Cantor Confusion
in [Math]
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From: Virgil on 16 Mar 2007 16:31 In article <1174055000.972069.261630(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 15:15, Carsten Schultz <cars...(a)codimi.de> wrote: > > mueck...(a)rz.fh-augsburg.de schrieb: > > > > > > > > > > > > > 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. > > > > I can imagine. > > > > > 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. The > > > domain is the same as the range, namely N. That is fact, not by claim > > > but by construction of the tree. That's why I constructed it. > > > > You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that > > really must have been fun. Ok, I agree on this. Now we know a property > > of the function > > > > f : N -> N > > n |-> 2^n. > > > > This does not tell us anything about 2^aleph_0. > > > > aleph_0 is not a natural number. Are you just discovering that? > > Don't mistake the infinite number of finite paths with infinite paths. An infinite path corresponds to an infinite set of finite paths in the which the nodes of each path in the set are included as nodes in every longer path in the set. There is one such infinite path for each such maximal infinite set of finite paths and vice versa. And the union of the infinite set f finite paths, as set of nodes will be the set of nodes of the infinite path. And that is using the legitimate meaning of union as in ZF or NBG. > In the union of all fiite trees every path has a finite length, given > by a natural number of nodes. Presently we are considering the union > of all such finite paths. (The union of all finite natural numbers is > an infinite union - nevertheless this union cotains only finite > numberrs.) The the mathematical union of the set of all finite paths is merely the set of nodes of the tree. But each infinite path requires a different infinite subset of the set of all finite trees, as described above. So WM is faced with something like the power set of his set of all finite paths, not merely the set of finite paths itself. And WM cannot handle it.
From: mueckenh on 17 Mar 2007 09:10 On 16 Mrz., 15:54, Carsten Schultz <cars...(a)codimi.de> wrote: > mueck...(a)rz.fh-augsburg.de schrieb: > > > Don't mistake the infinite number of finite paths with infinite paths. > > I don't. I am not sure. A set of finite paths is bounded by aleph_0. > > > In the union of all finite trees every path has a finite length, given > > by a natural number of nodes. Presently we are considering the union > > of all such finite paths. (The union of all finite natural numbers is > > an infinite union - nevertheless this union cotains only finite > > numberrs.) > > The set of all finite subsets of N is countable, the set of all subsets > of N is not, Alas, these subsets, as represented by indexes of paths, cannot be distinguished. The set of all finite paths is identical with the set of all finite and infinite paths. > and this will not change, regardless of the number of > reformulations that you give. As long as the paths exist as sets of nodes within the tree, they are countable. After having left the tree and lost any contact to nodes, the paths do no longer represent real numbers. And this will not change, regardless of the number of opposing believers. Regards, WM
From: mueckenh on 17 Mar 2007 09:24 On 16 Mrz., 16:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174039357.642800.20...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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: > ... > > > > 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. > > No, the finite is required here. Below you say that there are differences which are not in erms of nodes. > > > 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. > > Pray define "continuity". The function Card(paths) over n cannot jump from finite to uncountable. > > > > > 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. > > By what definition of limit? Pray, for once, provide a definition. Read it: "going on without end". > > > > > 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. > > Eh? So there are not even infinitely many paths? Why not? There are. Infinite is "going on without end". > > > 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. > > Yes, there is a difference. The difference being that in one set all paths > are finite and in the other set all paths are infinite. > > > In the union of all finite trees there are only finite paths. > > Why? The axiom of infinity states different. As the tree shows, this statement is self contradictory. > > > So you > > agree that in the union of all finite trees the number of paths is > > countable? > > No, because in the union there are also infinite paths. Or do you now > think that 0.010101... is *not* a path in that union? Is it in the union of all finite paths? However, as long as it consists of nodes it has only countably many co- 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? > > Not in terms of nodes, why should there be a difference in terms of nodes? What else can distinguish paths? Soul? Feeling? Belief? Regards, WM
From: mueckenh on 17 Mar 2007 09:40 On 16 Mrz., 16:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174041538.385009.91...(a)d57g2000hsg.googlegroups.com> mueck...(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. And it is easily proven that there are as many paths as nodes in the tree. You need only calclate the sum of the fractions. But, more obvious, there can never uncountably many paths exist in the tree, because there are never uncountably many nodes within oner level. > > > > > > > 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. It holds as long as there are nodes occupied by paths. Even if there is not finish. There is no jumping from finite to uncountable. > > > 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. This is true as long as the paths occupy nodes. But you seem to claim, though you do not explicitly state it (as far as I remember): The number of paths / paths-bundles becomes uncountable before the paths leave the tree. Regards, WM
From: mueckenh on 17 Mar 2007 09:51
On 16 Mrz., 21:06, Virgil <vir...(a)comcast.net> wrote: > In article <1174042350.684943.146...(a)l75g2000hse.googlegroups.com>, > > > > > > mueck...(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. Not if the function is continuous as is the number of paths-bundles in the tree. Between uncountable and finite, there is mor than a factor of two. Regards, WM |