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.
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.
Of course you are not allowed to say: All n in N! 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. (And if you did so, then one could also say: All r in R. The contracdiction would be of same size.)