2007
DOI: 10.1090/s0002-9939-07-08871-5
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Turing incomparability in Scott sets

Abstract: Abstract. For every Scott set F and every nonrecursive set X in F , there is a Y ∈ F such that X and Y are Turing incomparable.

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Cited by 15 publications
(11 citation statements)
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“…H. Friedman and McAllister independently asked the following question: if S is a Scott set and X ∈ S is not computable, does there necessarily exist a Y ∈ S such that X | T Y ? Kučera and Slaman [10] have recently given a positive answer to this question using Theorem 2.1.…”
Section: Introductionmentioning
confidence: 99%
“…H. Friedman and McAllister independently asked the following question: if S is a Scott set and X ∈ S is not computable, does there necessarily exist a Y ∈ S such that X | T Y ? Kučera and Slaman [10] have recently given a positive answer to this question using Theorem 2.1.…”
Section: Introductionmentioning
confidence: 99%
“…K-trivial sets provide an injury-free solution to Post's problem. A more surprising application was given by Kučera and Slaman [25], who used K-trivial sets in their proof that no Scott set of Turing degrees is "hourglass-like". That is, for every noncomputable real in the Scott set there is a Turing incomparable real.…”
mentioning
confidence: 99%
“…For instance, K-trivial reals allow us to solve Post's problem "naturally" (well, reasonably naturally), without the use of a priority argument, or indeed without the use of requirements. Also, Kučera and Slaman [19] have used this class to solve a longstanding question in computable model theory, namely given a noncomputable Y in a Scott set S, there exists in S an element X Turing incomparable with Y .…”
Section: Introductionmentioning
confidence: 99%