"Tim Little" <email@example.com> wrote in message news:firstname.lastname@example.org... > On 2010-06-20, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <email@example.com> wrote in message >>> However, there is no way that you can then prove the existence of a >>> finite algorithm accepting only the *finite* input n and producing the >>> n'th antidiagonal digit. >> >> Of course I can prove the existence of a finite algorithm. I can produce >> it. >> >> To produce the nth decimal of the anti-diagonal, look at the first n >> items on the purported list of all computable Reals. > > In what way is that "a finite algorithm accepting only the *finite* > input n"?
Its finite, and its an algorithm.
>It either must embed the list within the algorithm (thus > rendering it not finite),
No, it embeds the first n digits of the first n terms to calculate the first n decimal places, and does for all n.
That a finite process can be used to determine a number to any required degree of accuracy makes it computable.
> or accept the list as additional ainput > (thus violating the condition that it accept only the finite input n). > > >> Well, I think I am making a pretty good hand of my case. But yes, I >> know what I am saying is not accepted. I am genuinely confused by >> this - the fact that my opinions are unique, and I would expect >> therefore to be almost certainly wrong (with probability 1). On the >> other hand my argument seems pretty solid to me. > > Read through very carefully the definition of a computable real, > perhaps also reading some of the other known results on computable > reals and following their proofs. At some point you will hopefully > see why the "construction" in Cantor's diagonal proof does not qualify > as a finite algorithm in the sense of the definition of computable > real. >
Its an explicit construction. I can even do it by hand to tell you the anti-diagonal to any required degree of accuracy in a finite number of steps. That makes it computable.