Obections to Cantor's Theory (Wikipedia article)
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From: stephen on 21 Jul 2005 11:15 In sci.math Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > stephen(a)nomail.com wrote: >> Why are you bringing physics into this? Whether black >> holes exist or not has nothing to do with set theory. > Theoretically: no. In practice: yes. Because some consequences of set > theory have invaded into physics by the fact that mathematics becomes > somewhat _applied_ there, huh ! Geez ... > Han de Bruijn Are you claiming that set theory is directly applied in General Relativity? Do you actually have evidence of that? Ignoring mathematical foundations, which physicists, and most everybody else, typically do, it seems that the limit of 1/sqrt(1-v^2/c^2) as v approaches c is infinite with or without set theory. Stephen
From: Robert Low on 21 Jul 2005 12:06 stephen(a)nomail.com wrote: > Are you claiming that set theory is directly applied in General > Relativity? Do you actually have evidence of that? I have seen a paper in which transfinite induction was used. And in the existence of maximal solutions to the initial value problem, appeal is (at least sometimes) made to Zorn's lemma.
From: malbrain on 21 Jul 2005 12:23 Virgil wrote: > In article <MPG.1d4863d52071fde5989f51(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > and that I was trying to prove > > things about sets, not numbers, which is also bullshit, since I was > > proving a property regarding a set DEFINED by a natural number, which > > is ultimately a property of that number. > > But the set N is not defined by any one natural number Under Tony's theory, the number representing N is defined by a string of an infinite number of ones. Yes, more than one Turing machine can produce this. karl m
From: Jeffrey Ketland on 21 Jul 2005 12:38 Helene.Boucher(a)wanadoo.fr > > Jeffrey Ketland wrote: > >> It depends upon how you formulate PA. >> >> They are equivalent over the base theory PA-, which has >> (a) axioms stating that + and x are associative, commutative, and satisfy >> the distributive law, >> (b) axioms stating that 0 is an identity for + and a zero for x, and that >> 1 >> is an identity for x; >> (c) axioms saying that < is linear order. >> (d) axioms saying that + and x respect order; >> (e) if x < y, then (Ez)(x+z = y) >> (f) that 0 is the least number and that < is discrete. >> See Kaye 1991, p. 16 and pp 45-6. >> Richard Kaye then defines PA as PA- plus the induction scheme (p. 43). > >> >> Other authors define PA as the six axioms for successor, plus and times >> plus >> the induction scheme. >> > > I don't have Kaye's article handy, but it certainly seems a > non-standard definition of PA. Indeed I don't think it's warranted to > call this particular axiomatization "PA" unless some kind of warning is > given ! This subtheory is nice because it relates PA to the theory of the (non-negative part) of a discretely ordered ring. It includes all the obvious algebraic and order-theoretic properties of (N, 0, 1, +, x, <). (Of course, the notion of successor is not treated as primitive here: Sx may be defined as x+1 or as the smallest y > x.) > <snip> > >> >> If by "PA minus the induction scheme", you mean the six usual axioms for >> successor, + and x, > > Yes that's what I (and I think most other people) mean. I agree. It's usually something like Q (with the definition of <), plus the induction scheme. So, your original questions are, I think, these: (a) Does Q + induction scheme imply the least-number principle? (b) Does Q + induction scheme imply the principle of total induction? The answer is yes. (See Hajek/Pudlak 1993, _Metamathematics of First-Order Arthmetic_, p. 35). Or did you mean to ask about the converse in (a), (c) Does Q + least number principle imply induction scheme? I'm guessing, since I haven't checked the proof, but I'd be surprised if that wasn't true as well. --- Jeff
From: Jeffrey Ketland on 21 Jul 2005 12:42
Robert Low > Jeffrey Ketland wrote: > > The least number principle does express that (N,<) is a well-ordering. > > > But in the weakened sense that it's only sets of naturals > which make some P(n) true that have to have a least element. For > example, the set of infinite elements of a non-standard model has > no least element, but that's OK because there's no way of expressing > 'n is infinite' in the language. Right. --- Jeff |