On 22/06/2010 7:34 PM, WM 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. >> >> 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. > > Regards, WM
The antidiagonal of the list of (A0, A1, A2,...) would only belong to the set if it is also the antidiagnoal of some Ln, which you haven't proved to be the case.