True, I forgot about that. I guess that would allow extending the notion of "computable well-ordering" to sets of size 2^aleph_0. Of course that doesn't necessarily mean any of them would actually be computable, and I expect none of them would. Well, I guess it has to be the case that none of them would, or else we ought to be able to prove "R is well-orderable" without choice, but there ought to be a simpler argument than that -- just as computable ordinals in the ordinary sense are downclosed, I would expect that these too ought to be downclosed, which would immediately imply that none of the uncountable ones are computable. Now I guess I should actually check that they are downclosed...