Axiom:Axiom of Choice/Formulation 3

Axiom
Let $\mathcal S$ be a set of non-empty pairwise disjoint sets.

Then there is a set $C$ such that for all $S \in \mathcal S$, $C \cap S$ has exactly one element.

Symbolically:
 * $\forall s: \left({ \left({\varnothing \notin s \land \forall t, u \in s: t = u \lor t \cap u = \varnothing}\right) \implies \exists c: \forall t \in s: \exists x: t \cap c = \left\{{x}\right\} }\right)$

Also see

 * Equivalence of Versions of Axiom of Choice