Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 25 Jul 2005 22:57 In article <dc472m$s66$1(a)lust.ihug.co.nz>, Barb Knox <see(a)sig.below> wrote: > In article > <ITSnetNOTcom#virgil-0F5C2E.14021925072005(a)comcast.dca.giganews.com>, > Virgil <ITSnetNOTcom#virgil(a)COMCAST.com> wrote: > [snip] > > >If the naturals satisfy the Peano properties, then any non-empty subset > >of the naturals will have a smallest (first) member. > > > >If the set of infinite naturals is not empty, it must have a smallest > >member. Subtract one and that must be the largest finite natural. > > But note that the 1st-order Peano axioms do allow non-standard models > with "infinite naturals", such as these 2 lines together: > 0 1 2 3 ... > ... w-2 w-1 w w+1 w+2 w+3 ... > > Note that there is no smallest "infinite natural" here. I guess I always think 2nd order. > > This doesn't violate 1st-order Peano well-foundedness because the set of > just the "infinite naturals" (those with a 'w') can not be defined in > the first plase using 1st-order logic (on account of compactness). > > >Unless TO can produce evidence either of a largest finite natural or a > >smallest infinite natural, he argues in violation of the Peano > >properties. > > Not necessarily.... What about in 2nd order, where I believe that every non-enpty set of naturals must have a first member? > > >So that whatever TO's "numbers" are, they are not the Peano naturals. > > Note that if TO actually took the trouble to learn some axiomatic > mathematics he would likely find non-standard Peano models interesting. > But of course as it is all he can do is handwave and bluster. Actually, what he is doing is more like trolling than mere bluster.
From: malbrain on 25 Jul 2005 23:06 Barb Knox wrote: > In article <1122338688.718048.162860(a)g47g2000cwa.googlegroups.com>, > malbrain(a)yahoo.com wrote: > > >Barb Knox wrote: > >> In article <MPG.1d4ecd45545679a8989f6b(a)newsstand.cit.cornell.edu>, > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> [snip] > >> > >> >keep in mind that > >> >inductive proof IS an infinite loop, so that incrementing in the loop > >> >creates > >> >infinite values, and the quality of finiteness is not maintained over those > >> >infinite iterations of the loop. > >> > >> Using your computational view, consider the following infinite loop > >> (using some unbounded-precision arithmetic system similar to > >> java.math.BigInteger): > >> > >> for (i = 0; ; i++) { > >> println(i); > >> } > >> > >> Now, although this is an INFINITE loop, every value printed will be > >> FINITE. Right? > > > >Not so fast. The behaviour of incrementing i after it reaches INT_MAX > >is undefined. > > Not so fast yourself. You missed the part about "unbounded-precision > arithmetic system", which has no max. Sorry, but the C standard admits no such system. INT_MAX must be declared by the implementation. There is no room for exceptions. karl m
From: imaginatorium on 25 Jul 2005 23:12 Tony Orlow (aeo6) wrote: > Daryl McCullough said: > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > >> "Bigger" in the sense of no surjection from the "smaller set to the > > >> "larger", is one thing, "bigger" in the sense of having the "smaller" > > >> set as a proper subset is different. While these two measures happen to > > >> coincide for finite sets, they do not coincide for infinite sets, as the > > >> definition of infinite for sets should hint to you. > > >> > > >gee, they coincide for finite sets AND infinite sets, under Bigulosity > > >but I don't suppose you consider that extra consistency any sort of > > >progress. > > > > No, that's no progress at all. You can prove that for finite sets > > if there is a bijection between set A and set B, then they have > > the same Bigulosity. But that fails for infinite sets. So Bigulosity > > is an inconsistent notion of "size". > > > > -- > > Daryl McCullough > > Ithaca, NY > > > > > Huh? Bigulosity doesn't rely on bijections. That why it DOES work. Remind us how you determine the bigulosity of the set { 1, 1/2, 1/4, 1/8,...} Brian Chandler http://imaginatorium.org
From: malbrain on 25 Jul 2005 23:13 stephen(a)nomail.com wrote: > In sci.math malbrain(a)yahoo.com wrote: > > Barb Knox wrote: > >> In article <MPG.1d4ecd45545679a8989f6b(a)newsstand.cit.cornell.edu>, > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> [snip] > >> > >> >keep in mind that > >> >inductive proof IS an infinite loop, so that incrementing in the loop creates > >> >infinite values, and the quality of finiteness is not maintained over those > >> >infinite iterations of the loop. > >> > >> Using your computational view, consider the following infinite loop > >> (using some unbounded-precision arithmetic system similar to > >> java.math.BigInteger): > >> > >> for (i = 0; ; i++) { > >> println(i); > >> } > >> > >> Now, although this is an INFINITE loop, every value printed will be > >> FINITE. Right? > > > Not so fast. The behaviour of incrementing i after it reaches INT_MAX > > is undefined. > > There is no INT_MAX for an unbounded-precision arithmetic system. The C language is defined by the C standard, as defined by ISO. There are no "unbounded" standard types in the C language. karl m
From: imaginatorium on 25 Jul 2005 23:24
Tony Orlow (aeo6) wrote: > Martin Shobe said: <big snip> > > BTW, there is a caveat on convergence. You have to assume the > > standard topology. In other topologies, that sequence can converge. > You mean with a ring? That's really not what we're talking about, unless you > agree that the number line is a circle, and even then it's not relevant. In > pure quantitative terms, a sum of infinite 1's is infinite. Tony, could you please clarify: when you use the word "ring", what do you mean? (a) The algebraic structure known by mathematicians as a ring (b) Something else (in which case please call it a T-ring) (c) You're sure it is (a), but cannot actually sketch the axioms for a ring (a) If you select (c), please confirm you really meant (a) by sketching the axioms. If you select (a), please suggest why you think the name "ring" is used. (Since I really have no idea, I'd be interested in informed comments on the last question.) Brian Chandler http://imaginatorium.org > > So, please make up your mind. Do we increment to get a successor an infinite > number of times, or only a finite number of times, to get N? > > > > Martin > > > > > > -- > Smiles, > > Tony |