On 18 Jun., 14:17, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > Newberry <newberr...@gmail.com> writes: > > On Jun 15, 9:46 am, 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. > >> That means that d is not on the list. > > > How do you know that it does not prove that an anti-diagonal does > > exist i.e. that an antidiagonal is a contradiction in terms? > > Because every infinite sequence of digits represents a real number? And > the antidiagonal is one such sequence?
No. An infinite sequence of digits does not represent a number. In general it does not even converge. In order to have convergence, you need the powers of 10 or 2 or so. But without a finite definition there are no infinite sequences at all, neither with nor without powers.