Virgil
Posts:
9,012
Registered:
1/6/11


Re: Endorsement of Wolfgang Mueckenheim from a serious mathematician
Posted:
Jan 31, 2013 4:31 AM


In article <6595939006814d8a8e3c7b1e8dad94b1@w7g2000yqo.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 31 Jan., 01:58, Virgil <vir...@ligriv.com> wrote: > > In article > > <397a90fb2e3e411fae402365cadd1...@b11g2000yqh.googlegroups.com>, > > > > WM <mueck...@rz.fhaugsburg.de> wrote: > > > I have developed a new prooftechnique, namely proof by ignorance > > > > That is hardly new for Wolkenmuekenheim, but has been the ONLY standard > > there for years. > > Thinks are easy if things are easy. > Consider a box with a dozen different pralines. If I take three out of > the boxh and return it to you, you can decide which I did not take out > by looking at the remaining ones. > > Consider a Binary Tree. When I remove or colour some paths, you can > decide which are remaining by looking at the uncoloured paths.
> > And finally consider a Binary Tree with uncountably many paths. When I > colour a countable set of paths, then you have to decide by looking at > the remainings which paths have survived. While WM may include every node in his countable set of paths, he does not, and cannot, simultanteously include every path.
For example, you can obviously include every node in the set of "binary rational" paths (those that have only finitely many left branchings or only finitely many right branchings) but this omits the vast majority of paths which branch infinitely many times in each direction.
This corresponds in set theory with counting subsets of the actually infinite set N. If one counts all the finite subsets of N , and all the cofinite subsets of N, one still has not touched that vast majority of subsets of N which are neither finite nor cofinite in N.
And note that once WM claims to consider a COMPLETE INFINITE BINARY TREE, as he does above, he is conceding the existence of infinite sets so cannot later argue that they do not exist. 

