Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 19 Jul 2005 09:27 Alec McKenzie <mckenzie(a)despammed.com> writes: > David Kastrup <dak(a)gnu.org> wrote: >> Alec McKenzie <mckenzie(a)despammed.com> writes: >> >> > 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. > > I am not suggesting there is anything 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. > > 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. Well, it is now about a hundred years later, and millions of mathematicians have checked those 7 lines. It is not like they are difficult to comprehend or something. > Otherwise one is proclaiming infallibility. > > 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. And that's what you call "with no justification that I can see". I mean, look up "justification" in a dictionary of your choice. It would be hard to find anything _more_ justified. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David C. Ullrich on 19 Jul 2005 09:51 On Tue, 19 Jul 2005 13:43:55 +0100, Alec McKenzie <mckenzie(a)despammed.com> wrote: > David C. Ullrich <ullrich(a)math.okstate.edu> wrote: > >> On Tue, 19 Jul 2005 09:34:45 +0100, Alec McKenzie >> <mckenzie(a)despammed.com> 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. >> >> I once had a person tell me the following, with a straight face: >> >> (*) "You can't say for sure there's no such thing as a square >> circle! I mean just because they haven't found one yet doesn't >> mean they won't discover one tomorrow." >> >> Please choose one of the following replies: >> >> (i) No, (*) is nonsense. If it's square then _by definition_ >> it's not a circle. So they will _never_ find a square circle. >> >> (ii) Hmm, good point. >> >> You really should choose one of (i) or (ii), so people know >> how to reply to your post. The point: If you say (ii) then >> we know that there's no point worrying about anything you >> say. Otoh if you say (i) then there's hope - you agree that >> we're _certain_ they will never find a square circle, now >> we just have to convince you that our assertions about >> enumerations of subsets of N are just as certain, for >> entirely similar (although slightly more complicated) >> reasons. >> >> So which is it, (i) or (ii? > >It is (i), of course. Fabulous. >But you seem to be suggesting that the >proof in question is flawless for similar reasons to its being >so _by definition_. That I cannot see. Ok, here's another question. Suppose that we want to prove that A implies B. Suppose that we have an completely flawless proof that A implies C, and a completely flawless proof that C implies B. Does the union of those two proofs constitute a flawless proof that A implies B? I imagine you'll say yes to that as well. But the proof of the theorem in question really does involve nothing more than statements which are true by definition and statements which follow from previous statements by "if A implies C and C implies B then A implies B" arguments. ************************ David C. Ullrich
From: Alec McKenzie on 19 Jul 2005 09:59 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. > 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. -- Alec McKenzie mckenzie(a)despammed.com
From: Ross A. Finlayson on 19 Jul 2005 10:01 Infinite sets are equivalent. In one sense that's constructively illustrable via induction, with well-ordering and transfer. That's basically a three line proof that infinite sets are equivalent. Another notion is that it leads to the conclusion of the dual representation of an ur-element as the null element and the universal element. In a von Neumann universe, or rather ubiquitous ordinals, the powerset is successor is order type. In the bijection f(x)=x+1, S(N)={}, S({})=N, and the direct sum of infinitely many copies of N is the empty set. With regards to the naturals and a segment of the reals, nested intervals reinforces the notion that a well-ordering of the reals shows the structure of the reals to be a contiguous sequence of points. It's "Post-Cantorian." Particularly with regards to bijections between the naturals and reals, there are a variety of useful analytical results that do exist. Consider Vitali's reasoning why there exist unmeasurable sets, and why the existence of variously an infinitesimal or double infinitesimal, or "continuous" and "discrete" infinitesimal, on a one-dimensional line, leads to a ready explanation why it is not so. This way of reasoning leads to perhaps more applicable results than the transfinite cardinals, which in general have no utility. Skolemize, your model is countable. Via induction, the order type of all ordinals would be an ordinal. Quantify. V = L anyways. Appealing to constructibility, or finitudinous, finity, does not avoid variously these consequences, with infinity, and the infinite universe is the infinite constructible universe. Obviously I suggest the null axiom theory. ZF is inconsistent. Ross -- "Also, consider this: the unit impulse function times one less twice the unit step function times plus/minus one is the mother of all wavelets."
From: Jesse F. Hughes on 19 Jul 2005 09:55
David C. Ullrich <ullrich(a)math.okstate.edu> writes: > Ok, here's another question. Suppose that we want to > prove that A implies B. Suppose that we have an > completely flawless proof that A implies C, and > a completely flawless proof that C implies B. > Does the union of those two proofs constitute > a flawless proof that A implies B? Am I the only one suffering from flashbacks of Achilles and the Tortoise? Fortunately published back when copyrights really did expire. <http://www.ditext.com/carroll/tortoise.html> -- "I am the barbarian at the gates, raw creative force, willpower, and the will to fight for the truth no matter what, no matter who stands against me, no matter how many of you band [...] together in your weakness to fight against the math." -- James S. Harris |