Paul
Posts:
258
Registered:
7/12/10
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Re: Division without the axiom of choice
Posted:
Jan 10, 2013 6:18 AM
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On Thursday, January 10, 2013 10:48:46 AM UTC, William Elliot wrote:
> On Thu, 10 Jan 2013, pepstein5@gmail.com wrote:
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> > Let A and B be sets. Assume ZF without assuming choice. Then, for all
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> > non negative integers n, I believe (correct me if I'm wrong) that, if n
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> > x A is equipotent to n x B, then A is equipotent to B.
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> It's false for n = 0.
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> If A is infinite, the so is B and
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> A equipotent n x A equipotent n x B equipotent B
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> If A finite, then so is B and for n > 0, if n|A| = n|B| in N.
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> then |A| = |B|.
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> > There's a famous Conway/Doyle paper which proves this for n = 2 and n =
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> > 3.
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> What do you mean this for n = 2 and n = 3, when this is about all n >= 0?
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> > However, it doesn't seem rigorous or clear and I have trouble
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> > understanding it.
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> I don't uunderstand what you're talking about.
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> > Does anyone know a more axiomatic treatment? (I don't have access to a
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> > university, and I'm not in the market for maths purchases, so only free
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> > references would be helpful.)
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> How about an explicit statement of "this for n = 2 and n = 3"?
You are correct that my statement is false for n = 0 -- apologies for the error.
The requested explicit statement is:
Let n be an integer such that either n = 2 or n = 3.
Let A and B be sets such that a bijection exists between n x A and n x B.
Prove without using the axiom of choice that a bijection exists between A and B.
As for your comment about A being equipotent to n x A when A is infinite, there are 3 (exhaustive but not mutually exclusive) possibilities. Possibility 1 is that you are assuming the axiom of choice. Possibility 2 is that your asssertion is not even true without the axiom of choice. Possibility 3 is that your assertion is true without assuming the axiom of choice, but is highly non-trivial.
I am interested in extending the above to the case where n > 3.
Thank You,
Paul Epstein
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