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 $S$ is a mapping $f: \mathbb S \to \bigcup \mathbb S$ defined as:
 * $\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$

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

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

The codomain of $f$ is $\bigcup \mathbb S$, that is, the union of all the sets which comprise $\mathbb S$

Axiom of Choice
It is allowed that, given any set $S$ one can select an element from it.

However, what is debated is whether, given a set of sets $\mathbb S$, one can construct a mapping $f$ such that every set in $\mathbb S$ can be an element of the domain of $f$.

So it's not:
 * "Can I choose an element from any given set (in $\mathbb S$)?"

as much as:
 * "Can I construct a (mechanistic) procedure that will always return some element of any set (in $\mathbb S$)?"

The Axiom of Choice (AoC) is a philosophical position which states that a set of nonempty sets always has a choice function.

Non-Acceptance of Axiom of Choice
If one does not accept the AoC, then a choice function can be proved to exist for the following categories of $\mathbb S$:

A Choice Function Exists for All Finite Sets
If $\mathbb S$ is finite, we can construct a choice function on $\mathbb S$ by picking one element from each member of $\mathbb S$.

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

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


 * The Well-Ordering Theorem is Equivalent to the Axiom of Choice, which demonstrates the truth of the converse of the Well-Ordering Theorem.