Cantor Confusion
in [Math]
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From: Virgil on 15 Mar 2007 20:31 In article <1173954799.919385.61730(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 13 Mrz., 14:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173724460.248046.46...(a)s48g2000cws.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > Terminating paths. > > *Passing* path-bundles. Path-bundles are just another of WM's irrelevancies invented to obfuscate the obvious. > > > Those are *terminating* paths. (A path-bundle can be seen as a terminating > > path: it is a set of nodes containing a finite number of nodes.) > > A path bundle splits off into two bundles which pass said node. > Therefore every set of path-bundles in the tree has a finite cardinal > number > > This, in the "limit", may yield an infinite number, but certainly not > an uncountable number without having an intermediate countably > infinite number. For even binary trees ( where even here means all paths are of equal length), the number of paths increases exponentially with number of levels (lengths of a path). Adding 1 to the number of levels doubles the number of paths. > > The tree is continuous because its nodes are connected by paths. That is a distinctly non-standard meaning for "continuous" in mathematics. And nodes are connected by edges, not paths. > There > is never more than the factor 2. There are no interruptions possible > and no jumps from "finite" to "uncountable". Your claim would require > that. When one takes what WM calls the 'union' of all his finite binary trees, one has one such tree of n levels for each member n of N, which N is itself not finite, so that the union cannot be finite either. And one can easily see that the number of paths in any finite tree is the same as the power set of non-root levels in all his trees, of finite or infinite level. > PS: What about the review of chapter 10? When Chapter 9 is already shown to be so corrupt, why bother?
From: mueckenh on 16 Mar 2007 06:02 On 15 Mrz., 16:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 13 Mrz., 14:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1173724460.248046.46...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > Terminating paths. > > > > *Passing* path-bundles. > > > > > Those are *terminating* paths. (A path-bundle can be seen as a terminating > > > path: it is a set of nodes containing a finite number of nodes.) > > > > A path bundle splits off into two bundles which pass said node. > > Therefore every set of path-bundles in the tree has a finite cardinal > > number > > As long as path-bundles are finite sets of nodes that is right. You should say: As long as path-bundles are sets of nodes that is right. There are only countably many nodes. Unless the number of path-bundles first gets countably infinite it cannot become uncountably infinite. Reason: Continuity of the tree. > > > This, in the "limit", may yield an infinite number, but certainly not > > an uncountable number without having an intermediate countably > > infinite number. > > I do not know. What do you understand under "limit"? In this case nothing else but: "going on without end". But I will agree to any other definition which you might supply. The limit of U[k<n]{1,2,3,...,k} is the union of all finite initial segments for example. > > > The tree is continuous because its nodes are connected by paths. There > > is never more than the factor 2. There are no interruptions possible > > and no jumps from "finite" to "uncountable". Your claim would require > > that. > > My claim requires nothing of the sort. The number of finite paths is > countable. The number of paths of every set of finite paths is finite. The number of nodes of every set of finite paths is finite > The number of infinite paths is uncountable. There is no > jump from countable to uncountable, there is a jump from finite paths > to infinite paths. You seem to imply a difference between the set of all finite paths (which is countable) and the set of infinite paths. In the union of all finite trees there are only finite paths. So you agree that in the union of all finite trees the number of paths is countable? The set of all finite paths covers the whole tree with all its nodes. What is the difference, in terms of nodes, between the set of all finite paths and the set of infinite paths? Regards, WM
From: mueckenh on 16 Mar 2007 06:38 On 15 Mrz., 16:36, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173953509.704127.222...(a)b75g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > You try to make a distinction between the results of infinitely many > > countable parentheses and of infinitely many levels of the tree. This > > distinction is wrong, because both situations are identical. > > I do not make that distinction. In the harmonic series there are *no* > paths defined, so using it in connection with paths is wrong. It is very easy to define such paths. The following, for instance, is a path: 1/2 1/4 1/8 1/16 .... As it does never end, it is an infinite path. Do you agree? Do you claim that there are uncountably many of such paths? > > > > > An uncountable set of paths cannot exist other than outside of the > > > > tree, i.e., outside of mathematics. > > > > > > Why? A proof, please. > > > > Inside of the tree the number of paths is restricted to the cardinal > > number of levels (which easily can be understood as cross sections). > > Why? A proof, please. Do you need a proof that 2^n is finite for finite n? If so, is it sufficient to show that n is finite for finite n, or is a proof required that 2 is finite too? The cross section of a finite tree (i.e. the number of nodes in its basis) is finite. The union tree U(T(n)) consists of levels all of which are basis levels of finite trees and, therefore, are finite and have a finite cross section. As long as a path runs within U(T(n)) there is no chance for him to get more than the countable infinity of companions, i.e., a number which is growing without end but at any level n this number can be counted by natural numbers. Regards, WM
From: mueckenh on 16 Mar 2007 06:52 On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: > In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, > For even binary trees ( where even here means all paths are of equal > length), Only those are under discussion here. > the number of paths increases exponentially with number of > levels (lengths of a path). Adding 1 to the number of levels doubles the > number of paths. > > > The tree is continuous because its nodes are connected by paths. > > That is a distinctly non-standard meaning for "continuous" in > mathematics. It shows, however, that the number of paths cannot jump from finite to uncountable. >And nodes are connected by edges, not paths. Correct. Nodes ae connected by edges. But as these edges are elements of paths, nodes are also connected by paths. > > > There > > is never more than the factor 2. There are no interruptions possible > > and no jumps from "finite" to "uncountable". Your claim would require > > that. > > When one takes what WM calls the 'union' of all his finite binary trees, - meanwhile you should have learned this definition - > one has one such tree of n levels for each member n of N, which N is > itself not finite, so that the union cannot be finite either. As very element of the union is finite, every path is finite. "Infinite" means here only "finite numbers of nodes and finite numbers of separeted paths growing from level to level without end". > > And one can easily see that the number of paths in any finite tree is > the same as the power set of non-root levels in all his trees, of > finite or infinite level. As easily it can be seen that the number of paths within the tree, i.e., where there are nodes, remains countable. Every path-bundle splits at a node. The number of path-bundles is countable. As long as there are nodes, the path-bundles are countable. What happens beyond? Regards, WM
From: Carsten Schultz on 16 Mar 2007 09:35
mueckenh(a)rz.fh-augsburg.de schrieb: > On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: >> In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, > >> For even binary trees ( where even here means all paths are of equal >> length), > > Only those are under discussion here. > >> the number of paths increases exponentially with number of >> levels (lengths of a path). Adding 1 to the number of levels doubles the >> number of paths. >> >>> The tree is continuous because its nodes are connected by paths. >> That is a distinctly non-standard meaning for "continuous" in >> mathematics. > > It shows, however, that the number of paths cannot jump from finite to > uncountable. Using a word does not constitute proof. And indeed sup_{n<aleph_0} 2^n = aleph_0 < 2^aleph_0, so in this sense the function kappa |-> 2^kappa is not continuous. If you can prove (not claim!) by using your tree that it is, then you will finally have succeeded in showing that ZF is inconsistent. Have fun, Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. |