Virgil
Posts:
8,833
Registered:
1/6/11


Re: The uncountability infinite binary tree.
Posted:
Dec 16, 2012 5:42 PM


In article <348b10c6a5e04b218a0cb6d6733cfa16@4g2000yqv.googlegroups.com>, WM <mueckenh@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. If that were true than a Complete Infinite Binary Tree could not be defined purely in terms of its nodes
There is a Complete Infinite Binary Tree defineable by its nodes that has uncountable many paths. Each path in it is defined by a subset of the set of positive natuals, N, by requiring the path to branch left from each level in the subset of N and branch right at all other levels. This clearly establishes a bijection between paths and subsets of N.
So that WM must also be claiming that he can list all subsets of N.
It is equal to all finite initial strings of digits that > can be applied in a Cantorlist. But every path contains infinitely many branchings, any of which can be to the laft or to the right, so no finite string of 0forleft or 1forright digits can determine all of them.
(Every digit changed there has a > finite index.)
Meaning that all infinitely many digits have finite indices. > > Actually infinite paths like that of 0.010101... = 1/3 (in binary) > cannot be defined by nodes. But they can by infinite sequences of nodes each identified as either a left child or right child of its parent node.
> No infinite sequence has ever been defined > by its terms.
How about the sequence of all 0's and the sequence of all 1's? Or your own example of 0.010101... = 1/3 (in binary) ?
Besides which, there is no necessity for a path to be accessible to be a memeer of the set of paths any more than for a real number to be accessible in order to be a real number. Most of them aren't! 

