"WM" <firstname.lastname@example.org> wrote in message news:email@example.com... > On 21 Jun., 15:05, Sylvia Else <syl...@not.here.invalid> wrote: > > On 21/06/2010 7:51 PM, WM wrote: > > > > > > > > > > > > > On 20 Jun., 22:18, Virgil<Vir...@home.esc> wrote: > > >> In article > > >> <f92c169d-ee85-40c2-aa82-c8bdf06f7...@j4g2000yqh.googlegroups.com>, > > > > >> WM<mueck...@rz.fh-augsburg.de> wrote: > > >>> On 20 Jun., 17:51, "Jesse F. Hughes"<je...@phiwumbda.org> wrote: > > > > >>>> Cantor was this utterly insane freak who chose not to accept Newberry's > > >>>> word for it, and instead *prove* that there was no list of all real > > >>>> numbers. Obviously, his proof is nonsense, because, after all, Newberry > > >>>> said there was no list. > > > > >>> His proof is nonsense because it proves that a countable set, namely > > >>> the set of all reals of a Cantor-list and all diagonal numbers to be > > >>> constructed from it by a given rule an to be added to this list, > > >>> cannot be listed, hence, that this indisputably countable set is > > >>> uncountable. > > > > >> That is a deliberate misrepresentation of the so called "diagonal proof". > > > > > But this proof can be applied to this countable set and shows its > > > uncountability. > > > > Let's see. How might that work... > > > > OK, by way of example, take the set of rationals. It's countable, so we > > can list them. > > > > Now construct an anti-diagonal. It's clearly not in the list, so... > > > > ... um, well what, exactly? > > 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 sequnce 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.
> This set is certainly not countable, because you can prove that > there is always a diagonal number not in the list.
What set? It seems you're thinking there is some "limit list" for the sequence of lists (L0, L1, L2, ...), or do you mean a specific list Ln for some n?
If you mean some kind of limit list, please define exactly what this is.
> On the other hand, > the set is countable by construction. What now?
That depends on how you answer my previous question!