Axiom:Axiom of Choice for Finite Sets

Axiom
If $\mathcal S$ is a non-empty family of finite, non-empty sets, then there exists a choice function for S.

Remark
The axiom of choice for finite sets is trivially implied by the axiom of choice, but it is strictly weaker.

In fact, it is among the weakest forms of choice.

Proof from the Ordering Principle
By the Axiom of Union, $\mathcal S$ has a union.

Let $U = \bigcup \mathcal S$.

By the Ordering Principle, there is a total ordering $\preceq$ on $U$.

For each $S \in \mathcal S$, $S$ is a chain in $U$.

By Finite Subset of Totally Ordered Set, each $S \in S$ has a minimum.

Let $f:\mathcal S \to U$ be defined by:
 * $f(S) = \min S$

Then $f$ is a choice function for $\mathcal S$.