```Date: Mar 17, 2013 4:38 PM
Author: Virgil
Subject: Re: Matheology � 224

In article <077d0447-e3a0-4154-b23c-6c9b8c910cba@gp5g2000vbb.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:> On 17 Mrz., 12:07, William Hughes <wpihug...@gmail.com> wrote:> > >> > If the question is "Can a findable line be necessary"> > the question of whether a findable line is needed is> > certainly relevant.-> > Stop!> > I do not force anybody to accept my model. I cannot and I do not wish> to.> > The question in current mathematics was and is only this: Can a set of> more than one lines in the given list contain more than a single line?> In other words: Can more than one line be necessary?> > Regards, WMIf those lines are FISONs. any finite set of them is limited to the finitely many naturals contained in the largest of them, so one needs a non-empty set of FISONs that does not have a largest FISON, which can only be achieved but having an actually infinite set of FISONs.Which means that in WM's world, it can never happen at all.######################################################################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.--
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