm stuck at proving $F \subseteq \overline{A \cap B}$. Can you help me?

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  • $\begingroup$ A little observation, to simplify your argument: the inclusion you got follows from $\overline{A\cap B} \subseteq \overline{A}\cap\overline{B}$, which holds for any $A$ and $B$, in any topological space, not just $\mathbb{R}^n$ (because $X\subseteq Y$ implies $\overline{X}\subseteq\overline{Y}$). The other inclusion is harder because it is false if you don't use your extra assumptions. $\endgroup$ – user144221 Oct 6 '16 at 1:00
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Fix $p\in F$ and $\varepsilon>0$. Since $F=\overline A$, there exists some $q\in A$ such that $\|p-q\|<\varepsilon$. Let $$\xi\equiv\varepsilon-\|p-q\|>0.$$ Since $A$ is open and $q\in A$, there exists some $\delta>0$ such that $B(q,\delta)\subseteq A$. Also, given that $q\in A\subseteq \overline A= F=\overline B$, there exists some $r\in B$ such that $\|q-r\|<\min\{\delta,\xi\}.$

On the one hand, you have that $r\in B(q,\delta)\subseteq A$, so that $r\in A\cap B$.

On the other hand, you have also that $$\|p-r\|\leq\|p-q\|+\|q-r\|<\|p-q\|+\xi=\varepsilon,$$ so that $r\in B(p,\varepsilon)$.

Hence, $B(p,\varepsilon)\cap A\cap B$ is not empty for any $\varepsilon>0$, which implies that $p\in\overline{A\cap B}$.

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    $\begingroup$ Great! I had tried that idea before, but failed in choosing a good $\epsilon$. Thanks a lot! $\endgroup$ – Guido A. Oct 6 '16 at 0:45

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