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, WM


If 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.




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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.
--