In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 12 Mrz., 23:20, Virgil <vir...@ligriv.com> wrote: > > > > > WM has frequently claimed that a mapping from the set of all infinite > > binary sequences to the set of paths of a CIBT is a linear mapping. > > In order to show that such a mapping is a linear mapping, WM must first > > show that the set of all binary sequences is a vector space and that the > > set of paths of a CIBT is also a vector space, which he has not done and > > apparently cannot do, > > The field of real numbers (|R, +, *) should satisfy your wishes. > Written in the form of a tree with the decimal point common to all > paths that stretch from oo to -oo you get the same space as a decimal > tree. And if you translate that into binaries, you have the desired > fields. > > You can here without any limits add and subtract and multiply and > divide. > > Here is a sketch of the resulting Binary Tree extended to a complete > space: > > ... > 0 1 0 1 > \/ \/ > 0 1 > \ / > . > / \ > 0 1 > /\ /\ > 0 1 0 1 > ... > > Are you really too stupid to create such a simple structure by > yourself?
I am wise enough to know that neither the set of binaries nor the set of paths have been shown to have the structure of a linear space, or even of a subset of a linear space, and the obvious bijection between them does not have the necessary properties to make it into the linear mapping which WM has so often and so ignorantly claimed it to be. > > Or do you only want to distract the reader from the well known > contradiction?
What I want is for WM to act like a mathematician and either prove his too often made claim that the mapping from binary sequences to paths IS a linear mapping, or to retract his claim. --