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(Previously asked at MSE.)

Let the determinacy number, $\mathfrak{g}$ (for "game"), be the smallest cardinal such that for every (two-player, perfect-information, length-$\omega$) game on $\omega$ at least one of the following holds:

  • There is a set $\Sigma$ of strategies for player $1$ such that $\vert\Sigma\vert\le\mathfrak{g}$ and, for every strategy ${\bf t}$ for player $2$, there is some ${\bf s}\in\Sigma$ such that ${\bf s}\otimes{\bf t}$ is a win for player $1$.

  • There is a set $\Sigma$ of strategies for player $2$ such that $\vert\Sigma\vert\le\mathfrak{g}$ and, for every strategy ${\bf s}$ for player $1$, there is some ${\bf t}\in\Sigma$ such that ${\bf s}\otimes{\bf t}$ is a win for player $2$.

In $\mathsf{ZF+AD}$ we have $\mathfrak{g}=1$, and in $\mathsf{ZFC}$ we have $\aleph_1\le \mathfrak{g}\le 2^{\aleph_0}$. A bit less trivially, $\mathsf{ZFC+MA}$ implies $\mathfrak{g}=2^{\aleph_0}$. I'm curious whether a low determinacy number is consistent with choice:

Is $\mathsf{ZFC}$ + $\mathfrak{g}<2^{\aleph_0}$ consistent?

I strongly suspect the answer is negative, but I don't see how to prove it; the possibility of $\kappa<2^{\aleph_0}<2^{\kappa}$ breaks every construction of a "hard-to-cover" game I can think of.

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    $\begingroup$ It seems to me that as long as $2^{\aleph_0}$ is regular, you can show $\mathfrak{g} = 2^{\aleph_0}$ by a diagonalization argument. I'm not sure about the case when $2^{\aleph_0}$ is singular though. $\endgroup$ Sep 23, 2022 at 21:32
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    $\begingroup$ This shows that $\mathfrak{g}$ does not behave much like other cardinal characteristics (which you can usually separate from $2^{\aleph_0}$ in models in which $2^{\aleph_0} = \aleph_2$). $\endgroup$ Sep 23, 2022 at 21:33
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    $\begingroup$ The diagonalization argument Patrick presumably alludes to shows that $\text{cof}(2^{\aleph_0})\leq\mathfrak{g}$. Namely, enumerate all the strategies, and then create a game such that for every bounded collection of the strategies in the enumeration, there is a constant-play strategy defeating those strategies. $\endgroup$ Sep 26, 2022 at 7:56
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    $\begingroup$ @JoelDavidHamkins Yes, that's the argument I was referring to. Another thing that might be worth mentioning is that if $2^{\aleph_0}$ is singular then there is no single game witnessing that $\mathfrak{g} = 2^{\aleph_0}$: for every $A \subseteq \omega^\omega$, either player 1 has a winning set of strategies of size $\text{cof}(2^{\aleph_0})$ or player 2 has a winning set of strategies of size strictly less than $2^{\aleph_0}$. $\endgroup$ Sep 26, 2022 at 8:36
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    $\begingroup$ The proof of this is pretty easy. First pick a length $2^{\aleph_0}$ enumeration of player 2 strategies, $\{\tau_\alpha\}_{\alpha < 2^{\aleph_0}}$ and a cofinal sequence $f \colon \text{cof}(2^{\aleph_0}) \to 2^{\aleph_0}$. Then try to pick a winning set of strategies for player 1. On step $\alpha$, look for a player 1 strategy which defeats all player 2 strategies enumerated before $f(\alpha)$. If no such player 1 strategy exists then you have found a winning set of strategies for player 2 of size $f(\alpha)$. $\endgroup$ Sep 26, 2022 at 8:40

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