Let $(X,d)$ be a complete metric space and suppose that $A$ and B are both dense and $G_δ$ subsets of X, show that $A\cap B\neq \emptyset$.

Question

A set is said to be $G_\delta$ set if it can be expressed as countable intersection of open sets.
Let us suppose on the contrary that $A\cap B=\emptyset$
Now, choose $x\in B$. Then there exists a seq $(x_n)$ in $A$ such that $x_n\to x$. (as $\bar{A}=X$)
Now, no $x_n\in A$ for all $n\in\Bbb{N}$.
But for any $\epsilon>0,\exists\ N\in\Bbb{N}$ such that $x_n\in B(x,\epsilon)\ \forall n\ge N$.
Since $B=\bigcap B_k$ where each $B_k$ is open. $x\in B_k\ \forall n$
So, for all $x\in B_k, \exists \epsilon_k>0$ such that $B(x,\epsilon_k)\subset B_k$. Now, for each $\epsilon_k,\exists N_k\in\Bbb{N}$ such that $x_n\in B(x,\epsilon_k)\ \forall n\ge N_k$.
So, $x_n\in B_k\ \forall n\ge N_k$
From this I cannot proceed further, basically the maximum of $\{N_k|k\in\Bbb{N}\}$ exists (say $N$), then I can say $x_n\in B_n\ \forall n\ge N\implies x_n\in B\ \forall n\ge N$, which is a contradiction.
But this maximum will not exist in most of the cases. But I even have not used the dense nature of $B$ and $G_\delta$ property of $A$, and not the completeness of $X$.
Can anyone give me an idea about the solution? Thanks for assistance in advance.

Answer 1

It follows from: $$\left (\bigcap _n A_n \right ) \bigcap \left (\bigcap _m B_m \right ) = \bigcap_{n,m} A_n \cap B_m $$ and the fact that the intersection of two dense open sets is also a dense open set. By Baire's Category Theorem the right hand side is more than just not empty: it's dense in $X$.