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Let $A$ be a set of non-empty sets, then $\bigcup A$ is a set. Furthermore, $(\bigcup A)^{A}$ is a non-empty set. Besides let $P$ be a binary predicate such that for all $X\in A$ there is a unique $x \in X$ satifies $PXx$. $f:=\{(X,x)\in A \times \bigcup A|PXx\}$ is a functional class and $f \in (\bigcup A)^A$. Hence $f$ is also a set.

According things discussed above, $f$ is indeed a choice function of $A$ and is a set. So does this fact means AC is naturally holds?

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    @AsafKaragila Okay, I see.2012-11-10

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If such $P$ exists then by the replacement schema you have defined a choice function on $A$.

If this $P$ has the property above for all non-empty collections of non-empty sets, then $P$ is actually a function and in fact it is a global choice function.

In either case the predicate $P$ implies that the axiom of choice holds (at least for $A$) and that one can define a canonical choice.

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    @Popopo: It is$a$subset of $A\times\bigcup A$. If it **exists** then it means that this is a set, because only sets exist. If it was a class it was a class bounded by a set and therefore a set itself.2012-11-10