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An $\infty$-Borel set is a set $X\subseteq\mathbb{R}$ which has an $\infty$-Borel code - a set $r$ coding the construction of $X$ via open sets, complementation, and well-ordered unions (see http://en.wikipedia.org/wiki/Infinity-Borel_set). Under $ZFC$, every set of reals is $\infty$-Borel; however, this is not provable in $ZF+AD$ (it follows from a stronger axiom, $AD_+$).

Consider the following statement:

$$\text{$(*)$ Every $\infty$-Borel set (viewed as a game in the usual way) is determined.}$$

Clearly $(*)$ is independent (assuming consistency of large cardinals) of $ZF$, and is implied by $AD$. My first question is whether this is sharp:

(1) Is $(*)$ strictly weaker than $AD$?

(I'm mostly asking this for actual implication - that is, whether $ZF+(*)\vdash AD$ - but I'm also interested in the question of their relative consistency strengths.)

EDIT: Andres' comments below completely answer this question. However, I'll leave the following paragraph from the pre-comments question, for the record:

I suspect the answer is "yes"; however, the only approach I see is to start with a model of $AD$, and add an undetermined set of reals without adding any new $\infty$-Borel sets of reals. For example, if the $\infty$-Borel codes in a model of $ZF+AD$ could be reduced to individual reals, then starting with $V\models ZF+AD$ and adding an undetermined game via countably closed forcing would do the trick; but I don't see how to justify the hypothesis, and without it I don't know what to do.

Assuming the answer to (1) is "yes," I also have the following question:

(2) Is $(*)$ an interesting principle on its own?

Obviously this is somewhat vague; I am asking both about the principle $(*)$ itself, and the theory $ZF+(*)$ (the difference being, maybe $ZF+(*)$ happens to be some well-known theory but the specific sentence $(*)$ is not particularly interesting).

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    $\begingroup$ It is open whether $\mathsf{AD}$ and $\mathsf{AD}^+$ are equivalent; in particular, it is open whether $\mathsf{AD}$ proves that every set of reals is $\infty$-Borel. $\endgroup$ – Andrés E. Caicedo Apr 16 '14 at 6:33
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    $\begingroup$ If every $\infty$-Borel set is determined, then every set of reals in $L(\mathbb R)$ is determined, so at least we have equiconsistency. $\endgroup$ – Andrés E. Caicedo Apr 16 '14 at 6:35
  • $\begingroup$ I didn't know that it was still open whether $AD$ proved "all sets of reals are $\infty$-Borel," and it didn't occur to me that all sets of reals in $L(\mathbb{R})$ are $\infty$-Borel; your two comments (as far as I'm concerned) completely answer question 1. $\endgroup$ – Noah Schweber Apr 16 '14 at 7:11
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    $\begingroup$ I believe that my answer to this question is relevant for question 1. $\endgroup$ – Paul Larson Aug 28 '14 at 19:46
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    $\begingroup$ Sorry, I just saw this. Forcing with $\mathcal{P}(\omega)/fin$ adds a nonprincipal ultrafilter on $\omega$. Such an ultrafilter cannot have the Baire property (as a subset of the space $\mathcal{P}(\omega)$), so the existence of such an ultrafilter contradicts AD. $\endgroup$ – Paul Larson Feb 22 '15 at 22:00

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