Date: Mar 8, 2013 5:17 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots

In article 
<1df692af-a0d7-47da-a854-2db09999d100@14g2000vbr.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 8 Mrz., 22:14, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <6b761426-151b-43b7-adbb-3841e48fe...@y9g2000vbb.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:

> > > On 8 Mrz., 11:05, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > To make a change: Do *you* agree with the statement: It is silly to
> > > claim the existence of a set of natural numbers that has no first
> > > element?

>
> > every NON-EMPTY subset
>
> An empty set does not contain natural numbers. Therefore it is not a
> subset of natural numbers but at most a subset of the set of all
> unicorns.


According to what definition?

The definition accepted by everyone outside Wolkenmuekenheim is:

set A is a subset of set B if and only if
for all x, if x is a member of A then x is a member of B,
Which is true for any set B when A is empty.
>
> < of it is also

> > well ordered, and thus has a first element,
>
> Fine. Why do you sometimes appear to have forgotten this elementary
> wisdom?


What appears to WM is much of the time only delusion caused by the
astigmatism inherent in WMytheology.
> >
> > And where is WM's proof that some mapping from the set of all binary
> > sequences to the set of all paths of a CIBT is a linear mapping?

>
> I told you already EOD with respect to this isomorphism because you
> are too stupid to understand this fact.


You have yet to show ANYONE that either the set of binary sequences or
the set of paths is a linear space, and unless both of them are, there
is no possibility of having a linear mapping between them.

And where is WM's proof that some mapping from the set of all binary
sequences to the set of all paths of a CIBT is a linear mapping?
WM several times claimed it but cannot seem to prove it.
--