Cantor Confusion
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From: mueckenh on 9 Jan 2007 08:02 Virgil schrieb: > In article <1168271417.069985.286650(a)51g2000cwl.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > That L_D does not exist. > > > > > > Assume a line that contains any element that can > > > be shown to be in the diagonal exists. > > > > Otherwise the diagonal could not exist, because it is, by definition, > > the union of line ends in the EIT. > > > > > Call > > > this line L_D. > > > > > > A: L_D is a line, therefore L_D has a largest > > > element. > > > > > > B: L_D contains any element that can be shown to exist > > > in the diagonal, therefore L_D does not have a largest > > > element. > > > > > > Contradiction. Therefore L_D does not exist. > > > > Therefore the union of line ends including the end of L_D does not > > exist. > > It is any L_D which can not exist since it would be required to have > both a last member and also the successsor to its last member. > > > Fine. There are no infinite sets. There is only potential > > infinity. Every line including the diagonal > > The diagonal, by not having a last member, is not a line, all of which, > by definition, must have last members. But the diagonal consists of last members. If there was no line for each member of the diagonal, which included this and every smaller member, then the diagonal would not exist. Therefore, the diagonal contains only such elements with indexes {1,2,3,...n} which are simultaneously in a line - in one line. All together! Regards, WM Regards, WM
From: mueckenh on 9 Jan 2007 08:13 Virgil schrieb: > In article <1168290342.751050.317770(a)51g2000cwl.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > No one, except possibly WM, is claiming that an infinite binary tree is > > > > > the same as the union of all binary rational, and therefore finite, > > > > > trees. > > > > > > > > > What else could it be? > > > > > > It could be, and is, the union of infinitely many finite trees. > > > Since the union of two trees is not a tree, whyever should the union of > > > more than two trees be a tree? > > > > The union of two trees is a tree. > > Then it will have nodes which are simultaneoulsy terminal nodes and not > terminal nodes. One can embed one tree into a larger tree, but that > operation is not union. > > > >That is the same as the union of two > > initial segments of natural numbers which is an initial segment of > > natural numbers. > > > > The union of the one-level tree > > > > 0. > > / \ > > 0 1 > > > > and the two-level tree > > > > 0. > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > > > is the two-level tree > > > > 0. > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > > > similar to the union of the segment {0,1} and the segment {0,1,2} which > > is the segment {0,1,2}. > > That is merely the embedding of the smaller into the larger, it is not a > union. Embedding is a union. The union given above is simply the union {(0,0), (1,0), (1,1)} U {(0,0), (1,0), (1,1),(2,0a), (2,1a), (2,0b), (2,1b)} > > > > In the union of all finite paths there is no final node. > > But in a union of trees one does not get a union of paths. Why not? The union of of initial segments of N is not N? > > > > In the unuion of all last levels of the set of finite trees there is no > > last level. ! Regards, WM
From: mueckenh on 9 Jan 2007 08:27 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > The finite union of two or more finite trees is a finite tree. An > > infinite union of finite trees is the infinite tree. > > What is an infinite union? (Please give a definition) An infinite union of trees is the union of all (or nearly all) finite trees which reach from level 0 to level n where n eps N. (By the way the result is the same when taking only the last levels of all finite trees.) It is easy to see that these unions result in the infinite tree, i.e., in the tree containing all levels n for n eps N. Therefore this tree contains all paths which represent sequences of bits which can be indexed by natural numbers. There is no such path or part of such path outside of the union of all finite trees. Since the countable union of finite sets is countable, the result is that all paths in the infinite tree belong to a countable set. This is already clear when considering the countablility of nodes and edges, because there can be no more paths than edges. But now we can be absolutely sure: Only persons with very limited logic capabilities can claim that an infinite tree conaining all edges and all nodes and all levels which can be enumerated by natural numbers, will and simultaneously will not contain all possible paths (infinite sequences of nodes which can be enumerated by natural numbers). Regards, WM
From: William Hughes on 9 Jan 2007 08:31 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > A: L_D is a line, therefore L_D has a largest > > > > element. > > > > > > > > B: L_D contains any element that can be shown to exist > > > > in the diagonal, therefore L_D does not have a largest > > > > element. > > > > > > > > Contradiction. Therefore L_D does not exist. > > > > > > Therefore the union of line ends including the end of L_D does not > > > exist. > > > > The union of line ends is not a line. > > But it is supposed to exists as the complete, finished diagonal. [The claim is that the line L_D does not exist. Consider each of your comments in light of this claim.] No. No such claim was made. The claim was that L_D does not exist. Knowing whether a complete finished diagonal exists will not tell you whether L_D exists. This comment says nothing about whether L_D exists. > > > L_D does not exist > > so the union of line ends including the end of L_D does not exist. > > Therefre a complete, finished diagonal does not exist. Irrelevent. The diagonal is not a line. This statement says nothing about whether L_D exists. > > > > > > > Fine. There are no infinite sets. There is only potential > > > infinity. Every line including the diagonal > > > > The diagonal is not a line. The diagonal is a potentially infinite > > set. > > Correct. Therefore it does not contain all natural numbers (as indexes) > because then it would be actually infinite. > > > A line is not a potentially infinite set > > Correct. A line is or is not. But the system of lines is potentially > infinite. A system of lines is not a line. This statement says nothing about whether L_D exists > > > > > has a greatest element > > > until the existence of a greater one is shown. > > > > However, and contrary to your repeated claim, there is no > > single line which contains every element that can be shown > > to be in the diagonal. > > Every element that can be shown to be in the diagonal does exist as the > end of a line. Yes? > But not every end of a line does belong to a line? That is a surprising > aspect of existence. Every end of a line belongs to a line. However, these lines can be different for different line ends So this statement can be true whether or not L_D exists. This statement says nothing about whether L_D exists. > > By showing that an element of the diagonal exists, you show that the > line contaning it and every smaller element exists. However, "the line containing it" can be different for different elements. This statement can be true whether or not L_D exists This statement says nothing about whether L_D exists. You have made no statement relevent to the question of whether L_D exists, nor have you addressed the contradiction which shows that L_D does not exist. L_D does not exist. - William Hughes
From: mueckenh on 9 Jan 2007 08:42
Dik T. Winter schrieb: > In article <1168291077.860958.62580(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1168107015.838378.71940(a)51g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > > > different systems of paths, i.e., strings of nodes and edges? > > > > > > > > > > Yes, it entirely depends on how you find your paths. > > > > > > > > There are no paths to find. Paths do exist in the tree. > > > > Ok? Every infinite tree, which contains all levels enumerated with > > natural numbers, contains all possible paths and, therefore, contains > > the representations of all real numbers of [0, 1]. > > In that case the set of paths in the infinite tree is not the union of the > sets of paths in the finite trees. > The union of all finite trees (= union of all levels) of the tree automatically contains all paths, because there is no path or part of a path outside of this union. > > > > The union of all finite initial segments {1,2,3,...n} is the infinite > > initial segment N. So the infinite tree is the union of all finite > > trees. > > Yes, so what? I repeat that I am talking about the union of the sets of > paths. The union of all finite trees (= union of all levels) of the tree automatically contains all paths, because there is no path or part of a path outside of this union. > > > > So no infinite set N emerges from the union of all finite initial > > segments {1,2,3,...n} ? I agree! But, I am afraid, you do not. > > No, because I do not state that. I only state (I repeat): > there is no infinite number in the union of initial segments > and that means that N does not contain an infinite number. It contains > only finite numbers (but infinitely many). And the union of all finite > initial segments *is* N. Correct. As the union of all finite paths turning always right is 0.111... . > > > > How can that path be the union if none of the constituents is that path? > > > The union of the sets {0.1}, {0.11}, {0.111}, etc. would be the set > > > {0.1, 0.11, 0.111, ...}, but 0.111... is not in it. If it is in it is > > > must be in at least one of the constituent sets of the union. And I > > > have no idea how you come to the "union" of numbers when they are rational. > > > > Yeah. You are beginning to see why the union of all finite numbers n > > canot yield an infinite set. > > U{n is natural} {n} = N. Because every k in N is in one of the sets used > in the union, namely {k}. {0.111...} is *not* in > U[ {0.1}, {0.11}, {0.111}, ...} ] because it is in *none* of the sets > used in the union. But {0.111...} is {{0.1}, {0.11}, {0.111}, ...} (see below). > > > > > Similarly the union of all > > > > initial segments of a well-ordered countable set is this countable set. > > > > > > That is, eh, slightly different. > > > > Why? > > The union of {0.1}, {0.11}, {0.111}, ..., is *not* {0.111...}, it is: > {0.1, 0.11, 0.111, 0.1111, ...}. Remember: unions are defined for *sets*. > I do not know any definition of union for numbers as these. > And what is the set {0.1, 0.11, 0.111, 0.1111, ...}, or, translated to digits, the set {1,2,3,...}? Is it not {0.111...} or, translated to digits, N (or omega)? > > > So we go a bit abstract already. So consider the circles below. Their meaning should be unique. > > > > > They will understand, at least by experiment, > > > > > > > > oo > > > > ooo > > > > _____ > > > > ooooo > > > > > > You apparently have not done the experiment. No, they will not understand, > > > and such experiments have been done. > > > > Depends on their intelligence, which I don't know. > > It has nothing to do with intelligence. In their culture they do not use > *any* abstraction. They do everything with concrete things only. In above circles there is no abstraction. Regards, WM |