Axiom:Axiom of Choice/Formulation 3

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

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

Symbolically:
 * $\forall s: \paren {\paren {\O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O} \implies \exists c: \forall t \in s: \exists x: t \cap c = \set x}$

That is, there exists a transversal $C$ for $\SS$.

Also see

 * Definition:Transversal


 * Equivalence of Formulations of Axiom of Choice