Cantor Confusion
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From: Virgil on 16 Jan 2007 14:40 In article <1168958258.416263.123350(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > The union of all such restricted infinite binary trees will be a subtree > > of the compete infinite binary tree. > > > > So far, I hope WM and I agree. > > Of course. > > > > The issue between us is whether that union will be a proper subtree of > > the complete infinite binary tree or will be the whole tree. > > This question can easily be decided by finding a node which is in the > whole tree but not in the union. Should that be impossible, then a > property of the subtree is that it is not a proper subset but the > complete tree. > > > > Since every path in every one of the restricted infinite binary trees in > > the union is eventually constant, every path in the union of all those > > trees will also be eventually constant. > > That would be correct for a finite union of finite trees. What is > correct for the finite case need not be correct for the infinite case. If every path in every tree in a union of trees has a certain property then every path in the resulting tree has that property. In this case the property is that of being eventually constant. > > > > So that, for example, the union will not contain any path which > > alternates between branching left and branching right. > > That would be correct for a finite union of finite trees. What is > correct for the finite case need not be correct for the infinite case. So that WM claims that there are members of a union of sets which are not members of any of the sets being unioned. Not hardly! > > > > Further, it will not contain any path which has infinitely many > > branchings in both directions. > > That would be correct for a finite union of finite trees. What is > correct for the finite case need not be correct for the infinite case. We have infinitely many finite sets of eventually constant paths. We form the set-union of all those sets of paths. And WM requires that the union of those sets contain something not in any of the separate sets. What does WM think a "union" of sets is? The proper definition of a union of sets is a set that contains x as a member if and only if at least one of the sets being unioned contains x as a member. Whether the union is of finitely many sets or infinitely many makes no difference to that definition. So whatever WM's idea of "union" is, it is not the mathematical union of sets.
From: Virgil on 16 Jan 2007 14:49 In article <1168960204.698395.278050(a)m58g2000cwm.googlegroups.com>, 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. Are we expected to accept such garbage? > > > > > > > > > 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? Apples and oranges. The union of the finite trees will allow at most countably many of the uncountably many possible infinite paths to occur. We start by assuming that all trees and all paths issue from a common root node. We may embed every such finite binary tree in a complete infinite binary tree with the same root node by extending each finite path infinitely in the direction of its last branching, so each such infinite path is "eventually constant", in that past some point all its branchings are in the same direction. Then each finite binary tree extends to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree . (Where in an incomplete tree not every node has the maximal allowable number of child nodes while in a complete tree every node does have the maximum number of child nodes). The every path in every finite subtree corresponds to a unique infinite path in the complete tree by this extension. The 'union' of all such incomplete infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the entire tree. Since every path in every one of the incomplete infinite binary trees in the union is 'eventually constant', every path in the union of all those trees will also be eventually constant. But only countably many of the uncountable paths in the complete binary tree are eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then easily show that the set of paths in the union will be countable, but the set of paths in the complete infinite binary tree will be uncountable.
From: Virgil on 16 Jan 2007 14:56 In article <1168961119.168661.39130(a)38g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > 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. For such finite path, there are uncountably many infinite paths having the finite path as an initial segment. > 2) The union of all finite trees contains all nodes. Irrelevant. > 3) You cannot add another tree to the set of all finite trees without > adding at least one node. Irrelevant. > (The paths are subsets of the set of nodes. Two different sets must be > distinguished by at least one element.) But each such path so far is a finite set of nodes, whereas every infinite path is an infinite set of nodes. The set of finite subsets of a countably infinite N is countable. The set of all subsets of a countably infinite N is uncountable. So counting only the finite subsets doesn't count. > 6) We know that the set of all paths contained in his union is > countable. WM is remarkable for the number of things he "knows" that aren't so.
From: Virgil on 16 Jan 2007 15:01 In article <1168961641.904508.35010(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It is not by single nodes but by sets of nodes that one determines a path. So that WM's argument by nodes is irrelevant. The set of all finite subsets of any infinite but countable set is again countable, but the set of all subsets is not.
From: Virgil on 16 Jan 2007 15:06
In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > There are more limits than sequences? There are certainly more limits than finite 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. Why do we need falsehoods? > All possible nodes are in the union > tree. Nodes are irrelevant. > > > > 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? How many different subsets are there to an infinite set of nodes? For finite subsets, countably many , for all subsets, uncountably many. |