On 22 Jun., 01:29, "Mike Terry" <news.dead.person.sto...@darjeeling.plus.com> wrote: > "WM" <mueck...@rz.fh-augsburg.de> wrote in message >
> > Correct. Therefore the set of antidiagonals appears to be uncountable > > = unlistable. But it is countable. > > I don't see why you say it appears to be uncountable. (See below) >
> > The set of antidiagonals that can be constructed (by a given > > substitution rule) from an infinite sequence of lists. > > Ah, OK. > > So we have (L0, L1, L2, ...), and corresponding to each Ln we have an > antidiagonal An. So we have a sequence (A0, A1, A2, ...). > > 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 because the antidiagonal of the supposed list would belong to the set but not to the list. Therefore it is not countable.
