"WM" <mueckenh@rz.fh-augsburg.de> wrote in message news:41a66545-c0bb-44e3-bb41-1fdaf0c83d71@j4g2000yqh.googlegroups.com... > On 24 Jun., 01:02, "Mike Terry" > <news.dead.person.sto...@darjeeling.plus.com> wrote: > > > > a) Does this list contain the anti-diagonal of > > > (..., An, ... A2, A1, A0, L0)? > > > > This is not a list of numbers. L0 is not a number, it is a list. > > > > Therefore (..., An, ... A2, A1, A0, L0) does not have an anti-diagonal. > > You are wrong.
I am obviously right.
What you have written is not a list of real numbers.
Anti-diagonals (as used in Cantor's proof) require as "input" a list of real numbers.
Therefore, what you've written does not have an anti-diagonal in the sense of Cantor's proof.
> > The symbols above abbreviate the sequence of lists > > An > ... > A0 > L0
Which I'll take that you're trying to say that that (..., An, ... A2, A1, A0, L0) abbreviates the SEQUENCE of lists (L0, L1, L2...)
(since by your construction, (An, ...A0, L0(0), L0(1),...) is the list L_(n-1).)
NOTE: a sequence of lists does not have an antidiagonal, so your earlier question:
> > > a) Does this list contain the anti-diagonal of > > > (..., An, ... A2, A1, A0, L0)?
is nonsense, as I pointed out.
> > Each of them has an antidiagonal
... yes, the antidiagonal of Ln is An..
> that either is not in the list
... sorry, what list? Do you mean the list of sequences (L0, L1, ...) or the list Ln, or something else?
> (then > the set of all of them is unlistable) or is in a list. Then Cantors > argument is wrong. > > Regards, WM
