The Wikipedia article on the Axiom of Determinacy (AD) claims:

Equivalent to the axiom of determinacy is the statement that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined.

Is this claim true?

AD is usually stated for the Gale-Stewart game $G(S)$ with payoff set $S\subseteq \omega^\omega$, in which the players build a sequence in $\omega^\omega$ by alternately choosing integers, and Player I wins if this sequence is in $S$.

In the Banach-Mazur game $BM(X)$ with payoff set $X\subseteq \mathbb{R}$, the players build a descending sequence $U_0\supseteq U_1\supseteq U_2\supseteq \dots$ by alternately choosing nonempty open sets, and Player II wins if $\bigcap_{n\in \omega}U_n\subseteq X$.

AD implies the determinacy of $BM(X)$ for all $X\subseteq \mathbb{R}$, since $BM(X)$ can be replaced by an equivalent game in which the players are additionally required to play open intervals with rational endpoints, which can be coded by natural numbers.

After a bit of Googling, I couldn't find any information about the converse (aside from the claim on Wikipedia). Of course, this might just be because it's very obvious. But on the other hand, the fact that $BM(B)$ is determined when $B$ is Borel (or even just has the Baire property) is an easy theorem, while the corresponding fact for $G(B)$ is a very hard theorem ("Borel determinacy"). So I can believe that the converse might be false.

Edit: In light of the answer to this question, I've edited the Wikipedia page, replacing the (false) equivalence with the (true) implication:

The axiom of determinacy implies that for every subspace X of the real numbers, the Banach–Mazur game BM(X) is determined.


The claim is false. The Banach-Mazur game, also known as the $**$-game shows (and is equivalent to) that every set of reals has the Baire property. What is true, as you've noted, is that if one has a pointclass $\Gamma$ which is adequate and closed under Borel ($\Delta^1_1$) substitutions then we have $$Det(\Gamma) \rightarrow\forall A\in \Gamma, G^{**}(A)\text{ is determined}$$ This can be shown by noting that the payoff set of the Banach-Mazur Game $G^{**}(A)$ is in $\Gamma$

However the Baire property alone is not enough to imply determinacy for all sets of reals. The Baire is much weaker in consistency strength than the assertation that the axiom of determinacy holds in $L(\mathbb{R})$ for example. By Work of Shelah, the assertation that the Baire property holds for all sets of reals ordinal definable from a real has same consistency strength as $ZFC$. In other words, one may "take away" the inaccessible Solovay used in his construction. However, by work of Woodin the theory $ZF+AD$ has same consistency strength as $ZFC+\omega$-many Woodin cardinals.

On the other hand, $AD$ is actually equivalent to Turing determinacy in $L(\mathbb{R})$, or also equivalent to cofinally many below $\Theta$ cardinals having the strong partition property.

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    $\begingroup$ About "the Baire property alone is not enough to imply determinacy for all sets of reals": There's even a big difference in consistency strength. $\endgroup$ – Andreas Blass Apr 3 '18 at 21:54
  • $\begingroup$ I don't think we know the result about Turing determinacy. Only in, say, $L (\mathbb R) $. $\endgroup$ – Andrés E. Caicedo Apr 3 '18 at 22:07
  • $\begingroup$ Thanks. I see now that what I was looking for was exactly a model of ZF in which every set of reals has the Baire property (so we have determinacy for Banach-Mazur games) but AD fails. And indeed Shelah showed that one can get models of ZF+DC+"every set of reals has BP" without any large cardinal assumptions. Since AD implies "every set of reals is measurable", and this alone has higher consistency strength (requiring an inaccessible), I guess it should be possible to ensure that AD fails in Shelah's models. Right? $\endgroup$ – Alex Kruckman Apr 3 '18 at 22:53
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    $\begingroup$ @AlexKruckman: AD fails in Shelah's model because there is an $\omega_1$ sequence of reals. That's how he shows that not every set is Lebesgue measurable. $\endgroup$ – Asaf Karagila Apr 3 '18 at 23:26
  • $\begingroup$ @AndrésE.Caicedo Thanks for the correction, I edited the post. $\endgroup$ – Rachid Atmai Apr 4 '18 at 4:59

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