Minimal Covers and Arithmetical Sets1

If a and b are degrees of unsolvability, a is called a minimal cover of b if b<a and no degree c satisfies b<c<a. The degree a is called a minimal cover if it is a minimal cover of some degree b. We prove by a very simple argument that 0" is not a minimal cover for any n. From this result and the axiom of Borel determinateness (BD) we show that the degrees of arithmetical sets (with their usual ordering) are not elementarily equivalent to all the degrees. We also point out how this latter result can be proved without BD when the jump operation is added to the structures involved. Our notation is standard. We use N to denote the set of all natural numbers and "l.u.b." to abbreviate "least upper bound." To prove that 0" is not a minimal cover for any n, it is convenient to prove a somewhat stronger result, both for the sake of extra corollaries and in order to have a sufficiently strong inductive hypothesis in the proof. Theorem 1. If a 2^0" and a is a minimal cover of b, then 6 3:0". Proof. The proof is by induction on n. The theorem is trivial for n = 0. We now assume the theorem for n = k and prove it for n = k +1. Suppose a 3:04+1 and a is a mini nal cover of b. Let c = l.u.b. [o*+1, b}. Clearly b^cf^a and so c = b or c = a. If c = b, then 0k+1^b as required. Now suppose c = a. We claim that then a is r.e. in b. Since a 3:0*, it follows from the induction assumption that fo3;0*. Thus 0*+i js r e m Jj)j since it is r.e. in 0*. Hence a is the l.u.b. of two degrees r.e. in b and so must be r.e. in b. By relativizing the theorem of Friedberg and Muchnik that no r.e. degree is minimal [6, p. 66, Corollary l], it now follows that a is not a minimal cover of b, contrary to hypothesis. Corollary 1. If a^0n and a is r.e. in 0", then a is not a minimal cover. Received by the editors October 21, 1969. AMS Subject Classifications. Primary 0270, 0277.