Cantor Confusion
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From: Virgil on 9 Jan 2007 13:04 In article <1168349267.595403.120050(a)i15g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The finite union of two or more finite trees is a finite tree. An > > > infinite union of finite trees is the infinite tree. > > > > What is an infinite union? (Please give a definition) > > An infinite union of trees is the union of all (or nearly all) finite > trees which reach from level 0 to level n where n eps N. (By the way > the result is the same when taking only the last levels of all finite > trees.) That assumes that for each pair of trees there is a tree of which both are subtrees, which need not be the case. > > It is easy to see that these unions result in the infinite tree, i.e., > in the tree containing all levels n for n eps N. Not if the sets of nodes of the two trees are disjoint. > Therefore this tree > contains all paths which represent sequences of bits which can be > indexed by natural numbers. There is no such path or part of such path > outside of the union of all finite trees. Not when all the trees being unioned are disjoint.
From: Virgil on 9 Jan 2007 13:10 In article <1168350148.472024.53880(a)s80g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > > The union of all finite trees (= union of all levels) of the tree > automatically contains all paths, because there is no path or part of a > path outside of this union. Not unless every pair of trees have a tree of which they are both subtrees, and even then, your "union" does not necessarily imply that the union of two trees is a tree. > > The union of all finite trees (= union of all levels) of the tree > automatically contains all paths, because there is no path or part of a > path outside of this union. Such a union can only contain the paths that are already in some tree, but not paths that are not already in some tree. Note that the union of all finite ordinals does not contain anything except finite ordinals. So why should the union of all finite trees be different?
From: Virgil on 9 Jan 2007 13:24 In article <1168352732.164451.123280(a)m30g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > But it is supposed to exists as the complete, finished diagonal. > > > > [The claim is that the line L_D does not exist. > > Consider each of your comments in light of this claim.] > > > > No. No such claim was made. The claim was that L_D does not > > exist. Knowing whether a complete finished diagonal > > exists will not tell you whether L_D exists. > > You are wrong. No, it is you who are wrong. In ZFC, for example, the set of finite ordinals model the lines and the fist limit ordinal models the diagonal, and in that model there does not exist any L_D which contains every last member of a finite ordinal an d has itself a last member. > If the complete diagonal exists, then the complete line > L_D exists. If the complete diagonal does not exist, then the complete > line need not exist. That is just my arguing. And your arguing is false in ZFC or NBG or any number of other models for set theory. > Of course an actually > existing complete line is nonsense Agreed, so why do you keep trying to sell your "gold brick"? > but it results from the assumption > of a complete diagonal, which, unfortunately, is not considered > nonsense by set theorists. Since they have an explicit model for what WM claims cannot exist, why should they? > Your assertion is wrong. There are never different lines required for > different line ends. Different line ends are always elements of one > single line. How can a line have so many ends? > > > > L_D does not exist. > > Of course does it not exist, because th diagonal does not exist. Wrong! The ZFC model exists in which every line is finite but the diagonal is not.
From: Virgil on 9 Jan 2007 13:26 In article <1168353438.093402.58300(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >> L_D does not exist > > >> so the union of line ends including the end of L_D does not exist. > > > > > > Therefre a complete, finished diagonal does not exist. > > > > As far as I can see is the diagonal "only potentially infinite". Nobody > > claimed that it is "complete". Nonetheless, the conclusion that a > > single line L_D does not exist remains valid. > > If the diagonal is not complete, then it has a largest element. A line > containing this element (and all smaller ones) does exist. If it turns > out, that the diaogonal has a larger element, then it turns out that a > line containing this (and all smaller ones) does also exist. Non-sequitur. In ZFC the finite ordinals as lines and the first limit ordinal as diagonal disprove all of WMs claims here.
From: Virgil on 9 Jan 2007 13:29
In article <1168354048.433381.269130(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> mueckenh(a)rz.fh-augsburg.de wrote: > > >> > > >> > The finite union of two or more finite trees is a finite tree. An > > >> > infinite union of finite trees is the infinite tree. > > >> > > >> What is an infinite union? (Please give a definition) > > > > > > An infinite union of trees is the union of all (or nearly all) finite > > > trees which reach from level 0 to level n where n eps N. > > > > 1. Please give a definition of "union of all finite trees". > > Definitin: Denote the nodes of the tree by > > (0,0) > (1,0) (1,1) > (2,0a) (2,1a) (2,0b) (2,1b) And what if two trees have no nodes in common? E.g., A D / \ / \ B C E F What is the union of these two trees? |