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October 19th, 2010, 11:07 AM   #1
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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?
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October 19th, 2010, 01:22 PM   #2
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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.
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October 19th, 2010, 06:48 PM   #3
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Re: Cardinality

You will need axiom of choice for the general case.
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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?
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October 20th, 2010, 06:45 PM   #5
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Re: Cardinality

I want to make sure I know what you mean before I respond.
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October 20th, 2010, 06:57 PM   #6
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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).
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October 21st, 2010, 09:20 AM   #7
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Re: Cardinality

You cannot. If S^2=S for all infinite cardinals, then the axiom of choice holds.
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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?
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October 21st, 2010, 12:31 PM   #9
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Re: Cardinality

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

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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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