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Topic: Division without the axiom of choice
Replies: 10   Last Post: Jan 12, 2013 4:51 PM

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Paul

Posts: 384
Registered: 7/12/10
Division without the axiom of choice
Posted: Jan 10, 2013 5:15 AM
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Let A and B be sets. Assume ZF without assuming choice. Then, for all non negative integers n, I believe (correct me if I'm wrong) that, if n x A is equipotent to n x B, then A is equipotent to B.
There's a famous Conway/Doyle paper which proves this for n = 2 and n = 3.
However, it doesn't seem rigorous or clear and I have trouble understanding it.

Does anyone know a more axiomatic treatment? (I don't have access to a university, and I'm not in the market for maths purchases, so only free references would be helpful.)

Thank you very much for your help.

Paul Epstein



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