# Axiom:Axiom of Choice for Finite Sets/Proof from Hall's Marriage Theorem

Jump to navigation
Jump to search

## Theorem

Suppose that Hall's Marriage Theorem holds for all sets of finite sets.

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

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

## Proof from Hall's Marriage Theorem

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

This article, or a section of it, needs explaining.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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.

$\blacksquare$