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Let $(X,d)$ be a metric space and let $A, B \subset X$. Prove that: If B is open, then $$\overline{A} \cap B \subset \overline{A \cap B}$$ where $\overline{S}$ indicates closure for some set $S$.

For proof I was only able to change the prove part slightly(not sure if this is a correct start):

Since $B \subset \overline{B} \implies \overline{A} \cap \overline{B} \subset \overline{A \cap B}$

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Given a set $E \subset X$ you have $x \in \overline E$ if and only if every neighborhood of $x$ intersects $E$.

Let $x \in \overline A \cap B$. Then $x \in B$ and every neighborhood of $x$ intersects $A$.

Let $U$ be a neighborhood of $x$. Since $x \in B$ and $B$ is open, $U \cap B$ is a neighborhood of $x$ too.

But every neighborhood of $x$ intersects $A$, so that $U \cap B$ contains a point of $A$. If you denote this point by $z$, you find $z \in U \cap (A \cap B)$. Thus $U$ contains a point of $A \cap B$.

Since $U$ was an arbitrary neighborhood of $x$, it follows that $x \in \overline{A \cap B}$.

Consequently $\overline A \cap B \subset \overline{A \cap B}$.

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Let $x\in {\overline A}\cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)\subseteq B$. Let $0 < s < r$. Then $B_s(0) \cap B \not=\emptyset$. Since $x\in{\overline A}$, $B_s(x)\cap A \not = \emptyset$. Thus $B_s(x)\cap A \cap B \not= \emptyset$.

We have just shown $x\in \overline{A \cap B}.$

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  • $\begingroup$ Why you assume 0 is included in the set? $\endgroup$ – Pumpkin Dec 6 '18 at 11:16

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