The theory of numberings: open problems

Positive undecidable numberings in the Ershov hierarchy

“…We prove something similar for all levels Σ −1 a of the Ershov hierarchy, where a is the ordinal notation of any nonzero computable ordinal: in particular we show that if a is notation of an infinite computable ordinal, and A is an infinite family of Σ −1 a sets, containing some set A which belongs to some finite level of the Ershov hierarchy, then A has infinitely many positive undecidable numberings, which are pairwise incomparable with respect to Rogers reducibility. As a consequence, the family of all Σ −1 a sets has positive undecidable numberings, verifying Conjecture 15 of [2] for all levels of the Ershov hierarchy. (Of course, for finite levels this conjecture had been verified by Talasbaeva's theorem).…”

“…(Of course, for finite levels this conjecture had been verified by Talasbaeva's theorem). A straightforward observation, derived as a consequence of Ospichev's theorem on the existence, at all levels, of families without Friedberg numberings, allows us to show also that at every level there exist families with positive numberings but without Friedberg numberings, answering negatively Question 17 of [2].…”

“…Conjecture 15 of [2] states that for every finite n ≥ 1, the family of all Σ −1 n sets has positive undecidable numberings. This conjecture was verified in [9].…”

Families Without Minimal Numberings

“…Therefore, it is natural to raise questions as to which properties of computable numberings that hold in the classical sense will also hold for computable numberings in the Ershov hierarchy. In this regard, an interesting problem is whether or not there exist families of sets without minimal computable numberings [7,Question 11]. The main result of the present paper is proving the existence of such families in each class of sets from all levels, whether finite or infinite, of the Ershov hierarchy.…”

“…is active, and for all z / ∈ {2x, 2x + 1} if d 0 (x, i) is not active, and go to step (7). Otherwise, go to step (10).…”