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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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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Division without the axiom of choice
Posted: Jan 10, 2013 10:49 PM
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On Jan 10, 6:59 pm, Butch Malahide <fred.gal...@gmail.com> 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 <= kn then m <= n. ***TYPO CORRECTED***
>
> 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.


Sorry about the typo in the statement of Theorem 2!



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