On 23 Jun., 06:47, Sylvia Else <syl...@not.here.invalid> wrote:
> > Rather than argue that VMs proposition fails on that point, I wanted to > address the flaw, in order to find a more substantial objection.
My initials ar WM.
There is no flaw in the argument: Your bijection either contains all constructed antidiagonals. Then a list contains also its antidiagonal, because every list has an antidiagonal and your bijection (that is only a permutation of my list) contains all lines and antidiagonals. I.e., there is no missing antidiagonal of a "limit" list outside of the bijection.
Or there is a last diagonal of the limit list that does not belong to your bijection (and to my list). Then there is a countable set (the set that includes this last diagonal) that is not listable.
Think over that only a little bit. It is not difficult to understand.