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 February 16th, 2013, 05:07 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Discrete Mathematics - Set Theory Hi all f is a function from R to N (from the real numbers, to the natural numbers) we are asked to prove there is a value k in N so that $card(f^{-1}({k}))=aleph1$ anyone has an idea?
 February 16th, 2013, 06:58 AM #2 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: Discrete mathematics – set theory We have $\mathbb{R}=\bigcup_{k=1}^{\infty}f^{-1}(k)$. Since $\mathbb{R}$ is uncountable, at least one of the $f^{-1}(k)$ must be uncountable (otherwise we would have a countable union of sets that are at most countable, which would be at most countable).
 February 16th, 2013, 07:19 AM #3 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Discrete Mathematics - Set Theory I thought so to, but when I asked my professor, he told me that if i add aleph0+ aleph0 an infinite amount of times, it's not sure at all that i will get aleph0. he said there is another solution that bypasses this problem but i cant find it.

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