
Re: Division without the axiom of choice
Posted:
Jan 10, 2013 3:51 PM


On 20130110, William Elliot <marsh@panix.com> wrote: > On Thu, 10 Jan 2013, pepstein5@gmail.com wrote:
>> 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.
> It's false for n = 0.
> If A is infinite, the so is B and > A equipotent n x A equipotent n x B equipotent B
Without the axiom of choice, it is not necessarily the case that A infinite implies it is equipotent with 2 x A. That is not true in ZF, but it is weaker than choice.
> If A finite, then so is B and for n > 0, if nA = nB in N. > then A = B.
>> There's a famous Conway/Doyle paper which proves this for n = 2 and n = >> 3.
> What do you mean this for n = 2 and n = 3, when this is about all n >= 0?
>> However, it doesn't seem rigorous or clear and I have trouble >> understanding it.
> I don't uunderstand what you're talking about.
>> 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.)
> How about an explicit statement of "this for n = 2 and n = 3"?
 This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558

