Hall's Marriage Theorem/Finite Set

Theorem
Let $\mathcal S$ be a finite set of finite sets.

Then the following are equivalent:


 * $(1)\quad$ $\mathcal S$ satisfies the marriage condition: for each subset $F$ of $\mathcal S$, $|F| \le \left| \bigcup F \right|$.


 * $(2)\quad$ $\mathcal S$ has an injective choice function.

$(1)$ implies $(2)$
Write $\mathcal S$ as $\{ S_0, \dots \S_n \}$.

Since $\mathcal S$ is a finite set of finite sets, there is a $\mathcal T = \{ T_0, \dots, T_n \}$ such that:


 * $\forall k \in \N_n: T_k \subseteq S_k$
 * $\mathcal T$ satisfies the marriage condition.
 * $\mathcal T$ is minimal in this regard: replacing $T_k$ with a proper subset of $T_k$ will produce a set that violates the marriage condition.

An injective choice function on $\mathcal T$ will then induce an injective choice function on $\mathcal S$.

Thus we can assume WLOG that $\mathcal S$ is minimal in this way.

Then: for each $T \in \mathcal S$ and each $x \in T$, $\mathcal S \setminus \{ T \} \cup \{T \setminus \{x\} \}$ violates the marriage condition.

We will show that each element of $\mathcal S$ has exactly one element.

Suppose for the sake of contradiction that $x, y \in T$, $T \in \mathcal S$, and $x ≠ y$.

Then there exists $\mathcal M \subseteq \mathcal S \setminus \{ T \} \cup \{ T \setminus \{ x \}\}$ such that $|\mathcal M| > |\bigcup \mathcal M|$.

Since $\mathcal S$ satisfies the marriage condition, $\mathcal M \nsubseteq \mathcal S$, so $T \setminus \{ x \} \in \mathcal M$.