On 3 Jun., 22:53, Virgil <virg...@nowhere.com> wrote:
> > > 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.
From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Reihenform]: E1, E2, ?, Ev, ? otherwise we would have the contradiction, that a thing [Ding] E0 would be both an element of M, but also not an element of M.
Can you see the word "contradiction" in that text? > > > 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.
Then define a list as a mapping from N --> R and everything is fine. > > 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).
If you give me a number n then there are n^n^n^...^n numbers not given. Therefore your claim is nonsense. > > > > > Of course you are not allowed to say: All n in N! > > All n in N, without the factorial symbol, though, is quite legitimate.
All fractions required for the Oresme proof 1 + 1/2 + (1/3 + 1/4) + ... of the divergence of the series SUM(1/n) make up a number of 2^aleph_0, iff there are aleph_0 parentheses. > > > 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
but not every n^n^n^...^n
> WM must > show that there is some n for which n = n is false to falsify my > argument.
The same could be claimed for a list of all real numbers. Here it is: r = r.