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?
Asked
Active
Viewed 462 times
1
-
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
-
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
-
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
-
Does a countable union of closed sets have to be countable?? – Asaf Karagila Apr 22 '14 at 21:10
-
@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