Date: Feb 25, 2013 8:43 PM
Author: Virgil
Subject: Re: WM's Mytheology � 222 Back to the roots

In article 
<f9a5bd78-68d5-4fa9-a2bc-a877605ff7bb@k8g2000yqb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 24 Feb., 22:22, Virgil <vir...@ligriv.com> wrote:
>
>
>

> > There is an equally forceful necessity that, for every and any member
> > of your list, unless your list has a fixed last member, the diagonal is
> > necessarily longer than any given member.

>
>
> And the list is longer then every FIS of lines l_1 to l_n. But what
> does the remaining part consist of if not of lines?


Each and every line of WM's list is exceeded in length by d.
>
>
>

> > So just what "isomorphism", other than mere bijection, is WM
> > imagining in his tiny mind?

>
> First let us record that your original assertion is wrong - as in most
> cases of your writings.
>
> Second, since the paths are nothing but another notation of the binary
> strings, we have identity. There remains nothing to prove.


Since WM's claimed linearity mapping does not require identity but does
require certain other structures which are absent in mere bijection,
WM is WRONG,
AGAIN,
AS USUAL!
>
>
>

> >> Simplest logic. Try to find a set that contains its number if it does
> >> not contain its number. Isn't that simple?

>
> > How does that apply to, say, the set of von Neumann natural numbers
> > in ZF?

>
> It applies to Hessenberg's "proof". The set of all subsets of |N that
> are not containing that natural number which is mapped upon them,
> simply does not exist. It is not predicably defined.


Actually, for finite domains, D, and for any given f:D -> P(D), one can
explicitly find THE "Hessenberg set",
H_f = {x | x in D and x no in in f(x)} of D sets in the image of f.

Proceedure:
List the members of D followed by the image under f of that member and
then a 'yes' or ;no' according to whether that element is in that set.

The the set of those members for which you wrote "no" is the (possibily
empty) Hessenberg set for f.

The same process works for any listable set, D, and any function from D
to P(D).




>
>
>

> > > In mathematics we have variables which can assume values.
>
>
>

> > But variables do not take arguments like A taking A_1.
>
> That depends on what you let them take. A variable can take value, but only one.

You need a function (a sequence being one form of function) to take on
a sequence of values.
>
> > What we think is that all your lines are in d, which is, in a sense,
> > only the union of all your infinitely many lines.

>
> Therefore it cannot be more than all lines. And each one is finite,
> wih no regard how many there are.


It cannot be more that the "collection" of all lines, but in WMytheology
one is not allowed to collect them all.
>
>

> >> And certainly you don't claim that you can
> >> find more than one line that would be required to contain what one
> >> line contains?

>
>

> > No one of your lines contains its own successor so no one of your lines
> > can contain the line d which contains all successors.

>
> There are not *all* successors in potential infinity but only every
> one up to every n.



Whenever there is "NOT ALL" of something, at least outside of
WMytheology, there is at least one exception.
--