Prove that if $A$ is open at $(X, d)$ and $B$ is any subset of $X$, then $\overline{(A\cap \overline{B})}=\overline{(A \cap B)}$ Hello friends could you help me with the following please:
Prove that if $A$ is open at $(X, d)$ and $B$ is any subset of $X$, then
$\overline{(A\cap \overline{B})}=\overline{(A \cap B)}$
I have tried to do it using that $A\cap \overline{B} \subset \overline{A \cap B}$ But that property has not let me see anything, I do not see what relationship it may have
 A: Clearly $A\cap B\subseteq A\cap\operatorname{cl}B$, so $\operatorname{cl}(A\cap B)\subseteq\operatorname{cl}(A\cap\operatorname{cl}B)$, but we still have to show that $\operatorname{cl}(A\cap\operatorname{cl}B)\subseteq\operatorname{cl}(A\cap B)$. Note, though, that $A\cap\operatorname{cl}B\subseteq\operatorname{cl}B$, which is a closed set; what does this tell you about $\operatorname{cl}(A\cap\operatorname{cl}B)$?
A: The closure $\overline{A}$ of a subset $A$ of a metric space $(X,d)$ is the smallest closed subset of $X$ containing $A$. Hence, $\overline{A\cap B}$ is the smallest closed subset of $X$ which contains the intersection $A\cap B$. Here smallest means that every subset of $X$ which is closed and which contains $A\cap B$, also contains the closure $\overline{A\cap B}$. In particular, $\overline{A\cap\overline{B}}$ is a closed subset of $X$ which contains $A\cap B$, hence you get one inclusion $\overline{A\cap B}\subset\overline{A\cap\overline{B}}$. For the opposite inclusion, let $x$ be a point of $\overline{A\cap\overline{B}}$. Then for every positive real number $\delta>0$, the open sphere $B_{\delta}(x)=\{y\in X:d(y,x)<\delta\}$ intersects both $A$ and $\overline{B}$. If you can prove that $B_{\delta}(x)\cap\overline{B}$ non-empty implies $B_{\delta}(x)\cap B$ non-empty, then you may conclude that $x$ is indeed an accumulation point for $A\cap B$. 
