In article <bde842ea-7152-4476-aa85-3234c67f63a0@x21g2000yqa.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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?
What have "diagonals" to do with anything? > > > > 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.
In standard set theories, countable but not listable is impossible.
And until WM can give us a complete axiom set (like FOL+ZFC) for his version of set theory, we need not pay any attention to its oddities.
If it is listable, however, then it > has no unlisted diagonal.
