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


Re: Poor Endorsement of Wolfgang Mueckenheim from a serious nonmathematician
Posted:
Feb 1, 2013 6:44 PM


In article <29e0263a02f947a69fb5d163d8c9d601@l9g2000yqp.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 1 Feb., 12:25, forbisga...@gmail.com wrote: > > On Friday, February 1, 2013 12:12:59 AM UTC8, WM wrote: > > > On 1 Feb., 06:01, forbisga...@gmail.com wrote: > > > > Yes it does, under the assumption that the continuation > > > > of all of your initial segments is not the repeating sequence > > > > [01]. If it is then I will > > > > > If you could determine it by nodes or digits, you need not make > > > provisions. So you cannot determine by nodes whether 1/3 it there or > > > not. QED. > > > > I can determine at the finite initial segment 0.01 that either > > 1/3 is not there or 1/4 is not there because one has the continuation > > [01} and the other [00]. > > It is not the question what is "not there", but the only question is > what is there! In any Complete Infinite Binary Tree, every one of the uncountably many infinite paths must be there.
If any infinite path were "not there", it could only be because of some finite initial segment of that infinite path not being there , i.e., the only thing that can stop an infinite path from existing is the absence from the set of nods of a finitely positioned node at the end of a finite initial segment.
So if WM's tree contains all finitely positioned nodes, it must also include all uncountably many infinite paths.
> A Cantor does not construct a diagonal, unless every line is defined. > But then also the diagonal is defined and, therefore, belongs to a > countable set.
Actually, that argument would only prove that the set of definitions is NOT countable. 

