Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 19 Jul 2005 14:36 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? 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 may all pat each other on the back and dismiss "anti-Cantorians" as cranks, but that is only with a diligent lack of attention to every other area of mathematics and logic. -- Smiles, Tony
From: Stephen Montgomery-Smith on 19 Jul 2005 14:35 Alec McKenzie wrote: > "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. As best I can see from your other posts, you are making one of two points: 1. There is such an enumeration, because set thoery is inconsistent. Yes, I cannot be sure that cannot happen, but it would not invalidate the proof (because the theorem would be both true and not true simultaneously). If that is your problem, I think your issue is with proof by contradiction, not with Cantor's argument us such. 2. There really is a flaw in the proof, but mathematicians have somehow simply not seen it. While one cannot totally discount this possibility, the chances of this being the case is so extraordinarily small that for all practical purposes it is just not the case. We are talking probabilities like that of a Monkey sitting at a typewriter and dashing off a Shakespeare play. In principle, yes it can happen, in reality, you should worry more about UFO's abducting you. Stephen
From: Tony Orlow on 19 Jul 2005 14:39 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. > > It is perfectly reasonable to assert that no such flaw exists (given > the axioms used in the proof). Indeed, why would anyone entertain any > doubts when he can confirm the correctness of each and every step of > the proof? > > In all actuality, the flaws in various proofs and assumptions in set theory have been directly addressed, and ignored by the mainstream thinkers here. 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. Is this the same as the proof concerning the "uncountability" of the reals? -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 14:50 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. 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. Still, discussing these things with Cantorians is like trying to discuss evolution with an evangelical christian. -- Smiles, Tony
From: David Kastrup on 19 Jul 2005 14:59
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? > 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. Let's take the number representing the set of numbers dividable by three. Is this number dividable by three? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |