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Call $X$ very hyperlow if $\mathcal{O}^X \le_T \mathcal{O}$, where $\mathcal{O}$ is your favorite $\Pi^1_1$-complete set. Note: Turing reducibility, not hyp-reducibility. Observe that this is a (Turing) degree invariant notion.

Are the very hyperlow Turing degrees closed under join?

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I don't think so. There is an $\mathscr{O}$-recursive sequence $\{D_i\}_{i}$ of dense open sets so that for any real $g$ meeting every member of the sequence, $\mathscr{O}^g\leq_T g\oplus \mathscr{O}$. Now it is simple to construct two such reals $g_1$ and $g_2$ so that $g_1\oplus g_2\equiv_T \mathscr{O}$.

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  • $\begingroup$ Nice, thank you. $\endgroup$
    – Dan Turetsky
    Commented Mar 13, 2020 at 3:07

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