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The question whether Turing determinacy implies $AD$ is a well-known open problem. I was wondering if anything is known about the following analogous question:

Let $\Gamma$ be a regularity property (e.g. Lebesgue measurability, Baire property, $\mathbb P$-measurability for a reasonable arboreal forcing $\mathbb P$, etc). Assuming $ZF+DC+"\Gamma$ holds for all Turing-invariant sets of reals$"$, does it follow that $\Gamma$ holds for all sets of reals?

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    $\begingroup$ I'm expecting not a whole lot more is known than can be deduced from Bjørn Kjos-Hanssen's paper, but I'm hoping for better answers in some cases. $\endgroup$
    – François G. Dorais
    Commented Jul 28, 2019 at 4:47
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    $\begingroup$ @FrançoisG.Dorais The paper looks interesting, but its relevance to the question is not immediately clear to me. Could you please elaborate on this? $\endgroup$
    – Haim
    Commented Jul 29, 2019 at 4:36
  • $\begingroup$ What about $\Gamma$ being all Turing-invariant sets? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 2, 2019 at 9:59
  • $\begingroup$ Have you thought about degrees generated by a locally countable Borel quasi-partial order? $\endgroup$
    – 喻 良
    Commented Aug 2, 2019 at 11:54
  • $\begingroup$ @喻良 Not yet. Is there a candidate for such a quasi-partial order that might make the problem more easy to approach? $\endgroup$
    – Haim
    Commented Aug 4, 2019 at 10:01

1 Answer 1

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Here is not an answer but an example:

Let PSP be the statement that every uncountable set of reals has a perfect subset and TPSP that every uncountable set of Turing degrees has a perfect subset.

Then PSP and TPSP are equivalently consistent over $ZF+DC$.

Clearly over $ZF+DC$, PSP implies TPSP but I don't know whether the inverse is true.

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  • $\begingroup$ Indeed, once you consider the consistency strength version of the question, it is my conjecture that the answer should be affirmative (in the case of Lebesgue measurability, I worked with Shelah to show that $\Sigma^1_3$-Lebesgue measurability is equiconsistent with the Turing invariant version of the statement, though we never finished writing the paper as it turned out to be too similar to the original "Solovay's inaccessible"). However, for the original question, a counterexample won't be too extremely surprising. $\endgroup$
    – Haim
    Commented Aug 2, 2019 at 4:33

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