Date: Jan 13, 2013 4:16 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 191
On 12 Jan., 23:03, Virgil <vir...@ligriv.com> wrote:

> > You are invited to "discern" another path from the countable bunch of

> > infinite paths that I used to construct the Binary Tree.

>

> Until you list the ones that you used, there is no way to "discern"

> another, but any list you provide also provides a nonmember.

Please do that work for me, if you like to have such a list. I told

you already that I appended every definable tail.

But your idea of listing is not very usefu in this context. Consider

the list of all Cantor-lists that are possible (i.e., definable) and

append all anti-diagonals that are possible (i.e., definable), then

you get a list that cannot be diagonalized because it contains its

anti-diagonal already. And if you would try to diagonlize the

countable union of all this countable stuff you must fail, because

this stuff already contains all anti-diagonals of infinitely many

steps - and more are not feasible.

There remain two possibilities:

Either there is no list of all lists including all their anti-

diagonals. This case is tantamount to there being not all steps to

infinity.

Or there is this countable list containing all its antidiagonals. Then

Cantor's argument fails because this list cannot be diagonalized.

In both cases there is nothing uncountable.

> One reason that no man can have a list of all reals is that any such

> list provides the proof of its own incompleteness.

So where and in which form *exist* all reals?

Regards, WM