2
$\begingroup$

Recall that a set $A$ is c.e.a. (computably enumerable in and above) if there is some $X<_T A$ such that $A$ is $X$-c.e.

I am interested in degrees (specifically $\Delta^0_2$ degrees) that are not only not c.e.a., but do not even compute any c.e.a. degree. Equivalently, for all in $B,C\leq_TA$, $B$ does not enumerate $C$ unless $B$ computes $C$.

Clearly no c.e. degree has this property, as they are trivially c.e.a. Less trivially, so are 1-generics, so nothing that computes a 1-generic or c.e. set has this property. Minimal degrees do have this property, however.

Is anything known about such degrees?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Have you looked at the results of Mingzhong an Richard? They show that a degree is $REA$ above closed if and only if it is $ANR$. $\endgroup$
    – 喻 良
    Commented Apr 4, 2022 at 10:00
  • $\begingroup$ Ooh, I hadn’t seen those, thanks! $\endgroup$
    – David J. Webb
    Commented Apr 4, 2022 at 20:08
  • 1
    $\begingroup$ One more point. Since every cea degree computes a 1-generic and every 1-generic real is cea. So a degree computes a cea iff it computes a 1-generic. But this might not be what you want. $\endgroup$
    – 喻 良
    Commented Apr 7, 2022 at 1:11
  • $\begingroup$ It is what the kind of thing I want! The paper you mentioned led me to one by Wei Wang with mentioned that first result, so thank you again $\endgroup$
    – David J. Webb
    Commented Apr 8, 2022 at 2:31

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .