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 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: 6,607 Thanks: 616 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 non-countable 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
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Re: Cardinality

Quote:
 Originally Posted by dman315 You will need axiom of choice for the general case.
You can still prove <= without, right?

 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
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Re: Cardinality

Quote:
 Originally Posted by dman315 You cannot. If S^2=S for all infinite cardinals, then the axiom of choice holds.
What does the second sentence have to do with the first?

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 Cantor-Bernstein holds independent of AC and |S|<=|SxS|.
October 21st, 2010, 12:42 PM   #10
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Re: Cardinality

Quote:
 Originally Posted by dman315 Cantor-Bernstein holds independent of AC and |S|<=|SxS|.
I agree with the first part, but can you prove |S| <= |S x S| without AC?

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