In article <2af854b2-1a9a-423f-8f6a-f831e78f584f@t10g2000yqg.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 22 Jun., 18:27, "Mike Terry" > <news.dead.person.sto...@darjeeling.plus.com> wrote: > > "WM" <mueck...@rz.fh-augsburg.de> wrote in message > > > > news:b71db24f-d637-4afd-a717-d2b5055f4fbf@a30g2000yqn.googlegroups.com... > > > > > On 22 Jun., 11:49, Sylvia Else <syl...@not.here.invalid> wrote: > > > > > > The antidiagonal of the list of (A0, A1, A2,...) would only belong to > > > > the set if it is also the antidiagnoal of some Ln, which you haven't > > > > proved to be the case. > > > > > Ah, noew I understand. You want to make us believe that there is a > > > limiting list but no limiting antidiagonal. > > > > > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > > > if its antidiagonal is the antidiagonal of some Ln. > > > > L0 is not a number. It's a list, and so is not eligible for belonging to a > > list of numbers. (I.e. this response makes no sense.) > > > > ----------------- > > > > So this is where I believe this sub-thread has got to as of Tue 16:00 UTC: > > (to stop us going round and around) > > > > WM has defined a sequence of lists (L0, L1, ...) with corresponding > > antidiags (A0, A1, ...). There is a claim that "the set of antidiagonals" > > appears to be uncountable, but it was not clear which set this was. > > > > WM seemed to suggest he meant the set (A0, A1, ...). > > No, I added the A's to a list. > > > > I pointed out this was obviously countable. > > > > WM agreed but said it can't be written as a list, since it's antidiag is in > > the list. > > > > Sylvia pointed out why this is obviously wrong. > > No, she misunderstood. > > > > WM has now suggested an alternative list (L0, A0, A1,...) but that is not a > > valid list of numbers. > > I have, from the beginning, used the following list > > Ln = > > An > ... > A2 > A1 > A0 > L0 > > where L0 is thelist of all rationals. > > This list Ln contains a countable set of numbers but the set of its > diagonals is not listable, because An is not in the list. > > > > It is true none the less that the image of L0 is countable, and if we append > > all the A0, A1,... it is still countable, so the combined set IS listable. > > Suppose LW lists this new set. Of course LW has a NEW antidiagonal which is > > not in the list LW, so this isn't going anywhere. > > All possiblke diagonals of this set of lists Ln belong to a coutable > set, but there is no list of all of them.
Sure there is.
Let L_0 = {q_0, q_1, q_2 ...} be a listing of the rationals. Let a_0 be an antidiagonal of L_0. Let L1 = { a_0, q_0, q_1, q_2 ...} be a new list.
Extend this recursively with a_n being an antidiagnal to L_n, and L_(n+1) = L_n with a_n prepended to it = { a_n, a_(n-1), ...a_0, q_0, q_1, q_2, ...}
If this is alt all correct then WM is totally incorrect. The set {a_n : n in N} is already a list as it is indexed by N.
And it is still the case in standard mathematics that listability is equivalent to countability. > > It is the same with all Cantor-list. All diagonals of all Cantor-lists > belong to a countable set.
Actually not. To start with, there exist uncountably many "Cantor-lists".
And WM's definition of countability is fatally flawed. > > Regards, WM
