"Tim Little" <firstname.lastname@example.org> wrote in message news:email@example.com... > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> So Cantor's proof when applied to computable Reals proves exactly >> what in your opinion? > > That there is a real not on the list. > > >> I might remind you that as the form of proof is identical to that >> used by Cantor for all Reals, whatever you believe that Cantor's >> proof applied to computable Reals proves, his proof applied to all >> Reals must prove the same thing. > > It does prove exactly the same thing: in both cases, all such lists > omit at least one real. > > >> Its the same proof, after all, except limitting the set to just >> computable Reals. > > No, there is a slight difference: the antidiagonal is a real number. > It does not necessarily produce a computable real number. >
Of course it is computable. Cantor provides a simple construction for the number.
> Why, in your opinion, does the construction fail for rational numbers? > >
Because the anti-diagonal may not be rational.
It is however computable. It can be specified to any arbitrary degree of accuracy. Cantor provides an explicit construction. You give me a list of all Rationals, I can compute the anti-diagonal of this list explicitly. Indeed, you give me any list of Real numbers, I can compute the anti-diagonal explicitly.