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 $f: \mathbb S \to \bigcup \mathbb S$ 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 $\mathbb S$.

Axiom of Choice
The Axiom of Choice (AC) is the following statement: All $\mathbb S$ as above have a choice function.

This set theoretic axiom has been controversial, but is nowadays generally accepted. It can be shown that it does not follow from the other usual axioms of set theory, and that it is relative consistent to these axioms (i.e., that AC does not make the axiom system inconsistent, provided it was consistent withaout AC).

Note that for any given ("fixed") set $S\in \mathbb S$, one can select an element from it (without using AC). AC guarantees that there is a choice function, i.e., a function that "simultaneously" picks elements of all $S\in\mathbb S$.

AC is needed to prove statements such as "all countable unions of finite sets are countable" (for many specific such unions this can be shown without AC), and AC is equivalent to many other mathematical statements such as "every vector space has a basis".

In some situations, AC is not needed to get a choice function:

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
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 AC 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 AC. This is related to the fact that generally we need AC to show that, e.g., 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


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