Cantor Confusion
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From: Franziska Neugebauer on 18 Jan 2007 03:27 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> You can read what _mathematically_ relevant in >> >> ,----[ http://en.wikipedia.org/wiki/Simple_path ] | In graph theory, a path in a graph is a sequence of vertices such that | from each of its vertices there is an edge to the next vertex in the | sequence. > No. Just what is mathematically relevant, is not yet given there. As I have mentioned in (one of) my last posts: You have not given a definition of a union of all finite trees. F. N. -- xyz
From: mueckenh on 18 Jan 2007 04:07 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > T1, the union of all finite trees, contains only finite paths. It would be nice if you could show the existence of at least one finite path in this union. Can you? Where does a path end? > If you define a tree to be a set of paths, then > two different trees can have the same nodes. My trees are defined to be a set of nodes and of edges. Every uninterrupted sequence of edges is called a path. Therefore my finite trees and T1 and T2 are trees. > > > > > What must be done to make Cantor's matrix non-node-complete (i.e., to > > remove the diagonal without changing the hard ware)? Or what has to be > > done to make any tree non-node-complete, for instance a complete tree > > or a tree which contains 0.010101... but not 0.010110110...? How can we > > recognize whether a tree is (partially) node-complete? > > We cannot. Nonsense. If there are all digits whicgh are indexed by natural numbers, then we have an infinite sequence, i.e., a real number. > > If you define a tree to be a set of paths then > just looking at the nodes will not tell you which paths are > in the tree. But I do not define a tree to be a set of paths ! > If you define a tree to be a set of nodes, and of edges > and > any path through the nodes is in the tree, any path consisting of edges. > then looking > at the nodes will tell you what paths are in the tree > but T1 is not a tree. > Why not? > However, this is not T1. Using your definition of tree, > the union of all finite trees is not a tree. You are wrong. Look at my definition. > > > But it does not contain the complete path 0.111... (Otherwise Cantor's > > proof of the uncountability of the real numbers was contradicted). > > Correct. A union of finite paths does not contain an infinite > path. The *finite* union of all finite paths does not contain an infinite path. The union of all finite initial segments {1,2,3,...,n} contains only finite initial segments. Nevertheless it is said to be an infinite segment. > Note, however, that the union of finite paths, does contain > every node of the infinite path. And who prohibits me to follow the sequence of nodes without end? "Without end" is "infinitely". > > > 0 0. > > 1 1 > > 2 1 > > 3 1 > > ................. > > > > As noted the EIT contains every node needed to make > up the diagonal. Whether the EIT contains the diagonal > is a matter of definition. No. The EIT contains every node. Nobody cuts off the path crossing each one. It is infinite, by definition. > > > > > Are the different notations "line" and "level" the reason for the > > difference? > > > > No. The difference is whether you say I prefer to prove it. > that a tree must > contain any path defined by its nodes. There are things one must see to believe them, and there are other things one must believe to see them. Regards, WM
From: mueckenh on 18 Jan 2007 04:09 Franziska Neugebauer schrieb: > As I have mentioned in my last post: You have not given a definition of > a union of all finite trees. The union of two finite trees T(m) and T(n) with m and n levels, respectively, where m < n, is the tree with n levels. This definition unites sets of nodes (and sets of edges, respectively) and it is valid for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). The union of all finite trees is the union of all trees with n levels where n is a natural number: UT = T(1) U T(2) U T(3) U ... The paths in a tree are completely defined by the sequences of nodes (or edges) which can be followed to an end in CT or without an end in their union as well as in WWT. Regards, WM
From: mueckenh on 18 Jan 2007 04:12 Carsten Schultz schrieb: > mueckenh(a)rz.fh-augsburg.de schrieb: > > Virgil schrieb: > > > >> In article <MPG.2016e6f7746f9f07989b49(a)news.rcn.com>, > >> David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > >> > >>> mueckenh(a)rz.fh-augsburg.de wrote: > >>>> 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 this union is > >>>> countable. > >>> How does 6 follow from 1-5? > >> It doesn't, > > > > It follows from the theorem that a countable union of finite sets is > > countable. > > Every finite tree contains only a finite set of paths. > > The union of finite trees is countable. > > > Let N_i be the node sets of the subtrees, which we will denote by T_i), > and let us assume that they are all induced subtrees. Let us also > denote the set of paths in T_i by P_i. Now let p be a path in the big > tree. Then the following statements hold: Is your "big tree" the union of all subtrees? Or is it more (or less)? > > 1) The path p is a path in the union of the trees T_i (ie the induced > subtree on the union of the N_i) if and only if every node that p visits > is in some N_i. > > 2) The path p is an element of the union of the P_i if and only if > there is an N_i that contains every node that p visits. You assume that the union P_i of paths contains more paths than can be constructed from finite initial segments? > > Now you might think that these conditions are the same and derive from A tree with a unique set of nodes and edges would not contain two different sets of paths. > the countability of the union of the P_i that there are only countably > many paths in the union of the T_i. Is Cantor's diagonal longer than the union of all of its initial segments? Is N more than the union of all of its initial segments {1, 2, 3, ..., n}? Yes? Then N is certainly also more than the union of all ends {n} of its initial segments? No? The union of ends is more than the union of segments? Next time, please switch on your brain before posting. > But that would not prove this > statement There are things one must see to believe them, and there are other things one must believe to see them. Seeing different paths in identical trees belongs to the latter. Your arguing is an example of the first. Regards, WM
From: mueckenh on 18 Jan 2007 04:40
William Hughes schrieb: > I will switch to your defintion of a tree. A tree consists > of a set of nodes, and contains all paths that > the nodes define. > > > Of course not. The complete tree T1 contains all paths. > > T1 is not a tree. Here are some examples and the definition. According to this definition, T1 is a tree. This is a tree of type Weeping Willow with n = 1 level before getting constant. >>> > 0. >>> > / \ >>> > 0 1 >>> > / \ / \ >>> > 0 1 0 1 >>> > | | | | >>> > 0 1 0 1 >>> > | | | | >>> > 0 1 0 1 >>> > .......................... This is a cut tree with one level >>> > 0. >>> > / \ >>> > 0 1 The union of two finite trees T(m) and T(n) with m and n levels, respectively, where m < n, is the tree with n levels. This definition unites sets of nodes (and sets of edges, respectively) and it is valid for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). The union of all finite trees is the union of all trees with n levels where n is a natural number: UT = T(1) U T(2) U T(3) U ... The paths in a tree are completely defined by the sequences of nodes (or edges) which can be followed to an end in CT or without an end in their union as well as in WWT. > > > > > > > T1, the union of all finite trees, contains only finite paths. You > > > cannot make > > > T1 contain an infinite path by playing with > > > definitions. > > > > Then show me which node of the path 0.010101... is missing in T1. > > No node is missing. The path is missing. T1 is not a tree. > The fact that the nodes exist does not mean that the path > exists. The fact that following the edges one does never have to stop means hat an infinite path exists. In the union of all cut trees this is the case. > > > > > > > > > > > Given any tree (set of paths) we can add paths to > > > > > it to make it node-complete. > > > > > > > > > How can we add? > > > > How can we add a path??? If it is not in the tree, but no node is missing therein? > > > > > > > > > T1 is not node-complete. T2 is node-complete. > T1 does not have the property that given a set of nodes > in T1, the path through those nodes is in T1. The path through those nodes *is* the (ordered) set of nodes. The order is coming in by the indexes of the levels. > > > > > > > If you define a tree to be a set of paths, then > > > two different trees can have the same nodes. > > > > Not if all paths go downwards as regularly as in my tree. > > If you define at tree as a set of paths, then it is > trivial to find two different trees with the > same nodes. If you change the definition of tree to have > a tree defined by its nodes, That is the definition of my trees. Path are nly subsets which are in the tree or not. This can be determined by the nodes and edges. > then two different trees > cannot have the same nodes (DUH!). In fact? That do even you agree to? > However, once > you change the definition, T1 is not a tree. Call it as you like. T1 is a set of nodes and edges. And all the paths through connected edges can be followed without end. That means the paths are infinite. Further it can be shown that no path is missing. Do you really think that the union of all initial segments of the number 0.010101... is not this number? Where would it end? > > > > > We cannot recognize from a string of digits, what number they > > represent? > > We can tell what number they represent. However, > we cannot tell whether this string of digits is > a member of the tree (old definition). > Using your definition of a tree there is no such > thing as a tree that is not node-complete, but > on the other hand, T1 is not a tree. What would you call it? What would be required to complete it to be a tree? > > It is a tree. > > No. Any tree that contains T1 must contain an infinite > number of nodes. Any tree that contains an infinite number > of nodes must contain an infinite path. T1 does not > contain an infinite path. T1 is not a tree. What would be required to complete it to be a tree? > > > It is an infinite complete tree. It cantains all paths > > which exist outside of the world of Harry Potter. > > No, the tree that contains all paths is T2. T2 is > a tree, but T2 is not T1. > > > > > > > > > > > > Try to find the difference by following simple construction: > > > > > > > > The Equilateral Infinite Triangle (EIT) contains all lines which can be > > > > enumerated by natural numbers: > > > > > > > > 1 0.1 > > > > 2 0.11 > > > > 3 0.111 > > > > ........... > > > > > > > > In addition it contains the complete diagonal 0.111... (Otherwise, > > > > Cantor's diagonal proof would fail.) > > > > > > > > The union of all finite binary trees contains all levels which can be > > > > enumerated by natural numbers: > > > > > > > > > > Correct > > > > > > > 0 0. > > > > / \ > > > > 1 0 1 > > > > / \ / \ > > > > 2 0 1 0 1 > > > > ............................... > > > > > > > > > > However, this is not T1. Using your definition of tree, > > > the union of all finite trees is not a tree. > > > > Look at the original definition: > > > > 0. > > / \ > > 0 1 > > / \ / \ > > 0 1 0 1 > > | | | | > > 0 1 0 1 > > | | | | > > 0 1 0 1 > > ........................... > > This is a 1-level tree. > > So? My point is that T1 is not the tree that you showed. Why is > showing another tree that is not T1 relevant? > > > > > > > > But it does not contain the complete path 0.111... (Otherwise Cantor's > > > > proof of the uncountability of the real numbers was contradicted). > > > > > > Correct. A union of finite paths does not contain an infinite > > > path. Note, however, that the union of finite paths, does contain > > > every node of the infinite path. > > > > What is then missing to make the infinite path exist? > > The path. Only if we have a tree can we conclude > that the path exists from the fact that the nodes exist. > We do not have a tree. I will not further reply to this silly arguing. Rgards, WM |