Cantor Confusion
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From: Virgil on 25 Jan 2007 14:53 In article <1169714020.350577.110980(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > I am beginning to understand. You equate a sequece with a function > > f: N -> objects > > and talk about the domain and range of that function. > > Yes. The domain defines the length of the path. Since the domain, N, is not finite, neither are such paths. > > The tree is a collection of sets, namely the collection of levels L(n) > as well as the collection of subtrees like T(n) = L(0) U L(1) U ... U > L(n). The tree is also a set. Its elements are nodes. A path is a > subset with the special property that it contains all nodes which go > from top to the end or, in case there is no end, continue to go on. Paths are determined by the set of edges of the tree, not by the set of nodes. For any set of more that two nodes there are many different sets of edges making that set of nodes into a binary tree, so without the set of edges one does not have a tree at all. > > > > > 3) The union of all finite trees includes the union of all nodes and, > > > with it, the union of all such subsets which are paths (because every > > > path is a well defined subset of the set of nodes if the structure of > > > the tree is well defined). > > > > If a tree consists (amongst others) of the set of nodes in it, the union > > of two trees indeed consists (amonst others) of the union of the sets of > > nodes. > > I agree. But this union may not be intermingled with pair-axiom. The > union of two or more paths is a set of nodes which can but need not be > a path. > > In the tree T(1) > > a > /\ > b c > > There are two paths, namely p1 = {a, b} and p2 = {a, c}. By pair axiom > we get a set {p1, p2} containing two paths as elements. The union p1 U > p2 = {a, b, c}, however, is not a path. WM ignores the fact that. in each pair of nodes forming an edges, one node is parent and the other child, so one needs an /ordered/ pair of nodes, not simply a set of two nodes. Mathematically speaking, which WM is not able to do very well, {p1,p2} = {p2,p1}, but that won't serve for edges. If (p1,p2) represents the ordered pair, so that when p1 =\= p2 one has (p1,p2) =\= (p2,p1), that represents an edge appropriately. > > I need not specify and do not specify how the paths in T(oo) come in. I > merely express that some paths like 0.010 belong to one set of paths > (namely those of tree T(3)) but do not belong to the set of path in > T(oo). > T(oo) contains no paths which are only initial segments of paths of > T(oo). Then T(oo) must consist only of endless paths. The set of all endless paths in maximal infinite binary tree will be uncountable, as proved originally by Cantor. So that if T(oo) contains only countably many paths, WM must have some way of excluding most of the potential paths from T(oo). > > Yes. But due to the schism "infinite set of finite numbers" there are > two answers possible (but not more). Wrong, as usual. WM presumes that all paths have an end, but that is not possible when one can find endless chains of edges which link child node of one edge to parent node of the next. A path can only end in a leaf node, but T(oo) has no leaf nodes. For every T(n), the leaf nodes of its edge set become non-leaf nodes in the edge set of T(n+1), and since there is no last T(n), every node becomes a non-leaf node. > 1) If we identify path-lengths with natural numbers That presumes what it is trying to prove. In mathematics and logic, that is called the fallacy of the circular argument. > No. When unioning the trees we union also their subsets, including > their paths. WM has declared that the set of paths is derived from the set of edges and the set of nodes, giving all possible paths so derivable. In which case the "union" of all finite binary trees gives all possible endless paths, contrary to WM's claim that one gets only the union of path sets. > > A subset of a tree is a path if and only if all its elements (nodes) > are connected by edges. False! It must also have only one child edge for any parent edge and it must be maximal in the sense that no further edges can be added without violating the child-parent restriction above. WM does not seem to understand what "if and only if" means. > The union of two paths is a path if and only if all its nodes are > connected by edges. (See the examples above.) Wrong again. WM is batting 0.000 so far. The "union" of two paths, where one is from a subtree of the tree of the other, is a path if and only if the set of edges of one is a subset of the set of edges of the other (equivalently, the set of nodes of one is a subset of the set of nodes of the other). > > The set of paths in T(oo) consists of unions of paths. It cannot > contain contain more paths than were unioned, i.e.,countably many. If the sequence of trees in the "union" is finite then each path in the "union" tree is the union of a finite /sequence/ of paths and will be the last path in that /sequence/. If the /sequence/ of trees is endless then each path in the "union" tree is the union of an endless /sequence/ of paths and there will be NO last path in that sequence. So that the "number" of paths for WM's T(oo) will be the same as the "number" of such endless sequences, which is NOT countable. > > But the union of all elements of the powerset is N. U(U P(N))) = N. So > the union of two or more paths are nodes. For a finite tree, there is a bijection between paths and leaf nodes, but T(oo) and T have no leaf nodes, so you can't use them to count paths. > > The set of all paths in T(oo) is not the union but only a subset of the > countable union of countable sets of paths. > > > > > Going on, we can say: > > > > > > 6) T(oo) = T contains only finite paths. > > > 7) T(oo) = T contains all paths including all infinite paths. But you cannot say either truthfully! > > > > By your definition (paths are specific subsets of nodes), T contains > > infinite paths. Conclusion (6) is false. On the other hand, the > > union of the sets of paths contains only finite paths. And they are > > countable indeed. > > If T(oo) = T concerning the set of nodes and edges, then they are > identical with respect to paths to. There is no further parameter to > fix. If you do not accept this identity, then further discussion is not > meaningful. Honest discussion with WM has long been impossible, since he pays no attention to what anyone else says unless they agree with him, and merely repeats arguments shown to be fallacious. What we attempt is to show up his frequent fallacies for what they are.
From: Virgil on 25 Jan 2007 15:04 In article <1169716417.932690.29940(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169626572.550075.102890(a)l53g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > > > > > You have to choose whether you want to see a tree as > > > > > > > > a) the standard tree, T = (M, E) > > > > b) the "Virgil"-tree. T = (set of paths having certain properties). > > > > > > > > Which definition do you choose? > > > > > > I defined the trees which I use by nodes and edges. > > > > > > 0. > > > / \ > > > 0 1 > > > / \ / \ > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > . . . . > > > . . . . > > > . . . . > > > > > > > > > This is a tree of type weeping willow, T(1), with edges 0 and 1. It is > > > easy to see how the general definition of T(n) looks like but it is > > > tedious to write it down here. However, who is unable to understand it, > > > will be unable to follow my thoughts and should refrain from trying so. > > > > > > > > > The corresponding cut tree T(1) is > > > > > > 0. > > > / \ > > > 0 1 > > > > > > Why not > > > > 0. > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > ? > > That is a complete binary tree of depth 2 within you WW tree. > > No. The bottom nodes have not two child nodes but only one. They are both clearly subtrees of the same tree > > > > And your Weeping Willow Trees are essentially the same as my eventually > > constant trees, to which you objected. > > If they are the same as mine, then I will not object to yours, but I > think we should concentrate on not too much definitions. In both mathematics and logic, clarity of definition is essential to understanding and to finding truth. Those who choose to ignore precision of definition, and careful adherence to definitions as given, eventually lose track of truth. > > > > > > We see that the union of two trees is the larger one. We see further > > > that the union is the set theoretic union. And by enumerating the nodes > > > > > > 0 > > > 1 2 > > > 3 4 5 6 > > > ... > > > > > > we prove that the union of all nodes and, threfore, of all levels and > > > of all finite trees does exist in the original set theoretic meaning. > > > > But since in each WW tree every path is eventually constant, in the > > union of all such trees every path is still eventually constant, and no > > path in that union can have both infinitely many 0 branchings and > > infinitely many 1 branchings. > > > > Thus WM's "union" of WW trees is not the complete binary tree he claims, > > and the countability of WW paths does not imply the countability of the > > set of apths of the complete infintie binary tree. > > > > So WM fails again! > > > > If the infinite union of finite trees, T(oo), is the complet tree, T, Which, by WM's construction, it is not. > concerning the set of nodes and edges, then both are identical with > respect to all properties including all paths too. There is simply no > further parameter to fix. If you do not accept this identity, then > further discussion between us about this topic is not meaningful. Discussion with WM has long been impossible, as he pays no attention to anything that does not support his creed. Then > you may maintain ZFC and may say that it is free of contradictions, but > you must tolerate that a unique structure may yield different results, > i.e. is not unique. i do not tolerate it. But it may be a matter of > personal taste. It is a matter of logic that WM asserts conclusions unjustified by anything in any definition of trees and their "unions".
From: Virgil on 25 Jan 2007 15:08 In article <1169717449.703191.72140(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I am not willing to > waste my time with exlicit nonsense. Then stop posting so much of it.
From: Virgil on 25 Jan 2007 15:25 In article <1169718468.573403.117170(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The structure of my trees is given. Therefore even the edges are not > necessary, but a guide for the eye. Given the set of nodes {George, William, Ralph}, which is the root node? Which is the left child of the root node? Unless the set of edges is inherent in the NAMING of your nodes, and thus is invisibly included in you set of nodes, your claim is nonsense. > Hence, the union of trees is the > union of ordered sets of nodes. What is it that provides that "order", actually a partial order for most trees, if it is not the set of edges? > If you accept this, then we may > continue. If you do not accept this, then leave it as it is. Why should we accept what is patently false? The "order", which WM sneaks in, does not exists in vacuo, but must be provided, usually as a set of edges, but equivalently as a set of paths, but it must be provided. > > >It is not the aim of any axiomatic theory to rely on what you see. > > Maybe someone can explain the rules of the game to you. > > I am not interested in a game which is as stupid as you report it. It is a better game than the silly one you are playing, proselytizing for your silly religion. > I am interested whether there are irrational numbers. In mathematical "reality", they are as actual as triangles or natural numbers. They are all mere ideas without any physical reality. > If T(oo) as defined by me is T as defined by me, concerning the set of > nodes and edges, then they are identical with respect to paths too. WM's T(oo) and T have no more existence than do such irrational numbers as sqrt(2). If WM accepts the existence of his own "T(oo)", he can have no reason to object to a similar existence for sqrt(2)
From: Virgil on 25 Jan 2007 15:41
In article <1169718794.583774.316240(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169639256.391493.236600(a)s48g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > On 23 Jan., 13:35, Franziska Neugebauer > > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > > > > > > ** The std-union of the set of sets of paths of every finite binary > > > > std-tree is not identical to the set of paths of the infinite > > > > binary std-tree G. ** > > > > > > Summary > > > > > > 1) Every complete infinite binary tree T (containing all nodes and > > > edges) contains all paths. > > > 2) The union tree T(oo) of all finite trees is well defined (as I have > > > shown elsewhere) > > > > What WM has constructed is not a union of trees. > > > Call it as you like. Concerning nodes and edges it is the complete > tree. Meaning that it is maximal with respect to both nodes and edges for binary trees with a given root node, there being no way to add another node or edge to get a larger binary tree? And is it also maximal with respect to paths, so that every set of nodes or every set of edges which is in the form of a path is a path? If so, then there is an obvious bijection between the set of paths and the set of all endless binary strings, which set of strings Canor, et al, have proven uncountable. > > The union of finite trees cannot contain an infinite path. Then it cannot contain an infinite set of nodes or an infinite set or edges, unless there is some way to count an infinite set as finite. > > > 5) A countable union of countable sets is a countable set (according to > > > ZF with AC). But each path in T, or T(oo), is formed as a /sequence/ of ever longer paths, whose limit, or union is an endless path. So the paths of the infinite tree are not merely the union of the path sets but the set of all endless sequences of paths. > > All the paths of T(oo) are paths from a finite tree and, hence, from a > finite set of paths. This lie don't fly! |