In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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.
That depends of WHICH set of "antidiagonals" one is talking about. the set of only those constructed from diagonals will be countable but the set of all numbers not in the list will be uncountable
> > > > 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
But it HAS been written as a list (A0, A1, A2, ...), dingbat!