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A Borel partial order is the partial order corresponding to a Borel preorder of some Polish space. For example, the Turing and enumeration degrees, $\mathcal{D}$ and $\mathcal{E}$ respectively, are Borel partial orders. Say that a Borel partial order $\mathcal{X}$ has the Borel cone property iff whenever $A\subseteq\mathcal{X}$ is Borel then there is some ${\bf x}\in\mathcal{X}$ such that $\mathcal{X}_{\ge{\bf x}}$ is either contained in or disjoint from $A$.

Martin showed that $\mathcal{D}$ have the Borel cone property. Meanwhile, $\mathcal{E}$ obviously does not: consider the total degrees (or any of a number of other examples). However, since the total degrees are the image of an embedding of $\mathcal{D}$ into $\mathcal{E}$, it's not quite true that $\mathcal{E}$ is devoid of coney behavior; we just have to turn to substructures of the whole degree structure.

I'm curious whether there are any other interesting substructures of $\mathcal{E}$ with the cone property. Specifically:

Is there a Borel substructure $\mathcal{A}$ of $\mathcal{E}$ disjoint from the total degrees such that

  • every enumeration degree is below some element of $\mathcal{A}$, and

  • $\mathcal{A}$ (as a partial order in its own right) has the Borel cone property?

(Note that the first bulletpoint prevents disjoint cones; I'm also interested in examples of structures satisfying merely "$\mathcal{A}$ is unbounded in $\mathcal{E}$," but in that case we should add the hypothesis that $\mathcal{A}$ is directed.)

Noah Schweber
  • 18,932
  • Incidentally, we could work in ZF+AD+DC+??? instead, and replace "Borel" with "any"; this might make things better. However, since I don't see any immediate benefit, I've elected to stay in ZFC. If someone has an answer specific to the determinacy case, though, I'd be happy to accept it! – Noah Schweber Dec 29 '16 at 18:15

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