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