Cantor Confusion
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From: mueckenh on 17 Jan 2007 06:34 Dik T. Winter schrieb: > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1168955262.910310.48680(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > 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 > > ........................... This finite tree I call: Trauerweide > > So it contains infinitely many nodes? You are again shifting your paradigm. > And I would not think about that as a finite tree, only as a tree that > splits at finitely many places. This was my first definition. The property "finite" belongs to the tree because there are splittings only between level 0 and level n. This shows that the union of paths can really be fomed. So your recent objections are removed (therefore I do not answer your recent contribtions.) > > > > 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. > > No answer to this? The answer is: We can use sets of levels, sets of edges, sets of nodes or sets of paths. Look at the "Trauerweide". > > > > > 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? > > None. But all sets of diagonals of finite squares contain only finite > diagonals, so there can not be an infinite diagonal in their union > (which is a set of infinite size). So there cannot be an infinite diagonal at all? 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: 0 0. / \ 1 0 1 / \ / \ 2 0 1 0 1 ................................ But it does not contain the complete path 0.111... (Otherwise Cantor's proof of the uncountability of the real numbers was contradicted). Now, the path 0.111... in fact would look like a diagonal 0 0. 1 1 2 1 3 1 .................. Are the different notations "line" and "level" the reason for the difference? > > > There are more limits than sequences? > > Eh? Why do you think so? Each sequence has only a single limit (when the > limit is properly defined and when it exists). Correct. > > What we need is: All paths > > in the union tree are countable. All possible nodes are in the union > > tree. > > Yes, you desperately need it. I do not need it, but you do. The countablity of all paths in the union of finite trees is a theorem of set theory. > > > 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?). > > What is the argument here? I think you are back to the old argument, that > is not valid either. > (1) For each pair of paths there is an edge were they differ. in the union of all finite trees. Accepted. > (2) For each path there is no edge where it differs from all other paths. in the union of all finite trees. Accepted. > (And do not come up with the finitistic view of 10^100 * pi. No. It would make this discussion useless. Regards, WM
From: Franziska Neugebauer on 17 Jan 2007 06:43 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > William Hughes schrieb: >> > The trees are sets of nodes. >> >> Trees usually consist of a set of nodes _and_ a set of *edges*. The >> set of paths in a particular tree is defined only if _both_ of the >> sets are given. > > Every edge is related to a node. Every edge is actually related to _two_ nodes (vertices). > The set of nodes is the set of ends of edges united with the set of > root nodes (which is only one). The "set of ends of edges" (united with a set of roots or not) does not _constitute_ the tree. Nonetheless this union may perhaps _equal_ the set of vertices. > Edges can but need not be used. Whether they "need" to be "used" is irrelevant. Relevant is: ,----[ 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. `---- Simple message: No paths without edges. > One can use only nodes as well as one can use only edges. As said: There is already a well-established notion of tree in mathematics. A tree is an ordered pair of a set of vertices and a set of nodes _by_ _definition_. F. N. -- xyz
From: Franziska Neugebauer on 17 Jan 2007 06:50 mueckenh(a)rz.fh-augsburg.de wrote: >> > 0. >> > / \ >> > 0 1 >> > / \ / \ >> > 0 1 0 1 >> > | | | | >> > 0 1 0 1 >> > | | | | >> > 0 1 0 1 >> > ........................... > This finite tree I call: Trauerweide Armutszeugnis is more appropriate. F. N. -- xyz
From: mueckenh on 17 Jan 2007 07:48 Franziska Neugebauer schrieb: > > > > What I defined as the union of finite trees is the union in the sense > > of set theory. > > I remember that you refused to use usual graph theoretical approach > where a tree is a directed graph. A graph is an ordered pair of a set of > nodes _and_ a set of edges. Please show that in general a union of > trees (ordered pairs of sets of nodes and sets of edges) is a tree. I do not assert that this is so in general. Therefore I will not prove it. What I said is that the union of the trees with n levels and the tree with m >= n levels is the tree with m levels. This is valid for the levels, for the nodes, for the edges, and for the paths of trees like the following with n = 2 levels. >> > 0. >> > / \ >> > 0 1 >> > / \ / \ >> > 0 1 0 1 >> > | | | | >> > 0 1 0 1 >> > | | | | >> > 0 1 0 1 >> > ........................... > > > > The Equilateral Infinite Triangle (EIT) contains all lines which can > > be enumerated by natural numbers: > > > > 1 0.1 > > 2 0.11 > > 3 0.111 > > ........... > > This is simply a(n infinite) list (sequence) of unary representations. The diagonal has an infinite number of digits. > > (Otherwise, Cantor's diagonal proof would fail.) > 2. Cantor's proof does not "fail" in that case cause 0.1... is not > _in_ the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, > n e N). I said *otherwise* Cantor's proof would fail. "Otherwise" means: If there is no infinite diagonal. If there is an infinite diagonal, however, then there is no reason for the path 0.111... of the union of finite trees to be finite. Regards, WM
From: mueckenh on 17 Jan 2007 07:54
Franziska Neugebauer schrieb: > > Every edge is related to a node. > > Every edge is actually related to _two_ nodes (vertices). And sometimes they are drawn with black ink while the nodes are red. Sometimes colours are reversed or even different colours are used. Relevance? > > > The set of nodes is the set of ends of edges united with the set of > > root nodes (which is only one). > > The "set of ends of edges" (united with a set of roots or not) does not > _constitute_ the tree. Nonetheless this union may perhaps _equal_ the > set of vertices. > > > Edges can but need not be used. > > Whether they "need" to be "used" is irrelevant. Relevant is: that definition which I gave for that kind of trees which I use in my proof. > > One can use only nodes as well as one can use only edges. > > As said: There is already a well-established notion of tree in > mathematics. A tree is an ordered pair of a set of vertices and a set > of nodes _by_ _definition_. As I said: Relevant is only that definition which I gave for that kind of trees which I use in my proof. Regards, WM |