Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 19 Jul 2005 16:59 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > 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. > > "Larger" is a synonymon for "can't be surjected onto from" in set > theory. And "uncountable" is a synonymon for "larger than the set of > naturals". It is not a leap or an assumption, but simply a > definition. > > >> > 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, > > Quite so. But there is a requirement that subset number x _either_ > include x as a member _or_ not include x as a member. Only one of > those two statements can be true. And then Q _either_ not includes x > as a member _or_ does include it, respectively. > > You are free to choose your mapping as you want to. But once you have > chosen your mapping, each subset number x _either_ includes x as a > member _or_ it doesn't. Whether it does, can be chosen independently > for every x. But once you are through, for every particular x, x will > be in f(x), or it won't. And depending on that, x won't be in Q, or > it will. Actually, as I think about it, given this natural mapping of the naturals to the subsets of naturals, subset number x will ONLY include x as a member for subsets 0, 1 and 2. Beyond that, subset x will NEVER contain x. So, you have non-empty Q for the null set, and the singletons {1} and {2}. So, what does that prove? > > -- Smiles, Tony
From: david petry on 19 Jul 2005 17:02 Jesse F. Hughes wrote: > This isn't about > "anti-Cantorians", whatever the hell that might mean. This is about > Petrians, a well-defined group in which there is no dissension at all > (since there is only one member). Uh, not quite. For example, in this discussion, Han deBruijn seems to understand and agree with almost everything I say.
From: Robert Low on 19 Jul 2005 17:05 Stephen J. Herschkorn wrote: > 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? That depends on what it says about those groups of people :-)
From: Chris Menzel on 19 Jul 2005 17:02 On Tue, 19 Jul 2005 16:00:53 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > Chris Menzel said: >> 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? > > I was asked that before, and never got around to fully analyzing the > axioms for lack of time, I suggest you do so, though your reluctance is understandable, as the realization that all of the axioms used in the proof of the theorem are intuitively true would of itself constitute a proof that your arguments against diagonalization are flawed. Look at the positive side though -- this may well be the cure for what ails you. > but the diagonal proof suffers from the fatal flaw of assuming that > the diaginal traversal actually covers all the numbers in the list. > Any complete list of digital numbers of a given length, even a given > infinite length, is exponentially longer in members than wide in terms > of the digits in each member. Therefore, the diagonal traversal only > shows that the anti- diagonal does not exist in the first aleph_0 > terms. Of course, the entire list is presumed to be aleph_1 long, > being a list of the reals, To say that a list of the reals is aleph_1 in length assumes the continuum hypothesis; is that what you intend? (And if so, why?) Moreover, granted, aleph_1 is omega_1 in pure set theory, but one should use ordinal numbers rather than cardinals when talking about such things as the length of a list, as there are many lists of different ordinal lengths that are the same size. Pretty clearly, you aren't terribly well-educated in set theory. Don't you think you should understand a field before you try to point out its flaws? And don't you see that it makes you look rather silly? Would you consider attacking superstring theory without understanding basic quantum mechanics? > and the antidiagonal simply exists on the list, below the line of > diagonal traversal. Aside from your tenuous grasp of basic transfinite arithmetic, the critical terms in your argument -- "digital number", "length", "exponentially longer" (cardinal or ordinal exponentiation?), "width", "diagonal traversal", "below the line of diagonal traversal" -- are much too vague for your argument to be evaluated. For all we know, you might have some genuine insights. But currently, your argument is smoke and mirrors; it hasn't been expressed as mathematics. > Cantorians seem to think infinity is simply infinity, Hard to fault them on that score. The fact that you have doubts about whether infinity is infinity (what does "simply" add here?) says quite a lot -- though *perhaps* if you were to try to distinguish different senses of "infinite" with any sort of mathematical precision there might be a point there. Hit the books, Junior; you've got some work to do. Chris Menzel
From: david petry on 19 Jul 2005 17:05
> There is no mention of one historical or living figure who is > anti-Cantorian, what their objections were Hmm, not quite. I did mention Kronecker. Did you miss that? |