If $A \subset X$ is dense and $B \subset X$ is open, then $B \subset \overline{A \cap B}$

Question

I would like to prove the following:

Let $A$ and $B$ be subsets of a metric space $(X,d)$.

Prove that if $A$ is dense in $X$ and $B$ is open in $X$, then $B\subset \overline{A\cap B}$.

Could you give me a hint?

Answer 1

Let $x\in B$. We seek to show that $x\in\overline{A\cap B}$. If we show this, we are done.

By the properties of openness, $\exists$ a neighborhood of size $\epsilon>0$ around $x$ such that all points in that neighborhood are in $B$. Now, in order for $A$ to be dense, it must be arbitrarily close to $x$. So, we form a sequence of points $a_1,...\in A$ such that $d(a_n,x)<\frac\epsilon n$. Note that $\forall n\in\mathbb{N}\;a_n\in B$, so $$\forall n\in\mathbb{N}\;a_n\in A\cap B$$Hence, we found a sequence of points, which get arbitrarily close to $x\in B$ and are in $A\cap B$. Hence, $x\in\overline{A\cap B}$ and we are done.

Answer 2

Hint. If $x\in B$ and $U$ is a neighborhood of $x$, then $U \cap B$ is a neighborhood of $x$. By density of $A$, $U \cap B$...