Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Virgil on 19 Jul 2005 22:39 In article <MPG.1d47266568fcba95989f2e(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Robert Low said: > > Tony Orlow (aeo6) wrote: > > > David Kastrup said: > > > > > I still do not get this. You have a set of naturals {0,1,2,3...}, and a > > > set of > > > binary numbers {0,1,10,11,100,101,110,111,....}. Surely there is a > > > bijection > > > between these two sets. > > > > Of course. But that isn't the issue. The issue is the existence > > of a bijection between either of these sets and the power set > > of the set of naturals. So, which binary number does your > > procedure associate with the set of *all* natural numbers > > divisible by 3? > > > Again, that infinite set is denoted by the infinite whole number > 100100100...100100100. That "number" would require that ther eexist infintiely large natural numbers. But ALL naturals except the first, are acheivable by adding 1 to previous naturals, all the way back to the first one. So that somewhere TO is claiming that adding 1 to a finite natural produces an infinite natural. What is that finite natural? If TO cannot produce it, he cannot have infinite naturals.
From: Virgil on 19 Jul 2005 22:46 In article <MPG.1d4726e11766660c989f2f(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > David Kastrup said: > > >> Alec McKenzie <mckenzie(a)despammed.com> writes: > > >> > > >> > David Kastrup <dak(a)gnu.org> wrote: > > >> > > > >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: > > >> >> > It has been known for a proof to be put forward, and fully accepted > > >> >> > by the mathematical community, with a fatal flaw only spotted years > > >> >> > later. > > >> >> > > >> >> In a concise 7 line proof? Bloody likely. > > >> > > > >> > I doubt it had seven lines, but I really don't know how many. > > >> > Probably many more than seven. > > >> > > >> It was seven lines in my posting. You probably skipped over it. It > > >> is a really simple and concise proof. Here it is again, for the > > >> reading impaired, this time with a bit less text: > > >> > > >> Assume a complete mapping n->S(n) where S(n) is supposed to cover all > > >> subsets of N. Now consider the set P={k| k not in S(k)}. Clearly, > > >> for every n only one of S(n) or P contains n as an element, and so P > > >> is different from all S(n), proving the assumption wrong. > > > > > I still do not get this. You have a set of naturals {0,1,2,3...}, > > > and a set of binary numbers {0,1,10,11,100,101,110,111,....}. Surely > > > there is a bijection between these two sets. > > > > Fine. > > > > > So, what is the problem if one interprets the binary numbers (with > > > implied leading zeroes) as being a map of each subset, where each > > > successive bit represents membership in thesubset by each successive > > > natural number? > > > > Ok, so let's construct P. It is actually easy enough, since n<2^n, > > and so P=N. So what number corresponds to N itself in your mapping? > > > > > An infinite string of 1's: 1111.....1111. This is (2^aleph_0)-1. > > Infinite whole numbers are required for an infinite set of whole numbers. Not by anyone who understand the consequences of such a requrement. Every natural except the first is the result of adding 1 to a previous natural, and the set of all naturals is the smallest set with this inductive property. If TO's assumprtions were actually the case, there would have to be a finite natural so large that adding 1 to it would produce an infinite natural. But TO cannot produce either a largest finite nor a smallest infinite, so the set of all finite naturals is already big enough.
From: Dik T. Winter on 19 Jul 2005 22:45 In article <mckenzie-27AD7F.22290719072005(a)news.aaisp.net.uk> Alec McKenzie <mckenzie(a)despammed.com> writes: .... > In 1879 a proof of the four-colour map theorem was published by > Arthur Bray Kemp, a member of the London Mathematical Society. > He became a member of the Royal Society in consideration for his > achievement. > > It was not until eleven years later, in 1890, that the fatal > flaw in the proof was pointed out by Percy John Heawood in a > paper in the Quarterly Journal of Mathematics. Was it an invalid conclusion, or was it a wrong assumption? In the first case, mathematicians of that time should have noticed. Conclusions that do not follow from their antecedents are in many cases easily recognised. It is more problematical with wrong assumptions, they indicate a gap that can be surmounted or not, depending on the actual assumption. The latter kind is much more difficult to find. Anyhow, Cantor's proof consists solely of logical conclusions from antecedents. As two of the anti-Cantorians have stated to me, yes, each step is right but the proof is wrong... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 19 Jul 2005 22:53 In article <MPG.1d473242aff1e303989f30(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > You have not shown such a thing, and of course it would be > > inconsistent with the Peano axioms defining the naturals. > That is simply not true. There is nothing in Peano's axioms that states > explicitly that all natural numbers are finite. The fifth axiom, defining > inductive proof, is used to prove this theorem, but it is a misapplication of > the method. I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. That "proof" only applies to "initial segments", and the set of all naturals is not an initial segment, so the "proof" proves nothing about the set of all naturals. > Just be aware that anti-Cantorians are sick of > being called crackpots, and the day will soon come when the crankiest > Cantorians will eat their words, and this rot will be extricated from > mathematics. It is the standard delusion of crackpots that "their day will come".
From: Virgil on 19 Jul 2005 23:00
In article <MPG.1d4733ca67be65e3989f31(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > David Kastrup said: > > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > >> > > >> > Now, I am not familiar, I think, with the proof concerning > > >> > subsets of the natural numbers. Certainly a power set is a larger > > >> > set than the set it's derived from, but that is no proof that it > > >> > cannot be enumerated. > > >> > > >> Uh, not? > > > > > Yes, not. "Larger" is not a synonym for "uncountable" except in > > > Cantorland, and that is a leap and an assumption. > > > > "Larger" is a synonymon for "can't be surjected onto from" in set > > theory. And "uncountable" is a synonymon for "larger than the set of > > naturals". It is not a leap or an assumption, but simply a > > definition. > > > > >> > Is this the same as the proof concerning the "uncountability" of > > >> > the reals? > > >> > > >> It's pretty similar. > > > Figures. > > >> > > >> Assume a set X can be put into complete bijection with its powerset > > >> P(X) such that we have a mapping x->f(x) where x is an element from X > > >> and f(x) is an element from P(X). Now consider > > >> Q = {x in X|x not in f(x)}. Clearly, for all x in X we have > > >> Q unequal to f(x), since x is a member of exactly one of f(x) and Q. > > >> So Q is missing from the bijection. > > >> > > >> > > > Again with the "Clearly". You might want to refrain from using the > > > word, and just try to be clear, without hand-waving. > > > > > > There is no requirement that subset number x include x as a member, > > > > Quite so. But there is a requirement that subset number x _either_ > > include x as a member _or_ not include x as a member. Only one of > > those two statements can be true. And then Q _either_ not includes x > > as a member _or_ does include it, respectively. > > > > You are free to choose your mapping as you want to. But once you have > > chosen your mapping, each subset number x _either_ includes x as a > > member _or_ it doesn't. Whether it does, can be chosen independently > > for every x. But once you are through, for every particular x, x will > > be in f(x), or it won't. And depending on that, x won't be in Q, or > > it will. > Actually, as I think about it, given this natural mapping of the naturals to > the subsets of naturals, subset number x will ONLY include x as a member for > subsets 0, 1 and 2. Beyond that, subset x will NEVER contain x. So, you have > non-empty Q for the null set, and the singletons {1} and {2}. So, what does > that prove? The mapping from X to P(X) is not "natural", it is 'arbitrary', meaning that it can be anything and cannot be assumed to have any special properties. In particular, f(0) = f(1) = f(2) = X is possible, whenever {0,1,2} is a subset of X, . |