I can prove that $\mathbb{Q}$ is not a $G_\delta$ set in $\mathbb{R}$. I was applying the same Baire space argument to show that $\mathbb{Q} \times \mathbb{Q}$ is not a $G_\delta$ set.
I was thinking like this:
We can write $(\mathbb{Q} \times \mathbb{Q})^c= \displaystyle{\bigcap_{(p,q)\in \mathbb{Q} \times \mathbb{Q}}\{(p,q)\}^c}= \displaystyle{\bigcap_{n=1}^{\infty}V_n}$ where each $V_n$ is open in $\mathbb{R}^2$. If we can write $\mathbb{Q} \times \mathbb{Q}= \displaystyle{\bigcap_{n=1}^{\infty}U_n}$ where each $U_n$ is open in $\mathbb{R}^2$. Then $\displaystyle{\bigcap_{n=1}^{\infty}(U_n \cap V_n)}= \phi$ is dense in $\mathbb{R}^2$-----which is a contradiction.!
Am I thinking correctly ?
Please help. Thanks in advance.
The argument for $\Bbb Q \times \Bbb Q$ in $\Bbb R^2$ is exactly the same as for $\Bbb Q$ in $\Bbb R$, in essence:
If $(X,d)$ is a complete metric space with no isolated points then any countable dense set $D$ is not a $G_\delta$:
Suppose $D=\bigcap_n O_n$, where all $O_n$ are open. It follows that the $O_n$ are dense too as they all contain the dense set $D$.
Also, $X\setminus \{d\}$ is open and dense for any $d \in D$ as there are no isolated points.
But then $$\emptyset= D \cap D^\complement = \bigcap_n O_n \cap \bigcap_{d \in D} (X\setminus \{d\}$$
contradicts the Baire theorem for the complete $(X,d)$: a countable intersection of dense open sets with empty (not dense!) intersection.
$\Bbb Q \times \Bbb Q$ is still countable and dense and $\Bbb R^2$ is complete metric without isolated points so the same proof applies.