The intersection of a closure with an open set I have been stuck on this problem : 
Prove :If $A$ is Open , then $A\cap \overline B \subset \overline {A \cap B}$.I have tried to use the previously proven fact that $\overline {A \cap B}\subset \overline{A} \cap \overline{B}$ , but I cannot seem to get anywhere ...In particular, I do not see how A being open is relevant here.......... 
 A: Let $x \in A \cap \overline{B}$. So $x \in A$ and $x \in \overline{B}$.
Let $O$ be any open set that contains $x$. Then, as $A$ is open, $O \cap A$ is an open set that contains $x$ as well. Because $x \in \overline{B}$, $(O \cap A) \cap B \neq \emptyset$. But this also says that $O \cap (A \cap B) \neq \emptyset$.
As $O$ was arbitary open with $x \in O$, this shows that $x \in \overline{A \cap B}$, thereby showing the inclusion.
A: Let's start with a definition for closure which I think is suitable for this task (see e.g. https://en.wikipedia.org/wiki/Closure_(topology)): Let $B\subset X$
$$\overline{B}=\{x\in X:\ \exists (x_n)_{n\in\mathbb{N}}\subset B\mbox{ such that }x_n\rightarrow x\}.$$
Now let $x\in A\cap\overline{B}$ be arbitrary. Since $A$ is open there exists an $\varepsilon>0$ such that $B_\varepsilon(x)\subset A$ and there also exists a sequence $x_n\in B$ with $x_n\rightarrow x$. Hence there exists an $n_0\in\mathbb{N}$ such that for every $n\geq n_0$ we have $x_n\in B_{\varepsilon}(x)$. Hence $x_n\in A\cap B$ and $x_n\rightarrow x$. Therefore $x\in \overline{A\cap B}$.
A: Let $a \in A\cap \overline B$. Because, $A$ is open so for every open set $V$ containing $a$ there is a open set $U \subset V$ containing $a$ and $U$ is completely in $A$. $a \in \overline B$ so for every open set containing $a$ like $V$ there is open set $U \subset V$ completely in $A$ such that there is $z \in A$ and $z \in B$ such $z \in U$. Now we have $a \in \overline {A \cap B}$.
