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"? It either must embed the list within the algorithm (thus rendering it not finite), 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.