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
> 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