On 23 Jun., 08:39, Sylvia Else <syl...@not.here.invalid> wrote:
> > Given that "ultimate list" {...a2,a1,a0,q0,q1,q2,...} merely rearrange > > it to {q0, a0, q1, a1, q2, a2, ...} and you have comprehensive list of > > everything with a findable antidiagonal which is not listed.
With or without all diagonals? > > I thought that was what I did.
It would not help, because then the set of all lines (of my construction) including the antidiagonal of this construction is a countable set but not listable. If it is listable, however, then it has no unlisted diagonal.
