In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 22 Jun., 11:49, Sylvia Else <syl...@not.here.invalid> wrote: > > > > 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. > > Ah, noew I understand. You want to make us believe that there is a > limiting list but no limiting antidiagonal.
There is certainly a list of successive anti-diagonals that is an endless list, but infinite sequences do not have "limits" without some sort of convergence rules, which is clearly not the case here. > > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > if its antidiagonal is the antidiagonal of some Ln.
Assuming that L_(n+1) is the antidiagonal to endless list (L_n, ..., L_0, A_0, A_1, A_2,...), there clearly is always, for each n in N, a finite list (L_0, L_1, L_2, ..., L_n). And the union of all such finite lists forma an infinite list.
> > But if (L0, A0, A1, A2,...) is not a list Ln, then something must have > been happened in between that was incompatible with the process of my > proof.