In article <ec965aaf-77ee-4190-846a-4acf52b68dbc@z8g2000yqz.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 15 Jun., 18:46, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > > > > > >On 15 Jun., 16:32, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > >> The proof does not make use of any property of infinite lists. > > >> The proof establishes: (If r_n is the list of reals, and > > >> d is the antidiagonal) > > > > >> forall n, d is not equal to r_n > > > > >As every n is finite, it belongs to a finite initial segment of the > > >infinite list. > > > > I'm not sure what you are saying. The fact is, we can prove > > that for every real r_n on the list, d is not equal to r_n. > > Of course. Every real r_n belongs to a finite initial segment of the > list.
And an infinite terminal segment, neither of which is relevant.
> That does not yield any result about the whole list
It yields a result about each and every member of that list being different from d.
> Look here: We can prove for any finite segment > {2, 4, 6, ..., 2n} > of the ordered set of all positive even numbers that its cardinal > number is surpassed by some elements of the set.
What you cannot prove is that that statement is in any way relevant to the argument you are trying (and failing) to justify.
