Cantor Confusion
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From: Virgil on 21 Jan 2007 16:31 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. > > > > > > Proof of existence of the union tree by proofs of A, B, and C: > > > A) Proof of the existence of one infinite path by induction over the > > > indexes > > > {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n}... U ... > > > is the same as the union of the ends of the initial segments > > > {1} U { 2} U {3} U... U {n}... U .. > > > which is N which exists. > > > B) Proof of the existence of all infinite paths: See proof of the > > > existence of all real numbers in [0, 1]. > > > C) Proof of the existence of all paths in the tree being infinite:: All > > > paths of a tree have same length by definition. > > > > First of all, the standard definition of finite binary trees does not > > require that all paths in such a tree be of the same length. > > But the definition given by me requires it. So it holds for both types > of my trees, the cut tree as the weeping willow. So? > > > It is only > > in our special examples here that we are making such a requirement, but > > it is not "by definition". > > > > Secondly, there are lots of different infinite binary trees in which > > every path is of infinite length. The one of major interest is that one > > having one path corresponding to every possible binary sequence, which > > we have been calling the complete binary tree. It has uncountably many > > paths. > > > > But there are also infinite binary trees containing the same set of > > nodes and the same set of edges but having only countably many paths. > > You assert so, yes, we know. But how can a tree prohibit a possible > path to exist? All the nodes and edges guiding the path are there. What > hinders the path to exist? Is it the spirit required to understand set > theory? One can allow or disallow infinite paths if one is only concerned with nodes and edges and still have a tree. 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. And none of WM's wigglings have persuaded me otherwise. What property of "treeness" does such a tree lack, WM? > > > > For example, the infinite binary tree having only "eventually constant" > > paths (all branchings are 0's or all 1's from some node on) has exactly > > the same set of nodes and exactly the same set of edges as the complete > > binary tree. > > > > So the range of such infinite binary trees includes both ones with only > > countably many paths and ones with uncountably many paths. > > The first type of trees is certainly existing. But how can we conclude > that the second type does exist? Isn't it the conclusion that the > representation of a number with N digits is the representation of a > real number? The incomplete infinite binary tree as described corresponds only to the set of binary rationals in [0,1], those whose fractional form has denominator a power of 2, the complete tree to the set of all reals in [0,1].
From: Franziska Neugebauer on 21 Jan 2007 16:33 David Marcus wrote: > Franziska Neugebauer wrote: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > V* is not in the union (as omega is not in the union over all >> > natural numbers or over all finite intial segments of omega). >> >> I want you to define >> >> T(1) U T(2) U ... > > Do you think WM can actually produce a definition? His knowledge of > math is rather limited. Until he does I have _no_ clue what his writing is really about or whether it is about sth. at all. F. N. -- xyz
From: Virgil on 21 Jan 2007 16:52 In article <1169396771.216800.71250(a)38g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > Where is your union of all finite trees? You have not yet shown: > > It is interesting: > William knows that T1 and T2 exist, but T1 is different. > Virgil is also accepting T1 but, like William, sees only finite trees > therein. What I do not see is why in an infinite tree, WM claims that the set of nodes and the set of edges is sufficient to define the tree, particularly when I have examples proving otherwise. What I have done is to embed each finite binary tree isomorphically into a proper subtree of the complete infinite binary tree. Then the union of those images, already all subtrees of a single tree, makes sense. For every finite binary tree with at least two branches from the root, one can match each finite path with an infinite path by appending infinitely many branches in the opposite direction from the finite path's last branching, i.e., if the last branching is left, one appends an infinite string of right branchings and vice versa. Now every finite tree, T has an image tree f(T) as a subtree of the complete infinite binary tree. And the unions of the f(T) trees are well defined as trees, since the union of their nodes sets is the node set of their union, the union of their edge sets is the edge set of their union, and /now also/ the union of their path sets is the path set of their union. That last property, the union of their path sets being the path set of their union, is one which WM has not been able to esatblish holds for an infinite union of finite sets, but is easy to show for the image trees in the complete infintie binary tree. > Dik knows that T1 contains infinite trees, but doubts that its paths > are the union of the paths of the finite trees. > You doubt the existence of T1 (seeing that otherwise set theory is > contradiced?). > > Couldn't you get to a consensus? It would spare me a lot of work. 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.
From: David Marcus on 21 Jan 2007 16:55 Franziska Neugebauer wrote: > David Marcus wrote: > > > Franziska Neugebauer wrote: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > >> > V* is not in the union (as omega is not in the union over all > >> > natural numbers or over all finite intial segments of omega). > >> > >> I want you to define > >> > >> T(1) U T(2) U ... > > > > Do you think WM can actually produce a definition? His knowledge of > > math is rather limited. > > Until he does I have _no_ clue what his writing is really about or > whether it is about sth. at all. I think it is quite clear that it is not about anything. If you want to play with a child, you might as well play a game that is on their level. Otherwise, there isn't much fun in the game. -- David Marcus
From: Virgil on 21 Jan 2007 17:18
In article <MPG.201d8bd1f8c7b013989bc1(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > It is interesting: > > William knows that T1 and T2 exist, but T1 is different. > > Virgil is also accepting T1 but, like William, sees only finite trees > > therein. > > Dik knows that T1 contains infinite trees, but doubts that its paths > > are the union of the paths of the finite trees. > > You doubt the existence of T1 (seeing that otherwise set theory is > > contradiced?). > > > > Couldn't you get to a consensus? It would spare me a lot of work. > > Perhaps it would help if you look at the following and see if it is what > you are saying. If it isn't, please note any corrections. > > An edge is an ordered pair of nodes. A tree T is an ordered pair (M,E) > where M is a set of nodes, E is a set of edges. Assume that any tree > we mention is a complete binary tree. > > Define a path P in T to be a function P: X -> M where X is either > {0,...,n} or N, P(0) is the root node, (P(i),P(i+1)) is in E for all i > such that P(i) and P(i+1) are both defined, and if the domain is > {0,...,n}, then P(n) is a leaf node. Denote by Q(T) the set of paths > in T. > > Define a nested sequence of trees to be a sequence {T_i} where T_i = > (M_i,E_i), M_i is a subset of M_{i+1}, and E_i is a subset of E_{i+1}. > Define the union of these trees to be T* = (M*,E*) where M* = U{M_i} > and E* = U{E_i}. > > Claim: |Q(T*)| <= |U{Q(T_i)}|. 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. |