Cantor Confusion
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From: mueckenh on 23 Jan 2007 06:11 Virgil schrieb: > In article <1169472355.486545.309380(a)q2g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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? > > Non sequitur. What I believe is that complete infinite trees always have > a set of paths which is actual but unlistable. > We talked about Cantor's diagonal method. You said one could apply it in the complete tree. I ask you if this mehod supplies a real number not represented in the tree or not. If it supplies one, then the complete tree cannot be complete. If not, what do you mean by "can be applied"? Regards, WM
From: mueckenh on 23 Jan 2007 06:24 Virgil schrieb: > In article <1169473008.561623.314970(a)v45g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > The minimal set of paths consistent with a tree being infinite is > > > countable but the maximal one is not. > > > > The minimal set is the maximal set. > Specifically the set of all paths which are "eventually constant" is a > proper subset of the set of all possible paths, but it still uses every > node and every edge. Correct. This shows that there are no other paths. It is the core of my contradiction of set theory. Everything else claimed by you is belief in ghost paths (vibrations of vacuum? small black holes in a green tree?), but has nothing at all to do with mathematics. > > > > > > 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. > > > > Both sets are the same countable set. > > As per my example above, the set of "eventually constant" paths is > countable (as can be easily shown) but the set of all paths is not (as > Cantor proved with his original diagonal argument). How did he prove that? Did he really advance to infinity? Did he really surpass the nodes of the union of trees AND the paths of the union of trees? Did anybody, two weeks ago, believer, that there are different sets of paths in the same tree, coming and going like sweet little birds, perhaps even whistling and twittering? No, he said that for a_nn we can put b_nn and claimed that this is valid for every n. I claim that a path touching a node will also touch the next one and that all possible combinations are realized in an infinite tree with all nodes, Why do you believe that in Cantor's matrix there is only one fixed kind of diagonals? According to your arguing, there are many choices. Regards, WM
From: Franziska Neugebauer on 23 Jan 2007 06:27 mueckenh(a)rz.fh-augsburg.de wrote: > 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. Of course. This in this special case (a = a U a) you are right. If A /= B then the A U B is not a tree. Rule: If you talk about "union of trees", where the latter mean sets of nodes and sets of edges, you are not talking about the set-theoretical union. Hence the axiom of union does not apply. >> > 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). Then we have to forget about the usual graph theoretic definition of tree. Then please do not call the objects trees. >> It is an equivocation. > > No. The set theoretic union of the trees (edges are implied by the > general type of tree considered) Please take a decision: 1. Use standard-trees (T = (N, E)) then you cannot use the standard-union. To get a meaningful non-standard-union of all finite trees you must define T_omega to be identical with my G defined elsewhere. The union of all sets of paths of all finite trees is not identical with the set of paths of G. (the former only contains finite paths). 2. Use Virgils trees (T = set of paths having certaing properties) then you can use the standard union. Please do not call the object trees. Since this union is a union of sets of paths of all finite trees it is different from the set of paths of G. 3. Use David Marcus' std-tree together with a non-standard union defined by tree-union V := (std-union set-of-sets-of-nodes(V), std-union set-of-sets-of-edges(V)), which is indeed a tree. You get the same result as under 1. but you do not have to explicitly define the union. I would prefer David Marcus' definition. We can than examine the sets of paths separately from the trees. F. N. -- xyz
From: mueckenh on 23 Jan 2007 06:33 Virgil schrieb: > In article <1169489199.691793.259120(a)m58g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > > > > > Max n,m is not defined either, > > > > > > Aha > > > > > > ,----[ WM in <1169111380.377993.67320(a)l53g2000cwa.googlegroups.com> ] > > > | 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. > > > `---- > > > > > > So you mean m < n is not defined? Then it makes no sense at all to write > > > about trees? > > > > > Sorry, this should read: Max (n,m) is not defined *other* (than for > > finite m and n). The union of m and n is the maximum of both. > > Nevertheless the union of all natural numbers exists as well as the > > union of segments {1,...,n} and {1,..., m} and the infinite union of > > all segments. > > The union of any finite number of finite segments is a finite segment, > but the union of infinitely many finite segments is not a finite > segment, so that WM's argument fails. Sorry, that IS my argument. The union of all finite trees is not a finite tree. Therefore it has no finite paths. But also: All finite trees contain only finite paths. Therefore the union of all finite trees contains no infinite paths. Result: There is no infinite union of elements of any kind. Regards, WM
From: mueckenh on 23 Jan 2007 06:37
Virgil schrieb: > In article <1169489693.112797.92660(a)a75g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > WM is not competent to pass judgement on mathematicians. > > > > That could be decided by mathematicians only. > > Right! And WM is not one of them. That could be decided by mathematicians only. Regards, WM |