Virgil
Posts:
7,021
Registered:
1/6/11


Re: Matheology � 223: AC and AMS
Posted:
Mar 14, 2013 7:47 PM


In article <d3115286b9e44d87b3c7198f5a2ed762@z3g2000vbg.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 14 Mrz., 23:16, fom <fomJ...@nyms.net> wrote: > > > > > > "... an element of T is not a set..." > > Let T = {{a}, {b,c}, {c,d,e,f}} > then T has three elements, each of which is a set.
That there are set all of whose members are sets does not forcibly eliminate being able to have sets not all of whose members are sets.
At least not outside such special set theories as ZF which do not provide for any nonset objects.
There are set theories having urelements which are not themselves sets.
That WM seems to be totally ignorant of such theories is only to be expected of someone of his wide ranging ignorance.
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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. 

