On necessary and sufficient condition for a set to be open

Question

Let $G \subset X$ then $G$ is open if and only if for all $A \subset X$, $\overline{G \cap A} = \overline{G \cap \overline{A}}$.\ I can prove one inclusion but not the other... Not sure how to use the fact that G is open also... Any tips?

Answer 1

Let cl K be $\overline K.$

Prove the important lemma if U is open, then U $\cap$ cl A $\subseteq$ cl (U $\cap$ A).

Thus for open G, cl (G $\cap$ A) $\subseteq$ cl (G $\cap$ cl A)
$\subseteq$ cl cl (G $\cap$ A) = cl (G $\cap$ A) and equality follows.

For the converse, set A = X - G in the equation.