# Is the Banach game quantifier "intractable"? (Becker's guess)

(This is a revised version of the original question. Below I work in $$\mathsf{ZF+DC+AD}$$, but I would be happy to add further axioms if appropriate: $$\mathsf{ZF+DC+AD_\mathbb{R}}$$, for example, seems a natural framework as well.)

The Banach game with target set $$A\subseteq\mathbb{R}$$, $$G_\mathit{Ban}(A)$$, is played as follows. Players $$1$$ and $$2$$ alternately play positive reals, each smaller than the previous real; player $$1$$ wins iff the sum of the reals in the play (exists and) is in $$A$$. Despite being a game on reals, $$G_{\mathit{Ban}}$$ is somewhat tamer than one might expect in terms of axiomatic power (it's equivalent to $$\mathsf{AD}$$, rather than the obvious upper bound $$\mathsf{AD_\mathbb{R}}$$. Nonetheless it seems extremely complicated structurally. In particular, following a remark of Becker (Determinacy of Banach games), I'm interested in whether there is a precise sense in which the Banach game is "particularly intractable" to analyze:

The (rather vague) problem which Banach posed was to characterize those sets $$A$$ for which I (II) has a winning strategy in $$G_{\mathit{Ban}}(A)$$ . . . This paper will not provide the reader with any answer to Banach's question. I know of no nontrivial way to characterize when player I (or II) wins, and I suspect there is none.

[Emphasis/notation mine]. Here's one way this might be made precise - or rather two ways, depending on whether we care more about detecting player $$1$$ winnability or player $$2$$ winnability:

• Suppose player $$1$$ wins $$G_{\mathit{Ban}}(A)$$. Must $$A$$ have a Borel subset $$B$$ such that player $$1$$ wins $$G_{\mathit{Ban}}(B)$$?

• Suppose player $$2$$ wins $$G_{\mathit{Ban}}(A)$$. Must $$A$$ have a Borel superset $$B$$ such that player $$2$$ wins $$G_{\mathit{Ban}}(B)$$?

Of course we can replace "Borel" with any other pointclass, or go further and ask how high up the Wadge hierarchy we have to go to find such "sufficiency" sets, but this seems a good starting point. Note that the trivial upper bound (Wadge-hierarchy-wise) is less trivial than it may first appear: while there is a way to code Banach game strategies by reals (Freiling, Banach games, Theorem $$1.3$$), this coding mechanism is rather complicated (and there is no obvious reason why "topologically simple" strategies should be good enough in general), and so "$$\Sigma_1^1$$ in a strategy" is at first glance much worse than "$$\bf\Sigma^1_1$$."

• Since you put no conditions on $\mathcal F_+$ beyond just consisting of Borel sets, your first question can be formulated in a simpler-looking (to me) way: If $B(X)$ is a win for player I, must $X$ have a Borel subset $Y$ such that $B(Y)$ is also a win for I? Note also that the answer becomes an easy yes if we replace "Borel" with "analytic", since the set of all outcomes of plays in which I uses a specific strategy is analytic. Sep 30, 2017 at 14:37
• @AndreasBlass I agree with the first part. I'm not sure, though, that it becomes trivial if we replace "Borel" with "analytic" - while strategies for Banach games can be coded by reals (Thm. 1.3 of Freiling's paper), this coding is quite complicated; it's not clear to me that "the set of allowed plays by $\Sigma$" is in fact analytic, since while it is $\Sigma^1_1$ in $\Sigma$ I don't see why $\Sigma$ is boldface $\Sigma^1_1$ as a map from $\mathbb{R}^{<\omega}$ to $\mathbb{R}$. Sep 30, 2017 at 15:00
• I think you're right; winning strategies can be more complicated than I assumed. Sep 30, 2017 at 18:31