In article <2000e81b-7c5a-41be-b6af-98f96f2fb630@w31g2000yqb.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 22 Jun., 21:34, Virgil <Vir...@home.esc> wrote: > > > > > But (A0, A1, A2, ...) is obviously countable. Above you say it's > > > > "certainly > > > > not countable", but it is. > > > > > The set is certainly countable. But it cannot be written as a list > > > > But it HAS been written as a list (A0, A1, A2, ...), > > Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, > L0)?
Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why should there be any antidiagonal for it?
On the other hand, if (A0, A1, A2, ...) is a list of reals, then it will have an antidiagonal.
>Then Cantor's argument fails. Does it not? Then a list can not > list all anti-diagonals that belong to the countabe set constructed in > my argument.
Outside of WM's world a recursive construction such as he suggests does allow an anti-diagonal to any list infinite anti-diagonals.
