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 $\displaystyle Y_F = \bigcup_{k \mathop \in F} S_k$.

Let $Y = Y_I$.

Then the following are equivalent:


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


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

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

Then:
 * $\left\vert{P}\right\vert \not \le \left\vert{Y_P}\right\vert$

Thus 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:
 * $\left\vert{I \setminus \left\{{e}\right\} }\right\vert = n - 1 < n$

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:


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

g \left({k}\right) & : k \ne e \\ y & : k = e \end{cases}$

Then $f$ is an injection satisfying the requirements.

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

Then there exists $P \subseteq I \setminus \left\{{e}\right\} \subsetneqq I$ such that:
 * $\displaystyle \left\vert{P}\right\vert > \left\vert{\bigcup_{i \mathop \in P} \left({S_i \setminus \left\{ {y}\right\} }\right) }\right\vert$

Note that $P$ cannot be empty.

Since $\mathcal S$ satisfies the marriage condition:
 * $\left\vert{P}\right\vert \le \left\vert{Y_P}\right\vert$

so $y \in Y_P$.

By Cardinality Less One
 * $\left\vert{Y_P \setminus \left\{ {y}\right\} }\right\vert = \left\vert{Y_P}\right\vert - 1$

Thus:
 * $\left\vert{P}\right\vert > \left\vert{Y_P}\right\vert - 1$

Thus:
 * $\left\vert{P}\right\vert = \left\vert{Y_P}\right\vert$

We have for $\left\vert{P}\right\vert < n$ that $P \subsetneqq I$.

So by the inductive hypothesis there is an injection $g: P \to Y_P$ such that:
 * $\forall k \in P: g \left({k}\right) \in S_k$

It is to be shown that $\mathcal T = \langle S_k \setminus Y_P \rangle_{k \in I \setminus P}$ satisfies the marriage condition.

Let $Q \subseteq I \setminus P$.

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

Thus:
 * $\left\vert{P}\right\vert + \left\vert{Q}\right\vert = \left\vert{P \cup Q}\right\vert \le \left| Y_{P \cup Q} \right| = |Y_P| + |Y_Q \setminus Y_P|$

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

As $\left\vert{P}\right\vert \ne 0$, $\left\vert{I \setminus P}\right\vert < n$.

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

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