On 22 Jun., 11:49, Sylvia Else <syl...@not.here.invalid> wrote:
> >> 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.
> > 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.
There is a sequence of lists including a sequence of diagonals. Every list Ln includes the diagonals A0 to A(n-1). There is no limit. Not every sequence has a limit, in particular the sequence 1, 2, 3, ... does not have a limit. But as the list is and remains countably infinite, there is no problem if you would like to have a limit. There is none, but if there was one, the list would maintain its same structure.
