Date: Jan 30, 2013 6:32 AM
Author: William Hughes
Subject: Re: Matheology § 203
On Jan 30, 12:21 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 30 Jan., 12:02, William Hughes <wpihug...@gmail.com> wrote:

>

> > Summary. We have agreed that

>

> > For a potentially infinite list L, the

> > antidiagonal of L is not a line of L.

>

Do you agree with the statement

For a potentially infinite list, L,

of potentially infinite 0/1 sequences

the antidiagonal of L is not a line

of L

?

> Yes, that is unavoidable. Up to any n, the diagonal differs from every

> antry.

>

> But if you assume that all terminating decimals can be enumerated and

> written in one list, then it is impossible that the antidiagonal

> differs from all of them at finitely indexed digits, because every

> finitely indexed digit belongs to a terminating decimal. And they are

> all in the list by definition.

>

>

>

> > The question is:

> > Does this imply

>

> > There is no potentially infinite list

> > of potentially infinite 0/1 sequences, L,

> > with the property that

> > any potentially infinite 0/1 sequence, s,

> > is one of the lines

> > of L.

>

> What means "there is" with respect to potential infinity?

> In my opinion potential infinity means an evolving process.

> Look at thís sequence:

> 1) 0.1

> 2) 0.11

> 3) 0.111

> ...

> where we can calculate ever line n. But we cannot calculate all lines,

> because then we had all n, i.e., the actually infinite set |N (in the

> first column). And with it we had the old problems.

>

> Moreover we had all natural indices in the columns without having all

> indices in one line (the last one, but that is not existing). So we

> have all indices in this triangle. But we know that all indices that

> are in two lines, also are in one of them. By induction we can prove

> that for every line, since every line has a finite number ...

>

> No actual infinity is untenable. And with it every "there is" with

> respect to infinity.

>

> Regards, WM