In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 3 Jun., 05:34, Virgil <virg...@nowhere.com> wrote: > > In article <KKn5F7....@cwi.nl>, "Dik T. Winter" <Dik.Win...@cwi.nl> > > wrote: > > > > > In article > > > <5df917e3-517f-4a38-b28b-363843496...@t21g2000yqi.googlegroups.com> WM > > > <mueck...@rz.fh-augsburg.de> writes: > > > > You drop the completeness condition in certain cases but you assume it > > > > in case of Cantor's proof. That is cheating. > > > > > You again misunderstand the proof completely. There is an assumption that > > > a complete list is provided and that is proven false. > > > > As I understand the Cantor diagonal proof, the only assumption is that > > whenever one is provided with a list then that list has to omit at least > > one sequence. I do not think it was, in its original form, an indirect > > proof as your statement seems to indicate. > > Cantor understood it as a proof by contradiction. "da wir sonst vor > dem Widerspruch stehen würden, daß ein Ding E0 sowohl Element von M, > wie auch nicht Element von M wäre." But probably you know better.
A quote without a source cited is not evidence.
There is certainly no necessity for it to be cast as an indirect proof when the direct proof is even simpler. > > The only thing that is interesting here is that the same holds for the > list of all natural numbers. Give me a list of natural numbers and I > will show you that it is incomplete.
A list in the context of the Cantor diagonal argument means a mapping from the set of all naturals to the members of the listed set, so that the function, f: N --> N: x |--> x is a list of natural numbers.
Now WM must shown how and why THAT listing is incomplete (in the sense of missing some member of N either as argument or as value of the function f). > > Of course you are not allowed to say: All n in N!
All n in N, without the factorial symbol, though, is quite legitimate.
> That would be a > contradiction, because there is no last n and consequently no chance > to check whether your list would contain all n in N.
Since my listing clearly includes every n in N for which n = n, WM must show that there is some n for which n = n is false to falsify my argument.