In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 20 Mrz., 17:18, YBM <ybm...@nooos.fr.invalid> wrote: > > Proof, in the Mückenheim way, that an dog with no legs has two legs. > > It is a pity that you have no idea of what set-inclusion means. But I > am not surprised.
It is a pity that WM has no idea of what a linear mapping means to claim that he can map the set of all infinite binary sequences through the field of real numbers to the set of all paths of a Complete Infinite Binary Tree and have that mapping linear.
WM has frequently claimed that a mapping from the set, B, of all infinite binary sequences to the set, P, 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, and then show that his mapping satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of a field of scalars and x and y are vectors in B and f(x) and f(y) are vectors in P.
By the way, WM, what are a, b, ax, by and ax+by when x and y are binary sequences?
If a = 1/3 and x is binary sequence, what is ax ? and if f(x) is a path in a CIBT, what is (1/3)*f(x)?
Until these and a few other issues are settled, WM will still have failed to justify his claim of a LINEAR mapping from the set (but not yet proved to be vector space) of binary sequences to the set (but not yet proved to be vector space) of paths ln a CIBT.
Just another of WM's many wild claims of what goes on in his WMytheology that he cannot back up.
Though for someone knowing more mathematics than WM, a linear mapping from B to P is quite possible --