Cantor Confusion
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From: mueckenh on 23 Jan 2007 04:52 Virgil schrieb: > In article <1169396052.939963.194070(a)q2g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > Therefore, by induction, all finite n-lever trees for all n in N exist. > > > But that is all that standard induction allows one to conclude. > > > > It is enough. Or should there one level be missing? Please specify > > which remains to be included. > > Anything that concludes anything about infinite trees. Conclusions are not members or subsets of trees. > One can allow or disallow infinite paths if one is only concerned with > nodes and edges and still have a tree. How do you manage that? > > The issue is what constitutes an infinite tree. I claim that an infinite > binary tree in which every path is "eventually constant" is in every > respect an infinite binary tree with the same set of nodes edges and the > same set of edges as the complete infinite binary tree. Of course. And the same set of paths too. =================== > Why should sparing you work be any sort of goal for us as you do nothing > by way of serious work or logical thought to spare us work. > In fact, you seem to go out of your way to misunderstand and > misrepresent what we do and say. You say that paths are not defined by nodes and edges alone. That is not serious. Regards, WM
From: mueckenh on 23 Jan 2007 04:57 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. Every path in a finite tree ends at a node. T1 is infinite and has no finite paths. > The fact > that you can continue a path does not mean that you must continue > a path. > Whenever there is an edge leading a path, this path is there. There is no question of continuing as an arbitrary process. Process? There are no processes in sets and subsets. They simply exist or do not exist. Your arguing is refused. > Regards, WM
From: mueckenh on 23 Jan 2007 05:20 Dik T. Winter schrieb: > > > If the union of singletons {1} U {2} U {3} U... U {n} U ... is an > > infinite number omega, then the union of domains of sequences {1} U {1, > > 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an > > infinite seqeunce {1, 2, 3, ...}. > > Pray explain what you understand under "domain of", and what you understand > under "union of domains". That is complete non-standard terminology. The domain of a sequence is a natural number. The domain of an infinite sequence is omega. > But > whatever, whenever in a collection of sets A_k, none of the sets A_k contains > a particular element E, it is also not in the union. Therefore there is no uncuntable set of path in the union tree. But there are all paths in the union trees. > > How could a path be in the union of finite trees if it were not a path > > of a finite tree? > > Ah, that is the key question. Clearly 0.010101... is not a path in any > of the finite trees. But it is because by one of your definitions of path > and tree such a path does exist. You never stop to rigurously define what > a tree and a path is. I have a few times attempted to define a tree, but > each time you skipped my definition and resorted to intuitive, bad-defined > concepts. Once you stated that a tree was a set of nodes. That is > obviously false. No, it is correct, provided the edges are determined by the construction of the tree. I use only this special type of trees (i.e., every node has two child nodes). > A set of nodes is nothing more than a set of nodes > until you introduce edges. When you introduce edges you get a graph. > But, equally obviously, not every graph is a tree. A tree needs a > set of nodes and a set of directed edges (in a particular fashion). > I did grant you all that and gave (I think) a rigorous definition of a > tree, and even how to produce unions of trees. It is clear how to produce the union of the finite trees to all who discuss here. Virgil, William, and you agree that this union contains all nodes and all edges and all levels. Franzisca also knows and undertands that but refuses to admit it, because she also knows the consequences. > > > Two trees which are identical by all nodes and edges are identical by > > all paths. > > Right. > > > T1 as the countable union of all finite sets of finite paths contains > > only a countable set of finite paths. > > No. It *is* the countable set of finite paths. Right. It is the union of all nodes (and edges and levels and paths). > > > And T1 = T2. > > No. T1 is a set of paths, it is not a tree. T1 does not contains nodes > or edges as elements. Only paths. The nodes can be enumerated in various ways. Here is a very simple method: 0 11, 12 21, 22, 23, 24, .... n1, n2, ..., n2^n This is the finite tree (of the kind defined by m) with n levels . The union of trees is the ordered union of their nodes. The union of above elements exists for any set of n in N. This union is a tree, i.e., an ordered set of nodes as given above. > > > > Here it is (for trees of the kind weeping willow): Taking the union of > > two trees corresponds to taking the union of their sets of paths. > > Why? The union of the sets of paths is a set of paths, not a tree. The union of trees is a union of their nodes. The paths are merely subsets which are defined by the nodes and the special kind of tree. > > > Then > > we have:The set of paths in the union is the set of paths in the larger > > tree. > > If you define a tree by the set of paths in it. Yes. > > > This holds if it holds for domains of finite seqeuences. The domain of > > two finite sequences is the domain of the longer sequence. The domain > > of all finite sequences is omega. > > But using this kind of defining and uniting trees, the complete tree will > not have 0.010101... as a path. Do you see how it crucially depends on > how you define things? If you define trees by nodes and edges, and paths > as a consequence of that, you get a different result. (And I still wonder > what you mean with "domain of".) The domain of the path 1,0,1 is 1,2,3. A path is a sequence. A sequence is defined as a mapping f from a natural number or omega into the reals. The set of reals (here only 0 and 1) is called the range. Abbreviations are usual: dom f and ran f. > > > > > The tree is an (ordered) set of nodes. The nodes can be used to form > > paths, i.e., subsets of the tree. > > As a tree is a set of nodes, any subset of it is a set of nodes. Odf course, therefore the subsets which are paths must exist in the tree if the subsets exist there. >Not > every subset of it is a path. But whatever, the number of subsets of > the final tree is not countable. The union of the sets of paths of > the finite trees is not the set of paths of the complete tree. If all nodes and edges are in the union, then all subsets are in the union too. Regards, WM
From: mueckenh on 23 Jan 2007 05:30 hagman schrieb: > 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, Where can we find this edge? Ijn my opinion you are at the edge either if you begin to count 1,2,3,... --- ore nowhere. > 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, The union of all finite trees has no terminating path because the union of all finite segments {1,2,3..., n} (with in N) does nowhere terminate. Regards, WM
From: mueckenh on 23 Jan 2007 05:38
Franziska Neugebauer schrieb: > 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. unless M_A, E_A and M_B, E_B have same elements at all levels they have in common. > > > 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. Then forget about the graph theoretic union for the moment. Then we are doing soemthing new (but not really). > It is an equivocation. No. The set theoretic union of the trees (edges are implied by the general type of tree considered) 0 11, 12 21, 22, 23, 24 .... n1, n2, ..., n2^n and 0 11, 12 21, 22, 23, 24 .... m1, m2, ..., m2^m is the larger one of both. The same definition applies to the union of natural numbers or initial segments of them. And simiarly we can take the union over all levels or over all finite trees. Regards, WM |