Date: Mar 17, 2013 3:46 PM
Author: Virgil
Subject: Re: Matheology � 224

In article 
<4f0d5742-ef15-4427-95e4-9285e1171c66@h14g2000vbe.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 17 Mrz., 00:30, Virgil <vir...@ligriv.com> wrote:
>

> > > > > We call it unfindable or unfixable
> > > > > because as soon as we have found it, it is no longer the last line.

> >
> > > > If finding it makes it not what it is supposed to be, the how does one
> > > > prove that any such thing exists?

> >
> > > Simply by observing that otherwise, there must be a set with at least
> > > two natural numbers, both of which do not belong to the set.

> >
> > Non Sequitur, at least outside WMytheology.
> >
> > Where actually infinite set of naturls is allowed, nothing like what WQM
> > demands is needed or even possible.

>
> In mathematics the assertion of existence of a non-empty set of
> natural line-numbers implies that there is a least line-number.


Which, while true, is, as usual and as expected, irrelevant!

While there may be natural numbers which are, in some sense or other,
unfindable, there is nothing which is simultaneously a natural number
and has no successor natural number.

At least not outside WOLKENMUEKENHEIM.




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



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.
--