Let $\mathbb{R}$ be equipped with the standard topology.
Let $E$ be the set of irrational numbers.
How do i prove that $E$ is not a countable union of closed subsets, using Baire Category Theorem?
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What property does a closed subset of $E$ have, and what does Baire's theorem say about countable unions of such sets? – Daniel Fischer Apr 22 '14 at 20:51
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That isn't yet a contradiction. But a countable union of nowhere dense sets is meagre (of the first category), and the irrationals are non-meagre. – Daniel Fischer Apr 22 '14 at 21:05
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You have to use Baire? Not enough then to use a version of Cantor's diagonal proof that the members of E are not countable? – Travelling Salesman Apr 22 '14 at 21:07
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Does a countable union of closed sets have to be countable?? – Asaf Karagila Apr 22 '14 at 21:10
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@TravellingSalesman All basic set theory arguments are assumed, but i don't think that is necessary to prove this exercise.. – user140374 Apr 22 '14 at 21:23