Cantor Confusion
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From: mueckenh on 16 Jan 2007 10:25 Dik T. Winter schrieb: I every node is in the tree, then every node of 0.01010101... is in the tree too. > > > > > When all nodes are there , then all paths are there. > > > > > > All paths are in the complete tree, > > > > But the union of all finite trees is not the complete tree? > > Pray read what I always state: "the union of the sets of paths in the > finite trees is not the set of paths in the infinite tree". You even > quote it below: Irrelevant. See below. > > Cantor's diagonal can be represented as a (diagonal) path in an > > infinite matrix. This matrix is the union of finite matrices. It is the > > same as with the trees. > > The diagonal is *not* the union of the sets of diagonals of the finite > matrices, because that is a set with infinitely many elements. It is also > not in that union, because all elements of that union have finite length. > To use unions here you have to be careful about defining the sets you > are uniting. Consider the ordered (multi)-sets consisting of the initial n > digits of that diagonal for each n. Each set represents an initial > segment of the diagonal. The union of them can properly be defined and > represents the diagonal. (The definition of union in this case would be > something along: > The union of two ordered (multi)-sets is defined if for every > n either both have an n-th element and they are equal, or at most > one has an n-th member. > You can extend this to an arbitrary collection.) Like the paths in a tree, for instance. > > But when you try to prove that the set of paths is countable you are *using* > the union of sets of paths, and that is not the set about which you want > to prove something. No. What I am using is this: 1) The union of all finite trees contains the union of all finite paths and this set is countable. 2) The union of all finite trees contains all nodes. 3) You cannot add another tree to the set of all finite trees without adding at least one node. (The paths are subsets of the set of nodes. Two different sets must be distinguished by at least one element.) 4) But there is no node to add. 5) Therefore the union of trees contains all possible paths. 6) We know that the set of all paths contained in his union is countable. Regards, WM
From: mueckenh on 16 Jan 2007 10:34 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > No. The (potentially) infinite sequence of all digits is not > > > in the list, nor is it in the union of all finite binary trees. > > > > Simply draw the path of 1/3 with a slope of 45°: > > > > 0. > > 0 > > 1 > > 0 > > 1 > > ... > > > > Then you see that it is in the tree which contains all levels > > enumerated by all n in N, if and only if it is the diagonal of Cantors > > list enumerated by all n in N. > > > > > > > > > > > > Thus one cannot say > > > > > Cantor's diagonal "consists only of finite initial segments". > > > > > > > > One must say so. It is similar to saying N consists of finite numbers > > > > only. > > > > > > No. The statement > > > > > > All elements of N are finite numbers > > > > > > is true. The statement > > > > > > All initial segments of Cantor's diagonal are finite > > > > > > is false. There is one (potentially) infinite initial segment. > > > > in a list enumerated by all natural numbers? But there is no infinite > > path in a tree the levels of which are enumerated by all natural > > numbers? > > There are two such trees. > > T1: The union of all finite trees > > T2: The inifinite tree. > > Both T1 and T2 have levels which are enumeratred by all natural > numbers. > However, the trees are not the same. T1 does not contain an infinite > path. > T2 does contain an infinite path. The trees are sets of nodes. Please give a node which is in one but not in the other tree. Otherwise withdraw your assertion of T1 =/= T2 as wishful believing. Paths are sets of nodes too, they are subsets of trees. > > For Cantor's diagonal we can say: > > The (potentially) infinite set of all finite initial segments of > Cantor's > diagonal does not contain an (potentially) infinite initial > segment. > > Cantor's diagonal does contain an (potentially) infinite initial > segment. For the tree we can say: The infinite set T1 of all finite initial segments of paths does not contain an infinite path. T2 does contain all possible infinite paths. If you cannot distinguish T1 and T2 by finding a node which is in T2 but not in T1, then T2 = T1. Regards, WM
From: mueckenh on 16 Jan 2007 10:48 Dik T. Winter schrieb: > In article <1168955262.910310.48680(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1168869430.273702.199810(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > Depends on definition of "finite tree". If all finite paths are > > > > continued with sequences of 000... or 111..., then the union of paths > > > > of two finite trees is the set of paths of the larger tree. > > > > > > Wrong. You can name the paths anything you like, that does not make it > > > right. At level 2 the finite tree has only paths that go through two > > > nodes. At level 3 the finite tree has only paths that go through three > > > nodes. In the union of the sets of paths there are both paths that go > > > through two nodes and paths that go through three nodes. They are distinct. > > > > I meant the following: Accordng to my original definition, the tree > > from level 0 to level 1: > > > > 0. > > 0 1 > > > > has the paths 0.0000..., 0.0111..., 0.1000..., and 0.1111. (the tails > > 000... and 111... are always appended.) This set of four paths is also > > in the tree from level 0 to level 2 and in all greater trees. > > No, they are not. I see only paths that go through two nodes. There > are no such paths in higher levels. You may give them the same name, > that does not make it the same path. Please look at the paths in original definition: I'll try to draw it again somewhat more suggestive: The finite tree from level 0 to level 1 is: 0. / \ 0 1 / \ / \ 0 1 0 1 | | | | 0 1 0 1 | | | | 0 1 0 1 ............................ > > > But this is irrelevant. We do not unite sets of paths. > > You do when you want to prove that the set of paths is countable. > Remember what you wrote: > (1) In each finite tree the set of paths is finite. > (2) A countable union of finite sets is countable. > (3) So the set of paths in the complete tree is countable. > (3) is not justified because (2) is not about the set of paths in the > complete tree. Now try to do this proof *without* using sets of paths. > > > > > What is the difference > > > > between the union of all finite trees and the infinite tree in terms of > > > > nodes? > > > > > > See above. And ultimately, in the union of the sets of paths of all > > > finite trees, there are only paths that go through finitely many nodes. > > > There is *no* path that goes through infinitely many nodes. > > > In the union of all finite squares > > > > (11) > > > > (11), (12) > > (21), (22) > > > > etc. > > > > there is no infinite diagonal. > > Why not? I only state that that diagonal is not in the union of the sets > of diagonals of the finite squares. Which digit is missing? > > > > > If there is a difference, then it can exist also in Cantor's list. As > > > > the digit a_nn has always a finite distance from the first line it has > > > > also always a finite distance from the first column. You never cover > > > > the complete list. > > > > > > That is something different. Above we are talking about union. Here > > > you are talking about limit. > > > > Why do you refuse to talk about the limit in case of paths? > > In that case *you* have to define what you understand under limit. But > lets suppose that s_n is the set of paths in the finite tree of level n. > I concede that you might be able define a limit such that > lim{n -> oo} s_n = s > where s is the set of paths in the complete tree. But with this you > can show nothing about the cardinality of s because > lim{n -> oo} | s_n | = | s | > is not necessarily valid. There are more limits than sequences? Seems to be not very well defined. But let that be. We don't need it. What we need is: All paths in the union tree are countable. All possible nodes are in the union tree. A path can only then differ from all paths of the union tree if there is another node (or if it is running in other directions --- try that argument?). > > > > > > How than can you draw conclusions about the cardinality of the sets of > > > > > paths in the union from the cardinalities of the sets of paths in the > > > > > finite trees? > > > > > > > > A union cannot have more elements than are united. > > > > > > Indeed. But the union is not the complete collection. > > > > The union of finite trees contains all (initial) sequences (of paths). > > Their number is countable. > > As long as you allow only the *finite* initial sequences, they are > countable. Not if you allow more. That is because the union of the > sets of paths in finite trees does exist just of the set of finite > initial sequences. And you proof about sets of paths works for this. > But that set is not the set of all paths. But if there are other paths, then they should prove their being different by showing a node not in the union. Different sets differ by at least one element. Or not? Did you receive and enjoy my book which I sent off on 10. Jan.? Regards, WM > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 16 Jan 2007 11:06 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > No. The (potentially) infinite sequence of all digits is not > > > > in the list, nor is it in the union of all finite binary trees. > > > > > > Simply draw the path of 1/3 with a slope of 45°: > > > > > > 0. > > > 0 > > > 1 > > > 0 > > > 1 > > > ... > > > > > > Then you see that it is in the tree which contains all levels > > > enumerated by all n in N, if and only if it is the diagonal of Cantors > > > list enumerated by all n in N. > > > > > > > > > > > > > > > Thus one cannot say > > > > > > Cantor's diagonal "consists only of finite initial segments". > > > > > > > > > > One must say so. It is similar to saying N consists of finite numbers > > > > > only. > > > > > > > > No. The statement > > > > > > > > All elements of N are finite numbers > > > > > > > > is true. The statement > > > > > > > > All initial segments of Cantor's diagonal are finite > > > > > > > > is false. There is one (potentially) infinite initial segment. > > > > > > in a list enumerated by all natural numbers? But there is no infinite > > > path in a tree the levels of which are enumerated by all natural > > > numbers? > > > > There are two such trees. > > > > T1: The union of all finite trees > > > > T2: The inifinite tree. > > > > Both T1 and T2 have levels which are enumeratred by all natural > > numbers. > > However, the trees are not the same. T1 does not contain an infinite > > path. > > T2 does contain an infinite path. > > The trees are sets of nodes. Plus a structure on the nodes. The set of nodes in T1 is the same as the set of nodes in T2. The structure placed on the nodes in T1 is not the same as the structure placed on the nodes in T2. > Please give a node which is in one but not > in the other tree. No such node exists. The set of nodes is the same for each tree. However, a tree is more than its nodes. > Otherwise withdraw your assertion of T1 =/= T2 as > wishful believing. No. The can be other differences. > Paths are sets of nodes too, they are subsets of > trees. Just because a set of nodes is in T1, does not mean that the path represented by that set of nodes is in T1. > > > > For Cantor's diagonal we can say: > > > > The (potentially) infinite set of all finite initial segments of > > Cantor's > > diagonal does not contain an (potentially) infinite initial > > segment. > > > > Cantor's diagonal does contain an (potentially) infinite initial > > segment. > > For the tree we can say: The infinite set T1 of all finite initial > segments of paths does not contain an infinite path. T2 does contain > all possible infinite paths. Correct. > If you cannot distinguish T1 and T2 by > finding a node which is in T2 but not in T1, then T2 = T1. Incorrect. Two different trees can have the same nodes. - William Hughes
From: mueckenh on 16 Jan 2007 11:27
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > > > > > No. The (potentially) infinite sequence of all digits is not > > > > > in the list, nor is it in the union of all finite binary trees. > > > > > > > > Simply draw the path of 1/3 with a slope of 45°: > > > > > > > > 0. > > > > 0 > > > > 1 > > > > 0 > > > > 1 > > > > ... > > > > > > > > Then you see that it is in the tree which contains all levels > > > > enumerated by all n in N, if and only if it is the diagonal of Cantors > > > > list enumerated by all n in N. > > > > > > > > > > > > > > > > > > Thus one cannot say > > > > > > > Cantor's diagonal "consists only of finite initial segments". > > > > > > > > > > > > One must say so. It is similar to saying N consists of finite numbers > > > > > > only. > > > > > > > > > > No. The statement > > > > > > > > > > All elements of N are finite numbers > > > > > > > > > > is true. The statement > > > > > > > > > > All initial segments of Cantor's diagonal are finite > > > > > > > > > > is false. There is one (potentially) infinite initial segment. > > > > > > > > in a list enumerated by all natural numbers? But there is no infinite > > > > path in a tree the levels of which are enumerated by all natural > > > > numbers? > > > > > > There are two such trees. > > > > > > T1: The union of all finite trees > > > > > > T2: The inifinite tree. > > > > > > Both T1 and T2 have levels which are enumeratred by all natural > > > numbers. > > > However, the trees are not the same. T1 does not contain an infinite > > > path. > > > T2 does contain an infinite path. > > > > The trees are sets of nodes. > > Plus a structure on the nodes. The set of nodes in T1 is the > same as the set of nodes in T2. The structure placed on the nodes > in T1 is not the same as the structure placed on the nodes in T2. Ah, well. Dik had just propsed some paths running horizontally. > > > Please give a node which is in one but not > > in the other tree. > > No such node exists. Nice to hear. > The set of nodes is the same for > each tree. However, a tree is more than its nodes. Yes, some breath of God must be in there. > > > Otherwise withdraw your assertion of T1 =/= T2 as > > wishful believing. > > No. There can be other differences. I deeply deplore that you and others are allowed to call your religious beliefs mathematics. But it is clear now why there is no contradiction in ZFC and why there will never be such a contradiction. In fact, there can always be other things in there which cannot be determined but which invalidate an obvious contradiction. > > > Paths are sets of nodes too, they are subsets of > > trees. > > Just because a set of nodes is in T1, does not mean > that the path represented by that set of nodes is in T1. All paths run downwards. There is no other possibility than being longer or shorter. But this property is easily translated in having more nodes or less. And this leads to at least one node which is different. > > > > > > > For Cantor's diagonal we can say: > > > > > > The (potentially) infinite set of all finite initial segments of > > > Cantor's > > > diagonal does not contain an (potentially) infinite initial > > > segment. > > > > > > Cantor's diagonal does contain an (potentially) infinite initial > > > segment. > > > > For the tree we can say: The infinite set T1 of all finite initial > > segments of paths does not contain an infinite path. T2 does contain > > all possible infinite paths. > > Correct. If there is no difference in terms of nodes, why does T1 not contain all possible paths? What must be changed so that T1 gets T2? Concerning the hard ware (not the breath of God). > > > If you cannot distinguish T1 and T2 by > > finding a node which is in T2 but not in T1, then T2 = T1. > > Incorrect. Two different trees can have the same nodes. What must be changed in order to make T1 being T2? What must be changed in Cantor's matrix to remove the diagonal but leaving the number of lines infinite? Regards, WM |