Proof that the intersection of an open set with a not open set is not open in a metric space Let $(X, d)$ be a metric space.
We suppose that $A\subseteq X $ is open in $X$ and $B\subseteq X $ is a not open set in $X$, where $A,B \neq \emptyset $. Is  $A\cap B$  a not open set in $X$?
Ι have tried to prove it by considering that  $A\cap B$ is open set in $X$ but I can't find the contradiction.
We say that a subset $A$ is open in $X$ if, whenever $a \in A$, we can find a $δ_1 > 0$  such
that $ B(a,\delta_1)\subseteq A $.
On the other hand, $B$ is a not open set in $X$ if, whenever $\delta_2>0$, we can find a $b \in B $  such
that $ B(b,\delta_2)\nsubseteq B$.
I suppose that $A\cap B$ is open set in $X$.
So, whenever $z \in A\cap B $, we can find a $δ_3 > 0$  such
that $ B(z,\delta_3)\subseteq A\cap B $.
If anybody could help me, I would be grateful. 

That's wrong. Consider the reals with the usual metric, $A=(0,4)$ open and $B=(1,5]$ not open. Then $A \cap B=(1,4)$ is open.
