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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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 Herman Rubin Posts: 399 Registered: 2/4/10
Re: Division without the axiom of choice
Posted: Jan 10, 2013 3:51 PM

On 2013-01-10, 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 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

> 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)494-6054 FAX: (765)494-0558

Date Subject Author
1/10/13 Paul
1/10/13 William Elliot
1/10/13 Paul
1/10/13 Herman Rubin
1/10/13 Butch Malahide
1/10/13 trj
1/10/13 Butch Malahide
1/11/13 Paul
1/11/13 Butch Malahide
1/12/13 Paul
1/12/13 Butch Malahide