Why can no one in sci.math understand my simple point?
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From: Virgil on 18 Jun 2010 15:40 In article <88476fa9-bbc3-4ac1-9080-557f6a3b5f74(a)j8g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 18 Jun., 13:09, "Peter Webb" > > > The computable numbers are countable. > > > > And similarly Cantor's proof does not show that there are an uncountable > > number of Reals. > > What do you understand by "uncountable"? > > > It proves exactly what Cantor claimed it did, which is that > > you cannot list all Reals. > > Cantor said that there are 2^aleph_0 reals and aleph_0 rationals. And > he "proved" that 2^aleph_0 > aleph_0. And he said that there are an > uncountable number of reals because countable means listable. > > Regards, WM WM is right on this point!
From: Virgil on 18 Jun 2010 15:43 In article <4c1b5407$0$17174$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message > news:slrni1mcki.jrj.tim(a)soprano.little-possums.net... > > On 2010-06-18, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> Of course this number is computable; there > >> is a simple algorithm to compute it. > > > > I see you still haven't consulted a definition of "computable number". > > Umm, yes I have. > > > No worries, let me know when you have. > > > > > > - Tim > > There can be no list of all computable numbers. > > The proof is quite simple. Lets imagine that you have such a list of all > computable numbers. > > Lets say it starts off ... > > .111111... > .141592 ... > .71828 ... > > Take the 1st digit of the first number. If it is a "1", then make the first > digit of the diagonal number "2", otherwise make it a "1". Well, the first > digit of the first number is a "1", so the first digit of the diagonal > number is a "2". > > Now take the second digit of the second number and do the same substitution. > Its a "4", so the second digit of the diagonal number is "1". > > Now take the 3rd digit of the 3rd number ... its a "8", so the third digit > of the diagonal number is a "1". > > Continue in this fashion. > > The number that is produced is clearly "computable", because we have > computed it. Its also clearly not on the list. Therefore the list cannot > have contained all computable numbers. > > Exactly the same as Cantor's proof that the Reals cannot be listed. > > However, this does *not* mean that there are an uncountable number of them. > The computable numbers are countable. > > And similarly Cantor's proof does not show that there are an uncountable > number of Reals. It proves exactly what Cantor claimed it did, which is that > you cannot list all Reals. At least one definition of "countability" for infinite sets is "listability", i.e., existence of a surjection from N to the set in question. By what definition of countability is an infinite set which cannot be listed still regarded as countable?
From: Virgil on 18 Jun 2010 15:45 In article <b8f9104b-9daa-4892-9363-6ee43739c36a(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > No. An infinite sequence of digits does not represent a number. In > general it does not even converge. Given a finite set of n digits and interpreting the sequence as an n-ary proper fraction, it ALWAYS converges. So the USUAL interpretation of such an infinite sequence DOES converge ALWAYS.
From: Virgil on 18 Jun 2010 15:46 In article <87r5k41dwf.fsf(a)phiwumbda.org>, "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote: > WM <mueckenh(a)rz.fh-augsburg.de> writes: > > > On 18 Jun., 14:17, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> Newberry <newberr...(a)gmail.com> writes: > >> > On Jun 15, 9:46�am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >> > wrote: > >> >> WM says... > >> > >> >> >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> >> >> The proof does not make use of any property of infinite lists. > >> >> >> The proof establishes: (If r_n is the list of reals, and > >> >> >> d is the antidiagonal) > >> > >> >> >> forall n, d is not equal to r_n > >> > >> >> >As every n is finite, it belongs to a finite initial segment of the > >> >> >infinite list. > >> > >> >> I'm not sure what you are saying. The fact is, we can prove > >> >> that for every real r_n on the list, d is not equal to r_n. > >> >> That means that d is not on the list. > >> > >> > How do you know that it does not prove that an anti-diagonal does > >> > exist i.e. that an antidiagonal is a contradiction in terms? > >> > >> Because every infinite sequence of digits represents a real number? �And > >> the antidiagonal is one such sequence? > > > > No. An infinite sequence of digits does not represent a number. In > > general it does not even converge. In order to have convergence, you > > need the powers of 10 or 2 or so. But without a finite definition > > there are no infinite sequences at all, neither with nor without > > powers. > > Yes, very enlightening. What a unique and remarkable grasp of > mathematics. > > And what a shame that anyone allows you to teach. I concur!
From: Virgil on 18 Jun 2010 16:00
In article <86419e27-de46-4972-83e8-3c09d037f867(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Jun., 21:59, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > I never said a real number can be defined by an infinite sequence. > > All finite words belong to a countable set. If you exclude infinite > words (sequences) then there is no chance for uncountability. > > > > > A real number > > > can be defined only by a finite word. But there is no diagonalization > > > over finite words. > > > > Without even commenting on what you mean or whether it is true, it > > does not refute that the formalized argument is first order logic > > applied to axioms and incontrovertible > > Incontrovertible is religion. Because its adherents exclude > refutations from their perception. Something being "Incontrovertible" in FOL means you can't controvert it that while playing by the rules of FOL. AS far as I am aware, there are no definite set of rules of logic like FOL (first order logic) for religions. > > > You've not said what "wrong" assumption I've "started with". > > The possibility of an infinite sequence of infinite sequences that can > be completed in order to obtain a completed "anti-diagonal" sequence. That possibility is a consequence of an axiom set like FOL plus ZFC. And what is possible within such an axiom set in not constrained by whatever other axioms WM wishes to impose, nor even by WM's views on "reality". > > > All of this business of yours does not refute what is simply > > introvertible, that a formal proof exists in the manner I've > > mentioned. > > There may be a proof. But as the result is wrong the proof is not > worth much. Since it has not been proven wrong in FOL + ZFC, or whatever other system it was proved in, the proof stands. There is a form of pure mathematics which operates much like games, in that one sets rules and then plays within those rules. For this sort of mathematics, those like WM who insist on repeatedly breaking those rules are viewed as cheaters, and deserve to be. > > > > > The translation of these notions into your "incontrovertible" theory > > > is the weak point. > > > > NO, you did not listen to what I said. I did NOT say anything about an > > incontrovertible THEORY. Rather, I said it is incontrovertible that a > > certain finite sequence of finite sequences of symbols exists. > > But this finite sequence leads to the result that an uncountably > infinite sequence of infinite sequences exists. And that is wrong. WM's assertion of error conclusion is cheating. > > > > > > > Does ZFC not prove that all constructible numbers are countable? > > > > I don't know. What is the definition IN to set the rules for otherTHE LANGUAGE of ZFC of > > 'constructible number'? > > > > Anyway, I have no idea how you think that bears upon what I just > > wrote. > > So there seems to be a gap in ZFC. But it is easy to prove that in > fact there are only countably many constructible numbers. WM is free to set whatever rules he wants for his games, but is not free to override the rules that others have set for their games. And that form of "cheating" is exactly what he is forever trying to do. |