Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 19 Jul 2005 15:06 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. > 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? > Is this the same as the proof concerning the "uncountability" of the > reals? It's pretty similar. 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. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Robert Low on 19 Jul 2005 15:11 Tony Orlow (aeo6) wrote: > 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? So, you accept that the power set of the naturals is bigger than the set of naturals, but also think that the power set of the naturals can perhaps be enumerated. What do you mean by 'bigger than' in this context?
From: Chan-Ho Suh on 19 Jul 2005 15:11 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. People have espoused the view that "it does not get better than that". No one is denying that humans are fallible. Your memory of your entire life is distorted and possibly even a delusion. You cannot know for certain. If that was your point, it's such an obvious and silly point as to not bear making a big deal out of. And it's certainly irrelevant to bring it into this discussion. > 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. 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. And there certainly have been no results commonly found in modern undergrad textbooks, which have been verified by virtually every research mathematician and found to have a flaw only years later. It's rather irrelevant to this discussion to bring up any kind of proof that only a dozen people have verified. That's a totally different situation.
From: Chan-Ho Suh on 19 Jul 2005 15:11 In article <mckenzie-60C274.14595219072005(a)news.aaisp.net.uk>, Alec McKenzie <mckenzie(a)despammed.com> wrote: > 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. > Not really. But if you are trying to assert that this is a complicated proof, significantly more complicated than a proof of say, the irrationality of two, then you are mistaken. > > And that's what you call "with no justification that I can see". > > No, it is not -- what I called "with no justification that I can > see" was something else: > > It was the assertion that no flaw having been found in a proof > leads to a certainty that such a flaw cannot exist. I still see > no justification for that. Nobody asserted this. People have asserted that a particular given proof has no flaw. Of course, nothing is absolutely certain. Just as you cannot be absolutely certain you exist or that your memory of what you ate for lunch is correct. But if your point is the rather trivial point that nothing is absolutely certain, then there was no need to make it.
From: David Kastrup on 19 Jul 2005 15:11
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. _Depending_ on that, the constructed subset will either _not_ or _do_ contain the number, respectively. This constructed subset then does not correspond to _any_ natural number. > if this rests on the diagonal proof, Rather the other way round. It is more basic than the diagonal proof. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |