On Saturday, January 12, 2013 1:18:40 AM UTC, Butch Malahide wrote: > On Jan 11, 2:01 pm, pepste...@gmail.com wrote: > > > 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 <= 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. > > > > > > 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). > > > > Looks like one detail to me. Let me try to explain that one to you, so > > you can enjoy the proof. > > > > Hmm. The statement "[k]eep doing this until either all horses have > > been paired off, or else you are left with just one horse" might be > > ambiguous. I can see how someone might take it to meen that, after a > > FINITE number of steps, all or all but one of the horses will have > > been matched. No, the process is supposed to continue as long as > > necessary, an infinite number of steps if that's what it takes. The > > claim is that, when the process has run to completion, there will be > > at most one unmatched horse (in each connected component). That is, > > each horse (with at most one exception) gets matched after some finite > > number of steps, but that number varies from one horse to another, and > > is not necessarily bounded. > > > > Assume for a contradiction that two or more horses (in the same > > component) remain unmatched after an infinite sequence of steps. Call > > one of them Admiral, and call his nearest unmatched neighbor Biscuit. > > Every horse between Admiral and Biscuit got matched at some finite > > stage. Since there were only a finite number of horses between A and B > > to start with, there was some finite stage when *all* horses between A > > and B had been matched. (So A and B are horses of opposite colors, > > because there were initially an even number of horses between them.) > > Now it's clear that, after at most 4 more steps, A and B would have > > been matched with each other, unless one of them got matched with > > another horse first. Q.E.D.? > > > > An easy modification of this argument shows that 2m <= 2n implies m <= > > n.
This is a much clearer exposition than the Conway/Doyle one in www.math.dartmouth.edu/~doyle/docs/three/three.pdf and I now completely understand the proof. It might be worthwhile submitting your argument for publication (I'm guessing that it's your proof or your exposition).
Initially I missed the induction argument at the end of your most recent post. However, I did get one thing right initially. When I said "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," I meant that it didn't seem transparent to me that almost all horses have the property that they get matched at a finite stage. I did not mean to imply that I believed that some finite stage existed at which point almost all the horses were matched.