Obections to Cantor's Theory (Wikipedia article)
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From: Daryl McCullough on 20 Jul 2005 11:05 Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >>> 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. Yes. So we have: If S is a set of natural numbers, and S has a largest member N, then N >= the cardinality of S. How do you prove that every set of natural numbers has a largest member? -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 20 Jul 2005 11:29 Virgil said: > In article <MPG.1d472484f809e37c989f2c(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > The idea of uncountability as being equivalent to "larger than the set of > > naturals" is unfounded. > > The meaning of "uncountable" is a matter of definition. One may, as TO > does, object to claims of its being instanciated, but it is perfectly > well-defined. > > > > There is no reason to believe that larger sets cannot > > be enumerated. > > What does "larger than" mean, then? You ask this as if any set that is larger than any other set is "uncountable". Do you consider the two terms, "larger" and "uncountable" to be synonymous? > > There is no reason to believe that larger sets cannot > be enumerated. > > That TO does not understand the reasons does not invalidate them. Why don't you try explaining them? > > > the power set of the naturals can be enumerated and bijected > > with the naturals, as I described in another post, as long as infinite > > natural > > numbers are allowed. > > But to allow infinite naturals means that one must have a finite natural > so large that adding one to it gives an infinite result, since every > natural, except the first, is produced by adding 1 to a previous natural. > There is no way to specify any such number. There is no point at which the finite suddenly becomes infinite. To focus on this point that doesn't exist is a waste of time. If you want an infinite set of naturals, then you need infinite values in it. The "Twilight Zone", as you called it, is irrelevant to that fact. -- Smiles, Tony
From: David Kastrup on 20 Jul 2005 11:29 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Barb Knox said: >> In article <MPG.1d4726e11766660c989f2f(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> [snip] >> >> >Infinite whole numbers are required for an infinite set of whole numbers. >> >> Good grief -- shake the anti-Cantorian tree a little and out drops a >> Phillite. Here's a clue: ALL whole numbers are finite. Here's a >> (2nd-order) proof outline, using mathematical induction (which I >> assume/hope you accept): >> 0 is finite. >> If k is finite then k+1 is finite. >> Therefore all natural numbers are finite. >> >> > That's the standard inductive proof that is always used, and in > fact, the ONLY proof I have ever seen of this "fact". Is there any > other? I have three proofs that contradict this one. Do you have any > others that support it? > > Inductive proof proves properties true for the entire set of > naturals, right? For each of its members. > That entire set is infinite right? Therfore, the number of times you > are adding 1 and saying, "yep, still finite", is infinite, right? No. He is not adding 1 more than a single time, just to check that for n finite (which means that the set 0..n obeys the pigeon-hole principle) n+1 is still finite (the case of one additional pigeon-hole can be reduced to the case n if you check for the hole in position n+1 and in 0..n both before and after permutation). > So, you have some way of adding an infinite number of 1's and > getting a finite result? No, he is just checking that the conditions for the fifth Peano axiom hold. The whole point of the axioms is not to have to check an infinite number of steps in order to get a statement about the members of a particular infinite set, the naturals. > You might want to discuss this with your colleagues specializing in > infinite series. There is a very simple rules that says no infinite > series can converge to a finite value unless the terms of the series > have a limit of zero as n approaches infinity. Does this constant > term, 1, have a limit of zero? No it doesn't, and the infinite > series of constant 1's cannot converge, but diverges to > infinity. Can you actually deny this? If so, then Poincare was > right. Nobody is bothered about adding numbers. The Peano axioms do not even define addition. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 20 Jul 2005 11:34 Virgil said: > In article <MPG.1d4725fbdfa9bb24989f2d(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > There is a simple, demonstrably valid proof of Cantor's Theorem in ZF > > > set theory. So you must think the proof is unsound. Which axiom of ZF > > > do you believe to be false? > > > > > > Chris Menzel > > > > > > > > I was asked that before, and never got around to fully analyzing the axioms > > for > > lack of time, but the diagonal proof suffers from the fatal flaw of assuming > > that the diaginal traversal actually covers all the numbers in the list. > > If it misses any of them, it must miss a first one. Which is the first > one misssed? > > If there is not a first one missed, then none are missed. > The first one missed is the first one directly below the diagonal traversal. If there are more real numbers than digits, which is what the proof actually proves, then the diagonal traversal will always have numbers as wide as it, but below it, that it has not touched. If you want the specific first one missed when the diagonal has completed its march across aleph_0 digits and down aleph_ 0 numbers, it's number aleph_0+1, and the diagonal misses aleph_1-aleph_0 of the aleph_1 real numbers in the list. You antidiagonal is one of those, and there are so many of them, that just choosing one at random will almost surely get you a number not on the diagonal, even without constrction of your antidiagonal. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 11:38
Virgil said: > 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. > We have been through all this before. You lay these requirement on me, but when I say you cannot have a smallest infinite omega, or omega-1 would be finite, you say omega-1=omega. So, I can play your stupid game, and say that alpha is the largest finite, but alpha+1=alpha. Tada! I am as senseless as you! Isn't that great? Let me know when you want to get off the "largest finite" kick. I won't be responding to it any more. -- Smiles, Tony |