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)
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I claimed that that proof doesn't rely on the axiom of choice: is this really true?  --[[User:JBL|JBL]] 11:56, 7 September 2006 (EDT)
 
 
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)
 
 
 
Really?  Where can I find a proofThanks. &mdash;[[User:Boy Soprano II|Boy Soprano II]] 21:53, 26 August 2006 (EDT)
 
 
 
See the bottom of http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument --[[User:ComplexZeta|ComplexZeta]] 01:25, 27 August 2006 (EDT)
 
 
 
Thank you very much.  &mdash;[[User:Boy Soprano II|Boy Soprano II]] 15:04, 27 August 2006 (EDT)
 

Revision as of 10:56, 7 September 2006

I claimed that that proof doesn't rely on the axiom of choice: is this really true? --JBL 11:56, 7 September 2006 (EDT)