"WM" <email@example.com> wrote in message news:firstname.lastname@example.org... > 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. > > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > if its antidiagonal is the antidiagonal of some Ln.
L0 is not a number. It's a list, and so is not eligible for belonging to a list of numbers. (I.e. this response makes no sense.)
So this is where I believe this sub-thread has got to as of Tue 16:00 UTC: (to stop us going round and around)
WM has defined a sequence of lists (L0, L1, ...) with corresponding antidiags (A0, A1, ...). There is a claim that "the set of antidiagonals" appears to be uncountable, but it was not clear which set this was.
WM seemed to suggest he meant the set (A0, A1, ...).
I pointed out this was obviously countable.
WM agreed but said it can't be written as a list, since it's antidiag is in the list.
Sylvia pointed out why this is obviously wrong.
WM has now suggested an alternative list (L0, A0, A1,...) but that is not a valid list of numbers.
It is true none the less that the image of L0 is countable, and if we append all the A0, A1,... it is still countable, so the combined set IS listable. Suppose LW lists this new set. Of course LW has a NEW antidiagonal which is not in the list LW, so this isn't going anywhere.