On 2010-06-19, Virgil <Virgil@home.esc> wrote: > There is, however, some question in my mind about the existence of a > list of all and ONLY computable reals.
Why? The computable reals can be proven countable, as you already know, so there is a bijection between N and the set of countable 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.
That is true. It is also true that the existence of a surjection from N to X implies the existence of a bijection from N to X, even without the Axiom of Choice. (It is not necessarily true without AC if you replace N with an arbitrary set)