Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 19 Jul 2005 15:50 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. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Tony Orlow on 19 Jul 2005 15:54 Robert Low said: > 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? > I mean it has more elements. The singleton subsets are equivalent to the naturals themselves, and then there are many more subsets. A power set of a set with n elements has 2^n elements, which is always larger than n, for finite and infinite sets. The idea of uncountability as being equivalent to "larger than the set of naturals" is unfounded. There is no reason to believe that larger sets cannot be enumerated. the power set of the naturals can be enumerated and bijected with the naturals, as I described in another post, as long as infinite natural numbers are allowed. -- Smiles, Tony
From: Robert Low on 19 Jul 2005 15:57 Tony Orlow (aeo6) wrote: > Robert Low said: > >>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? >> > > I mean it has more elements. And what does it mean to say 'the power set of N has more elements than N', if it doesn't mean 'there is no surjection from N to its power set'? Which is, of course, exactly what 'the power set of N is uncountable' means... > the power set of the naturals can be enumerated and bijected > with the naturals, as I described in another post, as long as infinite natural > numbers are allowed. No, you can make a surjection from the set of (possibly infinite) natural numbers to the power set of the set of *finite* integers. If you also remember that you have to include those subsets including your infinite integers, you find that you still can't have a surjection from your set of (possibly infinite) integers to the power set of that set. You're trying to have your cake and eat it.
From: Tony Orlow on 19 Jul 2005 16:00 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? > > Chris Menzel > > I was asked that before, and never got around to fully analyzing the axioms for lack of time, 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, and the antidiagonal simply exists on the list, below the line of diagonal traversal. Cantorians seem to think infinity is simply infinity, even during the course of a proof that that is not the case. -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 16:02
Robert Low said: > Tony Orlow (aeo6) wrote: > > David Kastrup said: > > > I still do not get this. You have a set of naturals {0,1,2,3...}, and a set of > > binary numbers {0,1,10,11,100,101,110,111,....}. Surely there is a bijection > > between these two sets. > > Of course. But that isn't the issue. The issue is the existence > of a bijection between either of these sets and the power set > of the set of naturals. So, which binary number does your > procedure associate with the set of *all* natural numbers > divisible by 3? > Again, that infinite set is denoted by the infinite whole number 100100100...100100100. Notice, every third bit, from right to left, is a 1. Those are the multiples of 3. -- Smiles, Tony |