But the set of programs that compute real numbers is a subset of all programs, and the set of all programs is countable. Therefore the set of computable numbers is countable.
edit: I think maybe it wasn't clear that I'm talking about cardinality and measure? To be a little clearer - I'm saying that the computable numbers have (standard) measure 0 because they have countable cardinality. Any subset of the real numbers with countable cardinality has measure 0. And the set of all real numbers, of course, has uncountable cardinality.
> But the set of programs that compute real numbers is a subset of all programs, and the set of all programs is countable. Therefore the set of computable numbers is countable.
A constructivist is also not going to buy into your argument that a subset of a countable set is countable. Heck, the constructivists are not even going to buy into an argument that a subset of a finite set is necessarily finite (they have a term for such subsets: 'subfinite').
Yes, I know that this constructivism feels so bizarre that it cannot possible be coherent; the whole notion of cardinality appears to become useless. But you get used to it after a while.
Heck, even in classical mathematics, trichotomy of cardinally requires (or rather /is/) the axiom of choice. So cardinality wasn't really super well behaved to begin with.