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.

######################################################################

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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