October 19th, 2010, 11:07 AM  #1 
Newbie Joined: Oct 2010 Posts: 2 Thanks: 0  Cardinality
For any set S that has infinite elements, prove that S and SxS has the same cardinality. Can I prove this without using the continuum hypothesis? 
October 19th, 2010, 01:22 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591  Re: Cardinality
Continuum hypothesis is irrelevant. If S is countable, the proof is straightforward (like proving rational number set is countable). I am not familiar with the proof for noncountable sets.

October 19th, 2010, 06:48 PM  #3 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: Cardinality
You will need axiom of choice for the general case.

October 20th, 2010, 04:51 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Cardinality Quote:
 
October 20th, 2010, 06:45 PM  #5 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: Cardinality
I want to make sure I know what you mean before I respond.

October 20th, 2010, 06:57 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Cardinality
I'm saying that I think you can prove that S x S <= S for infinite S in ZF (that is, without AC).

October 21st, 2010, 09:20 AM  #7 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: Cardinality
You cannot. If S^2=S for all infinite cardinals, then the axiom of choice holds.

October 21st, 2010, 09:25 AM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Cardinality Quote:
My claim was that (I think that) "S x S <= S for all infinite sets S" in ZF. Your claim is that "S x S = S for all infinite sets S" is equivalent to AC in ZF. How does the second disprove the first?  
October 21st, 2010, 12:31 PM  #9 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: Cardinality
CantorBernstein holds independent of AC and S<=SxS.

October 21st, 2010, 12:42 PM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Cardinality Quote:
 

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