Cantor Confusion
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From: Han de Bruijn on 11 Dec 2006 10:29 stephen(a)nomail.com wrote: > Han de Bruijn <Han.deBruijn(a)dto.tudelft.nl> wrote: > >>stephen(a)nomail.com wrote: > >>>Nobody has said that time is a set. But time can be modelled using >>>sets. > >>Modelling time as implied with functions results in a vicious circle, >>as I have demonstrated in a previous poster. > > No you did not demonstrate any such thing. Welles, nietes, welles, nietes .. (= yes, no, yes, no .. ) >>>Apples are not sets either, nor are they numbers, but you >>>can reason about apples using sets and numbers. Your arguments are >>>becoming increasingly irrational. You really do not appear to have >>>any rational opposition to set theory, just emotional ones, >>>and your arguments boil down to Communist style propaganda slogans. > >>Well, you're quite close, after all. But I can also say that _your_ >>arguments boil down to Capitalist style propaganda slogans. Who is >>the hypocrite here? At least _I_ will not deny that my ideas _are_ >>influenced by the world outside. While I don't expect _you_ to admit >>that Set Theory is only "fundamental" because of its setting within >>the Capitalist Economic System. You think you are objective and free >>of emotion, but you are not. Your propaganda is as bad as mine. > > You were the one complaining about Set Theory restricting your freedom > within mathematics. You are now the one denying people people the > freedom to pursue mathematics you do not like. That is hypocrisy, > plain and simple. > > And one again, you demonstrate that you do not know what the word > "foundation" means. Set Theory is fundamental in the sense that > mathematics can be described using set theory. That is all anyone > is saying. You have turned this simple statement into some > political/spiritual/philosophical nonsense. You are just making > up positions to argue against, instead of actually responding to > what people say. Well, think I've said everything I had to say. > I suppose you think the fact that Turing machines are a foundation > of computing is just propaganda as well. I'm not fond of Turing machines, but I won't say such a thing. Han de Bruijn
From: William Hughes on 11 Dec 2006 10:37 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > Given that you find the words function, and bijection so > > distasteful, I have reworked my points to avoid using > > them in conection with potentially infinite sets. > > Please stop me at the first point you disagree with. > > > [...] > > > > -given two sets of natural numbers E and F where E is a > > potentially > > infinite set, and F has a largest element. there does > > not exist an equitransform between E and F > > > > Until here all is correct, as far as I see. > > > -the diagonal is the potentially infinite set of natural > > numbers. > > > > -every line L has a largest element > > > > -there is no equitransform between the diagonal and a line L > > And here we have the contradiction: No. Here we have a conclusion you do not like. You want to say -given two sets of natural numbers E and F where E is a potentially infinite set, and F has a largest element. there does not exist an equitransform between E and F -the diagonal is a potentially infinite set -a line L has a largest element. -there is an equitranform between the diagonal and L does This is a contradiciton. What you really want to say is: the result above conficts with the result of a completely different argument I have. Therefore, although I cannot see anything wrong with the argument above, I claim that it is wrong. This is the argument you present. > > 1) The set of lines is also the potentially infinite set of natural > numbers. > 2) Every element of the diagonal is in at least one line. > 3) Every initial segment of the diagonal is (in) a line. No, there is one initial segment of the diagonal that is not in a line. > 4) The limit (n -->oo, for the number n of elements) of the diagonal is > oo. Yes and it attains this limit (i.e the limit is a maximum as well as a supremum) > 5) The limit (m -->oo, for the number m of elements of a line) is less > than infinite. No. The limit is oo. (The limit cannot be any finite number) However, this limit is not attained (i.e. it is a supremum but it is not a maximum.) .. > 6) Therefore not every initial segment of the diagonal can be (in) a > line. > 7) Contradiction between (3) an (6). Since, both (3) and (5) are false, there is no contradiction. - William Hughes
From: mueckenh on 11 Dec 2006 10:53 Tony Orlow schrieb: > >> Repeating a falsehood does not make it any less false. > > > > Why then did you repeat always that here are no balls in the vase at > > time 0? Why do you repeat without reasonable arguments that there are > > more paths than edges in the binary tree, although no path springs off > > without its own egde? > > ========================= > > > > Again I agree. There are 9 n balls after n iterations, and there are > half as many paths as edges or nodes in the tree, despite supposed > bijection hat tricks. Banach-Tarski is ridiculous, and omega is a phantom. Correct. The function f(n) = 9n will never yield the value 0. > > >> Why he maintains this this somehow bijects the individual branches with > >> paths, is not clear. > > > > Because every branch consists of two edges, one for the each of the > > resulting paths. > > > > ========================= > > > > To put it another way, each right branch may be considered a > continuation of the same original infinite path, like adding a 0 to the > right of the decimal point - it doesn't change the value. The left > branch from every node produces a new path, creating a new value with > the concatenation of a 1 to the string. Therefore, for every two > branches, there is one additional path. Of course that is a possible way to construct a bijection. But set theorist will argue that by this constructuibn you get only those paths representing rational numbers. (That is even true, because there are no other numbers.) But they do not see that there are no paths without edge. They just believe that "in the infinite" their God will provide the missing paths. > > Is the diagonal longer than any line? Nope. By definition, it cannot exist, where no line is. > > Perhaps, with a countable number of bits. But, there are more than any > finite number of reals in the unit interval, and this is an infinite > number. Where *are* they? They canot be addressed. They cannot be imagined. They cannot be known. And, moreover, the proof proving their existence, is false. Quite a poor existence. Regards, WM
From: mueckenh on 11 Dec 2006 10:57 Virgil schrieb: > In article <1165759395.675493.173220(a)n67g2000cwd.googlegroups.com>, > Han.deBruijn(a)DTO.TUDelft.NL wrote: > > > David Marcus schreef: > > > > > Han.deBruijn(a)DTO.TUDelft.NL wrote: > > > > > > > > Ah, now don't act as if you didn't have those heated debates with some > > > > manifest opponents of set theory, i.e. Wolfgang Mueckenheim. > > > > > > If you've been reading the threads, then you should know that WM has so > > > far failed to present any mathematics at all. All he does is present > > > incorrect arguments that he insists follow from the standard axioms and > > > then proclaim: "Behold, standard mathematics is inconsistent." Big deal. > > > > Nonsense. The arguments presented by WM are quite reasonable. > > Not in [such areas of] mathematics where f(n) = 9n yields the limit 0. > And that is the only relevant arena. Not for people who know and love mathematics. Regards, WM
From: mueckenh on 11 Dec 2006 11:21
Virgil schrieb: > > > > > > > > Wrong. > > > > > > It is WM who is wrong > > > > Wrong. > > Yup, Wrong! Wrong!! > > > > Repeating a falsehood does not make it any less false. > > > > Why then did you repeat always that here are no balls in the vase at > > time 0? > > Because that is what the logic requires. That would be a poor funeral of logic. > > > Why he maintains this this somehow bijects the individual branches with > > > paths, is not clear. > > > > Because every branch consists of two edges, one for the each of the > > resulting paths. > > WM is confusing branches (edges) with branch points (nodes). This is a branch: 0 / \ and that is a branch: 1 / \ > > > > > ========================= > > > > > It is not the contents of a single line but the set of all lines that is > > > being quantified, so that the finiteness of lines is irrelevant. > > > > Wrong. We know from every line that it is finite. therefore it is > > uninteresting which line we consider. > > Then all lines must, in WM's eyes, be identical. That is your logic again? > > > > > You cannot imagine the integer [pi*10^10^100]. > > > > > Why not, you just did! > > > > No. I did not. > > If you say that you did not write "[pi*10^10^100]", then perhaps you > should lock your computer when not using it yourself, as someone did it > in your name. If you write sqrt(2) you have not imagined this (non existing) number. If you write or think [pi*10^10^100] you have not imagined this (non existing) number. That is only par of you arstrology falsely called logic. > > > > > Sqrt(2) has a directly approachable decimal numerical address. > > > > Yes. But there are only countably many addresses. > > But that does not identify which numbers have them. But it identifies the possible number of numbers. Regards, WM |