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Proof that the rationals are not a G_\delta set.

  1. Aug 18, 2010 #1
    I understand that showing [itex]\mathbb Q[/itex] is not a [itex]G_\delta[/itex] set is quite a non-trivial exercise, involving (among other things) an invocation of the Baire category theorem. Do any of you guys know it, or know where I can find it online? I'd really appreciate it. Thanks!
  2. jcsd
  3. Aug 18, 2010 #2
    I think I encountered this when I had to show that there does not exist a function from R to R that is continuous on the rationals and discontinuous on the irrationals.

    You could start with http://en.wikipedia.org/wiki/Gδ_set#Examples".
    Last edited by a moderator: Apr 25, 2017
  4. Aug 18, 2010 #3
    Thanks. I actually *did* start there, but was a little mystified by the following statement:

    "If we were able to write [itex]\mathbb Q = \bigcap_1^\infty \mathcal O_n[/itex] for open sets [itex]\mathcal O_n[/itex], each [itex]\mathcal O_n[/itex] would have to be dense in [itex]\mathbb R[/itex] since [itex]\mathbb Q[/itex] is dense in [itex]\mathbb R[/itex]."

    Why is this so? What, or who, says that if you write a dense set as an intersection of open sets, each of the sets in the intersection has to be dense?
    Last edited by a moderator: Apr 25, 2017
  5. Aug 19, 2010 #4
    Any set containing a dense set is itself dense: Since [tex]\mathbb{Q} \subseteq \mathcal{O}_n \subseteq \mathbb{R}[/tex], we have [tex] \mathbb{R} = \bar{\mathbb{Q}} \subseteq \bar{\mathcal{O}}_n \subseteq \mathbb{R}[/tex].
  6. Aug 19, 2010 #5
    Of course! I don't know why I didn't realize that! I guess my mind was inexplicably converting the [itex]\cap[/itex] into a [itex]\cup[/itex]. The proof on Wikipedia makes sense to me now. (The provision of the Baire category theorem that is violated, BTW, is that if [itex]\{ \mathcal O_n \}_1^\infty[/itex] is a collection of dense open sets, then so is [itex]\bigcap_1^\infty \mathcal O_n[/itex].)
  7. Aug 19, 2010 #6
    Minor pedantry: the BCT says that a countable intersection of dense open sets is dense; not necessarily open.
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