Cantor Confusion
in [Math]
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From: mueckenh on 14 Apr 2007 08:07 On 13 Apr., 19:54, Virgil <vir...(a)comcast.net> wrote: > In article <1176459754.005795.36...(a)q75g2000hsh.googlegroups.com>, > > mueck...(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. If there are countably many paths separated by one node, then there is a chance to get all paths separated by countably many nodes. But you argue that more paths are even easier to be separated? > > > 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". You know for sure, they must be somewhere, alas you can't find them, because all separated paths which are really present and separated, are countable. Regards, WM
From: Virgil on 14 Apr 2007 14:47 In article <1176551363.847980.257110(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 13 Apr., 19:13, Virgil <vir...(a)comcast.net> wrote: > > In article <1176455271.328396.146...(a)w1g2000hsg.googlegroups.com>, > > > > > > 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. In infinite trees those sets are necessarily uncountable. In finite trees they are necessarily finite. > > > > 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. Finally realized that, have you? > > > > > > > > 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. On each side of all but the endmost. > 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? I have no idea why WM would think that there could be fewer paths through some levels than through others. Paths in an infinite tree do not 'have' only one level, they all 'have' (pass through) all levels, and there are uncountably many of them through any node on any level, as many through any node as through the root node.
From: Virgil on 14 Apr 2007 15:15 In article <1176552133.184605.119430(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. False, at least for any level in an infinite tree. That would require that there be a level beyond which no path passes. > Or how do they manage to be separated but to be not separated? It may appear so to those who assume falsehoods, but it is no so in any mathematically defined complete infinite binary tree. Any two paths separate at some node, along with the uncountably many others also separating into two disjount uncountable sets of paths at that same node. WM seems to think that for each node there is only one pair of paths separating at that node, but even in finite trees that can only happen at parents of leaf nodes. > > > > 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? There always are in CIBTs, as has been conclusively demonstrated a number of times and refuted never. > > > > > > > > > > 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. O.K. Each proper binary rational in (0,1) has two separated paths. > > > > 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? For any finite subset of nodes in any infinite path in any CIBT, there will be infinitely many paths through all the nodes of that subset. Thus to 'separate' one such path simultaneously from all other paths requires an infinite set of nodes from that path as points of separation. No finite set of nodes will do. > > > > > 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? WM's question assumes conditions contrary to fact. If one were to require a tree to have only countably many paths, it could not be a complete infinite binary tree at all. At least not in as complete infinite binary trees are defined. > If you need an out-time, feel free to take it. As it is not I who assumes conditions contrary to fact, I do not need one. Do you? > > > > > 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? 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? By what definition of "separated paths" does WM claim this. Every two paths separate at some node, but that same node separates uncountably many paths branching left at that node from uncountably many branching right at that node. To separate any one path from all other paths takes infinitely many nodes.
From: Virgil on 14 Apr 2007 15:22 In article <1176552435.377253.129000(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 13 Apr., 19:54, Virgil <vir...(a)comcast.net> wrote: > > In article <1176459754.005795.36...(a)q75g2000hsh.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > 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. > > If there are countably many paths separated by one node, then there is > a chance to get all paths separated by countably many nodes. There are more than countably many separated by every node in every CIBT. And even then countably many nodes suffice. > But you > argue that more paths are even easier to be separated? Only in the sense that a CIBT with only countably many paths is impossible. And the possible is easier than the impossible. > > > > > 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". > > You know for sure, they must be somewhere, alas you can't find them, > because all separated paths which are really present and separated, > are countable. The paths, or at least sets of paths, that I can count are certainly countable, but that in no way proves that I have counted everything that is there. And there are quite adequate and valid proofs to the contrary, which WM has not refuted by any valid counterproofs of his own.
From: mueckenh on 15 Apr 2007 07:39
On 14 Apr., 20:47, Virgil <vir...(a)comcast.net> wrote: > > 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? > > I have no idea why WM would think that there could be fewer paths > through some levels than through others. "Separated" paths. > > Paths in an infinite tree do not 'have' only one level, they all 'have' > (pass through) all levels, and there are uncountably many of them > through any node on any level, as many through any node as through the > root node. Count separated paths only. There are never more than countably many in any level you investigate. Regards, WM |