Date: Feb 25, 2013 7:05 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

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?



> 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.



>> 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.



> > 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.


> 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.


>> 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.

Regards, WM