# Prove that If B is open, then $\overline{A} \cap B \subset \overline{A \cap B}$

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

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

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 $$x\in \overline{A \cap B}.$$

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