In article <e98313e8-07a4-4b4f-86a9-cfc392a36616@c33g2000yqm.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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.
Nonsense. Every permutation of a listing creates a list whose antidiagonals are different from those of the original.
> Then a list contains also its antidiagonal
That is more of WM's nonsense since for every list, its own antidiagonal is constructed so as to be not be a member of THAT list. And if one inserts it into its own list, one then has a different list.
So for WM to demand a list containing a nonmember of that list is foolish.
, > 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.
I can find one quite easily. Arrange the original list together with all the countably many antidiagonals one has generated into a list (which is quite simple) and take the antidiagonal of that list. this last antidiagonal will not be in the original or in any of the derived antidiagnoals. > > 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.
WM may find it unlistable, but I do not. Merely prepend that last antidiagonal to the list from which it was generated. > > Think over that only a little bit. It is not difficult to understand.
I understand it quite clearly, but WM seems to be having a ton of problems with it.
