Axiom:Axiom of Choice/Formulation 3

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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


Sources

  • 1908: Ernst ZermeloNeuer Beweis für die Möglichkeit einer Wohlordnung ("A new proof of the possibility of well-ordering") (Math. Ann. Vol. 65: 107 – 128)