On a Question of Sacks - A Partial Solution on the Positive Side

Let us say that a c.e. operator E is degree invariant on any given Turing degree a if X, Y ∈ a → E(X) ≡T E(Y ). In [4] we construct a c.e. operator E such that ∀X[X <T E(X) <T X ]. While we are unable to produce degree invariance everywhere, we are able to ensure that for every degree a there exists b such that a ∨ 0 = b ∨ 0 and E is degree invariant on b. What appears here is an abbreviated version of the material from that paper, stopping short of most technical details.

• Type: Conference
• Year: 2005
• Where: CIE
• Authors: Andrew E. M. Lewis
• Keywords: CIE 2005, Degree Invariant, Operator, Turing Degree