Closure of intersections

Question

Prove that $U$ is open if and only if $\bar{A} \cap U \subset \overline{A \cap U}$ for any subset $A$.

I know that closure of intersections is a subset of intersections of closures, but how do I show this?

Answer 1

'If' part: Take $A=U^{c}$. We see that $\overline {A}\cap U$ is empty which means $U \subset(\overline {U^{c}})^{c}$. Prove that the reverse inclusion is also true by taking complements in $U^{c} \subset\overline {U^{c}}$. We have now proved that $U^{c}$ is the closure of some set. Hence it is closed and its complement, $U$ is open.

'Only if' part: Let $x \in \overline {A} \cap U$. If $V$ is any open set containing $x$ then $U \cap V$ is also an open set containing $x$.Hence it must intersect $A$. Thus $A \cap U \cap V$ is non empty. This proves that $x \in \overline {A \cap U}$.