Rupert
Posts:
3,806
Registered:
12/6/04


Re: The uncountability infinite binary tree.
Posted:
Dec 16, 2012 3:35 PM


On Dec 16, 6:05 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > On 16 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > > > > > > > On Dec 16, 1:30 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > On 16 Dez., 11:54, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > On Dec 16, 11:32 am, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > On 16 Dez., 11:02, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > > > On Dec 15, 7:07 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > > On 15 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > > > > > On Dec 15, 12:35 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > > > > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > > > > > > > On Dec 15, 10:56 am, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > > > > > > Thus the Infinite binary tree is UNCOUNTABLE. > > > > > > > > > > > > Since it can be proved to be countable, there is a contradiction. > > > > > > > > > > > How do you prove that it is countable? > > > > > > > > > > First: A Tree that contains all nodes also contains all reals of the > > > > > > > > > unit interval. > > > > > > > > > > I constrcut the tree, i.e., all nodes, by countably many infinite > > > > > > > > > paths. > > > > > > > > > Tell us more about this construction. > > > > > > > > I use all finite paths. Every node of the Binary Tree is the end of a > > > > > > > path. Then I append these countably many paths by a tail according to > > > > > > > my choice. For instance I can use the tail > > > > > > > 000... > > > > > > > or > > > > > > > 010101... > > > > > > > or > > > > > > > the bit sequence of pi > > > > > > > or anything else, for instance a mix of many tails. > > > > > > > > In order to show you that you are dreaming if you think that infinite > > > > > > > paths could be identified by their nodes, I don't tell you what tails > > > > > > > I have used. If you don't sleep to deep, then you will wake up and > > > > > > > recognize that infinite paths are merely defined by finite > > > > > > > definitions, and hence belong to a countable set. > > > > > > > > Regards, WM > > > > > > > Okay, so you seem to be saying that you would take all the finite > > > > > > paths and append an infinite tail to each one, thereby obtaining a > > > > > > countably infinite collection of infinite paths. Is the claim then > > > > > > that this would be equal to the collection of all infinite paths? Or > > > > > > not? > > > > > > It is equal to the collection of all paths that are defined solely by > > > > > their nodes. It is equal to all finite initial strings of digits that > > > > > can be applied in a Cantorlist. (Every digit changed there has a > > > > > finite index.) > > > > > > Actually infinite paths like that of 0.010101... = 1/3 (in binary) > > > > > cannot be defined by nodes. No infinite sequence has ever been defined > > > > > by its terms. Those things have to be defined by finite definitions > > > > > like "0.010101..." or "1/3". > > > > > Okay. That all sounds fair enough. > > > > > So, you were going to show us that there are only countably many > > > > paths. When are you going to do that? > > > > First I showed you that there are only countably many paths that are > > > defined by nodes. You seemed to agree. Call them the set A. > > > > Well, the other "paths" cannot be defined by nodes. Call them the set > > > B. They need definitions by finite words. And there are only countably > > > many finite words. > > > > Now consider the union of two countable sets A and B. > > > How do you know that there exists a countable language such that every > > path can be defined in this language? > > I do know that in every language (as well as in all languages together > since aleph_0 * aleph_0 = aleph_0, and since there are no uncountable > languages) only countably many paths can be defined.
There are uncountable languages, but that's neither here nor there. We can restrict attention to countable languages.
However, you haven't justified your claim that there are only countable many countable languages, and in fact it can be shown that this isn't true.
> What cannot be > defined in any language is not a path.
This, too, is something which needs to be justified. A set theorist would not accept that assumption.
> Therefore I know that there are > not more than countably many paths  at least paths that can be > applied in mathematics and can be the result of any diagonalization. >
No. You don't know that.
> Regards, WM

