"WM" <email@example.com> wrote in message news:firstname.lastname@example.org... > On 21 Jun., 21:35, "Mike Terry" > <news.dead.person.sto...@darjeeling.plus.com> wrote: > > > > It is an irrational number. Add it at first position to the former > > > list L0, obtain L1. Now construct the diagonal of L1 (according to a > > > fixed scheme). It is certainly an irrational number. Add ist to the > > > list L1 at first position, obtain list L2. Now construct the > > > diagonal, ... and continue. I this way you get a countable set > > > consisting of all rationals and of all diagonal numbers of these > > > lists. > > > > No... What you get is an infinite sequence of lists (L0, L1, L2, ...) > > > > Each Ln in the sequence contains all rationals, and Ln contains the > > antidiagonals for L0,...L(n-1). Ln doesn't contain its own antidiagonal, > > and doesn't contain any antidiagonals for Lm with m>n. > > Correct. Therefore the set of antidiagonals appears to be uncountable > = unlistable. But it is countable.
I don't see why you say it appears to be uncountable. (See below)
> > > > > This set is certainly not countable, because you can prove that > > > there is always a diagonal number not in the list. > > > > What set? > > The set of antidiagonals that can be constructed (by a given > substitution rule) from an infinite sequence of lists.
So we have (L0, L1, L2, ...), and corresponding to each Ln we have an antidiagonal An. So we have a sequence (A0, A1, A2, ...).
But (A0, A1, A2, ...) is obviously countable. Above you say it's "certainly not countable", but it is.