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

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Discussion Overview

The discussion revolves around the proof that the set of rational numbers, \(\mathbb{Q}\), is not a \(G_\delta\) set. Participants explore the implications of the Baire category theorem in relation to this topic, touching on concepts of density in topology and continuity of functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that proving \(\mathbb{Q}\) is not a \(G_\delta\) set is complex and involves the Baire category theorem.
  • Another participant mentions a related problem regarding the existence of a function from \(\mathbb{R}\) to \(\mathbb{R}\) that is continuous on \(\mathbb{Q}\) and discontinuous on the irrationals.
  • A participant expresses confusion about a statement from Wikipedia regarding the density of open sets in the context of intersections, questioning why each open set in the intersection must be dense if the intersection results in a dense set.
  • One participant provides a clarification that any set containing a dense set is itself dense, supporting this with a topological argument involving closures.
  • A later reply acknowledges the previous clarification and expresses understanding, while also noting the specific provision of the Baire category theorem that is violated in this context.
  • Another participant points out a minor distinction regarding the Baire category theorem, emphasizing that it refers to countable intersections of dense open sets being dense, not necessarily open.

Areas of Agreement / Disagreement

Participants generally agree on the implications of the Baire category theorem and the properties of dense sets, but there are nuances in understanding and interpretation, particularly regarding the definitions and implications of \(G_\delta\) sets.

Contextual Notes

The discussion includes assumptions about the definitions of dense sets and \(G_\delta\) sets, as well as the application of the Baire category theorem, which may not be fully resolved or universally understood among participants.

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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!
 
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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".
 
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snipez90 said:
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".

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?
 
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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].
 
adriank said:
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].
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].)
 
Minor pedantry: the BCT says that a countable intersection of dense open sets is dense; not necessarily open.
 

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