Talk:Power set

I deleted the part that said that for no infinite set was there a bijection between the set and its power set. I am fairly certain that this is undecided. It certainly is known that the proposition $\displaystyle 2^{\aleph _{n} } = \aleph _{ n+1 }$ is undecidable, so I am very suspicious of a proposition that such a cardinality as $\displaystyle \aleph _{n>1}$ exists at all. Or are these cardinalities known to exist after all? If so, how are they defined? —Boy Soprano II 21:35, 26 August 2006 (EDT)

It is true (and decidable) that there is no bijection between a set and its power set. --ComplexZeta 21:45, 26 August 2006 (EDT)

Really? Where can I find a proof? Thanks. —Boy Soprano II 21:53, 26 August 2006 (EDT)

See the bottom of http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument --ComplexZeta 01:25, 27 August 2006 (EDT)

Thank you very much. —Boy Soprano II 15:04, 27 August 2006 (EDT)

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Talk:Power set