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How to show that $\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$?

$\bf AD$ is abbreviated for axiom of determinacy. $\bf AC_{\omega}(\mathbb{R})$ states that for each family $(X_i)_{i∈\omega}$, in which, each $X_i$ is a subset of $\mathbb{R}$, the countable product set $\prod_{i∈\omega}X_i$ is non–empty.

It is known that $\bf AD$ and $\bf AC$ are incompatible(See e.g. Theorem 6.5, Axiom of Choice(Herrlich)), but why $\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$?

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Let $X_i\subseteq\mathbb\omega^\omega$ non-empty for $i\in\omega$. We define $A$ to be the set $\{x\in\omega^\omega\mid x_{II}\notin X_{x(0)}\}$, where $x_{II}$ denotes the sequence of odd coordinates in $x$.

The game is played on $\omega$ and the winning set is $A$, namely $I$ wins if and only if the resulting sequence is in $A$. Clearly $I$ cannot win. Therefore $II$ has a winning strategy $\tau$.

Then given $n\in\omega$ we know that the sequences of moves made by $II$ using $\tau$ if $I$ plays $(n,0,0,0,0,0,\ldots)$ is in $X_n$ for all $n\in\omega$, and there's your choice function.

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Thank you very much. It's quite an ingenious construction. –  Metta World Peace Feb 7 '13 at 17:49
    
@Metta: Thanks. Obviously it's not my own... :-) –  Asaf Karagila Feb 7 '13 at 17:50

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