Obections to Cantor's Theory (Wikipedia article)
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From: Virgil on 20 Jul 2005 20:46 In article <MPG.1d483af164930cd4989f46(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1d4733ca67be65e3989f31(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > 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? > > > > The mapping from X to P(X) is not "natural", it is 'arbitrary', meaning > > that it can be anything and cannot be assumed to have any special > > properties. > > > > In particular, f(0) = f(1) = f(2) = X is possible, whenever {0,1,2} is > > a subset of X, . > > > I defined the mapping and it's as natural as it gets. Each succesive bit > represents membership of each successive element. It doesn't get any more > natural than that.n Every unique infinite string of bits represents a unique > subset of the naturals, so I don't know WHAT that last sentence means. > There is a 1-1 correspondence between infinite bit strings and > subsets. But there is no 1-1 correspondence between naturals and INFINITE bit strings (only with finite bit strings). This is just another instance of that same delusion that TO has that there exist naturals with more than finitely many naturals as predecessors.
From: Virgil on 20 Jul 2005 20:53 In article <MPG.1d483b6159f51ebe989f47(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > > 000...000.000...001 > > > > It is not a real number unless there is a finite number of zeroes > > specified for each ellipsis. and once that is done, it is not a smallest > > positive real, since that is a mythical beast. > > > True in ways. It is an infinitesimal unit, which can be considered as a > smallest real. AS it is, it is not a real number at all. If the number of zeros between the decimal point and the '1' is any a natural number then it can represent a real, but not a smallest real, as inserting another zero would give something smaller yet. If there were a smallest real, it would have to be a very large negative one, but there is no such thing in any form of real number system which as a smallest non-zero real either in regular value or in absolute value.
From: W. Dale Hall on 20 Jul 2005 20:58 Tony Orlow (aeo6) wrote: > The World Wide Wade said: > >>In article >><MPG.1d472484f809e37c989f2c(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> >> >>>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. >> >>Sort of like saying "if 0 = 1" is allowed. >> > > No, more like saying "if infinite digits are allowed", which is what is > required to have an infinite set of digital numbers using a finite base. That > was a useless comment. I shouldn't even be responding. This is untrue, with the possible exception of whether it's worthwhile to respond to these articles. It is definitely not necessary to have any infinitely long numbers to accommodate an infinite set of numbers. It is necessarily to have numbers of arbitrarily finite length, but no integer requires infinitely many digits. I'm sure I'm not the only one to suggest the following, but here goes: 1. Note that the successor function is a 1:1 function from the set of natural numbers to a proper subset. This establishes N as an infinite set, since it is in 1:1 correspondence with a proper subset. 2. Note that the induction axiom (one of the Peano axioms) states: If A is a subset of N that is closed under the successor function, and which contains 0, then A = N. The set of all finite natural numbers satisfies the requirements of the induction axiom. Therefore, the set of finite natural numbers is equal to the set of all natural numbers. In other words, every natural number is finite. 3. Note that the set of finite natural numbers is (without the induction axiom) already an infinite set by virtue of (1) above: the successor of every finite natural number is another finite natural number, and the set of all successor numbers is a proper subset. Whatever set you take to be the set of natural numbers, if it contains infinite natural numbers, is not the same set that mathematicians are referring to when they use the term "natural numbers". Perhaps you mean to be claiming that there is no model of Peano arithmetic? I'll also note that your "infinite series" argument is not germane. The issue is not whether one *can* formulate infinite series, but whether one *must* formulate these. Arithmetic by itself does not mandate such, nor does algebra. In fact, you nod to that point by raising the notion of convergence, a notion that is decidedly non-algebraic, but instead topological, in nature. The very act of defining convergence and of assigning numerical values to the formal sums that can be written involves an expansion of the notion of summation; if it were not so, then there would be no choice in the matter of assigning the value of sums. Instead, there is a great deal of choice in the matter, viz. p-adic numbers, being completions of the rationals wrt norms other than the usual one. In addition, note that the phenomenon of conditional convergence carries with it the ability to assign the value of a sum to any real value whatsoever by merely rearranging terms. Dale.
From: Virgil on 20 Jul 2005 21:18 In article <MPG.1d48436ba92140bf989f4b(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > > > > > David Kastrup wrote: > > > > > >> Oh good grief. Successor in interest to JSH, are we? > > > > > > JSH is not an anti-Cantorian. So this argument, again, doesn't make > > > sense and may only be useful for the purpose of insulting. > > > > JSH was the one who repeatedly threatened a day of reckoning when his > > opponents would be cast from the ranks of mathematicians. > > > > And that is exactly what Tony was doing here. I quote what you > > snipped: > > > > >>> I will have my web pages published before too long, so I am not > > >>> getting into a mosh pit with you again right now. Just be aware that > > >>> anti-Cantorians are sick of being called crackpots, and the day will > > >>> soon come when the crankiest Cantorians will eat their words, and > > >>> this rot will be extricated from mathematics. > > > > > I never claimed you would be cast from the ranks of mathematics, but > that you will see the errors that you are currently ignoring, and > that the rot of Cantorian cardinality will be removed from mainstream > thought and replaced with ideas that don't lead to absurdity like > Banach-Tarski. Banach-Tarski is not half as absurd as TO's "infinite naturals". > I do see the ramifications of this nonsense in many areas. TO cannot see his hands in front of his face in matters mathematical. > Until you > can demonstrate that the theory is really correct This has been done to the satisfaction of better minds that TO will ever have. That To is not satisfied merely means he cannot cross that pons asinorum. > I am well within my rights to disagree with your axioms and > conclusions, and if that right is challenged, I will continue to > defend it and challenge your theory. And we are equally within our rights to shoot you down whenever you produce delusions, like your infinite naturals for example. > It takes two to tango, and if you end up with people vowing > vengeance, well hell, you probably deserve it. If anyone deserves it, TO himself is high on the list. > Then again, maybe JSH > is mentally unstable, but then so was Cantor, and so was Godel. So that TO fits in nicely, except that Cantor and Goedel lacked TO's crippling quantifier dyslexia.
From: Martin Shobe on 20 Jul 2005 21:36
On Wed, 20 Jul 2005 10:57:58 -0400, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >Barb Knox said: >> In article <MPG.1d4726e11766660c989f2f(a)newsstand.cit.cornell.edu>, >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >> [snip] >> >> >Infinite whole numbers are required for an infinite set of whole numbers. >> >> Good grief -- shake the anti-Cantorian tree a little and out drops a >> Phillite. Here's a clue: ALL whole numbers are finite. Here's a >> (2nd-order) proof outline, using mathematical induction (which I >> assume/hope you accept): >> 0 is finite. >> If k is finite then k+1 is finite. >> Therefore all natural numbers are finite. >> >> >That's the standard inductive proof that is always used, and in fact, the ONLY >proof I have ever seen of this "fact". Is there any other? I have three proofs >that contradict this one. Do you have any others that support it? > >Inductive proof proves properties true for the entire set of naturals, right? Yep. >That entire set is infinite right? Yep. >Therfore, the number of times you are adding >1 and saying, "yep, still finite", is infinite, right? Yep. But be careful here, at *every* stage of this process, we have still only done it a finite number of times. > So, you have some way of >adding an infinite number of 1's and getting a finite result? Nope. You weren't careful. > You might want to >discuss this with your colleagues specializing in infinite series. There is a >very simple rules that says no infinite series can converge to a finite value >unless the terms of the series have a limit of zero as n approaches infinity. >Does this constant term, 1, have a limit of zero? Nope. > No it doesn't, and the >infinite series of constant 1's cannot converge, but diverges to infinity. Yep. > Can >you actually deny this? If so, then Poincare was right. BTW, there is a caveat on convergence. You have to assume the standard topology. In other topologies, that sequence can converge. Martin |