> Step n owns 2^(n-1) initial segments. Bruno, why are we discussing this? Sure, in finite time you can compute all initial segments of size n. In countable time you can compute one real, or a countable number of reals. But each of your steps needs more than twice the time required by the previous step. Therefore you need more than countable time to compute all reals.
> Now, could you give me a bitstring which is not generated > by this countably infinite process ? Sure. The output of the program "While TRUE print 1" won't be computed in countable time by your procedure. Juergen
