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



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> 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.
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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](http://en.wikipedia.org/wiki/Determinacy) 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. 



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> 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? 
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"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.



Martin's cone theorem and recursion theory
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.


