In article <dd5850eb-6401-43e5-a342-df3166644cab@a30g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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. > > Regards, WM
However the finite sublists of so far constructed antidiagonals does have a "limit", it being a list of all the antidiagonals buildable by this iteration.
WM's hang up is that in his world no infinite process is ever allowed to be completed so that he is, like Xeno, unable to move past what the rest of us move past easily.
