
Re: Division without the axiom of choice
Posted:
Jan 10, 2013 5:48 AM


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
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"?

