On 22 Jun., 14:34, Sylvia Else <syl...@not.here.invalid> wrote: > On 22/06/2010 9:28 PM, WM wrote: > > > On 22 Jun., 11:49, Sylvia Else<syl...@not.here.invalid> wrote: > > > 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 > > (L0), (A0, L0), (A1, A0, L0), (... , A2, A1, A0, L0) ... > > being respectively, > > L0, L1, L2, L3, ...
Correct. > > For (A0, A1, A2, ...) to be a list that should contain its own > anti-diagonal,
I do not argue about (A0, A1, A2, ...). The list that I consider has the form
An ... A2 A1 A0 L0
where L0 is an infinite list of all rational numbers. This list Ln does not contain its own antidiagonal.
> it has to be the same as an Ln, yet I can see no reason > to think it would be.
It is not, need not cannot and should not.
> The fact that a contradiction would arise seems a > powerful indicator that (A0, A1, A2, ...) would not be same as any Ln.
A contradiction arises for every Cantor list. Cantor thought that he had proved the uncountability of all definable binary sequences (because he used only definable binary sequences in his list and obtained a definable binary sequence as its antidiagonal). The contradiction arose when the definable sequences were recognized to be countable. That was the point - not the present discussion.