Basic set inclusion in a metric space

Question

Consider the sets $A$ and $B$ in a metric space $(E,d)$. Clearly $A\cap B \subseteq B$. Suppose $closure(A\cap B) \neq B$. Then $closure(A\cap B) \subset B$ or $closure(A\cap B) \supset B$. My intuition says the latter is not possible. How to argue about it?

*A closure point $x$ of $A$ is such that all open ball centered in $x$ has a nonempty intersection with $A$.

Thanks in advance!

Answer 1

You cannot argue about it, since it is false. Suppose that your metric space is $\mathbb R$, endowed with its usual distance. Take $A=[0,1]$ and $B=\mathbb Q$. Then $\overline{A\cap B}=A$ and therefore neither $\overline{A\cap B}\subset B$ nor $\overline{A\cap B}\subset B$ hold.