Axiom:Axiom of Choice/Formulation 3

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Axiom

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

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

Symbolically:

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


Also see

  • Results about the Axiom of Choice can be found here.


Sources