WM misunderstands what "proof by contradiction" means. Such a proof requires that one start with a contrary assumption, that the theorem in some way fails to be true and from that assumption derive a contradiction, which is not done here.
The proof itself merely assumes ANY list of sequences of m's and w's and shows that there is another which is not included.
The proof is quite complete before it is noted that the theorem's falsehood would produce any contradiction, so that contradiction remark is in no way essential to the proof, and the proof is not, in any sense, a proof by contradiction.
WM's ignorance of formal logic again trips him up.
> > > > > 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.
Since there is a unique injection f:N -> R in which each natural gets mapped to its real value, while there are reals unlisted there are no natural-valued reals unlisted.
The reals have a unique multiplicative identity, which to distinguish it from the 1 in N I shall designate by one. Then the mapping from N to R defined by f)1 = one and f(sucessor(n)) = f(n) + one, where "+" is the real number addition operator, is that injection mentioned above. > > > > 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.
To justify calling a claim nonsense, one must be given evidence of its falsity, which WM carefully does not attempt.
For what sets does WM argue that the identity mapping from that set to itself is not a bijection? He seems now to be claiming it for N.
> > > > > 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.
And the difference between a duck is that one leg is both the same. > > > > > 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
Then WM is claiming that n^n^n^...^n is not equal to itself? What strange natural numbers exist in WM's world. > > > 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.
But I presented a function listing (as an image of N) all members of N, whereas surjective images of R do not count as listings until after R has been shown to be a surjective image of N, which Cantors followers have proved impossible. > > Regards, WM
