Another AC anomaly?
in [Logic]
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From: YBM on 17 Dec 2009 16:02 WM a �crit : > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: >> In article >> <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com>, >> >> WM <mueck...(a)rz.fh-augsburg.de> wrote: >>>> No. Suppose we have the paths 0.000 and 0.100, what is their union? And >>>> is it a path? >>> No. But the union contains two paths. And an infinite union of that >>> kind may contain the path 0.111... >> So according to WM a union of sets may contain an object not contained >> in any of the sets being unioned. > > For paths and initial segments to contain and to be is the same. In > fact: > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > and also > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. As usual you confuse to be an element and to be a party of... > This is so according to set theory. Of course it is rubbish. I am glad > that you share my opinion. This is not what he said. You are not only a liar, an incompetent and a fraud, but a dishonest man.
From: Quaestor on 17 Dec 2009 16:07 In article <a737fea3-6dd2-4a1d-9506-2df7cb6d412a(a)v30g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > �> It is. I constructed a finite path from the root node to each other > > �> node. > > > > Yup, you constructed a finite path, and that does not represent 1/3. > > It 1/3 can be represented by a bit sequence, then it is represented in > my tree by the union of the node sets representing > 0.0, 0.01, 0.010, ... > > > > �> � � � Then I appended an infinite tail. > > > > Whatever that may be, it is *not* a path according to your explicit > > statement > > that the tree did not contain infinite paths. > > You have misread my construction. I use finite paths and then I append > an infinite tail. We have not misread the consequences of your construction. Attaching only one infinite tail , or even a countably infinite number of infinite tails, to each finite subpath does not produce all possible paths. As Cantor proved and WM cannot disprove. > Consider the following model for a complete infinite binary tree: The set of nodes is N = {1,2,3,...}, each such natural being a node. For each node n in N, its left child is defined to be 2*n + 0 in N and its right child is defined to be 2*n + 1 in N. A path is, by definition, a set of nodes which contains 1, and, for each node in it, contains exactly one child of that node. Until WM can produce some cogent reason why the above definition fails to be a binary tree, all his claims about such trees are proved wrong.
From: WM on 17 Dec 2009 16:19 On 17 Dez., 22:07, Quaestor <quaes...(a)qqq.com> wrote: > In article > <a737fea3-6dd2-4a1d-9506-2df7cb6d4...(a)v30g2000yqm.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > It is. I constructed a finite path from the root node to each other > > > > node. > > > > Yup, you constructed a finite path, and that does not represent 1/3. > > > It 1/3 can be represented by a bit sequence, then it is represented in > > my tree by the union of the node sets representing > > 0.0, 0.01, 0.010, ... > > > > > Then I appended an infinite tail. > > > > Whatever that may be, it is *not* a path according to your explicit > > > statement > > > that the tree did not contain infinite paths. > > > You have misread my construction. I use finite paths and then I append > > an infinite tail. > > We have not misread the consequences of your construction. > > Attaching only one infinite tail , or even a countably infinite number > of infinite tails, to each finite subpath does not produce all possible > paths. Please prove your assertion by showing which path I did not produce. But prove it by knowing only all nodes of all paths. Information exceeding the infinite sequence of nodes of each path is not acceptable in mathematics. > > Consider the following model for a complete infinite binary tree: > > The set of nodes is N = {1,2,3,...}, each such natural being a node. > > For each node n in N, > its left child is defined to be 2*n + 0 in N and > its right child is defined to be 2*n + 1 in N. > > A path is, by definition, a set of nodes which contains 1, and, > for each node in it, contains exactly one child of that node. Which path do you see in the tree that has not that property? > > Until WM can produce some cogent reason why the above definition fails > to be a binary tree, It does not fail. Every node has two child-nodes. Regards, WM
From: WM on 17 Dec 2009 16:24 On 17 Dez., 22:02, YBM <ybm...(a)nooos.fr.invalid> wrote: > WM a écrit : > > > > > > > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > >> In article > >> <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com>, > > >> WM <mueck...(a)rz.fh-augsburg.de> wrote: > >>>> No. Suppose we have the paths 0.000 and 0.100, what is their union? And > >>>> is it a path? > >>> No. But the union contains two paths. And an infinite union of that > >>> kind may contain the path 0.111... > >> So according to WM a union of sets may contain an object not contained > >> in any of the sets being unioned. > > > For paths and initial segments to contain and to be is the same. In > > fact: > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > and also > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. > > As usual you confuse to be an element and to be a party of... > I am concerned with numbers. Your absurd distinctions are irrelevant. Either N exists or not. If yes, then it is the union of all natural numbers as well as the union of all initial segments of the ordered set of all natural numbers as well as the union of all these and itself, because {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...} If you see that this equation is true, then you acknowledge my argument. If you cannot see that, then either you are a matheologian or you have too little intelligence to understand it or both. Regards, WM
From: Quaestor on 17 Dec 2009 16:47
In article <9dde10e2-2ab6-4714-9dd1-d3ebaea51757(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:07, Quaestor <quaes...(a)qqq.com> wrote: > > In article > > <a737fea3-6dd2-4a1d-9506-2df7cb6d4...(a)v30g2000yqm.googlegroups.com>, > > > > > > > > > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > It is. I constructed a finite path from the root node to each other > > > > > node. > > > > > > Yup, you constructed a finite path, and that does not represent 1/3. > > > > > It 1/3 can be represented by a bit sequence, then it is represented in > > > my tree by the union of the node sets representing > > > 0.0, 0.01, 0.010, ... > > > > > > > Then I appended an infinite tail. > > > > > > Whatever that may be, it is *not* a path according to your explicit > > > > statement > > > > that the tree did not contain infinite paths. > > > > > You have misread my construction. I use finite paths and then I append > > > an infinite tail. > > > > We have not misread the consequences of your construction. > > > > Attaching only one infinite tail , or even a countably infinite number > > of infinite tails, to each finite subpath does not produce all possible > > paths. > > Please prove your assertion by showing which path I did not produce. > But prove it by knowing only all nodes of all paths. Until you give some explicit way of telling which paths you do produce, ther is no way of telling which ones you miss. > Information exceeding the infinite sequence of nodes of each path is > not acceptable in mathematics. I do not accept your fiats on what is or is not acceptable in mathematics, having seen you to be so often wrong in such claims. > > > > > Consider the following model for a complete infinite binary tree: > > > > � �The set of nodes is N = {1,2,3,...}, each such natural being a node. > > > > � �For each node n in N, > > � � � its �left child is defined to be 2*n + 0 �in N and > > � � � its right child is defined to be 2*n + 1 in N. > > > > � �A path is, by definition, a set of nodes which contains 1, and, > > � �for each node in it, contains exactly one child of that node. > > Which path do you see in the tree that has not that property? In my tree, none. But in there are sets of nodes satisfying my condition of pathhood which do not occur in your tree. > > > > Until WM can produce some cogent reason why the above definition fails > > to be a binary tree, > > It does not fail. Every node has two child-nodes. Then WM concedes his errors. |