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The standard axiom of choice states that every set of non-empty sets has a choice function. And the axiom of global choice states that every class of non-empty sets has a choice function. But I’m wondering how strong a choice axiom is required to get a choice function for a set-sized collection of classes. Let me explain.

Let $X$ be a set, and let $\phi(x,y)$ be a formula such that for all $x\in X$, there exists a set $y$ such that $\phi(x,y)$. Then my question is, what is required to prove that there exists a set $Y$ and a function $f:X\rightarrow Y$ such that $\phi(x,f(x))$ for all $x\in X$?

Is the standard axiom of choice sufficient to prove this?

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Yes, the standard axiom of choice is sufficient: intuitively, just replace each class $C$ with the set $V_{\alpha_C}\cap C$, where $\alpha_C$ is the smallest ordinal such that that set is nonempty. (This is the same idea behind Scott's trick, although the latter refers more specifically to the "paring down" of equivalence classes.)

We could also phrase this in terms of shifting attention from the original formula $\varphi(x,y)$ to the formula $$\psi(x,y)\equiv\varphi(x,y)\wedge\forall z(rk(z)<rk(y)\implies\neg\varphi(x,z)),$$ where "$rk(a)$" is the rank of $a$ (that is, the smallest ordinal $\theta$ such that $a\in V_\theta$).

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