On Apr 10, 1:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > On 10 Apr., 08:59, William Hughes <wpihug...@gmail.com> wrote: > > > Next argument.- > > Consider a Cantor-list that contains a complete sequence (q_k) of all > rational numbers q_k. The first n digits of the anti-diagonal d are > d_1, d_2, d_3, ..., d_n. It can be shown for every n that the Cantor- > list beyond line n contains infinitely many rational numbers q_k that > have the same sequence of first n digits as the anti-diagonal d. > Proof: There are infinitely many rationals q_k with this property. > All are in the list by definition. At most n of them are in the first > n lines of the list. Infinitely many must exist in the remaining part > of the list. So we have obtained: > For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ..., > q_kn > > Every property that holds for all FISs d_1, d_2, d_3, ..., d_n of d, > also holds for all digits d_n of d that can be subject to digit- > reversal and digit comparison. > > Regards, WM
They have the same sequence of first n digits, but you'll never find a number in the list that has "all digits the same as the anti- diagonal" . Which is the point .
For any natural number k , the k'th member of your list will have its k'th digit different from the k'th digit of the antidiagonal . Therefore the antidiagonal itself is not in the list , even as arbitrarily large "partial sub-sequences" of the antidiagonal may appear on the list .