On 23/06/2010 2:27 PM, Virgil wrote: > 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.
Yes. > > 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, ...}.
Yes, but at some point VM made it clear that the issue wasn't that the diagonal wasn't present in A, but in the 'ultimate' L.
I was concerned that as it was then formatulated, VM's proposition could be attacked on the basis that the diagonal of the 'ultimate' L couldn't be constructed - you couldn't even start to do so, because the first element of the list wasn't defined. It was the 'last' element of an infinite sequence.
Rather than argue that VMs proposition fails on that point, I wanted to address the flaw, in order to find a more substantial objection.
