Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
William Elliot

Posts: 1,485
Registered: 1/8/12
Re: Division without the axiom of choice
Posted: Jan 10, 2013 5:48 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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 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"?



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.