In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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. This set is certainly not countable, because you can prove that > there is always a diagonal number not in the list. On the other hand, > the set is countable by construction. What now?
In what position in each new list is the anti-diagonal of the immediately previous list to be inserted?
While any finite number of such insertions will indeed create new lists, will an infinite number of insertions work?
For example, if the new entry is always placed first in the new list, the after infinitely such insertions there will be original members with infinitely many predecessors, so that one no longer has a list.
And one clearly cannot put any new element after all others, as that also destroys the "being a list" property.
Thus WM's arguments again fall flat and again demonstrate his inability to think things through.