Cantor Confusion
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From: hagman on 22 Jan 2007 08:40 mueckenh(a)rz.fh-augsburg.de schrieb: > Virgil schrieb: > > > Cantor, using his original diagonal argument, and anyone else who wants > > to emulate him, can show that any /list/ of paths in a complete infinite > > binary tree is incomplete. > > So you do not believe in complete trees? Somewhere in the mist of > infinity the paths cease to split, or become unobservable? No, there are, at the edge to infinty, an awful lot of new paths that are brought into light, namely non-terminating paths. Finite trees have absolutely zero (non-repeating) non-terminating paths, infinite trees have them (and in fact have continuum-many). Restricting counts to terminating paths starting at the root, they are in correspondence with the nodes and thus countable. Thus you have - finitely many terminating paths - zero non-terminating paths for finite trees and - countably many terminating paths - uncountably many non-terminating paths for infinite trees. While the countably many terminating paths are somewhat like a "limit" of the finitely many terminating paths in finite trees, the non-terminating paths are a feature that applies only to infinite trees, so their count is /not/ a limit of counts in finite trees.
From: mueckenh on 22 Jan 2007 08:59 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > The directions of the paths are the same in T1 and T2. If you insist on > > > > a difference, then it can only result from the length of the paths. > > > > > > Yes. T1 contains only finite paths. > > > > Which path is finite? Where does a path end? > > Every path in T1 ends at a node. The fact > that you can continue a path does not mean that you must continue > a path. There is no path which ends at a node which you can specify. Every path of the union tree has maximum length. 0.11 is not a path in the tree of three or more levels. Define "the maximum set" as the union of {1}, {1,2}, {1,2,3}, ..., {1,2,3,...,n} It is obviously nonsense to assert that every maximum set ends somewhere, because there is only one maximum set which ends at n. > > > > > > > T2 contains only > > > (potentially) infinite paths. > > > > > > > > > The paths in T1 are not determined by the nodes in T1. > > > > > > > > By its nodes and edges every path is determined. > > > > > > Yes, but the fact that these nodes and edges are in T1 > > > does not mean the path determined by these nodes > > > and edges in in T1. > > > > Even if the paths are in T1 this does not mean that the paths are in > > T1, I assume? > > You assume wrong. > > The statement was that even if the nodes and edges of > path P1 are in T1, this does not mean that path P1 is in T1. This statement is wrong. According the definition of a path in the trees defined by me, the nodes and edges determine it completely. If you don't believe it, take it as a definition. > > More formally. T1 is the union of all finite paths. Let M be the > set of nodes in T1, (that is if a node m is in M, then there is a > path p(m) in T1, such that the node m is in p(m)). > > Let p1 be a (potentially) infinite path. Then > > Every node in p1 is contained in some path > in T1. The infinite union of all finite trees can only contain nodes which are in a path. Otherwise the nodes could not enter the union. Nodes are ends of paths of finit trees. > > However > > There is no path in T1 that contains every node in p1 Then not every node of T1 is in T1. > > Proof: > > Let a path in T1 that contains every node in p1 be called L_D > > L_D is in T1 so L_D is finite. > > L_D contains every node in p1, so L_D is (potentially) infinite. > > Contradiction. Therefore L_D does not exist. Let P be in the union of all intial segments {1,2,3,...,n} of N. There is no initial segment P of N which contains all natural numbers. Therefore the union of all P (which is N) does not exist? You make the following error: A node cannot enter the union tree unless it belongs to a path. If there is not one finite tree containing all nodes, then not all nodes can be in the union of all finite trees. This implies that the complete infinite tree contains more nodes than the union of finite trees. The result is: There is no complete set N of all natural numbers unless it contains an infinite number (which, however, by definition cannot belong to N). Result: There is no complete set N of all natural numbers. Regards, WM
From: mueckenh on 22 Jan 2007 09:05 Virgil schrieb: > In article <1169413190.884269.174180(a)q2g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > The directions of the paths are the same in T1 and T2. If you insist on > > > > a difference, then it can only result from the length of the paths. > > > > > > Yes. T1 contains only finite paths. > > > > Which path is finite? Where does a path end? > > At its leaf node, of course. There are no leaf nodes in the unuion tree. > > > > > We do have "to create or make paths", in particular because you cannot > > define what we would have to do in order to make paths without changing > > nodes and edges. If there *are nodes* (connected by edges or my general > > pescription), then there *are the paths* corresponding to them. > > It may be false that every possible path trough a given node or edge > exists, At least in an infinite tree. No. Every possible path is possible, every existing path exists, every possible path exists, and every existing path is possible. > > > Here, we are interested in domains only (length of paths). > . > > > > The interval is a set of points. The union of two sets contains all the > > elements which are in at least one of the sets. This can be one > > interval. > > But need not be. It can be two intervals also, but not if they have one common point. Regards, WM
From: mueckenh on 22 Jan 2007 09:08 Franziska Neugebauer schrieb: > Virgil wrote: > > > The difficulty with that definition is that for "infinite" trees, > > the set of nodes and the set of edges is not enough to determine the > > set of paths, as it is in finite trees. > > > > The minimal set of paths consistent with a tree being infinite is > > countable but the maximal one is not. > > > > As the distinction between the set of paths being countable and > > uncountable is what WM is interested in, those sets of paths must be > > taken into consideration. > > Do you agree that a tree (finite and infinite) is completely determined > by the nodes and edges? > > Do you agree that the set of paths if kind of a "derived" property? > > I asked since I have the apprehension that you take the structure > (M, E, P) (M = set of nodes, E = set of edges, P = set of paths) for > the tree. Whereas I take (M, E) for the tree. That's it! Everything else is unjustified by mathematical thinking. Regards, WM
From: Franziska Neugebauer on 22 Jan 2007 09:24
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> Even if we now use >> >> T(m) U T(n) := T(sup(m, n)). (fin-u') >> >> and and try to define (inf-u) by >> >> T(1) U T(2) ... := T(omega) >> >> it is still left open what T(omega) shall mean. One has to take >> special care for the notation: The union symbol "U" does not mean the >> usual set theoretical union. > > It does. No, it does not. The usual set theoretical union is defined as A U B := { x | x e A v x e B } If A and B are trees (M_A, E_A) and (M_B, E_B) then the union is A U B = { x | x e {M_A, {E_A}} v x e {M_B, {E_B}} } = { M_A, {E_A}, M_B, {E_B} } This union is not a tree. > The nodes can be enumerated in various ways. Here is a very simple > method: > 1 > 11, 12 > 21, 22, 23, 24, A fomal version of this "numbering" I have given in <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net> > ... > The union of trees is the ordered union of their nodes. > The union of above elements exists. > > What is the problem? The problem is that this tree-union is _not_ the set or graph theoretical union. It is an equivocation. F. N. -- xyz |