Cantor Confusion
in [Math]
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From: mueckenh on 15 Apr 2007 07:48 On 14 Apr., 21:15, Virgil <vir...(a)comcast.net> wrote: > > 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. So in the tree never all paths are separated? > > > 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. That means that only uncountably many sets of paths and not individual paths can be addressed in the tree? > > > > > > 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. O, I must have missed this demonstration. At which level can we find these 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. > But the tree has infinitely many nodes and levels. > > > > > 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 consider an infinite tree which contains only the rational paths. Does it contain any node less than the infinite tree containing all real paths? > > To separate any one path from all other paths takes infinitely many > nodes.- But this is beyond the capability of the tree? Regards, WM
From: Virgil on 15 Apr 2007 15:28 In article <1176637182.881550.115690(a)y5g2000hsa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Separated from what? Every path is already separated from every other path somewhere in a CIBT, so there are no paths that are not separated in such a tree. > > > > 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. Separated from what? Every path is already separated from every other path somewhere in a CIBT, so there are no paths that are not separated in such a tree. > There are never more than countably many > in any level you investigate. As every path passes through every level, there are uncountably many paths through every level. Those uncountably many paths are separated at that level into finitely many separated_from_each_other uncountably large sets of paths, one uncountable set of paths through each node at that level. The only sort of node in any tree through which only one path passes must be a leaf node. So is WM requiring that every path in a CIBT must have a leaf node in order to exist as a "separated" path? "Separatedness" is irrelevant, a straw man that WM brings in only to cover the absence of his proof for claims and the lack of any valid counters for the proofs of what he object to..
From: Virgil on 15 Apr 2007 16:03 In article <1176637688.332542.149280(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 14 Apr., 21:15, Virgil <vir...(a)comcast.net> wrote: > > > > 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. > > So in the tree never all paths are separated? Pairwise they are, but in order for one path to be separated simultaneously from all others it must contain one node not in any other path, and such a node is necessarily a leaf node. So WM wants leaf nodes in CIBTs. Never happen! > > > > > 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. > > That means that only uncountably many sets of paths and not individual > paths can be addressed in the tree? In my CIBTs, at each level one gets finitely many sets of uncountably many paths each , NOT uncountably many sets of paths. There is a significant difference. > > > > > > > > > > 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. > > O, I must have missed this demonstration. At which level can we find > these separated paths? As paths in a CIBT pass through ALL levels and separation of any one path from all others requires ALL levels, or at least infinitely many of them, WM's question assumes a condition contrary to fact. Or in a different interpretation, one finds all those separated paths at any and every level, but not their "separations". The proof of the separation of a path from all others in a CIBT is established by any infinite subset of its nodes, and cannot be established by less. > > > > > > 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. > > > But the tree has infinitely many nodes and levels. Which means that such separations can and do occur. In a finite tree, one leaf node can separate one path from all others, but in the absence of leaf nodes, more is required, in fact, any infinite subset of the node set of a path will work, but nothing less will suffice. > > > > > > > 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 consider an infinite tree which contains only the rational > paths. It would then not be a CIBT, as having only "rational" paths would require the presence either of leaf nodes or nodes with a single child node, neither of which are allowed in CIBTs. > Does it contain any node less than the infinite tree containing > all real paths? No, but it would, for example, exclude the path 0.101001000100001... in which the nth 1 is followed by n 0s, which must occur in a CIBT. > > > > > To separate any one path from all other paths takes infinitely many > > nodes.- > > But this is beyond the capability of the tree? How so? It does contain infinitely many nodes. As sequences of binary digits, it takes infinitely many binary digits to distinguish any one proper binary rational from all others, so there is no increase of difficulty in going from the countable set of binary rational sequences to the uncountable set of all binary sequences So that if WM's rational binary tree requires infinitely many nodes to separate one binary rational path from all others, why does he object to the same thing for CIBTs? WM continually swallows camels and strains at gnats.
From: Dik T. Winter on 15 Apr 2007 20:19 In article <1176298903.609533.227830(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 11 Apr., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > Do you claim that all paths exist as separated entities? If so: how or > > > where do they exist? > > > > Philosophy? I would expect mathematics in this newsgroup. > > I think that numbers belong to mathematics. But the question "where do they exist" is not a mathematical question. > > > How are they related to separated paths? > > > > Each path is (in principle) a Cauchy-sequence. > > To be or not to be, that is the question. If yes, then the tree must > have a width of 2^aleph0. If not, then there are no irrational numbers > - nowhere, neither in the tree nor else in mathematics. Well, in that case you *first* have to define what you mean with "the width of the table". Width is not a set-theoretic concept. So, what do you mean with "the width of the table"? > > > Are the > > > numbers represented by paths which are completely within the > > > infinitely many levels of the tree or not? > > > > This sentence is quite difficult to pars. Yes, each real number is > > represented by a path which is completely within the infinitely many > > levels of the tree. > > But not all real numbers coexist in the tree? Only a countable number > of them is admitted simultaneously? Why? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Apr 2007 20:43
In article <1176300393.408149.165530(a)w1g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 11 Apr., 03:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175968689.240804.64...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > The mirror is not "just virtual" unless you claim that the infinite > > > paths exist only virtually. > > > > That is not what I state. In your mirrored sequence you have a first > > element on the right. Well if you wish to go with the right-left > > writing people, do so. But that does not change the sequence at all, > > Only the way you look at it. > > In the geometric series there is no last element. Nevertheless you > think the series has a limit value. RIght, there are precise definitions for that. > In a geometric series with last > but not first element, the things are different? Yes, because a series *does* have a first element. Pray use standard terminology. > Why is there not a > limit value for > > lim[n --> oo] SUM [1/2^n + ... + 1/4 + 1/2 + 1]? There is. By the very definition of limit. But I see nothing in there about ... 1/16 + 1/8 + 1/4 + 1/2 + 1. The limiting series is 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... . > > > By definition the final sequence is the result of the sequence of > > > interchanges. > > > > By *what* definition? Pray state the definition. And do not come up with > > the single definition: > > [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {..., 4, 3, 2, 1} > > because in that case I come up with the single definition: > > [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {2, 3, 4, 5, ...} > > Furthermore you should be able to prove that with your definition the > > definition does not change the sum. > > Here is a mathematical proof: > We know that the sum cannot be larger than 2. > Proof: SUM [1/2^n + ... + 1/4 + 1/2 + 1] = 2 - 1/2^n, and 1/2^n >= 0 > for every natural number n. > We know that the sum is larger than 2 - eps for any positive real eps. > Proof: Take any real eps > 0. Find n such that 1/2^n < eps. > What do you require further? I ask for a proof that the limit of the series of interchanges (1 n) is {..., 4, 3, 2, 1}, as you state. As you do not provide such a proof, I think you cannot. I think that with a reasonable definition of that series of interchanges you will get: {2, 3, 4, ...}. But you simply give proofs for things for which I do *not* ask a proof. > > > For every path there is a node assumed to contribute 1. If it turns > > > out that this node is shared by another path, then take the next node > > > and continue to go on. > > > > And you never will terminate searching because there is no last node. Now > > what? > > That is the same for searching the digits of pi. Nevertheless you say > pi exists. The situations are quite different. Given a number n there is the mathematical possibility to calculate the n-th digit of pi. (I may note that I talk about "mathematical" possibility. Whether it is feasible is not of concern to mathematics.) On the other hand, there is *not* the mathematical possibility to find the last node of an infinite path, i.e. the node that contributes 1 to the sum. It is worse, given any arbitrary value 2^(-n), there is no way to find the node in the infinite path that contributes that arbitrary value to the sum. > > > Wrong. It has been proven always only for a finite part of the list. > > > > You think so. In that case you should throw away *all* of mathematics, > > because there is (for instance) no direct proof that sqrt(n) is rational > > only if n is a square. When something is proven for an arbitrary element > > that means that is has been proven for all elements. > > Yes, it has been proven for all elements (which exist). But it has not > been proven for the sequence eventually constructed by these elements. What sequence constructed by what elements? The list given is a countable set of reals in decimal form. By assumption they all do exist, otherwise it is not a list. The number constructed (not sequence) is proven to be different from each element of the list, and so for all elements of the list. > So to say: It has not been proven for the diagonal number. Well, you have a serious misunderstanding about mathematics. > > > You cannot apply Cantor's proof to the whole diagonal. In particular > > > because the diagonal does not exist. (It does not exist as a path > > > separated from all other paths in the binary tree.) > > > > Again, your mantra. Proof, please. > > If all paths exist simultaneously, then there must exist uncountably > many in the infinite tree. But there are. > > So every finite part of the path does *not* > > mean all parts of the path. > > You did not talk about such strange parts before. And this explanation > does not fit my original statement: > "Every finite node of a path means the whole path because there are no > infinite nodes". Yes, I never contradicted *that*. > Nevertheless, we can agree upon these things if we > clarify their meaning. Does "Every finite part of the path" mean "the > whole path"? Does "Every finite part of the path" include or cover > also that special part you just invented? And now you switch again from "finite node" to "finite part of the path". As "every finite part of the path" would mean to me (without further definition) a set of parts of paths, I would state: no, because a path is not a set of parts of paths but (by your own definition) a set of nodes. And even as a set of parts of paths, "every finite part of the path" does *not* contain the path itself as an element (if the path is not finite). What you do mean with the word "cover" escapes me. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |