
Re: Why can no one in sci.math understand my simple point?
Posted:
Jun 23, 2010 7:38 PM


"Virgil" <Virgil@home.esc> wrote in message news:Virgil12E0AD.00311823062010@bignews.usenetmonster.com... > In article <88dhukFesiU1@mid.individual.net>, > Sylvia Else <sylvia@not.here.invalid> wrote: > > > 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 > > >>> <2000e81b7c5a41beb6af98f96f2fb630@w31g2000yqb.googlegroups.com>, > > >>> WM<mueckenh@rz.fhaugsburg.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 antidiagonal 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 > > >> antidiagonal of the An1, An2,...., 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 0origin 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. > > The "ultimate L", as presented by WM, does not exist as a standard list, > but only as a list open at both ends, so cannot have a standard > antidiagonal, though all sorts f equivalent nonmembers can be > constructed merely by rearranging that "ultimate L" into a list, which > may be done in many ways.
Yes, WM's "list" is simply not a list, and I think all this "trying to turn a clearly wrong statement into something meaningful on WM's behalf" ultimately just makes the thread go on and on without getting anywhere. *WM* has to say what he actually meant if he can, and if he can't then we can leave and go on to another thread where we can maybe achieve something (not sure what! hehe). It seems to me that although WM makes numerous mistakes in his wording, with persistence he actually does end up clarify what he's saying so that things can move forward.
But I had my own idea which doesn't really help WM, but neatly avoids the main problem at the moment...
We start with the list
L0 = (Q0, Q1, Q2, ...) [list of rationals as before]
and form it's antidiagonal A0 as before. BUT... we create our new list L1 by inserting A0 at position 2:
L1 = (Q0, A0, Q1, Q2, Q3, Q4, ...)
And call L1's antidiagonaly A1, which we stuff into L1 at position 4 to get L2:
L2 = (Q0, A0, Q1, A1, Q2, Q3, Q4, ...)
... and so on. (An is the antidiag for Ln, and is inserted at posiiton (2n+2) to form list L_(n+1).
The advantage is that there is now a well defined limit list containing ALL the antidiagonals, which I'll name Loo:
Loo = (Q0, A0, Q1, A1, Q2, A2, Q3, A3, Q4, A4, Q5, ...)
This seems to be as close to what WM is after as I can come up with. But it still doesn't work, as Loo has an antidiagonal which is not in Loo, and there is still no contradiction to be had.
Regards, Mike.
> > > > 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. > > One can still create equivalents of antidiagonals simply by reordering > that "final" list. Nthing about an "antidiagonal" requires taking the > elements of the base list in any particular order, as long as one can > take ALL of them one after another in in SOME order. > > > > Rather than argue that VMs proposition fails on that point, I wanted to > > address the flaw, in order to find a more substantial objection. > > > The flaw in WM's argument is that he claims that no nonmember can be > found for his "ultimate" open at both ends list, but he is wrong. > > Given that "ultimate list" {...a2,a1,a0,q0,q1,q2,...} merely rearrange > it to {q0, a0, q1, a1, q2, a2, ...} and you have comprehensive list of > everything with a findable antidiagonal which is not listed.

