Cantor Confusion
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From: mueckenh on 28 Oct 2006 17:06 Virgil schrieb: > > > But for infinite ones, you must do one > > > more transfinite step. > > > > No. > > Yes! In order to proceed from 1^/2^n to zero? In order to proceed in f(n+1) = 1*E + f(n)/2 to infinity? How could Cauchy calculate sqrt(2) by such a recursive procedure without knowing transfinite induction? > > Any real number has only finite digit positions, according to Dik > > And is Dik your "Authority"? No, but his answer shows that your party contradicts your own opinion. > > Do we need transfinite induction for all the > > reals? > > Not to establish their existence. But to establish the limit n --> oo? Representations of irrationals do not exist without that limit. Regards, WM
From: David Marcus on 28 Oct 2006 17:52 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > And once again WM deliberately confuses, "all positions are > > finite", with "there are a finite number of positions". > > There is no confusing! Every finite position belongs to a finite > segment of positions (indexes). If you don't believe that and assert > the contrary, then try to find a finite position which dos not belong > to a finite segment (of indexes). > > It is simply purest nonsense, to believe that "all positions are > finite" if "there are an infinite number of positions". Before any of us wastes more time on this, please pick one: 1. You are making a statement that you say is provable within some standard mathematical system, e.g., ZFC. 2. You are making a statement that is true within your own system. -- David Marcus
From: William Hughes on 28 Oct 2006 17:58 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > And once again WM deliberately confuses, "all positions are > > finite", with "there are a finite number of positions". > > There is no confusing! Every finite position belongs to a finite > segment of positions (indexes). Correct. But every position belongs fo a finite set of positions, does not mean there are only a finite set of positions. The first means: for every position N, there exists a finite set of positions A(N) such that N belongs to A(N) The second means: there exists a finite set of positions B, such that for every position N, N belongs to B You cannot simply exchange the quantifiers. >If you don't believe that and assert > the contrary, then try to find a finite position which dos not belong > to a finite segment (of indexes). This cannot be done. So what. The assertion that "every finite position belongs to a finite segment" implies "there are only a finite number of positions" is simple quantifier dyslexia. > > It is simply purest nonsense, to believe that "all positions are > finite" if "there are an infinite number of positions". > Why. Why does "there are an infinite number of positions" imply that "at least one position is infinite"? - William Hughes
From: William Hughes on 28 Oct 2006 18:16 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > > > > Within the *real* numbers the limit does exist. And a decimal number is > > > > nothing more nor less than a representative of an equivalence classes. > > > > > > So we are agian at this point: The real numbers do exist. > > > > Not according to you (at least not by the usual definition). > > According to > > your reasoning, there can only be a finite number of real numbers ever > > defined. This makes limits kind of weird, since you cannot get > > arbitrarily close. > > You cannot get arbitrarily close either. Can you calculate sqrt(2) to > 10^100 digits? No. But we can both get as close as necessary to obtain > correct results. > > > > > > > > > > > > For the real > > > numbers we have LIM 10^(-n) = 0. Therefore, i ther limit n--> oo tere > > > is no application of Cantor's argument. > > > > > > > > > Given that there are only a finite number of integers, what > > do you mean by n--> oo ? > > > > In these discussions I understand the usual by n-->oo, because I do not > believe that my private opinions are of great interest for the > majority. But if you are interested: I understand by, e.g., n --> oo > 1/n a function the argument of which gets as large as necessary in > order to make 1/n as small as desired, such that we cannot distinguish > it from 0 in any calculation. This is possible because we have the > tools to construct natural numbers as large as we want. Piffle. If there is a limit to how many numbers we can define there is also a limit to how large a number we can define. - William Hughes
From: Virgil on 28 Oct 2006 18:26
In article <1162067841.239007.74250(a)e64g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Please name the mathematicians that agree that your argument is correct. > > (Han doesn't count, since he says he is a physicist.) > > Every mathematician whose mind has not yet suffered the drill of set > theory you probably were exposed to knows that my argument is correct. "Mueckenheim" cannot speak for all those people. There are, no doubt, some of them that will come to understand how wrong "Mueckenheim" is even without the benefits of formal education. |