Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: MoeBlee on 6 Nov 2006 17:56 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > Here is my translation: It is allowed to understand the new number > > > omega as limit to which the (natural) numbers n grow, if by that we > > > understand nothing else than: omega shall be the first whole number > > > which follows upon all numbers n, i.e., which is to be called larger > > > than each of the numbers n. > > > > Okay. I don't see any problem with that. Would you please refresh the > > context by saying what point it is that you draw from that quote? > > Cantor's first book (general paper): Grundlagen einer allgemeinen > Mannigfaltigkeitslehre (Leipzig 1883). Foundations of a general set > theory. > § 11, showing how one is lead to the *definition* of the new > numbers... (Es soll nun gezeigt werden, wie man zu den Definitionen der > neuen Zahlen geführt wird und auf welche Weise sich die natürlichen > Abschnitte in der absolut-unendlichen realen ganzen Zahlenfolge, welche > ich Zahlenklassen nenne, ergeben.) I asked you what point you draw from the quote that you translated. MoeBlee
From: Virgil on 6 Nov 2006 18:02 In article <1162845651.473904.203110(a)h54g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > > Why then do you not understand how an edge of the binary tree can be > > > divided? > > > > Because you don't state clearly how you wish to divide the edges and how > > the division is used in the "relation" between edges and paths that you > > are defining. > > It is completely irrelevant how an edge is divided. It is clearly relevant how the correspondence between edges and paths is to be constructed. One can "name" each edge by the sequence of left-right branchings from the root node to the terminal edge of that edge. One can name each path by the infinite sequence of left right branching from the root node. Thus each edge bijects with a unique finite binary string of {L,R} branchings and each path with a unique infinite binary string of {L,R} branchings. Until WM can show a surjection from the set of finite strings, E for edges, to the set of infinite strings, P for pathes, he has not produced a satisfactory argument for there being as many edges, much less for there to be more. > If you divide a > cake of five pounds into 10 equal pieces then you can calculate the > weight of each piece without knowing the details of the division. Irrelevant. > > > If you knew what it means to divide a number, then I could explain > "whole number" as the opposite, namely an undivided number 1, like a > whole edge is an undivided edge. But, alas, I am afraid you cannot > understand. Certainly we cannot understand what WM is incapable of explaining, neither the content not the reason for WM's reluctance, unless that reason is WM's realization that he has no useful definition t provide. > > > > Please don't conclude from your own ignorance on mine. > > > > Wouldn't dream of it. > > I am sure. But I know what a whole number is. Apparently it is a secret.
From: MoeBlee on 6 Nov 2006 18:04 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > > > > > > For ordinals, > > > > > > > > x<y <-> xey > > > > > > > > where 'e' is the epsilon membership symbol. > > > > > > That is identical with my definition. Take for instance Zermelo's > > > definition of the naturals or Cantor's own (Collected works, p. > > > 289-290), then you can see it. Your definition has been simply > > > translated from Cantor's. Please don't conclude from your own ignorance > > > on mine. > > > > You wrote: > > > > "My question is : Do you maintain omega > n for all n e N? I know that > > modern set theory says so. If something can be larger than a number, > > then it must be a number." > > > > Given that "n < omega" stands for "n e omega", and given that N is > > omega, what point is there in asking anyone whether they maintain that > > n < omega for every n e N? You're asking someone whether he or she > > maintains that n e omega for every n e omega. What is the point of > > asking that? > > > If omega > n then we cannot have a diagonal with omega digits in a > matrix the lines of which have only n (= less than omega) digits, > because the digits of the diagonal are digits of the lines. That's just another reenactment of your dogmatic pronouncements. I gave you a proof in set theory, from axioms using only first order logic. You give an argument that does not use any recognizable logic and from no stated axioms. Please just show any step of my proof that is not justified by first order logic applied to the axioms, or show a proof of a sentence P and ~P in the language of set theory such that both sentences are theorems of set theory. > > > > > If it cannot be a fraction because ZF does > > > > > not yet know how to divide elements, > > > > > > > > In ZF we define various operations of division. As far as I know, there > > > > is not a dvision operation for sets in general. > > > > By the way, I do see that there are definitions of subtraction, > > division, and logarithms for ordinals. But they are not anything like > > your non-definition of ordinal division. > > > > > Why then do you not understand how an edge of the binary tree can be > > > divided? > > > > Haven't I asked you a few times already not to effectively put words in > > mouth? I never said I don't understand how an edge can be divided. I > > never said anything about "divided edges". > > So do you understand it now, how an edge can be divided in two pieces > and each one can be related to a path? So do you understand now, how > half an edge can be divided in two pieces and each one can be related > to a path? So do you understand now, how a fraction of an edge can be > divided in two pieces and each one can be related to a path? If you want to define a particular relation on a particular object that is proven to exist in set theory, then I'm all ears. > > You are ignorant of the basics of set theory. > > I know its basics. That does not mean that I have to accept them. No, you don't know basic set theory. You don't even have a SENSE (let alone a precise understanding) of mathematical definition. MoeBlee
From: Virgil on 6 Nov 2006 18:05 In article <1162845739.878615.214650(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > In article <1162472271.664965.315880(a)m7g2000cwm.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > Does WM choose to claim that a set of naturals which has no maximum > > > > > > must > > > > > still have a finite upper bound? > > > > > > 1 > > > 11 > > > 111 > > > ... > > > > > > The length of each column is omega. > > > > I assume that we are using ZFC as our logical system. > > > > Assuming by the "length of each column" that you mean the number of 1's > > in each column, then the number of 1's in each column has cardinality > > aleph_0. So, what you wrote is essentially correct. > > > > > The length of each line is less than omega. > > > > Each line has a finite number of 1's. Finite cardinality is less than > > aleph_0. So, what you wrote is essentially correct. > > > > > The length o the diagonal is less than omega. > > > > The number of 1's in the diagonal is aleph_0. So, what you wrote is > > false. In fact, I have no clue what you could possibly be thinking that > > would lead you to make such an obviously incorrect statement. The length > > of the diagonal is clearly the same as the length of the first column, > > and you just wrote above that the length of each column is omega. > > The length of the diagonal is clearly not more than the length of any > line. If every line is finite but at least of length equal to is line number, then even more clearly, the diagonal is longer that every line. So that WM could hardly be more wrong if he tried. > Nevertheless a mathematician would understand where maximum and > supremum occur in the above reasoning. Mathematicians, even amateur mathamaticians, do, but WM does not.
From: Virgil on 6 Nov 2006 18:13
In article <1162845887.720670.222930(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > Whatever your point, you won't be able to show that merely dropping the > > > axiom of rabbithood from the Z axioms entails that there are only > > > non-rabbit > > > sets. > > > > Which axiom of ZF or NBG does WM connect with rabbits? > > Which axiom of ZF - INF do you connect with infinity? At least "infinity" is cogent, but rabbits? Perhaps WM would be happier as a veterinarian. |