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

[AxiomOfChoice]

I understand that showing \mathbb Q is not a G_\delta 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!

 

[snipez90]

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".

 

[AxiomOfChoice]

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 \mathbb Q = \bigcap_1^\infty \mathcal O_n for open sets \mathcal O_n, each \mathcal O_n would have to be dense in \mathbb R since \mathbb Q is dense in \mathbb R."

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?

 

[adriank]

Any set containing a dense set is itself dense: Since \mathbb{Q} \subseteq \mathcal{O}_n \subseteq \mathbb{R}, we have \mathbb{R} = \bar{\mathbb{Q}} \subseteq \bar{\mathcal{O}}_n \subseteq \mathbb{R}.

 

[AxiomOfChoice]

Any set containing a dense set is itself dense: Since \mathbb{Q} \subseteq \mathcal{O}_n \subseteq \mathbb{R}, we have \mathbb{R} = \bar{\mathbb{Q}} \subseteq \bar{\mathcal{O}}_n \subseteq \mathbb{R}.

Of course! I don't know why I didn't realize that! I guess my mind was inexplicably converting the \cap into a \cup. The proof on Wikipedia makes sense to me now. (The provision of the Baire category theorem that is violated, BTW, is that if \{ \mathcal O_n \}_1^\infty is a collection of dense open sets, then so is \bigcap_1^\infty \mathcal O_n.)

 

[adriank]

Minor pedantry: the BCT says that a countable intersection of dense open sets is dense; not necessarily open.