On 2/6/2013 12:51 PM, WM wrote: > On 6 Feb., 19:16, fom <fomJ...@nyms.net> wrote: >> On 2/6/2013 8:46 AM, WM wrote: >> >> >> >> >> >>> On 6 Feb., 02:34, fom <fomJ...@nyms.net> wrote: >>>> On 2/5/2013 11:02 AM, WM wrote: >> >>>>> On 5 Feb., 17:06, fom <fomJ...@nyms.net> wrote: >>>>> This correspondence is as impossible, as I have shown above, as >>>>> finding a set of natural numbers with negative sum. >> >>>> No. >> >>> If you think no, then explain this: >> >>> Have you ever seen a Cantor-list where more than half of the >>> interesting sequences (a_j) of digits a_kj with k < j had infinite >>> length? Have you ever seen a Cantor-list with at least one of the >>> interesting sequences of digits having infinite length? No? Why the >>> heck do you believe that they play the crucial role in Cantor's >>> "proof"? >> >> It is called individuation. >> > Cantor proves: For every line a_n the inequality a_nn =/= d_n implies > a_n =/= d.
Before he proves. He asserts the presumption of the presumed claimant. That presumption involves a one-to-one correspondence. A one-to-one correspondence is not a one-to-many correspondence.
> > This implies: For *every* line a_n: (a_n1, a_n2, ...., a_nn) is > *finite*. >
Yes. Cantor identifies a terminating search for each line in order to make stepwise decisions on what symbol to use in extending the sequence of symbols that form the counter-example.
> In words: The proof unavoidably requires that the lines a_n, with no > exception, are finite up to the crucial point a_nn - and the rest is > silence.
No. The counter-example only requires a terminating search for each line. The proof requires a presupposition of individuated numbers named using a consistent naming algorithm based on a finite set of sequentially ordered symbols.
> The rest has *nothing* to do with the proof, and if the rest > is empty, the proof is not in the least changed. > No individuation.
And with the two words of this sentence, you identify your complete and total ignorance of...