"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message news:E4GdnSMEFu3jxoPRnZ2dnUVZ8m2dnZ2d@brightview.co.uk... > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message > news:4c1d8758$0$14086$afc38c87@news.optusnet.com.au... >> >> "Tim Little" <tim@little-possums.net> wrote in message >> news:slrni1qqsn.jrj.tim@soprano.little-possums.net... >> > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> >> Of course it is computable. Cantor provides a simple construction >> >> for the number. >> > >> > Only if the list itself is a recursively enumerable function. >> > Cantor's proof makes no such assumption. >> > >> >> Yes it does. It requires that the nth digit of the nth term can be >> calculated. > > Not so - the proof simply needs the n'th digit of the n'th term to EXIST.
Of course the nth digit of the nth term exists. Or are you claiming there are computable Reals which don't have decimal expansions?
> It is then possible to prove the EXISTENCE of the missing real number (via > the antidiagonal argument). His proof should not be read as instructions > for a "C" program or some such! :-) It's an existence proof, not really > an > "algorithm" at all in the sense I would use the word...
Of course it is an algorithm.
It accepts the first n digits of the first n terms on the list and computes the anti-diagonal to n decimal places for all n.
> > >> This is not quite as strong as being re, but it is close. In any >> event, it is exactly the same restriction as I place on the purported >> list >> of all computable Reals. > > No, you require the list itself to be computable, so that given m,n as > input > you can compute "the n'th digit of the m'th number". That's MUCH more > than > Cantor requires.
Cantor's proof uses the nth digit of the nth term to find a Real not on the list.
