Hall's Marriage Theorem/Finite Set

Theorem
Let $\SS = \family {S_k}_{k \mathop \in I}$ be a finite indexed family of finite sets.

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

Let $Y = Y_I$.

Then the following are equivalent:


 * $(1): \quad \SS$ satisfies the marriage condition:


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

$(2)$ implies $(1)$
Let:
 * $\exists P \subseteq I: \card P > \card {Y_P}$

Then:
 * $\card P \not \le \card {Y_P}$

By contrapositive of Injection implies Cardinal Inequality, 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 $\SS = \family {S_k}_{k \mathop \in I}$ be an indexed family of finite sets that satisfies the marriage condition.

Let $e \in I$.

Let $y \in S_e$.

Let $\SS_y = \family {S_k \setminus \set y}_{k \mathop \in I \setminus \set e}$.

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


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


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

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

By Cardinality Less One:
 * $\card {I \setminus \set e} = n - 1 < n$

Then by the inductive hypothesis, there is an injection $\ds g: I \setminus \set e \to \bigcup \SS_y$ such that for each $k \in I \setminus \set e$, $\map g k \in S_k \setminus \set y$.

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


 * $\map f k = \begin{cases}

\map g k & : k \ne e \\ y & : k = e \end{cases}$

Then $f$ is an injection satisfying the requirements.

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

Then there exists $P \subseteq I \setminus \set e \subsetneqq I$ such that:
 * $\ds \card P > \card {\bigcup_{i \mathop \in P} \paren {S_i \setminus \set y} }$

Note that $P$ cannot be empty.

Since $\SS$ satisfies the marriage condition:
 * $\card P \le \card {Y_P}$

so $y \in Y_P$.

By Cardinality Less One:
 * $\card {Y_P \setminus \set y} = \card {Y_P} - 1$

Thus:
 * $\card P > \card {Y_P} - 1$

Thus:
 * $\card P = \card {Y_P}$

We have for $\card P < n$ that $P \subsetneqq I$.

So by the inductive hypothesis there is an injection $g: P \to Y_P$ such that:
 * $\forall k \in P: \map g k \in S_k$

It is to be shown that $\TT = \family {S_k \setminus Y_P}_{k \mathop \in I \setminus P}$ satisfies the marriage condition.

Let $Q \subseteq I \setminus P$.

We have that:
 * $P$ and $Q$ are disjoint sets
 * $\SS$ satisfies the marriage condition.

Thus:

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

As $\card P \ne 0$, $\card {I \setminus P} < n$.

Thus there is an injection $h: I \setminus P \to Y_{I \setminus P} \setminus Y_P$ such that:
 * $\forall k \in I \setminus P: \map h k \in S_k \setminus Y_P$

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