# Is this cardinal characteristic trivial? (Number of strategies needed to guarantee at least one win)

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.

• 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. Sep 23, 2022 at 21:32
• 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 $2^{\aleph_0} = \aleph_2$). Sep 23, 2022 at 21:33
• 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. Sep 26, 2022 at 7:56
• @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 $2^{\aleph_0}$. Sep 26, 2022 at 8:36
• 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 $f(\alpha)$. Sep 26, 2022 at 8:40