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

  • 2
    $\begingroup$ Your first assumption is not correct as $\mathbb R\setminus\mathbb Q$ is a countable intersection of open sets and has no interior point. $\endgroup$ – Hagen von Eitzen Dec 19 '13 at 11:42
  • 1
    $\begingroup$ Mind that only finite intersections of opens are open (and thus contain interior points). This is in general false, e.g. $\{0\}=\bigcap_{n=1}^\infty(-\frac1n,\frac1n)$. $\endgroup$ – 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 $\Bbb R$.

  • $\begingroup$ Got it. Thank you Brian! $\endgroup$ – topsi Dec 19 '13 at 12:24
  • $\begingroup$ @Shir: You’re welcome! $\endgroup$ – Brian M. Scott Dec 19 '13 at 22:05
  • $\begingroup$ Am I crazy or.... The union when you define your scrip U should be an intersection, otherwise it's always equal to X. $\endgroup$ – user139779 Apr 2 '14 at 13:34
  • 1
    $\begingroup$ No, it's right. $\mathscr{U}$ is a family of open sets, and the intersection of all elements of $\mathscr{U}$ shall be empty. $\endgroup$ – Daniel Fischer Apr 2 '14 at 13:43
  • $\begingroup$ So, in any T1 infinite Baire space countable intersection of open dense sets is uncountable (and dense). $\endgroup$ – Mahbub Alam Dec 10 '17 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.