Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: William Hughes on 23 Nov 2006 12:51 Albrecht wrote: > > > > P.S. Any progress on the definition of complete > > I work on it. In the meantime I use yours. Unfortunately you need a definition of "complete" that does not need a context. Your repeated claim is that infinite sets cannot be complete. However, you never give a context. The definitions of complete I use all require a context. A set is said to be complete with respect to something. None of my definitions will work for you. - William Hughes
From: William Hughes on 23 Nov 2006 12:55 Eckard Blumschein wrote: > > Before I may decide I would like to clarify the meaning of the term set. > Cantor's definition of a set has been confessed invalid at least since > 1923 with no possiblity of correction in case of infinite sets. Why? > > Fraenkel confessed paradoxa due to this definition. The reason behind > them is ambiguity. Cantor's definition claims to provide each singe > element of the set as well as simultaneously the infinite set as an > entity. So it is a chimera. > Why? What is there about providing the infinite set as an entitiy which precludes providing each single element of the set? - William Hughes
From: mueckenh on 23 Nov 2006 13:50 Virgil schrieb: > > > I will accept that it can be connected if the "right" uncountable set is > > > removed, but I can think of uncountable sets of points whose removal at > > > least appears to totally disconnect the set of those remaining. > > > > Of course, for example if you remove all points. > > The remaining empty set is trivially connected. That is nonsense. If there is nothing to be connected then there is nothing connected! > > > > > > For example if one removes all the points in the Cartesian plane which > > > have either coordinate non-integral, what is left is about as > > > disconnected as one can imagine. > > > > Cantor wanted to suggest that the removal of a countable set leaves the > > plane connected while the removal of an uncountable set does not. > > Those who try to read Cantors mind often deceive themselves. I read his words. > > As an > > example he chose the algebraic numbers (obviously in order to > > distinguish them from the transcendental numbers). > > > > I proved that there is by far a simpler proof for the algebraic > > numbers. > > I proved that this proof can also be applied to the transcendental > > numbers. > > See the appendix of http://arxiv.org/pdf/math.GM/0306200 > > I have seen it. No serious mathematician would dare present such a > sloppy paper for publication. Nor would any well-referreed mathematical > journal publish it. > The alleged proofs, at least as far as I bothered > to read, are fatally flawed, In fact, this was the opinion of the mathematicians at Cantor's time already. Therefore he had problems to get his papers published. Therefore I wrote my paper. > and do not establish their claimed results. With other words, you did not understand them. But our question concerned the appendix only. The proof given there is very simple, so simple that even Mr. Bader asserted to have understood it (it seemed so to him, at least). You cannot confirm that my proof is correct? Regards, WM PS: You really did not see that Cantor's arguing in his first proof is valid for rational numbers as well as for transfinite numbers? It is the same situation as in the appendix.
From: mueckenh on 23 Nov 2006 13:56 Virgil schrieb: > In article <1164198530.509417.22070(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1164143519.034139.154960(a)h54g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Randy Poe schrieb: > > > > > > > > > > > > > What about all contructible numbers or all words? They cannot be > > > > > > mapped > > > > > > on N though they are a countable set. > > > > > > > > > > The set of all finite words, the set of polynomials, or any > > > > > countable set can be put in bijection with N. Where do you > > > > > get your view that it "can't be mapped to N"? > > > > > > > > The list of all finite words cannot be constructed. > > > > > > It can be constructed inductively, which means that in ZF or NBG, it can > > > be constructed. > > > > The natural numbers cannot be constructed inductively. Induction proofs > > do not cover N, I was told. > > Then you were told wrong. Induction yields infinitely many numbers? So you can do by induction infinitely many steps? So, if you would count these steps, you would get an infinite number? But if you would count the differences 1, you would not get an infinite number? > > > > The list of all > > > > constructible numbers cannot be constructed. > > > > > > > > > The list of all naturals can be constructed inductively, which means > > > that in ZF or NBG, it can be constructed. > > > > It can be proven inductively that all initial segments of the set of > > natural numbers are finite. So, if the induction proof concerns all n > > in N, then N is finite. > > WM claims that a quality of members of a set must be inherited by the > set itself. All that WM can prove is that every MEMBER of N is finite, > which says nothing about N itself. I can prove that every step generating the next member n+1 from the member n belongs to a finite set of steps. The set of steps is finite and remains so for all n. > Those whose "feelings" tell them that they are Napolean Buonaparte have > feelings as relevant to the structure of mathematics as EB's. You prefer to be Julius Caesar? Regards, WM
From: mueckenh on 23 Nov 2006 13:59
Virgil schrieb: > In article <1164197100.642614.53260(a)e3g2000cwe.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > The countability of the set W of all the words of a finite alphabet is > > > > proved by other means than the usual countability proof. The latter > > > > consists in constructing a bijection with N, i.e., in constructing a > > > > list. But it is impossible to construct a list of all words. It is > > > > impossible to construct a bijection between W and N. > > > > > > WRONG. It is a theorem that from the set of finite sequences of members > > > of a finite alphabet there is an injection into N. > > > > Nevertheless it is impossible to construct such an injection (as it s > > impossible to construct a well order of the reals). > > Actually, it is quite trivial, Give each character in the alphabet of n > characters a different integer value between 0 and n-1 inclusive, and > represent the members of N in base n, and the bijection associates each > word with the numeral, base n. > By induction, at least in ZF or NBG or any system allowing induction, > the construction is complete! Induction yields infinitely many numbers? So you can do by induction infinitely many steps? So, if you would count these steps, you would get a larger number than any finite number? As far as I know, induction shows that you always have done a finite number of steps. By induction one can even prove that always a finite number of words has been mapped on the natural numbers. Now, after you have "completed" this finite list of words, take it and construct a diagonal number. This is always a finite number, as the list is finite. Fine. So we have proved that there is a finite word which was missing in the "complete" list. But you think, the proof that we always have mapped a finite set of words, shows in fact, that we have mapped an infinite set of words? I know. I know what you think. Stay in your world. Regards, WM |