Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 19 Jul 2005 16:17 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > 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. And what number corresponds to the subset of N without the number 1111.....1111 in it? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: The World Wide Wade on 19 Jul 2005 16:20 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. There is no reason to believe that larger sets cannot > be enumerated. 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. Sort of like saying "if 0 = 1" is allowed.
From: Stephen J. Herschkorn on 19 Jul 2005 16:24 Should a reputable encyclopedia contain an entry devoted entirely to people who think the earth is flat? An entry only for those who think that sun revolves the earth? An entry devoted specifically to those who think that man never landed on the moon? To those who insist there is a smallest positive real number? -- Stephen J. Herschkorn sjherschko(a)netscape.net Math Tutor in Central New Jersey and Manhattan
From: Alec McKenzie on 19 Jul 2005 16:31 Chan-Ho Suh <suh(a)math.ucdavis.nospam.edu> wrote: > In article <mckenzie-3D5A6E.14153919072005(a)news.aaisp.net.uk>, Alec > McKenzie <mckenzie(a)despammed.com> wrote: > > > I quite agree that it does not get better than that, but I think > > one must allow some room for doubt, however small, for any > > proof. Otherwise one is proclaiming infallibility. > > No one has proclaimed infallibility. To allow no room whatever for any doubt is to declare it incapable of error, ie infallible. > > 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. > > I doubt this. "Fully accepted" means that the community either was not > paying attention or didn't care enough to check themselves, etc. Quite possibly. > I know of no proof in the modern literature that was verified correct by > a large number of experts and with a flaw only found years later. I did not say it was verified correct by anyone, let alone a large number of experts. -- Alec McKenzie mckenzie(a)despammed.com
From: Tony Orlow on 19 Jul 2005 16:53
David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > Alec McKenzie said: > >> >> "Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote: > >> >> > >> >> > Can anti-Cantorians identify correctly a flaw in the proof that there > >> >> > exists no enumeration of the subsets of the natural numbers? > >> >> > >> >> In my view the answer to that question a definite "No, they > >> >> can't". > >> >> > >> >> However, the fact that no flaw has yet been correctly identified > >> >> does not lead to a certainty that such a flaw cannot exist. Yet > >> >> that is just what pro-Cantorians appear to be asserting, with no > >> >> justification that I can see. > >> >> > >> > Even though every subset of the natural numbers can be represented > >> > by a binary number where the first bit denotes membership of the > >> > first element, the second bit denotes membership of the second > >> > element, etc? > >> > >> Well, what number will then represent the set of numbers dividable by > >> three? > > 100100...100100100100 > > > Of course, you will argue that this infinite value is not a natural > > number, since all naturals are finite, but that is clearly > > incorrect, as it is impossible to have an infinite set of values all > > differing by a constant finite amount from their neighbors, and not > > have an overall infinite difference between some pair of them, > > indicating that at least one of them is infinite. > > 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. So, which inductive proof do you believe? You cannot add 1 an infinite number of times to your maximal element, without it achieving an infinite value. I provided two other proofs that an infinite set of naturals must include infinite values, which were dismissed, but never refuted, or any flaw pointed out. I will have my web pages published before too long, so I am not getting into a mosh pit with you again right now. 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. > > >> Let's take the number representing the set of numbers dividable by > >> three. Is this number dividable by three? > > > > Why does it have to be? > > It does not have to be. But if it is a natural number, it either is > dividable by three, or it isn't. You claim that it is a natural > number. So what is it? Is it dividable by three, or isn't it? > > It must be one, mustn't it? The number is 100100...00100100. It's certainly even, and a multiple of 4. Is it divisible by 3? That can only be determined in a binary system with a finite number of digits, as far as I can tell. Infinite whole numbers aren't always as convenient as finite ones, but they still must exist for the set to be infinite, and they still can be used to represent infinite subsets of the naturals. If you divide this number by 3 (11) you find it is divisible or not, depending on whether you have an odd or even number of 100's in your infinite string. Of course, this question is not really answerable, so I don't have an answer for you. What do you think? What is aleph_0 mod 3? -- Smiles, Tony |