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(I feel like I might have to apologise in advance for this question, but oh well..)

I just rediscovered a comment from Asaf K here on MO that states that full Replacement is not needed for Borel Determinacy (BD), a fact which flies in the face of everything I feel I've been led to believe. The Wikipedia article on the set-theoretic niceties is stated perfectly correctly, but a shallow reading might miss (as I did) the fact there's wriggle room between ZC (Zermelo with Choice), which is insufficient, and ZFC, which is sufficient. For a set theorist, the gap might be silly, but not when it comes to models.

Some Replacement is needed, but not unboundedly-large instances, it seems. Asaf says that it only requires that $\beth_{\omega_1}$ exists. So is this my question: for which $V_\alpha$ does BD hold? Is it just those ordinals $\alpha$ so that $\beth_{\omega_1} \in V_\alpha$? Or is there a different characterisation? How about models of the form $H_\alpha$? Then one gets models of some set theory that is stronger than ZC, but weaker than ZFC, in which BD holds. Such information would be good to add to WP, for instance, but also to draw the sort of fine line that Reverse Mathematicians like to see.

As Noah points out in the comments, I'm making a bit of a muddle here. But I think what I want to know is how much replacement over ZC does one actually need to prove BD. François indicates that something like Replacement for functions with countable domain is sufficient, but not tight. I would be willing to countenance arguments of the sort that justify Dependent Choice in an otherwise choice-free setting, in order to extrinsically justify countable Replacement aside from knowing it proves BD.

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    $\begingroup$ BD holds in all $V_\alpha$ in which all relevant parameters exist: so $\alpha \geq \omega+5$ is enough for most encodings and $V_{\omega+\omega}$ is more than enough for everyone who is not deliberately acting silly. Similar (but slightly different) situation for the $H$ hierarchy are true. This is because these are defined in ZFC and $\beth_{\omega_1}$ does exist; these are not predicative definitions! $\endgroup$
    – François G. Dorais
    Commented Jun 16, 2021 at 3:31
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    $\begingroup$ Sorry for the wording, countable replacement is enough but it's a bit too much for reasons I explained. (Though the "too much" part is mostly visible in combination with other axioms.) $\endgroup$
    – François G. Dorais
    Commented Jun 16, 2021 at 4:18
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    $\begingroup$ @DavidRoberts There's an odd mix-up in this question between theories and structures here. The $V_{\omega+\omega}$ (for example) of any model $V\models\mathsf{ZFC}$ will satisfy BD, since BD is essentially just a statement about reals. But this has nothing to do with the amount of replacement needed for BD. Basically, "Which $V_\alpha$s (or similar) satisfy BD?" is just not the right question to ask. $\endgroup$
    – Noah Schweber
    Commented Jun 16, 2021 at 4:25
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    $\begingroup$ I believe it is not so much replacement that is needed, but rather existence of suitably large power iterated power sets that is. A reasonably precise statement of strength is given in Theorem 1.1 here - over set theory without power set, existence of $P^\alpha(\omega)$ is sufficient for $\Sigma^0_{1+\alpha+2}$-determinacy, but not $\Sigma^0_{1+\alpha+3}$ determinacy, at least in the presence of $\Sigma_1$-replacement. I'm not sure to what extent this answers your question. $\endgroup$
    – Wojowu
    Commented Jun 16, 2021 at 10:25
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    $\begingroup$ Also, although this may be a slightly silly point, it should not at all be surprising that BD doesn't require full replacement - BD is a single formula in the lanuage of ZFC, so it can't depend on an entire axiom schema. $\endgroup$
    – Wojowu
    Commented Jun 16, 2021 at 10:35

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To summarise discussion in the comments, particularly from Wojowu, the answer I was after is in Tony Martin's book, Determinacy of infinitely long games (author's draft pdf). At the cost of assuming $\Sigma_1$ Replacement, so that some standard constructions work (I think von Neumann ordinals and cartesian products), what is really needed is the existence of the power set $P^\alpha(\omega)$, for all countable ordinals $\alpha$. Replacement using functions on countable sets is sufficient (and, intriguingly, something Zermelo more-or-less considered as an axiom, in the 1920s). Without $\Sigma_1$ Replacement, it will still work (noted by Martin), but one must use different models for the ordinals, for instance.

But there is a much more nuanced post and discussion in the blog post by Tom Leinster, Borel Determinacy Does Not Require Replacement, the title of which is, I think, deliberately provocative.

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