On 23/06/2010 8:44 PM, WM 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. 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. > > Regrads, WM
Thing is, I just cannot see that you've constructed a countable set that should contain its own anti-diagonal when expressed as a list. All you've done is construct a countable set of countable sets, each of which contains the anti-diagonals of its predecessors, but none of which by construction can be expected to contain its own anti-diagonal.
You've then claimed, but without any kind of formal argument, that if you take this to the limit, the result is a countable set that should contain its own anti-diagonal when expressed as a list.