In article <91a57f95-54ee-4f0a-8341-b2a7dc2f11de@h13g2000yqm.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 16 Jun., 05:37, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote: > > Virgil <Vir...@home.esc> writes: > > > But until you can determine which of those 10 cases, how can you > > > compute the number? > > > > You can't. The number is computable nonetheless, in the sense that there > > exists an effective procedure for churning out its decimal expansion. > > > > As noted, computability is a purely extensional notion. Recall the > > classical recursion theory exercise, which we find, in some form or > > other, in pretty much any text on the subject: > > > > Let f : N --> N be a function such that > > > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > > > Is f computable? > > All computable, Turing-computable, nonetheless computable, definable > (in any useful, i.e., finite language), and by other means > determinable numbers form a countable set.
I am not sure that in standard mathematics that that description can "form a set" at all, much less a countable one. > > If Cantor's diagonal proof results in any such a number, then it > proves in effect the uncountability of a countable set.
If WM claims that every set of 'real' numbers is countable, meaning that one can construct a surjection from N to it, then the very constructability of such a surjection proves that set of numbers to be incomplete by the Cantor "anti-diagonal" construction.
