On Jan 1, 6:42 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > On 30 Dez. 2012, 23:31, George Greene <gree...@email.unc.edu> wrote: > > > YOU CAN'T PROVE IT. We by contrast HAVE EASILY > > proved that since for EVERY n, the nth position of the diagonal > > DIFFERS from the nth R on the list AT Rn's nth position, > > THE ANTI-DIAGONAL *IS*NOT*ON* the list. If it WERE on, it would > > have to be on it *AT* some row n. But the anti-diagonal IS NOT > > on the list at row n because Rn DIFFERS from the anti-diagonal IN > > POSITION n. > > But if you use a list of all finite sequences s(n) (of every finite > length n) then there is always a finite sequence s(n) that is > identical to the initial sequence d(n) of the diagonal. And as the > diagonal can only be investigated up to any finite sequence, comparing > s(n) with d(n), it is clear that Cantor's argument shows only one side > of the medal, namely there is no sequence s(n) that is identical to > d(n) for every n. The other side is that, by construction of the list, > there is for every n and every d(n) an s(n) = d(n) in the list.
Right! This is VERY TRIVIAL PROOF and points  and  are TRIVIALLY CORRECT!
segment = finite sequence of digits
 every segment of every finite size can be listed.
 every SEQUENCE of segments can be listed
by induction, a SEQUENCE of SEQUENCE OF DIGITS is a SEQUENCE of DIGITS!
Notice George just goes off on Tangents about definitions and ignores the topic.
GEORGE WHY DONT YOU PUT YOUR MONEY WHERE YOUR MOUTH IS