There is no low maximal d. c. e. degree – Corrigendum

Abstract

The purpose of this short paper is to clarify and correct the proof of the main result contained in [1]: namely, that there exists no low maximal d. c. e. degree. There we gave a simple proof obtained as an immediate corollary of the following posited extension of the Robinson Splitting Theorem ([1, Theorem 1.7]: For any c. e. set A,Denis Hirschfeldt (private communication) was the first to notice a problem with the particular application of the Recursion Theorem in the proof of this result, one which does not occur in the original Robinson proof [5].We present below the following reformulation of the use of the Recursion Theorem sufficient to correct the proof of our main result (the non-existence of a low maximal d. c. e. 1) ), via a degree-theoretic extension of the Robinson Splitting Theorem:Theorem For any d. c. e. degree l, any c. e. degree a, if l is low and l < a, then there are d. c. e. degrees a 0 , a 1 such that l < a 0 , a 1 < a and a 0 ∪ a 1 = a. P r o o f. Let L be a d. c. e. set of low degree. Let L = L 0 − L 1 for some c. e. sets L 0 , L 1 such that L 0 ⊃ L 1 . Let f be a 1 − 1 computable function such that L 0 = {f (x) : x ∈ ω}, and M = f −1 (L 1 ). Then M is c. e. and M ≤ T L. (M is called Lachlan's set for L.)Given a c. e. set A, and a d. c. e. set L, assume that L < T A and L has low degree. First we construct ω-c. e. sets A 0 , A 1 to satisfy the following requirements:where i = 0, 1, e ∈ ω, and {Φ e : e ∈ ω} is a standard list of all partial computable (p. c.) functionals Φ. Let a = deg T (A), l = deg T (L) and a i = deg T (A i ⊕ L) for i = 0, 1. Then a i is an ω-c. e. degree, for each i = 0, 1. By the R-requirement, a 0 ∪ a 1 = a, and by the S-requirements, l ≤ a i < a, i ∈ {0, 1}. Then by the S-and R-requirements we will have l < a i .We will construct ω-c. e. sets so that they are c. e. in Lachlan's set M for L. The theorem follows from the following result from [2].Proposition 1 (Arslanov, LaForte and Slaman [2]) . Let A and C be sets such that C is c. e., A is c. e. in C, C ≤ T A, and A is ω-c. e. Then deg(A) is d.

Combinatorial principles weaker than Ramsey's Theorem for pairs

“…For a map f and a subset X of its domain, we denote the image of X under f by f "X. (RT 2 2 ) Ramsey's Theorem for pairs: Every 2-coloring of [N] 2 has a homogeneous set. It is easy to show that the number of colors does not matter, in the sense that for each n > 2, Ramsey's Theorem for 2-colorings of pairs is equivalent to Ramsey's Theorem for n-colorings of pairs.…”

“…In fact, one of the standard proofs of Ramsey's Theorem for k-tuples works in ACA 0 (for any fixed k), and so for each k > 2, Ramsey's Theorem for k-tuples is equivalent to ACA 0 . Whether RT 2 2 itself also implies ACA 0 remained open for twenty years. Seetapun (see Seetapun and Slaman [1995]) proved a degree theoretic cone avoiding theorem (for every set Z, coloring f of [N] 2 computable in Z, and sets C i T Z, there is a homogeneous set not computing any C i ) that implies that RT 2 2 does not imply ACA 0 (even over WKL 0 ).…”

“…Whether RT 2 2 itself also implies ACA 0 remained open for twenty years. Seetapun (see Seetapun and Slaman [1995]) proved a degree theoretic cone avoiding theorem (for every set Z, coloring f of [N] 2 computable in Z, and sets C i T Z, there is a homogeneous set not computing any C i ) that implies that RT 2 2 does not imply ACA 0 (even over WKL 0 ). Another view of the proof theoretic strength of such theorems and systems is provided by the analysis of their first order consequences and conservativity results.…”

“…Thus, in particular, the first order consequences and consistency strengths of the two systems are the same. While it is a major open question whether RT 2 2 actually implies WKL 0 , RT 2 2 is known to be strictly stronger than WKL 0 in this proof theoretic sense. The relevant principle is Σ 2 -bounding.…”

“…Moreover, even in the first order case, BΣ n+1 is not Σ 0 n+2 -conservative over IΣ n (Paris [1980, p. 331]). Thus, as Hirst [1987] showed that RT 2 2 implies BΣ 2 , RT 2 2 is not even Σ 0 3 -conservative over RCA 0 . (By Friedman [1976], any model M of IΣ n can always be extended to a model of RCA 0 + IΣ n with the same first order part by taking the sets to be those with ∆ 0 1 definitions over M .)…”