In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 16 Mrz., 01:21, fom <fomJ...@nyms.net> wrote: > > > > > Page 201, in fact. > > Page 204 AXIOM VI (Axiom of Choice): If T is a set whose elements are > all sets that are different from 0 and mutually disjoint ... > > Now I am looking forward with interest how you will put it to show > that you are right and Zermelo said: an element of T is not a set. >
That Zermelo SOMETIMES has sets all of whose members are also sets in no way means that he requires all members always to be sets.
This is just another of WM's many forms of quantifier dyslexia.
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, 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 f(x) and f(y) are vectors in suitable linear spaces.
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 af(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. --