Elementary confusion about determinacy

Question

1. In ZF we can prove Borel determinacy.
2. In ZF we can prove that the determinacy for closed sets is equivalent to the Axiom of Choice.

Since closed sets are Borel, it follows that the Axiom of Choice is a theorem of ZF. I am aware this is wrong and very likely an entirely trivial confusion emerging from my lack of knowledge on these matters. However, I'd appreciate someone to shed some light on this, probably with a line pointing out what is the confusion.

Thanks!

Answer 1

Closed, or Borel, in what space?

$\mathsf{ZF}$ proves that the axiom of choice is equivalent to the following statement:

$(*)\quad$ For every set $X$, equip $X^\omega$ with the product topology coming from the discrete topology on $X$. Then every game on $X$ of length $\omega$, whose payoff set is closed as a subset of $X^\omega$, is determined.

But $(*)$ is an incredibly general principle! Every set $X$ has to be taken into account here. By contrast, when we talk about Borel determinacy we're (unless otherwise stated!) referring to specifically Borel subsets of $\omega^\omega$ (or equivalently for this context, $2^\omega$). While $\mathsf{ZF}$ proves "Every Borel subset of $\omega^\omega$ is determined," this is not enough to imply $(*)$ above.

Indeed, in general the phrase "$\Gamma$-determinacy" refers either to $\omega^\omega$ or $2^\omega$ specifically (and even this needs further clarification for $\Gamma$ sufficiently simple - closed games on $2$ are fairly boring but closed games on $\omega$ are extremely complicated). So the phrase "Over $\mathsf{ZF}$, choice is equivalent to closed determinacy" is misleadingly phrased; it should be something like "Over $\mathsf{ZF}$, choice is equivalent to the determinacy of all closed games on all sets."