In article <88d6j2Fqq8U1@mid.individual.net>, Sylvia Else <sylvia@not.here.invalid> wrote:
> On 23/06/2010 11:03 AM, Virgil wrote: > > In article > > <2000e81b-7c5a-41be-b6af-98f96f2fb630@w31g2000yqb.googlegroups.com>, > > WM<mueckenh@rz.fh-augsburg.de> wrote: > > > >> On 22 Jun., 21:34, Virgil<Vir...@home.esc> wrote: > >> > >>>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's > >>>>> "certainly > >>>>> not countable", but it is. > >>> > >>>> The set is certainly countable. But it cannot be written as a list > >>> > >>> But it HAS been written as a list (A0, A1, A2, ...), > >> > >> Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, > >> L0)? > > > > Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why > > should there be any antidiagonal for it? > > Ach! Let's scrap A0 - it's confusing. > > If we let L_n be the nth element in the list L0, and An the > anti-diagonal of the An-1, An-2,...., A1, L_1, L_2, L_3,... > > then > > L_1 > A1 > L_2 > A2 > L_3 > A3 > L_4 > ... > > is a list. I'm still thinking about that. > > Sylvia.
There are a lot of possible lists here.
One starts with, for example, same listing of the rationals indexed by the 0-origin naturals: L0 = {q0, q1, q2, ...}.
For that list one finds an antidiagonal, a0, not in L0, and with it forms a new list L1 = {a0, q0, q1, q2, ...} with the antidiagonal a0 prepended to the list of which it is the antidiagonal.
This process is clearly recursive, allowing us now, for example, to find an antidiagonal a1 to L1 which we can prepend a1 to L1 giving a new list L2 = {a1, a0, q0, q1, q2, ...}.
The process also clearly may be in theory repeated infinitely often, so that one can derive from it new sequence of the antidiagonals taken in the order of their derivation, A = {a0,a1, a2, ...}.
But as listability = countability everywhere except in WM's world, WM's arguments claiming that tone can have ne without the other fails.
