Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Franziska Neugebauer on 22 Jan 2007 04:54 Virgil wrote: [...] > My analysis is based on "embedding" each of the finite trees into your > infinite Q(T*) as a proper subtree and then showing that the union of > all those subtrees is far short of the maximal infinite binary tree. I think noone (except WM) doubts that the countably infinite union of the set of the sets of *paths* (not the trees!) of all finite binary trees is different from the set of all paths in the infinite binary tree. The latter contains infinitely long paths but the former does not. But the union of the set of the sets of paths P(n) of the finite binary trees trees T(n) n e omega is not a tree at all. It is a set of paths. So to meaningfully speak of "union of all finite trees" (WM) T(1) U T(2) ... = U { T(i) | i e N } you must either 1. (WM-1) re-define union and the union-U so one gets the binary tree having max({ depth(T(i) | i e N }) depth as "union". -> undefined since max(N) does not exist. 2. (WM-2) re-define union and the union-U so one gets the binary tree having sup({depth(T(i) | i e N }) depth as "union". -> undefined yet. But after extending T(n) to T(omega) meaning the infinite binary tree it may be meaningful though due to the use of a redefined notion of "union" this leads to faulty conclusions about what cardinality can be "reached" by countable "unions" of finite sets. 3. (Virgil?) Use a variant definition of tree. Possibly by defining every set of paths (having certain properties) to be a tree. Regardless of what you choose you must deviate from the vital mathematical definitions to achieve any reasonable result (if at all). F. N. -- xyz
From: Franziska Neugebauer on 22 Jan 2007 05:11 Dik T. Winter wrote: > In article <1169375886.670658.154840(a)q2g2000cwa.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: [...] > Once you stated that a tree was a set of nodes. That is > obviously false. 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. Do you have a message id for your definition of unions of trees? [...] > > 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. Correct. > > > > We see, the trees T1 and T2 are identical with respect to all > > > > nodes, all edges, and all paths (which would already have been > > > > implied by the identity of nodes and edges). But the set of > > > > all paths is countable in the tree T1 and uncountable in the > > > > same tree T2. > > > > > > It is not countable, you have not given a proof. > > > > 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. Correct. > > 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. Doesn't it depend on the defintion of "the union"? [...] > > > > Then it exists in the union tree - together with all paths of > > > > same length. > > > > > > But I am not contradicting that. Again, I state that it is not > > > in the > > > union of sets of paths in finite trees. There is a huge > > > difference. > > > > See the proof above. The sets of paths can be unioned as the trees > > can be unioned. > > Not so. If you define trees as sets of paths, they can be united as > you wish, but you will not have 0.010101... in the union tree. True. [...] > > There is a bijecton between the set of finite trees T(n) and the > > sets of paths P(n) in finite trees. > > Yes. They are both countable. Wording: WM claims a bijection between { T(n) | n e N } (*) and what? a) P(n) n e N <- what is this or does he mean "the set of the sets of paths P(n) in finite trees T(n)"? b) { P (n) | n e N }. There is trivially a bijection between (*) and b). [...] F. N. -- xyz
From: mueckenh on 22 Jan 2007 08:14 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. The nodes can be enumerated in various ways. Here is a very simple method: 1 11, 12 21, 22, 23, 24, .... The union of trees is the ordered union of their nodes. The union of above elements exists. What is the problem? Regards, WM
From: mueckenh on 22 Jan 2007 08:25 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? That is a possible point of view to avoid inconsistencies. An even better point of view would be not to believe in complete infinite sets of digits of real numbers and of complete diagonals of infinite lists. Somewhere in the mist of infinity ... Regards, WM
From: mueckenh on 22 Jan 2007 08:36
Virgil schrieb: > > 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 tree is defined by the set of edges and the set of nodes. (If only trees are in question which I defined as cut trees or as weeping willow trees, then even the nodes are sufficient, in fact even the levels are sufficient to determine the tree completely. ) It is impossible that there are different paths in the same tree, i.e., in a tree which is well defined by its nodes and edges, the set of paths is well defined. All other assertions are unmathematical. > > 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. > > 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. Regards, WM |