A Theorem on Productive Functions

Proof. Myhill has proved that any productive set is productive under a 1-1 total function. A corresponding result is easily established for any contraproductive set having at least one total contraproductive function. Letting h(x, y) be the recursive function of [2, Theorem 2.4, p. 69], consider the sequence s0 = h(0, 0), Sx — h(l, s0), s2 = h(0, Sx), Si = h(i, s2), Si = h(2, S3), and so on. This sequence gives Ko mutually disjoint effective enumerations (e.g., so, Sx, Si, • ■ • ) of indices of all the r.e. sets. Let \pi(x), 0^i<co, be recursive functions providing these disjoint enumerations. Then, if p(x) is any 1-1 productive (contraproductive) function for a, it is straightforward to verify that (i) each function p o \pi is also a productive (contraproductive) function for a, and (ii) the p o i/\-'s have mutually disjoint ranges. Added in proof. Since this note was submitted, the author has observed the following result: any contraproductive set admitting a total contraproductive function is actually productive. Thus, in particular, the hypothesis of the theorem does not cover the case of the complement of a simple or mesoic r.e. set; nevertheless, the conclusion of the theorem (for contraproductivity) does hold for such sets.