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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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William Elliot

Posts: 2,479
Registered: 1/8/12
Re: Division without the axiom of choice
Posted: Jan 10, 2013 5:48 AM
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On Thu, 10 Jan 2013, 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 n|A| = n|B| 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"?

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