In article <email@example.com>, Tim Little <firstname.lastname@example.org> wrote:
> On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: > >> The relevant point: the *only* input to the Turing machine in the > >> definition is n. The rest of the tape must is blank. Peter > >> apparently believes that a number is still computable even if the > >> Turing machine must be supplied with an infinite amount of input (the > >> list of reals). > > > > No. > > Oh? Then what leads you to believe that the antidiagonal of a (not > necessarily recursive) list of computable reals is computable? > > > > Do you agree that Cantor's diagonal proof when applied to a > > purported list of all computable Reals will produce a computable > > Real not on the list > > No, for the fifth time now. The antidiagonal of a list of all > computable real is not computable. How many more times would you like > me to repeat this simple and mathematically obvious statement?
There is, however, some question in my mind about the existence of a list of all and ONLY computable reals.
For countability of a set it is certainly sufficient to have a list containing all its members even if that list is allowed to contain other things as well.