# If a subset $A\subset X$ in $(X,\tau)$ satisfies $A=U\cap \overline A$ with $(A\subset )U$ open, then $A$ is open in $\overline A$

In a topological space $(X,\tau)$, let $A\in X$ and let open $U\subset X$ contain $A$ such that $A=U\cap \overline A$. Does it mean that $A$ is open in $\overline A\cap U$?

I try to understand the equivalence between different definitions of a locally closed set (the above, theoretical $A$ for the record) and the above-mentioned claim keeps showing up in a various of proofs, not thoroughly explained. I have been spending hours on this equivalence question, while I already have a big mass of exams to study for and I am frustrated. Could you make it any clearer to me?

$U \cap (U \cap \overline{A}) = U \cap \overline{A} =A$, so $A$ is the intersection of an open set $U$ of $X$ with the subset $U \cap \overline{A}$ so by definition of the subspace topology your claim has been shown.