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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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Paul

Posts: 400
Registered: 7/12/10
Re: Division without the axiom of choice
Posted: Jan 11, 2013 3:01 PM
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On Friday, January 11, 2013 12:59:39 AM UTC, Butch Malahide wrote:
> On Jan 10, 4:15 am, pepste...@gmail.com wrote:
>

> > Let A and B be sets.  Assume ZF without assuming choice.  Then, for all [positive] 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.
>
>
>
> Yes, that is correct. More generally, it is proved in ZF that, for any
>
> cardinals a and b, and any nonzero finite cardinal (i.e. natural
>
> number) n, the inequality na <= nb implies a <= b. (This implies the
>
> proposition you asked about, in view of the well known fact that a = b
>
> <-> a <= b & b <= a.)
>
>
>

> > There's a famous Conway/Doyle paper which proves this for n = 2 and n = 3.
>
> > However, it doesn't seem rigorous or clear and I have trouble understanding it.
>
>
>
> I don't know the Conway/Doyle paper, and I don't know a proof for n =
>
> 3. A proof for n = 2 has been posted in this newsgroup:
>
>
>
> http://groups.google.com/group/sci.math/msg/1e65b64fee74fe07?hl=en
>
>
>

> > 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.)
>
>
>
> The following summary and references are cribbed and paraphrased from
>
> p. 174 of Waclaw Sierpinski's book Cardinal and Ordinal Numbers,
>
> second edition revised, Warszawa, 1965. The theorems are theorems of
>
> ZF (standard set theory without the axiom of choice); m and n are
>
> arbitrary cardinals (i.e., if they are infinite, they are not
>
> necessarily alephs); k is a natural number. The theorems you are
>
> interested in are:
>
>
>
> THEOREM 1. If km = kn then m = n.
>
>
>
> THEOREM 2. If km <= n then m <= n.
>
>
>
> F. Bernstein, Untersuchungen aus der Mengenlehre, Math. Annalen 61
>
> (1905), 117-155. [Proves Theorem 1 for k = 2 and outlines a proof for
>
> general k.]
>
>
>
> W. Sierpinski, Sur l'egalite 2m = 2n pour les nombres cardinaux, Fund.
>
> Math. 3 (1922), 1-16. [Another proof of Theorem 1 for k = 2.]
>
>
>
> W. Sierpinski, Sur l'implication (2m <= 2n) -> (m <= n) pour les
>
> nombres cardinaux, Fund. Math. 34 (1947), 148-154. [Proof of Theorem 2
>
> for k = 2.]
>
>
>
> A. Tarski, Cancellation laws in the arithmetic of cardinals, Fund.
>
> Math. 36 (1949), 77-92. [Proof of Theorem 2 in general.]
>
>
>
> I guess Tarski's 1949 paper has what you're looking for. I don't know
>
> if it's available as a free etext; I'm inclined to doubt it, but I
>
> haven't looked. On the other hand, I bet your local public library can
>
> get you a copy at nominal cost by interlibrary loan.


Thank you for this very helpful reply. I'm afraid that I'm not convinced by the proof for the case n = 2 on the other thread. I suppose I could try to fill in the details but I do this type of maths for enjoyment, and I enjoy the process of reading through complete proofs more than I enjoy working out the details.
So what are these missing details I'm referring to? The answer is that it doesn't seem at all transparent to me that, in the infinite case, almost all the horses get matched at some finite stage. (The finite case is fine).

Paul Epstein





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