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I recently learned about three determinacy theorems, and I am curious about the reverse mathematical-strength of (the countable analogs of) these theorems.

These theorems are Zermelo's Theorem (Zermelo 1913), the Open Determinacy Theorem (Gale-Stewart 1953) and the Borel Determinacy Theorem (Martin 1974).

For reference, here are statements of each of these theorems. Let $A$ be any set, and let $G_A$ be any Gale-Stewart game on $A$. Endow $A$ with the discrete topology and $A^\omega$ with the product topology.

  • Zermelo's Theorem states that, if $A$ is finite, then $G_A$ is determined.
  • The Open Determinacy Theorem states that, if the payoff set for $G_A$ is an open or closed subset of $A^\omega$, then $G_A$ is determined.
  • The Borel Determinacy Theorem states that, if the payoff set for $G_A$ is a Borel subset of $A^\omega$, then $G_A$ is determined.

I understand that the reverse mathematics of these theorems has been studied to some extent. In particular, the Open Determinacy Theorem is equivalent to AC under ZF (Question on Gale-Stewart Theorem and Axiom of Choice) and that the Borel Determinacy Theorem requires "repeated use" of the Axiom of Replacement (https://en.wikipedia.org/wiki/Borel_determinacy_theorem) (Friedman 1971).

I am curious about where these theorems lie in the hierarchy of subsystems of second-order arithmetic, if we restrict these theorems to countable sets $A$. Does anyone know the answer to this question, or a source I could look to for more information?

Thank you!

Gavin Dooley
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  • You may want to start here: https://math.berkeley.edu/~antonio/papers/TDet21.pdf (The strength of Turing determinacy within second order arithmetic, A. Montalbán and R. A. Shore, Fund. Math., 232 (2016), 249--268.) – Andrés E. Caicedo Apr 26 '23 at 17:16
  • @AndrésE.Caicedo Thanks! That answers a lot of what I was curious about. – Gavin Dooley Apr 26 '23 at 17:47
  • In particular, that paper mentions the "classical" reverse mathematical fact that the Determinacy Theorem for "effectively open" (i.e., recursively enumerable or $\Sigma_1^0$) sets is equivalent to the system ATR${ }_0$. – Gavin Dooley Apr 26 '23 at 17:52

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One of the earliest results in reverse mathematics was Steel's $1972$ proof that each of clopen and open determinacy is equivalent to $\mathsf{ATR}_0$. This should in my opinion be surprising, since $(i)$ every computable clopen game has a hyperarithmetic winning strategy but $(ii)$ there are computable open games without hyperarithmetic winning strategies. I'm aware of three proofs of this result, one via $\Sigma^1_1$-Separation (this is the proof presented in Simpson's book), one via Harrison orders/overspill (I don't recall where this is written up), and one via a "tree of strategies" (see the end of this old answer of mine).

As we climb up the determinacy hierarchy we rise rapidly in strength. $\mathsf{Z}_2$ cannot even prove $\Delta^0_4$ determinacy; the best it can do is prove that finite Boolean combinations of $\Sigma^0_3$ games are determined. This latter principle is schematic, and so itself constitutes a hierarchy; the relationship between the $n\Sigma^0_3$-determinacy hierarchy and the usual $\Pi^1_k$-comprehension hierarchy is subtle and was established by Montalban/Shore 1, 2.

We can go below $\mathsf{ATR}_0$ if we replace games on $\omega$ by games on $\{0,1\}$ (that is, move from games in Baire space to games in Cantor space); for example, clopen determinacy in Cantor space is equivalent to $\mathsf{WKL}_0$. See e.g. Nemoto.

Noah Schweber
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  • I have a possibly stupid question: why do reverse mathematicians refer to determinacy for $\Sigma_1^0$ sets as “open determinacy”, when the $\Sigma_1^0$ sets are really just those that are effectively open? Is “open” simply shorthand for “effectively open”? Or does it have something to do with the fact that the $\Sigma_1^0$ sets are the sets that ATR${ }_0$ “knows” are open? – Gavin Dooley Apr 27 '23 at 14:13
  • OR: Am I understanding Steel’s result wrong? Is it really about all open sets and not just the $\Sigma_1^0$ ones? – Gavin Dooley Apr 27 '23 at 14:15
  • @GavinDooley I was being sloppy (although it's a common sloppiness): "$\Sigma^0_1$" should be "${\bf \Sigma^0_1}$", that is, all open sets. – Noah Schweber Apr 27 '23 at 15:47
  • No worries! Are all of the $\Sigma$'s, $\Pi$'s, and $\Delta$'s in your answer technically supposed to be the boldface versions? – Gavin Dooley Apr 27 '23 at 16:16
  • @GavinDooley Yes, that's correct. – Noah Schweber Apr 27 '23 at 16:56