Martin's cone theorem and recursion theory

Question

Martin's remarkable cone theorem in the theory of determinacy says the following:

Suppose $A\subseteq \omega^\omega$ is Turing invariant and determined. If $\forall x\exists y(x\le_T y\& y\in A)$ then $A$ contains a cone.

Let me explain what this means: $A$ is Turing invariant iff $\forall x\in A\forall y(x\equiv_T y\Rightarrow y\in A)$. Here, $\le_T$ is the relation of Turing reducibility and $\equiv_T$ is the corresponding equivalence relation.

"Determined" is in the usual sense of infinite games on integers.

A cone is a set of the form $$C_y=\{x\mid y\le_T x\}.$$ Clearly, cones are Turing invariant. We say that $y$ is the base of the cone $C_y$.

If $\forall x\exists y(x\le_T y\& y\in A)$ we say that $A$ is cofinal.

When Martin proved his theorem, he thought that it would be a quick way of showing determinacy fails (in ZF), by considering explicit sets coming from recursion theory. Instead, he found several results in recursion theory as a consequence.

Here are some examples:

• For every $x$ we have $x<_T x'$, where $x'$ is the Turing jump of $x$. This means that the set of jumps is cofinal. By Borel determinacy, it follows that there is a $y$ such that if $y\le_T x$, then $x\equiv_T z'$ for some $z$. Well known recursion theoretic results show that in fact we can take $y=0'$.
• Again by Borel determinacy, there is a real $x$ such that any $y$ with $x\le_T y$ is a minimal cover above some $z$. Again, recursion theoretic arguments show that we can take $x=0^{(\omega)}$.

I do not know many examples coming from recursion theory, but maybe somebody here does.

Are there (natural) examples of sets $A$ defined recursive theoretically that we know need to contain a cone, but for which we do not know of a (natural) base?

"Need to contain a cone" could be taken to mean, say, that they are Turing invariant and cofinal and Borel.

Naively, a negative answer would mean we have very strong abstract basis results. But I would be interested in natural candidates for a positive answer as well.

Answer 1

I would like to add another example.

Given a sentence $\phi$ from partial order language, then for any Turing degree $x$, either $D(\leq x)\models \phi$ or $D(\leq x)\models \neg\phi$. By the BD, there is a Turing degree $x_{\phi}$ so that either for all $y\geq_T x_{\phi}$, $D(\leq y)\models \phi$ or for all $y\geq_T x$, $D(\leq y)\models \neg \phi$.

Let $z$ be a Turing degree above all the $x_{\phi}$'s, then for every $y\geq_T z$, $D(\leq y)$ is elementary equivalent to $D(\leq z)$.

I don't know a natural base for this.

Answer 2

Let $$ A=\{x \mid x\ \equiv_T\ j(y)\quad \text{for some set } y\}, $$ where $j(y)$ is the $\Delta^1_{2n+1}$-jump of $y$, for $n\ge 1$.

Since $x\ <_T\ j(x)$ for all $x$, $A$ is cofinal in the Turing degrees. Hence assuming Projective Determinacy, $A$ must contain a cone in the Turing degrees.

Kechris showed a restriction on possible jump inversion theorems in the $\Delta^1_{2n+1}$-degrees (see Kastanas, The jump inversion theorem for $Q_{2n+1}$ degrees, Proc. AMS 1984). So I am guessing that no base for a cone contained in $A$ is known. But I guess this is closer to set theory than recursion theory.

EDITS: Changed the example since the $\omega$-jump or hyperjump do not work, by MacIntyre, Transfinite extensions of Friedberg's completeness theorem, J. Symbolic Logic, 1977; and added the assumption PD.