Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 18 Jan 2007 04:48 Virgil schrieb: > Since every path in every one of the incomplete infinite binary trees in > the union is 'eventually constant', every path in the union of all those > trees will also be eventually constant. Wrong. Every segment {1,2,3,...,n} is finite, but the union of all segments is not finite. Every finite segment of the path 0.010101... has only finitely many switches 0 to 1 and 1 to 0, but the union of al finite segments has infinitely many switches. > > Thus WM's argument that the union of finite binary trees generates a > complete infinite binary tree is falsified. By an invalid argument. Regards, WM
From: mueckenh on 18 Jan 2007 04:52 Virgil schrieb: > In article <1169030073.463923.305630(a)v45g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > > > > This is obviously not the union in the sense of set theory. Set > > > theoretically a tree is a directed graph which is defined as an ordered > > > pair (V, E). The union of two trees is in general not a tree at all. > > > > What I defined a sthe union of finite trees is the union in the sense > > of set theory. > > The set theory union of trees is not a tree at all, but with suitable > modifications one can construct something that works somewhat in the way > WM intends. Unfortunately for WM, it disproves his claims: 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. This definition unites sets of nodes (and sets of edges, respectively) and it is valid for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). The union of all finite trees is the union of all trees with n levels where n is a natural number: UT = T(1) U T(2) U T(3) U ... The paths in a tree are completely defined by the sequences of nodes (or edges) which can be followed to an end in CT or without an end in the union of all CTs as well as in the WWT and the union of all WWTs. > > 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.) > > So that WM's EIT is like the successor of omega, containing all the > members of omega together with omega itself. Wrong. The union of all natural numbers is N or omega. Regards, WM
From: mueckenh on 18 Jan 2007 04:57 Virgil schrieb: > In article <1169031719.407811.187320(a)a75g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > So far, I hope WM and I agree. > > > > > > The issue between us is whether that union will be a proper subtree of > > > the complete infinite binary tree or will be the entire tree. > > > > > > Since every path in every one of the incomplete infinite binary trees in > > > the union is 'eventually constant', every path in the union of all those > > > trees will also be eventually constant. > > > > Yes, this is valid for a finite union of finite trees. And perhaps, > > without giving a thought to it, may be claimed for the infinite union. > > And when one does give it more thought it is even more obviously valid. So the union of all finite segments {1,2,3,...,n} with n in N is finite? Think again! > > > > > > But only countably many of the uncountable paths in the complete binary > > > tree are eventually constant. > > > > > > So that, for example, the union will not contain any path which > > > alternates between branching left and branching right. > > > > Which alternation will be missing? > > For one, the path which repeats one left then one right. > For a second, the path which repeats one left then two rights. > For a third, the path which repeats one left then three rights. > And so on. And for the union tree, i.e. a tree with an ordinal number being transfinite? Regards, WM
From: mueckenh on 18 Jan 2007 05:04 Virgil schrieb: > In article <1169032407.857382.32680(a)51g2000cwl.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > Since every path in every one of the restricted infinite binary trees in > > > > > the union is eventually constant, every path in the union of all those > > > > > trees will also be eventually constant. > > > > > > > > That would be correct for a finite union of finite trees. What is > > > > correct for the finite case need not be correct for the infinite case. > > > > > > If every path in every tree in a union of trees has a certain property > > > then every path in the resulting tree has that property. > > > > If every finite initial segment of N has the property of being finite, > > then the union of all initial finite segments cannot be an infinite > > segment. N does not exist. > > False analogy. > > The correctly analogous statement is": > > if every member of every finite initial segemtn of the naturals > is finite then every member of the union of thsoe finite initial > segments is finite, i.e., every member of N is finite. Of course every member is finite. It is the union of infinitely many members which is an infinite set and an infinite tree and an infinite path in the infinite tree. > > > > > > We have infinitely many finite sets of eventually constant paths. We > > > form the set-union of all those sets of paths. And WM requires that the > > > union of those sets contain something not in any of the separate sets. > > > > > > What does WM think a "union" of sets is? > > > > I think the union of all initial finite segments {1,2,3,...,n} is the > > union of all natural numbers, namely N. > > Which, as above, has no infinite members. Correct. But the union is infinite. Of course no path in the infinite tree has an infinite initial segment. The path *is* the infinite iinitial segment. (Just like Canor's diagonal.) Regards, WM
From: Carsten Schultz on 18 Jan 2007 05:17
mueckenh(a)rz.fh-augsburg.de schrieb: > Carsten Schultz schrieb: > >> mueckenh(a)rz.fh-augsburg.de schrieb: >>> Virgil schrieb: >>> >>>> In article <MPG.2016e6f7746f9f07989b49(a)news.rcn.com>, >>>> David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >>>> >>>>> mueckenh(a)rz.fh-augsburg.de wrote: >>>>>> No. What I am using is this: >>>>>> 1) The union of all finite trees contains the union of all finite paths >>>>>> and this set is countable. >>>>>> 2) The union of all finite trees contains all nodes. >>>>>> 3) You cannot add another tree to the set of all finite trees without >>>>>> adding at least one node. >>>>>> (The paths are subsets of the set of nodes. Two different sets must be >>>>>> distinguished by at least one element.) >>>>>> 4) But there is no node to add. >>>>>> 5) Therefore the union of trees contains all possible paths. >>>>>> 6) We know that the set of all paths contained in this union is >>>>>> countable. >>>>> How does 6 follow from 1-5? >>>> It doesn't, >>> It follows from the theorem that a countable union of finite sets is >>> countable. >>> Every finite tree contains only a finite set of paths. >>> The union of finite trees is countable. >> >> Let N_i be the node sets of the subtrees, which we will denote by T_i), >> and let us assume that they are all induced subtrees. Let us also >> denote the set of paths in T_i by P_i. Now let p be a path in the big >> tree. Then the following statements hold: > > Is your "big tree" the union of all subtrees? Or is it more (or less)? It is a tree of which all the T_i are induced subtrees. Might contain more than the union, or it might not. Since it does not occur in the statements below, this does not matter. >> 1) The path p is a path in the union of the trees T_i (ie the induced >> subtree on the union of the N_i) if and only if every node that p visits >> is in some N_i. >> >> 2) The path p is an element of the union of the P_i if and only if >> there is an N_i that contains every node that p visits. > > You assume that the union P_i of paths contains more paths than can be > constructed from finite initial segments? I do not assume anything. I just note that being a path in the union of the T_i and being an element of the union of the P_i are a priori different things and that you would have to prove their equivalence in your setting should you claim this equivalence. >> Now you might think that these conditions are the same and derive from > > A tree with a unique set of nodes and edges would not contain two > different sets of paths. See how irrelevant (if it is meaningful att all) this statement is? >> the countability of the union of the P_i that there are only countably >> many paths in the union of the T_i. > > Is Cantor's diagonal longer than the union of all of its initial > segments? Can you stay on the subject? > Is N more than the union of all of its initial segments {1, 2, 3, ..., > n}? Are there subsets of N that are not subsets of a proper initial segment? > Yes? > Then N is certainly also more than the union of all ends {n} of its > initial segments? > No? The union of ends is more than the union of segments? > > Next time, please switch on your brain before posting. I should not be surprised that replying to cranks leads to abuse. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. |