How to prove this: $\overline{A\cap B}=\overline{A\cap \overline{B}}$?

Question

Let $A$ be an open set of $E$ an normed linear space, and $B\subset E$, then I have to prove that $$\overline{A\cap B}=\overline{A\cap \overline{B}}$$

(I'm stuck in the two $\subset$'s)

Any help will be appreciated.

Answer 1

One side is rather trivial, since the closure operator preserves inclusions: $$B \subset \overline{B} \implies A \cap B \subset A \cap \overline{B} \implies \overline{A \cap B}\subset \overline{A \cap \overline{B}}.$$

Now, for the other side. We have to use that $A$ is open. We do it like this: take $x \in \overline{A \cap \overline{B}}$ and $U$ an open neighbourhood of $x$. Then: $$U \cap A \cap \overline{B} \neq \varnothing$$

Since $A$ is open, $U \cap A$ is open. I claim that $U \cap A \cap B \neq \varnothing.$ Otherwise: $$(U \cap A)\cap B = \varnothing \implies (U \cap A)\cap \overline{B} = \varnothing.$$ So $U \cap A \cap B \neq \varnothing$ means that $x \in \overline{A \cap B}$ and we're done.