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 $F_\sigma$?
The irrationals, the Cantor Set, and $F_\sigma$ sets
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real-analysis
2 Answers
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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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0Great. Thank you very much! I'd been struggling with this question for hours, mostly because I failed to realize that category was relative. – 2011-11-18
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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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0Sorry, 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. – 2011-11-18