Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology � 223: AC and AMS
Posted:
Mar 17, 2013 2:49 PM


In article <f18d8aef68114993b4a5fe5f9639e899@z4g2000vbz.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 17 Mrz., 07:49, fom <fomJ...@nyms.net> wrote: > > On 3/16/2013 1:25 PM, WM wrote: > > > > > On 16 Mrz., 18:17, fom <fomJ...@nyms.net> wrote: > > > > > An additional remark: > > > > >> No. Zermelo's AC requires that one name can be written > > >> with certainty. > > > > > This statement is not Zermelo's original statement. It can be proven > > > to hold, iff it was possible to choose, in practice, one element from > > > every subset of T. If this was doable. > > > > Well, the critical investigation of the > > statement during the twentieth century > > resulted in taking it as an axiom. > > > > Its provability is not the criterion > > by which it is to be understood. > > > > > > > > > There have been many mathematicians criticizing Zermelo's axiom > > > (Borel, Peano, Poincaré and others). Zermelo discusses a lot of > > > objections in another 1908 paper. And the most amazing fact is, that > > > at that time none of the arguments aims at the fact, that there are > > > only countably many choices possible by theoretical reasons. > > > > > Zermelo agrees that the AC is not provable. He did not know, at that > > > time, that it is disprovable by theoretical mathematics. > > > > Disprovable by belief, perhaps. > > Zermelo created the axiom of choice because it was obvious to him that > is is correct, i.e., that his choice can be done, at least in > principle. Then he went on and "proved" from this axiom the well > ordering theorem. If he had known that the axiom of choice can be > disproved by proving that at most countably many choiced can be > executed, even in principle, why should he have used it?
Standard mathematics does not accept any such claimed disproof as valid!
Acceptance or rejection of the axiom of choice remains optional in real mathematics, even if not in WM's phony mathematics.
######################################################################
WM has frequently claimed that HIS 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 would first have to show that the set of all binary sequences is a linear space (which he has not done and apparently cannot do) and that the set of paths of a CIBT is also a vector space (which he also has not done and apparently cannot do) and then show that his mapping, say f, satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the field of scalars and x and y and f(x) and f(y) are arbitrary members of suitable linear spaces.
While this is possible, and fairly trivial for a competent mathematician to do, WM has not yet been able to do it.
But frequently claims already to have done it. 

