MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 (; 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.")

share|cite|improve this question
up vote 8 down vote accepted

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 Hugh has several proofs of this, and it remains one of the few unpublished results in the area.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.)

share|cite|improve this answer
Based on your answer, I googled around and found this:…. 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? – Noah Schweber Jun 16 '14 at 2:00
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. – Andrés E. Caicedo Jun 16 '14 at 2:01
(And, yes, you are right, it was on MSE that this came up.) – Andrés E. Caicedo Jun 16 '14 at 2:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.