Hall's Marriage Theorem/Finite Set

Theorem
Let $\mathcal S = \left\langle{S_k}\right\rangle_{k \in I}$ be a finite indexed family of finite sets.

For each $F \subseteq I$, let $\displaystyle Y_F = \bigcup_{k \in F} S_k$.

Let $Y = Y_I$.

Then the following are equivalent:


 * $(1): \quad \mathcal S$ satisfies the marriage condition: for each subset $F$ of $I$ such that $\left|{F}\right| \le \left|{Y_F}\right|$.


 * $(2): \quad$ There exists an injection $f: I \to Y$ such that $\forall k \in I: f \left({k}\right) \in S_k$.

$(2)$ implies $(1)$
Suppose that for some $P \subseteq I$, $\left|{P}\right| > \left|{Y_P}\right|$.

Then $\left|{P}\right| \not\le \left|{Y_P}\right|$, so there can be no injection from $P$ to $Y_P$.

Thus there can be no injection from $I$ to $Y$ satisfying the requirements.

$(1)$ implies $(2)$
The proof proceeds by induction on $n$, the cardinality of $I$.

If $n = 0$, then the empty set is the necessary injection.

Suppose the theorem holds for all $m < n$.

Let $I$ be a set with $n$ elements, where $n \ge 1$.

Let $\mathcal S = \left\langle{S_k}\right\rangle_{k \in I}$ be an indexed family of finite sets that satisfies the marriage condition.

Let $e \in I$.

Let $y \in S_e$.

Let $\mathcal S_y = \left\langle{S_k \setminus \left\{{y}\right\} }\right\rangle_{k \in I \setminus \left\{{e}\right\}}$.

By Law of Excluded Middle, one of the following must hold:


 * $(a): \quad \mathcal S_y$ satisfies the marriage condition.


 * $(b): \quad \mathcal S_y$ violates the marriage condition.

Case $(a)$
Suppose $\mathcal S_y$ satisfies the marriage condition.

By Cardinality Less One:
 * $\left|{I \setminus \left\{{e}\right\} }\right| = n - 1 < n$

Then by the inductive hypothesis, there is an injection $\displaystyle g: I \setminus \left\{{e}\right\} \to \bigcup \mathcal S_y$ such that for each $k \in I \setminus \left\{{e}\right\}$, $g \left({k}\right) \in S_k \setminus \left\{{y}\right\}$.

Let $f: I \to Y$ be defined thus:


 * $\displaystyle f \left({k}\right) = \cases {

g(k) &: $k \ne e$ \\ y &: $k = e$}$

Then $f$ is an injection satisfying the requirements.

Case $(b)$
Suppose $\mathcal S_y$ does not satisfy the marriage condition.

Then there exists $P \subseteq I \setminus \{ e \} \subsetneqq I$ such that $|P| > \left|{ \bigcup_{i \in P} (S_i \setminus \{ y \}) }\right|$.

Note that $P$ cannot be empty.

$\bigcup_{i \in P} (S_i \setminus \{ y \}) = \left({ \bigcup_{i \in P} S_i }\right) \setminus \{ y \} = Y_P \setminus \{ y \}$, so $|P| > |Y_P \setminus \{y\}|$.

Since $\mathcal S$ satisfies the marriage condition, $|P| \le |Y_P|$, so $y \in Y_P$.

By Cardinality Less One, $|Y_p \setminus \{y\}| = |Y_P| - 1$.

Thus $|P| > |Y_P| - 1$.

Thus $|P| = |Y_P|$.

Since $P \subsetneqq I$, $|P| < n$, so by the inductive hypothesis there is an injection $g: P \to Y_P$ such that $g(k) \in S_k$ for each $k \in P$.

We will show that $\mathcal T = \langle S_k \setminus Y_P \rangle_{k \in I \setminus P}$ satisfies the marriage condition.

Let $Q \subseteq I \setminus P$.

Then since $P$ and $Q$ are disjoint and $\mathcal S$ satisfies the marriage condition, $|P| + |Q| = |P \cup Q| \le \left| Y_{P \cup Q} \right| = |Y_P| + |Y_Q \setminus Y_P|$.

Since $|P| = |Y_P|$, $|Q| \le |Y_Q \setminus Y_P| = |\bigcup_{k \in Q} (S_k \setminus Y_P)|$.

As this holds for all such $Q$, $\mathcal T$ satisfies the marriage condition.

As $|P| ≠ 0$, $|I \setminus P| < n$, so there is an injection $h: I \setminus P \to Y_{I \setminus P} \setminus Y_P$ such that $h(k) \in S_k \setminus Y_P$ for each $k \in I \setminus P$.

Then $f = g \cup h$ is the injection we seek.