# How to prove that $\mathbb{Q}$ is not the intersection of a countable collection of open sets.

I am reading chapter 4 of Jech - Set Theory and trying to solve question 4.14, in which we are asked to show that:

$\mathbb{Q}$ is not the intersection of a countable collection of open sets. The clue is to use Baire's Category Theorem.

I can't see how to do that, and can't see the connection to Baire's category theorem. Intuitively, I agree that this seems reasonable, since any intersection of open sets is a set with non empty interior (am I right about that?). So, if $\mathbb{Q}$ were an intersection like that it would contain an open interval (again, hope I am right would like to know if not.). But, I guess this is not a proof, and, can't see the connection to Baire's theorem.

Thank you!! Shir

• Your first assumption is not correct as $\mathbb R\setminus\mathbb Q$ is a countable intersection of open sets and has no interior point. – Hagen von Eitzen Dec 19 '13 at 11:42
• Mind that only finite intersections of opens are open (and thus contain interior points). This is in general false, e.g. $\{0\}=\bigcap_{n=1}^\infty(-\frac1n,\frac1n)$. – Andrea Mori Dec 19 '13 at 11:42

One version of the Baire category theorem says that if $X$ is a complete metric space, and $\{G_n:n\in\Bbb N\}$ is a countable family of dense open sets in $X$, then $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$. $\Bbb R$ with the usual metric is complete.
Suppose that $\Bbb Q=\bigcap_{n\in\Bbb N}G_n$, where each $G_n$ is open in $\Bbb R$. $\Bbb Q$ is dense in $\Bbb R$, so each $G_n$ is dense in $\Bbb R$. For each $q\in\Bbb Q$ let $U_q=\Bbb R\setminus\{q\}$; clearly $U_q$ is a dense open set in $\Bbb R$. Let $$\mathscr{U}=\{G_n:n\in\Bbb N\}\cup\{U_q:q\in\Bbb Q\}\;;$$ then $\mathscr{U}$ is a countable family of dense open subsets of $\Bbb R$, so its intersection is non-empty. However, the construction clearly ensures that $\bigcap\mathscr{U}=\varnothing$. This contradiction shows that $\Bbb Q$ cannot be a $G_\delta$ in $\Bbb R$.
• No, it's right. $\mathscr{U}$ is a family of open sets, and the intersection of all elements of $\mathscr{U}$ shall be empty. – Daniel Fischer Apr 2 '14 at 13:43