Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 19 Jul 2005 15:34 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. > > > The only objection to this bijection between the natural numbers and > > the subsets of the natural numbers is the nonsensical insistence > > that every natural number in the infinite set is finite, which is > > mathematically impossible, given the fact that each additional > > element requires a constant incrementation of the entire range of > > values in the set. > > You are babbling. Anyway, let's assume just for kicks that infinite > numbers are part of the natural numbers, and lets take your numbering > scheme. I am not babbling, and that kind of insult is what leads to all this nasty nonsense. Please try to be civil. I notice that the Cantorians routinely accuse any opponent of babling, spewing nonsense being a crackpot or crank, or other personal insults, rather than replying with any form of logic. It's a convenient way of changing the subject. > > 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? There is no requirement that the number that represents a subset of numbers be a member of that subset of numbers, any more than it is a requirement that the natural number corresponding to a given rational number be either the numerator or denominator of that rational number, in Cantor's diagonal proof of the countability of the rationals. Does the rational number that is 10th in Cantor's enumeration need to have a 10 in it? No. That argument is so pointless, it can't possibly be accepted as any proof of anything. Is it really that bad? Have we sunk so low, that we prove things based on ideas that don't even have a semblance of sense? I concur with Poincare and Bouwer. > > -- Smiles, Tony
From: Chris Menzel on 19 Jul 2005 15:24 On Tue, 19 Jul 2005 14:50:33 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > ... > Is the above your 7-line proof? it makes no sense. There is no reason > to expect the natural number corresponding to the subset to be a > member of that subset. if this rests on the diagonal proof, there is > a very clear flaw in that proof which you folks simply dismiss as > irrelvant, but which is fatal. 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
From: David Kastrup on 19 Jul 2005 15:36 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. >> 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? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 19 Jul 2005 15:40 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > Jesse F. Hughes said: > >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> > >> > "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. > >> > >> Huh? > >> > >> The proof of Cantor's theorem is easily formalized. It's remarkably > >> short and simple and every step can be verified as correct. > > > > In all actuality, the flaws in various proofs and assumptions in set > > theory have been directly addressed, and ignored by the mainstream > > thinkers here. > > Guffaw. Again with the insulting remarks. Grow up. > > > 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. > > > 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, despite the fact that x is a member of an infinite number of subsets. Subset number 3, represented as ...00011, includes only the first and second elements. What does that prove? That 3 is not a element of subset 4 (...000100), which consists ONLY of 3? I can't even imagine why this is thought to prove anything. -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 15:47
David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> > >> > "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. > >> > >> Uh, what? There is nothing fuzzy about the proof. > >> > >> Suppose that a mapping of naturals to the subsets of naturals exists. > >> Then consider the set of all naturals that are not member of the > >> subset which they map to. > >> > >> The membership of each natural can be clearly established from the > >> mapping, and it is clearly different from the membership of the > >> mapping indicated by the natural. So the assumption of a complete > >> mapping was invalid. > >> > >> > Yet that is just what pro-Cantorians appear to be asserting, with no > >> > justification that I can see. > >> > >> Uh, where is there any room for doubt? What more justification do you > >> need apart from a clear 7-line proof? It simply does not get better > >> than that. > > > Is the above your 7-line proof? it makes no sense. > > If you don't get it. > > > There is no reason to expect the natural number corresponding to the > > subset to be a member of that subset. > > There is no such expectation. The only expectation is that _every_ > natural number is _either_ a member of its corresponding subset, or > not. Okay. > _Depending_ on that, the constructed subset will either _not_ or > _do_ contain the number, respectively. Redundant, but okay. > This constructed subset then > does not correspond to _any_ natural number. This is where it goes wrong. The consructed subset corresponds to the natural number denoted by the binary string constructed from the right to the left, where each successive bit is a 1 if the successive natural is a member, and 0 if it is not. Any subset can be denoted as a string of bits, and any string of bits can denote a natural number. The only reason to reject this bijection is if one clings to the idea that all natural numbers are finite, which is impossible. > > > if this rests on the diagonal proof, > > Rather the other way round. It is more basic than the diagonal proof. It is so basic, I cannot even quite see what erroneous assumption you are making. It seems like you are assuming subset number X must contain X as a member. If so, how do you justify this assumption? > > -- Smiles, Tony |