# Why is $\mathbb{Q}$ not a $G_\delta$ set? [duplicate]

A $$G_\delta$$ set is defined as the intersection of a countable family of open sets. If $$n \in \mathbb{N}$$ and $$x_j \in \mathbb{Q}$$, $$\mathbb{Q}$$ can be expressed as $$\bigcap\limits_{r=1/n}^{\infty} (\bigcup\limits_{j=1}^{\infty} B_r(x_j))$$. Union of (in this case countably many) open sets is an open set in $$\mathbb{R}$$, so why is $$\mathbb{Q}$$ not a $$G_\delta$$ set?

• What is your proof that $\mathbb Q$ is equal to that intersection?
– Pedro
Commented Jun 26, 2023 at 17:59
• @Pedro my intuition failed me here, it's a bit of a counterintuitive fact Commented Jun 26, 2023 at 18:02

Notice that $$\cup_{j=1}^\infty B_r(x_j) = \mathbb{R}$$ for all $$r>0$$. Hence your intersection is still $$\mathbb{R}$$.
Assume $$\Bbb{Q}=\bigcap_{n=1}^\infty U_n$$ with each $$U_n$$ open. Then. $$\emptyset = \left(\bigcap_{n=1}^\infty U_n\right)\cap\left(\bigcap_{x\in\Bbb{Q}}\left(\Bbb{R}\setminus\{x\}\right)\right)$$
For each $$n$$, $$\Bbb{Q}\subseteq U_n$$, so $$U_n$$ is an open dense set. Also, $$\Bbb{R}\setminus\{x\}$$ is an open dense set for each $$x$$. So a countable intersection of open dense subsets of $$\Bbb{R}$$ is empty - in contradiction to BCT.