Date: Dec 10, 2012 5:09 AM
Author: Zaljohar@gmail.com
Subject: Re: Mathematics in brief

On Dec 10, 12:17 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 10 Dez., 09:07, Virgil <vir...@ligriv.com> wrote:
>

> > In article
> > <5a2d9b2e-c558-446a-908f-1a5f24d3f...@r14g2000vbd.googlegroups.com>,

>
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 10 Dez., 06:32, Zuhair <zaljo...@gmail.com> wrote:
> > > > On Dec 10, 12:21 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > If so what is the proof that ALL reals belong to that tree?
>
> > > There is no proof because the Binary Tree contains all reals between 0
> > > and 1 by definition.

>
> > What definition is that? I know of no such definition.
>
> All real numbers that have binary representation starting with 0. are
> in the Binary Tree and are in the interval [0, 1]. No one is missing
> in any of these systems.
>

> > Every binary sequence is in any complete infinite binary tree.
>
> Of course. But you had forgotten that above?
>
>
>

> > > Cantor's proof (concerning the binaries with bits w and m) shows that
> > > not all that are in the tree can be in the list.

>
> > Right for once!
> > > This proof is given by constructing the whole Binary Tree by countably
> > > many actually infinite paths (i.e. finite paths with infinite endings
> > > like 010101... or 000... or 111... or 001001001... or any desired
> > > ending that I do not publish). My proof rests upon your inability to
> > > find out what paths are missing.

>
> > And Cantor's counterproof says that no such list contains all paths and
> > provides an unambiguous way of determining from any such list ( which
> > exists as a consequence of WM's claim of countability) any finite number
> > of the uncountably many  paths which are missing.

>
> But his proof is taken as evidence that the Binary Tree contains
> uncountably many paths *that are defined by nodes only* (as is every
> path in the Binary Tree). And this result is false.
>

> > > Find a path that I have not used!
>
> > List the ones you have used
>
> I have constructed all paths that can be defined by nodes in the
> Binary Tree, because no node is missing in my construction.
>

Which Binary Tree, Is it the binary tree on NxN grid? which indeed has
countably many nodes. If we hold that the number of paths cannot
exceed the number of nodes, then clearly it follows that this Binary
tree of yours is countable. But the question is:

Why must we believe that ALL reals can be represented in that
Countable binary tree of yours.
Just diagonalize the set of all infinite paths of your binary tree,
and then you'll get a path that is not in your tree and that of course
correspond to some real. So your tree provably cannot supply a
representation for each real.

Zuhair