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prove that $\mathbb{Q}$ can not be $G_{\delta}$ set in $\mathbb{R}$.

let $\mathbb{Q}$ is $G_{\delta}$ set. then $\mathbb{Q}=\cap_{i=1}^{\infty}\ O_i$.

for open sets $O_i$. now each open set in $\mathbb{R}$ is countable union of disjoints open intervels . then each open set $O_i$ containg $\mathbb{Q}$ must be of the form $\mathbb{R}/P_i,$ where $P_i$ contain countable many irrationals points.

i am not getting this last line that $O_i =\mathbb{R}/P_i,$ with $P_i$ contain countable many irrationals points.

any hint.thanks in advanced.

paarth
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    The standard proof of this uses the Baire category theorem, and shows that $\mathbb{R} \setminus \mathbb{Q}$ is $G_\delta$ also. – B. Mehta Feb 20 '18 at 19:37
  • In the ever-reliable Wikipedia, the article on $G_\delta$ sets proves this statement by appealing to the Baire CategoryTheorem. So this outsider to the area would guess that any proof of your fact would involve fairly deep mathematics. – Lubin Feb 20 '18 at 19:47

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