Axiom:Axiom of Choice for Finite Sets

From ProofWiki
Jump to navigation Jump to search


Let $\SS$ be a non-empty set of finite, non-empty sets.

Then there exists a choice function for $\SS$.


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

Proof from the Ordering Principle

By the Axiom of Union, $\SS$ has a union.

Let $U = \bigcup \SS$.

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

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

By Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, each $S \in S$ has a minimum.

Let $f: \SS \to U$ be defined by:

$\map f S = \min S$

Then $f$ is a choice function for $\SS$.


Proof from Hall's Marriage Theorem

Without loss of generality, suppose $\SS$ is pairwise disjoint.

We will show that $\SS$ satisfies the marriage condition.

That is, for each finite subset $\FF$ of $\SS$:

$\card \FF \le \card {\bigcup \FF}$

as follows:

Let $\FF$ be a finite subset of $\SS$.

By the Principle of Finite Choice, $\FF$ has a choice function $f_\FF: \FF \to \bigcup \FF$ such that for each $S \in \FF$, $\map {f_\FF} S \in S$.

Since $\SS$ is pairwise disjoint, $f_\FF$ is an injection.

Thus $\card \FF \le \card {\bigcup \FF}$.

By Hall's Marriage Theorem for sets of finite sets, $\SS$ has a choice function.