Cantor Confusion
in [Math]
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From: Dik T. Winter on 16 Mar 2007 11:15 In article <1174041538.385009.91590(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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? Yes, and yes, and that is easily proven by a slight adaption of Cantor's proof. > > > > > 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? No. > If so, is it sufficient to show that n is finite for finite n, or is a > proof required that 2 is finite too? No. What you need to prove is that that also holds for the *infinite* tree together with *infinite* paths. > 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. Yes. > 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. Makes no sense. The number of paths in U(T(n)) is the same regardless at what level we look at the paths. That number is *not* growing. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 16 Mar 2007 15:29 In article <1174039357.642800.20000(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Why should he say anything so foolish? > 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. There is no mathematical definition of continuity under which trees are continuous. If WM has some notion of his own, he must explain it before claiming it. So that, as yet, his "reasin" is unreasonalbe. > > > > > 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. That union does exist, but in what sense is it a limit? Absent any appropraite definition, no such thing exists. > > > > > > The tree is continuous because its nodes are connected by paths. Its nodes are connected by edges, not paths. Paths derive from nodes and edges so any connections of nodes must precede the notion of 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. You claim might but ours does not. Ours springs from the form of the CIBT (complete infinite binary tree) itself, which has a path set bijecting withe P(N), which is uncountable. > > > > 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 Something right for a change. > > > 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. Does WM declare that there is no difference between the set of all finite paths and the set of all infinite paths? That would require, among other idiocies, that every finite path be infinite and every infinite path be finite. > > 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? Provided one is only counting finite paths, but the set of infinite paths more nearly corresponds to the power set of the set of finite paths than to the set itself, since each infinite path derives from an infinite set of finite paths. > 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? Each infinite path corresponds to an infinite set of finite paths, and the set of infinite subsets of a countable set is uncountable. > > Regards, WM
From: Virgil on 16 Mar 2007 15:43 In article <1174041538.385009.91590(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? Of course! Every such path is an infinite subset of the countably infinite set of nodes, and thus is potentially uncountable. Every such path can be represented by a subset of the set of non-root levels, which is countable, so must have cardinality of the power set of a countable set. > > > > > > > > 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). Actually, the number of paths of a binary tree in which all paths are of level n is 2^(n-1): 1 level (root node only), 2^(1-1) = 1 path 2 levels (oot and one each left and right branch), 2^(2-1) = 2 paths. And with each added level, the number of paths doubles as each former path splits in two with the addition of a new left and right branch. > > > > Why? A proof, please. proof impossible as the claim is false. > > 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? Relevance? If one follows WM's undefined limit arguments, we would have to conclude that if lim n [as n -> aleph_0] = aleph_0 then lim 2^n [as n -> aleph_0] = 2^aleph_0 > aleph_0
From: Virgil on 16 Mar 2007 16:06 In article <1174042350.684943.146210(a)l75g2000hse.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It shown no such thing. If n can jump from finite to aleph_0, then 2^n can jump to 2^aleph_0. > > >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 - As what WM calls a "union" violates every mathematical definition of "union", I will still call is versiont "what WM calls 'union'". > > > 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. The every path, being ordered by the level of its nodes and being finite, must have both a first and a last node. So that ALL WM's paths, including any infinite ones, have last nodes. While in the CIBT, no path has a last node. So, whatever WM is playing his silly games with, they are not CIBTs (complete infinite binary trees). > "Infinite" means here only "finite numbers of nodes and finite numbers > of separeted paths growing from level to level without end". Except that WM claims and 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. WM's notion of a tree is then not a CIBT. > 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? Define the level of a node in a tree to be the number of edges in the chain of edges connecting it and the root node. Then the root is at level 0, its child nodes at level 1, and s on. Then for a finite binary tree with all paths being of level n, the number of paths is 2^n. And every path may be uniquely identified by its set of left branchings, wish may be identified by the level at which these branchings end. So that every path corresponds uniquely to a set of non-zero level and vice versa. Why WM should claim that this is no longer true when the set of levels is unbounded is not clear.
From: Virgil on 16 Mar 2007 16:18
In article <1174054064.244699.153490(a)n59g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 14:35, Carsten Schultz <cars...(a)codimi.de> wrote: > > mueck...(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, > > I had already quite a lot. > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > the sense that never a jump by more than a factor 2 can occur because > the nodes of the tree are connected by an untearable network. Following WM's argument, g:n |--> n is even more continuous in that it can never "jump" by a difference of more than 1, so can never become infinite at all. Thus WM's "untearable" network can never represent anything but strictly finite trees, unless he is wrong about them. > The > domain is the same as the range, namely N. The range of f: n |--> 2^n can never be the same as the domain, unless both are empty. > That is fact, not by claim > but by construction of the tree. That's why I constructed it. A construction which requires N for both the domain and range of f: n |--> 2^n is fatally flawed. And this is equally true whether one uses the definition of function favored by Moblee or that favored by me. |