Incompatible degrees $a,b$ s.t. $x < a$ implies $x \leq b$

Question

Are there incompatible Turing degrees $a,b$ s.t any degree computable in $a$ either computes $a$ or is computed by $b$?

Obviously, if $a$ was above $b$ then $a$ would be a strong minimal cover of $b$. But do incomparable such degrees exist and is there a name for them?

Or am I missing some obvious fact that implies such degrees can't exist? I feel like I should already know the answer but i don't seem to.

Edit: I meant to rule out the case where this is trivially possible so I should have added: and this relation holds for no $b$ below $a$.

Answer 1

This type of algebraic property is observed in the local structure of the enumeration degrees: there we call them Ahmad pairs. Ahmad (a student of Lachlan) showed that there are incomparable $\Sigma^0_2$ enumeration degrees $a$ and $b$ such that for all enumeration degrees $x$ if $x<a$ then $x<b$. She also showed that there are no symmetric Ahmad pairs. This property plays a key role in our attempts to analyze the 2-quantifier theory of the structure.

In the c.e. Turing degrees there are no Ahmad pairs because of Sacks splitting theorem (in an Ahmad pair the first element cannot be the join of two lower ones). In the $\Delta^0_2$ Turing degrees there are symmetric Ahmad pairs: any pair of distinct minimal degrees.

Answer 2

Every countable upper semilattice with least element is isomorphic to an initial segment of the Turing degrees (Lachlan/Lebeuf, Countable initial segments of the degrees of unsolv- ability). Consider the partial order $P$ consisting of a copy of $\omega$ together with "an incomparable pair of $\infty$s" (fine, plus a third $\infty$ to be their join); taking these top points as our $a$ and $b$ gives a nontrivial example of the situation.

Answer 3

Just in case anyone else is looking what was nagging me about having seen a construction for this it's essentially the result of the minimal upper bound construction.

Let $a$ be any degree that's not a minimal cover and use the method of recursively pointed trees to build a minimal upper bound of the degrees below $a$. Interleave with extensions to ensure that the set being built doesn't compute $a$. By fact that $a$ isn't a minimal cover there is no greatest element below $a$ so $b$ can't be below $a$ (and the solution isn't trivial) and $b$ doesn't compute $a$.

And no you don't need the trees. You can just build a set whose n-th column is a finite modification of the n-th element in our chain elements below (really finite joins of elements) $a$.