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

1. Aug 18, 2010

### 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!

2. Aug 18, 2010

### 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".

Last edited by a moderator: Apr 25, 2017
3. Aug 18, 2010

### AxiomOfChoice

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?

Last edited by a moderator: Apr 25, 2017
4. Aug 19, 2010

### adriank

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

5. Aug 19, 2010

### AxiomOfChoice

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 $\bigcap_1^\infty \mathcal O_n$.)

6. Aug 19, 2010

### adriank

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

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