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.

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