Definition:Choice Function

Definition
Let $\mathbb S$ be a set of sets such that:
 * $\forall S \in \mathbb S: S \ne \varnothing$

that is, none of the sets in $\mathbb S$ may be empty.

A choice function on $\mathbb S$ is a mapping $f: \mathbb S \to \bigcup \mathbb S$ satisfying:
 * $\forall S \in \mathbb S: f \left({S}\right) \in S$.

That is, for any set in $\mathbb S$, a choice function selects an element from that set.

The domain of $f$ is $\mathbb S$.

Use of Axiom of Choice
In some situations, AoC is not needed to get a choice function:

A Choice Function Exists for Set of Well-Ordered Sets
If every member of $\mathbb S$ is a well-ordered, then we can define a choice function $f$ by:
 * $\forall S \in \mathbb S: f \left({S}\right) = \inf \left({S}\right)$

Note that this only applies if we are given a well order for each $S \in \mathbb S$, more formally, if there is a function that maps $S \in \mathbb S$ to a well-order of $S$. If we just know that each $S \in \mathbb S$ is well-orderable, we generally do need AoC to get a choice function (to apply the proof above, we have to pick a well-order for each $S\in \mathbb S$, which requires AoC. This is related to the fact that generally we need AoC to show that, for example, the countable union of countable sets is countable.)

A Choice Function Exists for Well-Orderable Union of Sets
If the union $\bigcup \mathbb S$ is well-orderable, we can create a choice function for $\bigcup \mathbb S$.

Also see

 * Well-Ordering Theorem


 * Well-Ordering Theorem is Equivalent to Axiom of Choice