"Tim Little" <firstname.lastname@example.org> wrote in message news:email@example.com... > On 2010-06-20, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <firstname.lastname@example.org> wrote in message >>> No you couldn't, as it makes no assumption of any finite algorithm. >>> Only *your* alteration of Cantor's proof assumes any sort of finite >>> algorithm, as it is required for the definition of "computable real". >> >> I don't say anything about finite algorithms. I just ask for a list of >> computable Reals, in the same form a Cantor asks for a list of Reals. > > If you don't realize that "finite algorithm" is contained within the > definition of "computable real" that you use, there really is no use > in continuing this line of conversation any further. Goodbye. >
I do understand that. But I don't actually ask for a "finite algorithm" in my proof, I only require that I am provided a purported list of all comptuable Reals.
Yes, in practice, this means that each item is specified by a finite algorithm, but so what? I ask for a purported list of all computable Reals and show there is at least one missing. Exactly the same as in Cantor's original proof. Producing the purported list of all computable Reals is your problem. That you cannot do this is exactly my point, yes I know there is no finite algorithm for calculating every computable Real, that is what I am trying to prove - that there can be no list of exactly the computable numbers. Any more than there can be no list of all Reals. And its not because the computable Reals are uncountable ... they are enumerable but not recursively enumerable.