Date: Jan 30, 2013 6:21 AM
Subject: Re: Matheology § 203

On 30 Jan., 12:02, William Hughes <> wrote:
> Summary.  We have agreed that
> For a potentially infinite list L, the
> antidiagonal of L is not a line of L.

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

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