Condition for Open Sets

Question

How can one verify that $U$ is open iff $\overline{B} \cap U \subset \overline{B\cap U}$ for any subset $B?$. I have attempted, and I just want to know if I am going in the correct direction.

Proof: Let U be open, we have $B \cap U \subset \overline{B\cap U}$, since U is open, any limit point of $B$ contained in $U$ must also be a limit point of $B \cap U.$ Thus we have $B'\cap U \subset (B\cap U)' \subset \overline{B\cap U} $, which further implies $\overline{B}\cap U=(B \cap U)\cup(B'\cap U )\subset \overline {B\cap U}$

This is just one way, for the other direction, I am not quite sure how to show. Can anyone help me out further, and also if the above is not correct, could anyone help me show this problem?

Answer 1

Suppose $U$ is not open. Then $U^c$ is not closed, which means that $U^c$ is a proper subset of $\overline{U^c}$. Consequently $\overline{U^c}\cap U$ is nonempty. Since $\overline{U^c\cap U}=\emptyset$, we have that $\overline{U^c}\cap U \not\subseteq \overline{U^c\cap U}.$

Answer 2

Conversely set B = $U^c.$ Then
$\overline U^c \cap U = \overline {U \cap U^c}$ is empty.
Thus U subset $U^o$ which assures U is open.