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The axiom of Turing determinacy is a weakening of the full axiom of determinacy, $AD$, in which only games with payoff sets which are $\equiv_T$-invariant are demanded to be determined.

In "Turing determinacy and the continuum hypothesis" (published in 1989), Ramez Sami writes:

"The main question so far unsettled in this particular domain can be roughly put this way: is it true that for any "reasonable" pointclass $\Gamma$ we have: Turing-Det$(\Gamma)\implies$Det$(\Gamma)$? In particular is it the case that: [over $ZF+DC$, presumably] Turing $AD$ implies $AD$?"

My question is, what is the status of this question currently? Do we know whether Turing $AD$ is strictly weaker than $AD$? The only recent work I know of around Turing determinacy is from the reverse mathematical side (http://www.math.cornell.edu/~shore/papers/pdf/TDet21.pdf); I'm not at all familiar with the set theory on the subject.

(I vaguely recall that Turing determinacy implies that every Suslin set is determined, but I can't remember where I supposedly learned this "fact.")

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  • $\begingroup$ It seems even unknown whether $TD$ implies the nonexistence of a Bernstein set. $\endgroup$
    – 喻 良
    Apr 26, 2020 at 7:09
  • $\begingroup$ I think this is no longer unknown : arxiv.org/abs/1912.12558 $\endgroup$ Jul 29, 2020 at 19:22
  • $\begingroup$ @PaulLarson Unless I'm missing something, that's specifically about the perfect set property - I don't see how to lift it to an equivalence between general determinacy principles, and in the intro at a glance they only claim to answer a special case of question $2$, not question $1$ itself. $\endgroup$ Jul 29, 2020 at 19:23
  • $\begingroup$ @PaulLarson And in section IV they list that as a still-open problem. $\endgroup$ Jul 29, 2020 at 19:24
  • $\begingroup$ I should have read it more carefully. I believe that that paper is not up to date, but I may be misinformed. $\endgroup$ Jul 29, 2020 at 19:36

2 Answers 2

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This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) is the claim in your parenthetical remark at the end. This gives determinacy in $L(\mathbb R)$ using, for example, a reflection argument.

(I mentioned this a while ago somewhere on this site. Maybe that's where you heard of it? This can be used to prove that $\omega$-board determinacy is equiconsistent with determinacy. I seem to recall that's how the topic came up.)


Update (Jun 25/22): Hugh has just published a proof that Turing determinacy implies the determinacy of Suslin sets (which gives $L(\mathbb R)$-determinacy and more). He indicates in the paper that the general question remains open. Coordinates:

NEW ZEALAND JOURNAL OF MATHEMATICS Volume 52 (2022), 845–863, "Turing determinacy and Suslin sets" https://doi.org/10.53733/140

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  • $\begingroup$ Based on your answer, I googled around and found this: math.stackexchange.com/questions/656414/…. So that must be what I was remembering. It's probably too much to ask, but do you know if there is a version of the proof floating around? $\endgroup$ Jun 16, 2014 at 2:00
  • $\begingroup$ I don't think so, unfortunately. Hugh lectured on this at the seminar at Berkeley a few years ago, but I couldn't find anybody who had taken notes. $\endgroup$ Jun 16, 2014 at 2:01
  • $\begingroup$ (And, yes, you are right, it was on MSE that this came up.) $\endgroup$ Jun 16, 2014 at 2:02
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To show this in L(R) (or any well understood determinacy model) one apparently has to run a core model induction as in section 6.2 of https://ivv5hpp.uni-muenster.de/u/rds/core_model_induction.pdf which explains why it is open in the abstract.

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