Difference between revisions of "Talk:Power set"

 
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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 <math>\displaystyle 2^{\aleph _{n} } = \aleph _{ n+1 }</math> is undecidable, so I am very suspicious of a proposition that such a cardinality as <math>\displaystyle \aleph _{n>1} </math> exists at all.  Or are these cardinalities known to exist after all?  If so, how are they defined? &mdash;[[User:Boy Soprano II|Boy Soprano II]] 21:35, 26 August 2006 (EDT)
 
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 <math>\displaystyle 2^{\aleph _{n} } = \aleph _{ n+1 }</math> is undecidable, so I am very suspicious of a proposition that such a cardinality as <math>\displaystyle \aleph _{n>1} </math> exists at all.  Or are these cardinalities known to exist after all?  If so, how are they defined? &mdash;[[User:Boy Soprano II|Boy Soprano II]] 21:35, 26 August 2006 (EDT)
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It is true (and decidable) that there is no bijection between a set and its power set. --[[User:ComplexZeta|ComplexZeta]] 21:45, 26 August 2006 (EDT)

Revision as of 20:45, 26 August 2006

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)