Date: Mar 11, 2013 5:40 PM
Subject: Re: Matheology § 222 Back to the roots

On 11 Mrz., 21:22, William Hughes <> wrote:

> l is a line of G and hence findable.
> d_max is not findable and used ("for every n")

l_max is used too ("for every line").
> Do you agree with the statement
> If G is a subset of lines of L
> and G has a findable last element
> then there is no line, l, in G
> for which it is true that
>     For every n, the nth
>     FIS of d is contained in l

I agree with this statement:

For every findable line of L there is an identical findable FIS of the
diagonal up to that line. And for every findable FIS of the diagonal
there is an identical line. Same holds for the diagonal 1, ..., max of
L and the last line 1, ..., max of L.

I do not see any use in answering your questions which try to make a
difference between changing the FIS of the diagonal and changing the
due line. When changing the FIS of the diagonal you speak of the same
diagonal, but when changing the line, you speak of different lines.

This is unjustified. In order to see it, write the list in the form


where every line and the diagonal are written in one and the same
line. Does this answer your problems? If you have pleasure in
continuing to "prove" that there is a difference between line(s) and
diagonal, please go on, but leave me out of the play - since I do not
see a difference and will not change my mind in this respect.

And a last remark: You will never succeed in proving that pot. inf. is
the same as act. inf, since your unsurmountable obstacle is the
requirement that all natural numbers have to be in the list, but
cannot be in one line but must be in one line.

Meanwhile I am tired to answer your questions. 600 postings are
enough, and there are many further §§ of matheology waiting to be
published as soon as the current discussions will have ceased.

Regards, WM