Virgil
Posts:
6,993
Registered:
1/6/11


Re: Matheology � 223: AC and AMS
Posted:
Mar 18, 2013 3:52 PM


In article <e8ee197eb471494e9d0d0238f6eee4ff@h11g2000vbf.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 18 Mrz., 07:26, fom <fomJ...@nyms.net> wrote: > > > > You turn to an outdated strategy directed to > > a situation that no longer exists rather > > than do the hard work of grounding your > > claims. You do this to say that just > > because you do not believe a particular > > axiom, > > Wrong. I prove that the axiom is nonsense liek the axiom that a > triangle with four edges exists.
WM very rarely manages to proves anything to the satisfaction of anyone other than WM himself.
For example WM has now failed to prove a claim of linearity to the satisfaction of anyone other than WM himself. > > Of course I presume that in mathematics facts can be proven.
In standard mathematics, only statements can be proven, or disproven.
> A simple > statement of faith in the Cartesian product cannot be overcome. So be > satisfied, nobody will ever cause you to change your belief, if it is > only fimr enough. You may even include the belief that you are doing > mathematics or logic.
A belief that WM has, but isnot justified in having. > > > > > Yet, there is an > > established criterion for demonstrating > > that the axiom you do not believe > > is, in fact, in error. > > If the inhabitants of a mad house establish a criterion I am not > obliged to accept it.
In saner words, if mathematics says what WM does not like, he can at will reject it. > > > You say that > > you do not need to respect this > > criterion. Nor, do you elucidate > > an alternate criterion that others > > might consider. > > Hahaha. Every axiom, beginning with Euclid, established or formalized > a triviality. AC formalizes a counterfactuality.
Then do not use it. But since ZF does not use it either, you have presented no justification for your rejecting ZF.
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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. 

