On 22/06/2010 9:28 PM, WM 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.
Hadn't even crossed my mind.
> > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > if its antidiagonal is the antidiagonal of some Ln. > > 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. >
Strictly speaking, by your method of construction, the lists are
For (A0, A1, A2, ...) to be a list that should contain its own anti-diagonal, it has to be the same as an Ln, yet I can see no reason to think it would be. The fact that a contradiction would arise seems a powerful indicator that (A0, A1, A2, ...) would not be same as any Ln.
