Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Tony Orlow on 20 Jul 2005 16:45 Virgil said: > In article <MPG.1d4825defff809ee989f39(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > I proved it for all n in N, which I think you agree is an infinite number of > > naturals. > > But that does not prove that what is true for members of N is true for N > itself. Consider the set containing a single apple. By TO's argument, > that set must itself be an apple, since that is true for all its members. > > To uses a similar argument in reverse, that an infinite set must cinatin > infinite objects, so not that appple must be a set as well as the set > being an apple, at least by TO_logic. You are deliberately misrepresenting what I have said, Virgil, as you have done countless times. Stop it. You are lying. > > > I think you were perhaps one of those claiming that inductive proof > > only works for finite iterations, but then it wouldn't work for the infinite > > set of naturals, now, would it? I am not going in these stupid circles with > > you. > > TO makes up his own stupid circles, and goes round and round them > endlessly. So says the pot. > > > > Again this goes back to the Cantorian mantra, "no largest finite" > > It is not strictly Cantorian. To the best of my knowledge, even those > mathematicians who avoid infinities do not object to 'no largest finite > natural', though all mathematicians object to infinite naturals, at > least in standad models. I never objected to that fact either. It's just not an excuse for dismissing obvious proofs. > > If TO wants infinite naturals, he had better look up Abraham Robinson, > et al, but if he thinks Cantorian math is confusing, he will not get far > with non-standard analysis. > -- Smiles, Tony
From: Robert Low on 20 Jul 2005 16:54 Daryl McCullough wrote: > Robert Low says... >>Funnily enough, there is a similar sounding statement that >>is true in non-standard analysis: any set containing arbitrarily >>large finite integers must also contain an infinite integer. >>But in that game, the class of all finite integers isn't >>a set :-) > Actually, maybe Tony would be happier with nonstandard > analysis... As long as he doesn't think he'll get a smallest infinitesimal that way...
From: Daryl McCullough on 20 Jul 2005 16:43 In article <3k7o8jFt4lfgU2(a)individual.net>, Robert Low says... > >Daryl McCullough wrote: >> But the Peano axioms say that *all* nonempty sets of naturals have >> a smallest element. So if you say that there is no smallest infinite >> natural, then that implies that there are *no* infinite naturals. > >I'm not entirely sure what you're claiming here, but order isn't >even mentioned in the Peano axioms, so they certainly can't include >a statement that the naturals are well-ordered. I guess I was thinking in terms of second-order PA. If you are going to use induction to prove something about infinite sets, then you have to be working in second-order PA, because you can only use first-order induction to prove things expressed in the language 0, +, *, successor. -- Daryl McCullough Ithaca, NY
From: Robert Low on 20 Jul 2005 17:18 Daryl McCullough wrote: > In article <3k7o8jFt4lfgU2(a)individual.net>, Robert Low says... >>Daryl McCullough wrote: >>>But the Peano axioms say that *all* nonempty sets of naturals have >>>a smallest element. >>I'm not entirely sure what you're claiming here, but order isn't >>even mentioned in the Peano axioms, so they certainly can't include >>a statement that the naturals are well-ordered. > I guess I was thinking in terms of second-order PA. It's pretty unclear in a lot of the thread when people are talking about first order and when second order PA; but I don't see how that helps. The only difference is replacing the axiom scheme "If P is a one-place predicate, then P(0) together with \forall n (P(n)->P(n+1)) entails \forall n P(n)" with the axiom "If S \subset N, and 0 \in S, and S is closed under successor, then S=N". We still don't have an explicit mention of order, so the axioms can't just *say* than N is well-ordered, even if they imply it. But maybe part of the problem is that the entire conversation is taking place in an ill-defined context...
From: Daryl McCullough on 20 Jul 2005 17:02
Tony Orlow writes: > >Daryl McCullough said: >> Once again, if your claims had any merit whatsoever, then you >> would be able to rephrase them in a way that does not rely on >> unorthodox meanings of terms. Rephrase your claim without using >> the word "finite" or "infinite". Is that possible? >Are you talking to me or Dik? You. You're the one taking about infinite naturals. >No one else seems to have a definition for "infinite" as a >stand-alone word. Mathematicians have a definition for "infinite set", namely "a set such that there exists a bijection between that set and a proper subset of itself". >Finite means with an end I don't want more words. I want a *mathematical* definition. Give a definition in terms of the standard mathematical concepts such as "subset", "element of", etc. >If the leftmost possible significant digit is at position n, >then the largest number possible in base b is b^n. Who says that there is a leftmost possible significant digit? You are confusing two different things: (1) For every string, there is a finite number n such that the length of the string is less than n. (2) There is a finite number n such that for every string the length of the string is less than n. (1) is true, but (2) is false. >If b and n are known, then b^n is known. If b >and n are finite, then b^n is finite. This is just using the >standard definition of finite, independent of cardinality. There is no standard mathematical definition, other than the one you've rejected. -- Daryl McCullough Ithaca, NY |