Does every nonzero Turing degree adimit a nonzero incompatible degree?

Question

It is well known that every nonzero Turing degree $\bf x$ adimits an incomparable degree $\bf y$, that is a degree such that $\bf x\not\le_T\bf y$ and $\bf y\not\le_T\bf x$.

I was wondering whether this statement generalizes to the stronger following one: every nonzero Turing degree $\bf x$ adimits a nonzero degree $\bf y$ which is incompatible with it, meaning that the set $\{\bf x,\bf y\}$ has no nonzero lower bounds.

In case this statement were known to be false, is it known exactly what degrees provide counterexamples to it?

Answer 1

Yes, every degree has this property. Specifically, for every Turing degree d there is a minimal Turing degree a not below d (in particular, any sufficiently-Sacks-generic real will have this property, with "sufficiently" depending on d).

Alternatively, we can do a single forcing and produce a perfect set of Turing incomparable minimal degrees; since any Turing degree only computes countably many things, this immediately implies that for every degree d there is a minimal a not below d.

It's worth keeping in mind that the minimal degrees display far more variety and structure than one might guess. In particular, "minimal" does not really mean "small" - for example, for any d$\ge_T0'$ there is some minimal a such that a$'=$ d. (This is due to Cooper.) Similarly, for any degrees d, x there is some minimal a such that a $\vee$ d $=$ x (and this strengthens the affirmative answer to your question given above).

And I can't end without mentioning an interesting open questions about the minimal degrees (I believe each due to Yates). Say that a is a strong minimal cover of d if a $\ge_T$ d and every degree $<_T$ a is $\le_T$ d. (By contrast, a minimal cover simply satisfies $($d, a$)=\emptyset$.) Yates asked - and I believe this is still completely open - whether every minimal Turing degree has a strongly minimal cover. It is known meanwhile that every sufficiently large Turing degree has a strongly minimal cover, and that there are Turing degrees without strongly minimal covers, so really either answer is plausible.