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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?

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

2 Answers 2

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Notice that $\cup_{j=1}^\infty B_r(x_j) = \mathbb{R}$ for all $r>0$. Hence your intersection is still $\mathbb{R}$.

This fact is related to the Baire category theorem. See e.g. pg 131 of Carothers Real Analysis.

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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.

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