Cantor Confusion
in [Math]
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From: mueckenh on 13 Apr 2007 06:22 On 12 Apr., 23:48, Virgil <vir...(a)comcast.net> wrote: > In article <1176409858.797211.294...(a)p77g2000hsh.googlegroups.com>, > > > Why then do not all real numbers of [0, 1] exist side by side > > simultaneously in the so defined tree, while they exist in > > mathematics? > > And who says they do not? If all paths exist side by side, then there must exist uncountably many paths side by side, i.e., with different nodes. But we know that in the whole tree there can only exist countably many side by side. Regards, WM
From: Virgil on 13 Apr 2007 13:13 In article <1176455271.328396.146230(a)w1g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Apr., 23:48, Virgil <vir...(a)comcast.net> wrote: > > In article <1176409858.797211.294...(a)p77g2000hsh.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 12 Apr., 20:02, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1176397336.310678.13...(a)l77g2000hsb.googlegroups.com>, > > > There can be more things than names, but not more names than names. > > > Numbers are nothing unless being named. > > > > They are something if one wishes them to be. Whoever ruled that a number > > without a name does not exist? > > In what form does it exist? In what form can it be used? > > > > > > > > > In the tree we see that every creation of another sequence requires a > > > node. > > > > Only in finite trees do leaf nodes biject with paths. In infinite trees > > there are no leaf nodes. > > But nodes of separation. What is a "node of separation? At each parent node in any binary tree, one SET of paths branches left and another SET branches right. The only nodes that separate one path from all the rest must be such that one of the two sets is a singleton set, which only can occur when it is the parent of a leaf node. So in that sense, only a parent of a leaf node can be a "node of separation". So which nodes in a complete infinite binary tree does WM claim are "nodes of separation (parents of leaf nodes)"? > > > > > > > > Again, your mantra. Proof, please. > > > > > > > > > If all paths exist simultaneously, then there must exist > > > > > > > uncountably > > > > > > > many in the infinite tree. > > > > > > > > Which is precisely the case! At least in mathematics. > > > > > > > But the tree is not in mathematics? > > > > > > If mathematically defined trees, finite or infinite, do not exist > > > > within > > > > mathematics, why does mathematics bother to define them? And it does > > > > bother to!- > > > > > Why then do not all real numbers of [0, 1] exist side by side > > > simultaneously in the so defined tree, while they exist in > > > mathematics? > > > > And who says they do not? > > That one who says that there are less than all paths side by side in > the tree. Whether they are "side by side" depends on what is meant by "side by side". In the sense of having a nearest neighbor on either side, ,only a few of them ( the binary ratinals) do, but in the sense of having all the others on one side or the other , they all do. Whom does WM claim claims otherwise?
From: Virgil on 13 Apr 2007 13:54 In article <1176459754.005795.36550(a)q75g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Apr., 23:48, Virgil <vir...(a)comcast.net> wrote: > > In article <1176409858.797211.294...(a)p77g2000hsh.googlegroups.com>, > > > > > > Why then do not all real numbers of [0, 1] exist side by side > > > simultaneously in the so defined tree, while they exist in > > > mathematics? > > > > And who says they do not? > > > If all paths exist side by side, then there must exist uncountably > many paths side by side, i.e., with different nodes. The delusion that in a complete infinite binary tree the branching from a given node separates only finitely many paths from other paths seems to be what gives rise to WM's delusion that there must be as many nodes as paths, despite several clear disproofs that have been presented here. > But we know that > in the whole tree there can only exist countably many side by side. "WE" know no such thing. "It ain't what you don't know that hurts you, it's what you 'know' that ain't so". Mark Twain. "WE", meaning those who accept such set theories as ZF and NBG, know that in a complete infinite binary tree in such set theories, there are "more paths than can be counted by the naturals".
From: mueckenh on 14 Apr 2007 07:49 On 13 Apr., 19:13, Virgil <vir...(a)comcast.net> wrote: > In article <1176455271.328396.146...(a)w1g2000hsg.googlegroups.com>, > > > > > There can be more things than names, but not more names than names. > > > > Numbers are nothing unless being named. > > > > They are something if one wishes them to be. Whoever ruled that a number > > > without a name does not exist? > > > In what form does it exist? In what form can it be used? In what form does it exist? In what form can it be used? > > > > > In the tree we see that every creation of another sequence requires a > > > > node. > > > > Only in finite trees do leaf nodes biject with paths. In infinite trees > > > there are no leaf nodes. > > > But nodes of separation. > > What is a "node of separation? A node of separation separates two setsof paths. > > So in that sense, only a parent of a leaf node can be a > "node of separation". > So which nodes in a complete infinite binary tree does WM claim are > "nodes of separation (parents of leaf nodes)"? There are no leaves in the infinite binary tree. > > > > That one who says that there are less than all paths side by side in > > the tree. > > Whether they are "side by side" depends on what is meant by "side by > side". In the sense of having a nearest neighbor on either side, ,only > a few of them ( the binary ratinals) do, but in the sense of having all > the others on one side or the other , they all do. Then there must be uncoutably many of them on one side or the other. If we start at the path 0.000... all the other paths must be on the other (right hand) side. But if we count all paths in any level, there are not so many. How do you explain this discrepancy? Regards, WM
From: mueckenh on 14 Apr 2007 08:02
On 13 Apr., 19:44, Virgil <vir...(a)comcast.net> wrote: > In article <1176459546.678219.176...(a)o5g2000hsb.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 12 Apr., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > In article <1176409486.686541.279...(a)p77g2000hsh.googlegroups.com>, > > > > > I am claiming that there are as many points of separation as are > > > > separated paths. > > > > What does WM mean by a "separated path". > > > A separated path in the tree is a path which can be distinguished from > > every other path. For instance 0.000... or 0.111... or 0.010101... or > > 1/pi etc. > > In that sense every path is a separated path in every binary tree, as > every path can be distinguished from any other path by the fact that > there is some common node at which they branch in different directions. As two separated paths will never meet again, there must be a level where all paths are separated, if all paths exist as separated paths. Or how do they manage to be separated but to be not separated? > > According to your latest definition of "separated paths". there are > uncountably many separated paths and no non-separated paths, But there are never uncountably many separated paths in the tree? > > > > > > Until WM clarifies what he means by "separated paths" versus > > > "non-separated paths"I have no idea what he is asking. > > > A separated path in the tree is representing a real number, i.e., a > > number like 0.000... which can be uniquely addressed because it > > differs from any other real number. > > Bad definition, as each proper binary rational has two separated paths. Not the rational 0.000... in the binary tree containing the interval [0, 1]. A mathematician would know that. > > As each "separation" separates uncountably many paths from uncountably > many other paths, not merely one path from one other path, it only takes > countably many separations to isolate any one path from all the others. That is difficult logic for me. Could you explain it a bit further? > > > > > How can there be more separated paths than points of separation? > > If the branching from each node only separated finitely many paths from > finitely many others, as in finite trees, WM might have an argument, but > as that is not the case in infinite trees, WM's argument fails. So uncountably many paths are separated by less nodes than countably many paths? If you need an out-time, feel free to take t. > > > I am asking whether the path P does exist in the tree as a path which for > > every other path Q has a node different from that path Q. > > Each other path has some node not in path P, but it may be a different > node for different other paths, and one node will serve to differentiate > many other paths, so that the correspondence WM is trying to imply is > false. In fact? > > > If this is > > the case for every path P, then there must be as many nodes as paths. > > WRONG, unless each node can serve to exclude only finitely many other > paths from P, which is not the case. > > In fact, in every complete infinite binary tree with path P, at each > node of P, there are as may paths which separate from P as there are > paths in the whole tree. So there are no separated paths in the tree, after all? Regards, WM |