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Definability in the Recursively Enumerable Degrees

Published online by Cambridge University Press:  15 January 2014

André Nies ,
Richard A. Shore  and
Theodore A. Slaman
André Nies
Affiliation:
Department of Mathematics, University of Chicago, Chicago, IL 60637, USA.E-mail:nies@math.uchicago.edu
Richard A. Shore
Affiliation:
Department of Mathematics, Cornell University, Ithaca, NY 14853, USA. E-mail: shore@math.cornell.edu
Theodore A. Slaman
Affiliation:
Department of Mathematics, University of California, Berkeley, CA 94720, USA. E-mail: slaman@math.berkeley.edu
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§1. Introduction. Natural sets that can be enumerated by a computable function (the recursively enumerable or r.e. sets) always seem to be either actually computable (recursive) or of the same complexity (with respect to Turing computability) as the Halting Problem, the complete r.e. set K. The obvious question, first posed in Post [1944] and since then called Post's Problem is then just whether there are r.e. sets which are neither computable nor complete, i.e., neither recursive nor of the same Turing degree as K?

Let be the r.e. degrees, i.e., the r.e. sets modulo the equivalence relation of equicomputable with the partial order induced by Turing computability. This structure is a partial order (indeed, an uppersemilattice or usl)with least element 0, the degree (equivalence class) of the computable sets, and greatest element 1 or 0′, the degree of K. Post's problem then asks if there are any other elements of .

The (positive) solution of Post's problem by Friedberg [1957] and Muchnik [1956] was followed by various algebraic or order theoretic results that were interpreted as saying that the structure was in some way well behaved:

Theorem 1.1 (Embedding theorem; Muchnik [1958], Sacks [1963]). Every countable partial ordering or even uppersemilattice can be embedded into .

Theorem 1.2 (Sacks Splitting Theorem [1963b]). For every nonrecursive r.e. degreeathere are r.e. degreesb, c < asuch thatb ∨ c = a.

Theorem 1.3 (Sacks Density Theorem [1964]). For every pair of nonrecursive r.e. degreesa < bthere is an r.e. degreecsuch thata < c < b.

Type
Research Article
Information
Bulletin of Symbolic Logic , Volume 2 , Issue 4 , December 1996 , pp. 392 - 404
Copyright
Copyright © Association for Symbolic Logic 1996

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References

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