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I know that $\mathbb{R}\setminus\mathbb{Q}$ is not an $F_\sigma$ set. However, what about $\mathbb{R}\setminus\mathbb{Q}\cap$ Cantor set? Is that an $F_\sigma$?

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HINT: Let $C$ be the Cantor set. $(\mathbb{R}\setminus\mathbb{Q})\cap C=C\setminus\mathbb{Q}$, which in some ways is a lot like $\mathbb{R}\setminus\mathbb{Q}$. How do you prove that $\mathbb{R}\setminus\mathbb{Q}$ is not an $F_\sigma$? Can you use the same basic idea on $C\setminus\mathbb{Q}$?

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When proving that the irrationals were not an $F_\sigma$ I used the fact that any subset of the reals that is an $F_\sigma$ is either of the first category, or it contains an open interval. So, if I were to adapt that to this case, I'd say that $C\setminus\mathbb{Q}$ contains no intervals, but it's not of the first category, it can't be an $F_\sigma$. Is that argument good, or something along those lines? – FPP Nov 18 '11 at 12:22
@FPP: Yes, that works, provided that you know why $C\setminus\mathbb{Q}$ isn’t first category. – Brian M. Scott Nov 18 '11 at 12:48
I'm not sure why.... If C is of the first category, why would one of its subsets be of the second? Is that possible? – FPP Nov 18 '11 at 14:10
Is there something I'm missing? First of all.... Is $C$ of the first or second category? – FPP Nov 18 '11 at 14:17
@FPP: You beat me to it: I was just about to point out that category is relative to the ambient space. And yes: $C\setminus\mathbb{Q}$ is a dense $G_\delta$ in $C$, so it’s also a Baire space, and the Baire category theorem still applies. – Brian M. Scott Nov 18 '11 at 14:33

Do you know the properties of the Cantor set, i.e., is it closed, open, etc? Find out, and that will answer your question.

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Well, I do know that, but I guess I couldn't put it together? I know it's a closed set, I know it's perfect, totally disconnected, of the first category, compact, nowhere dense, etc. I feel like I have a lot of "fun facts" about it, but I can't reach a conclusion from them. – FPP Nov 18 '11 at 1:31
So $\mathbb R- \mathbb Q$ is the complement of an $F_{\sigma}$ , which can be expressed, thru De Morgan, in terms of closed sets, which you can then combine with the closedness of the Cantor set. – Alphonse Nov 18 '11 at 1:34
But after using DeMorgan, I'd end up with an intersection of open sets, the opposite of what I want, right? – FPP Nov 18 '11 at 1:39
Sorry, I meant, start by describing $\mathbb Q$ itself in terms of unions, intersections, etc. , after which De Morgan gives you a description of $\mathbb R- \mathbb Q$ , after which you can compose with the closed Cantor set, and determine what the whole expression is. – Alphonse Nov 18 '11 at 2:23

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