Cantor Confusion
in [Math]
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From: Virgil on 21 Apr 2007 13:57 In article <1177158654.185563.309460(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 19 Apr., 14:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > It is a set of nodes that separates a path, not a single > > node. There are an uncountable number of sets > > of nodes. So you do have enough nodes. > > First we should clear the notation: > There are bunches of paths like 0,1 which contains all paths from > 0,1000... to 0.111... > A searated path is a bunch of paths with only one element like > 0,111... > > In order to get two separateed bunches of paths, there is one bunch > required which contains all these paths and one node to split the > incomin bunch into wo bunches. > > The infinite tree contains all separated paths. Therefore there are as > many nodes (minus one) required in the tree. Non sequitur. There is no proof that in an infinite binary tree, at least as defined in mathematical terms, the cardinality of the set of nodes is as large as the cardinality of the set of paths. There are proofs that in an infinite binary tree, at least as defined in mathematical terms, the cardinality of the set of nodes is less than the cardinality of the set of paths. Such proofs have been presented here and have not been shown to be invalid, but every attempt by WM to prove that there are as many noes as paths has been invalidated, at least in mathematics. If WM wants things to work differently in his own private dreamworld, that is fine, as long as he does not try to insist that there is no other world where things are different.
From: Virgil on 21 Apr 2007 14:10 In article <1177158861.675889.121570(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Apr., 16:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176981294.085164.20...(a)n76g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 18 Apr., 16:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > Without the chance that uncountably many are separated in the > > > > > whole > > > > > tree? > > > > > > > > They are all separated from each other. I do not know why you need > > > > uncountably many separation points for that. > > > > > > Because one node separates only one (bunch of) path(s) from another > > > one. > > > > It makes sense only if you remove the parenthesis. When you remove the > > stuff between parenthesis it makes no sens. > > That depends on the notation. The "bunch of" makesw it somewhat > clumsy. So would WM rather have it neat but false than clumsy but true? And what is the advantage does WM claim for "bunch" over "set". Is not every "bunch" of paths a set of paths? > > There are bunches of paths like the bunch 0,1 which contains all paths > from 0,1000... to 0.111... > A searated path is a bunch of paths with only one element like > 0,111... > > In order to get two separateed bunches of paths, there is one bunch > required which contains all these paths and one node to split the > incomin bunch into wo bunches. But in a CIBT, all incoming and outgoing 'bunches", i.e., sets, of paths are uncountable as sets (or as bunches, unless bunches are a lot different than sets), > > The infinite tree contains all separated paths. Therefore there are as > many nodes (minus one) required in the tree. Often claimed but never justified. There is no way that one node (or unit or whatever you wish to call it) accomplishes the separation of one path from all others. It takes infinitely many of nodes or whatevers. So that it is the set of sets of nodes, not just the set of nodes, which has to be as large as the set of paths. > > Should no separated paths exist, then they would also not exist in > Cantor's list as entries or as diagonal number. Paths do exist, but not as WM envisions them to exist. What WM envisions can only be seen in his own private personal universe, which mathematicians, fortunately, need never access in order to do actual mathematics.
From: Virgil on 21 Apr 2007 14:31 In article <1177159491.684039.144700(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Apr., 17:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176981838.819590.80...(a)l77g2000hsb.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 18 Apr., 16:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > The mathematical question is: "are all paths available in the infinite > > > > tree", and the answer is: "yes". A follow-up question can be "is each > > > > path separated from all other paths", and the answer is, again, "yes". > > > > And a further question can be, "is there a level such that a > > > > particular > > > > path is separated from all other paths", and the answer to this is > > > > "no". > > > > "exist" is barely a mathematical term. > > > > > > Existence of separated paths is not possible without being separated. > > > But the set of occasions for separations (= nodes) is countable. > > > > In what way is that a contradiction? As each path separates at each of > > its elements (nodes) from uncountable many other paths, I see no > > contradiction. > > The tree must contain all separated path due to real numbers between 0 > and 1. For this sake we need as many nodes as separated paths. WM may need that, but no one else can be forced into that Procrustean bed. The set of nodes in a CIBT is countably infinite. Each path consists of, or is defined by, an infinite subset of that infinite set. While not every infinite subset of the set of nodes determines a path, enough do to give the set of paths a cardinality equal to the of the set of all subsets of the set of nodes. > Instead of "path" which seems to be misleading use "bunch of paths". Instead of "bunch" use "set", or does WM wish to assert that his bunches are not sets? of trichotomy.) > > Cantor's (first diagonalization) method allows to enumerate every > rational number. So each rational has its place in the sequence. But > this sequence does not enumerate *all rationals". Otherwise there must > be the largest rational less than 1. Deliberate conflation of two incompatible order relationships. In the standard rational ordering, there is no largest rational less than any rational. When one ennumerates the rationals, then, unless 1 as a rational is ennumerated first, there will be a last rational ennumerated before 1 is ennumerated, but this is not a problem, as it is quite independent of the standard rational ordering. > > The absence of this rational is a natural as he absence of the set > {N,k,f} in Hessenberg's proof. False. Given any non-empty set S and it power set P(S) of all subset of S, and any function f:S -> P(S), then H_f = {x in S : not x in f(x)} is a subset of S and therefore a member of P(S) but is not in the range of f. But the set of rationals does not surject to the power set of the naturals so WM's argument fails, as usual.
From: mueckenh on 21 Apr 2007 16:33 On 21 Apr., 15:00, William Hughes <wpihug...(a)hotmail.com> wrote: > On Apr 21, 8:30 am, mueck...(a)rz.fh-augsburg.de wrote: > > > On 19 Apr., 14:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > It is a set of nodes that separates a path, not a single > > > node. There are an uncountable number of sets > > > of nodes. So you do have enough nodes. > > > First we should clear the notation: > > There are bunches of paths like 0,1 which contains all paths from > > 0,1000... to 0.111... > > A separated path is a bunch of paths with only one element like > > 0,111... > > > In order to get two separateed bunches of paths, there is one bunch > > required which contains all these paths and one node to split the > > incomin bunch into wo bunches. > > So a single node does not separate a single path. However, > a set of nodes can. A single node increases the number of bunches of paths by 1. > > There are uncountable many sets of nodes. That is completely irrelevant. > > Therefore: > > There are enough nodes to separate uncountable many paths. No. A single node increases the number of bunches of paths by 1. > > We have to settle the question of whether there are enough nodes > to separate uncountable many paths, before we go on to > discuss the question of whether they in fact do separate uncountable > many paths. This question has been settled. Every single node increases the number of bunches of paths by 1. The number of bunches surpassing the number of nodes is given by 1 + 2-1-1+2-1-1+2-1-1... Regards, WM
From: mueckenh on 21 Apr 2007 16:35
On 21 Apr., 19:57, Virgil <vir...(a)comcast.net> wrote: > In article <1177158654.185563.309...(a)o5g2000hsb.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 19 Apr., 14:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > It is a set of nodes that separates a path, not a single > > > node. There are an uncountable number of sets > > > of nodes. So you do have enough nodes. > > > First we should clear the notation: > > There are bunches of paths like 0,1 which contains all paths from > > 0,1000... to 0.111... > > A separated path is a bunch of paths with only one element like > > 0,111... > > > In order to get two separateed bunches of paths, there is one bunch > > required which contains all these paths and one node to split the > > incomin bunch into wo bunches. > > > The infinite tree contains all separated paths. Therefore there are as > > many nodes (minus one) required in the tree. > > Non sequitur. > > There is no proof that in an infinite binary tree, at least as defined > in mathematical terms, the cardinality of the set of nodes is as large > as the cardinality of the set of paths. Every node increases the number of path-bunches by 1. Regards, WM |