$A,B \subset (X,d)$ and $A$ is open dense subset, $B$ is dense then is $A \cap B$ dense? I am trying to solve this problem, and i think i did something, but finally i couldn't get the conclusion. The question is:


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*Let $(X,d)$ be a metric space and let $A,B \subset X$. If $A$ is an open dense subset, and $B$ is a dense subset, then is $A \cap B$ dense in $X$?.


Well, i think this is true. We have to show that $\overline{A \cap B}=X$. Or in other words, $B(x,r) \cap (A \cap B) \neq \emptyset$ for any $x \in X$. Since $A$ is dense in $X$, so we have $B(x,r) \cap A \neq\emptyset$. That means there is a $y \in B(x,r) \cap A$. Which means there is a $y \in A$. And since $A$ is open we have $B(y,r_{1}) \subseteq A$ for some $r_{1} > 0$. Couldn't get any further. I did try some more from here on, but couldn't get it. Any idea for proving it or giving a counter example.

Let $(X,\tau)$ be a topological space. Let $A$ and $B$ be two dense subsets of $(X,\tau)$  with $A$ is an open set.
Let $G\in \tau$. Then since $A$ is dense in $(X,\tau)$, $G\cap A\neq\emptyset$.
Now since $B$ is dense in $(X,\tau)$, for open set $G\cap A$ which is a
nonempty set, $(G\cap A)\cap B\neq \emptyset$.
Hence $(A\cap B)$ is a dense set in $(X,\tau)$.
