In article <firstname.lastname@example.org>, WM <email@example.com> 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 You turn to an outdated strategy directed to 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.
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.