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

(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.

• 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. yesterday
• 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 $2^{\aleph_0} = \aleph_2$). yesterday