There Is No Degree Invariant Half-jump

We prove that there is no degree invariant solution to Post’s problem that always gives an intermediate degree. In fact, assuming definable determinacy, if W is any definable operator on degrees such that a <W (a) < a′ on a cone then W is low2 or high2 on a cone of degrees, i.e., there is a degree c such that W (a)′′ = a′′ for every a ≥ c or W (a)′′ = a′′′ for every a ≥ c. A striking phenomenon in the early days of computability theory was that every decision problem for axiomatizable theories turned out to be either decidable or of the same Turing degree as the halting problem (0′, the complete computably enumerable set). Perhaps the most influential problem in computability theory over the past fifty years has been Post’s problem [10] of finding an exception to this rule, i.e. a noncomputable incomplete computably enumerable degree. The problem has been solved many times in various settings and disguises but the solutions always involve specific constructions of strange sets, usually by the priority method that was first developed (Friedberg [2] and Muchnik [9]) to solve this problem. No natural decision problems or sets of any kind have been found that are neither computable nor complete. The question then becomes how to define what characterizes the natural computably enumerable degrees and show that none of them can supply a solution to Post’s problem. Steel [13] suggests that a natural degree should be definable and its definition should relativize to an arbitrary degree (and so, in particular, be defined on degrees independently of the choice of representative). Along these lines an old question of Sacks’ [11] asks whether there is a degree invariant solution to Post’s problem, i.e. a computably enumerable degree invariant operator W such that A <T W (A) <T A ′. (A function W : 2 → 2 is a computably enumerable operator if there is an e ∈ ω such that, for each set A, W (A) = W e , the e th set computably enumerable in A. Any function f : 2 → 2 is degree invariant if for every A and B, A ≡T B implies that f(A) ≡T f(B).) Such an operator would clearly be a candidate for a natural solution to Post’s problem. Received by the editors February 16, 1996 and, in revised form, May 9, 1996. 1991 Mathematics Subject Classification. Primary 03D25, 03E60, 04A15; Secondary 03D30.