Hall's Marriage Theorem/General Set

Theorem
Let $\SS = \sequence {S_k}_{k \in I}$ be an 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:


 * $(1): \quad \SS$ satisfies the marriage condition: for each finite subset $F$ of $I$, $\card F \le \card {Y_F}$.


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

$(2)$ implies $(1)$
Suppose that for some finite $F \subseteq I$, $\card F > \card {Y_F}$.

Then $\card F \not \le \card {Y_F}$.

By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $F$ to $Y_F$.

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

$(1)$ implies $(2)$
Let $\map \Phi x$ be the set of all elements of $y$ in the expression of $x$.

Let $\FF$ be the set of all partial functions $g$ from $I$ to $Y$ such that:


 * $g$ is one-to-one.
 * If $\tuple {k, y} \in g$, then $y \in S_k$.

Then for any $k \in I$, $\set {\map g x: g \in \FF \land k \in \map {\operatorname {Dom}} g}$ is finite.

$\FF$ has finite character:

Let $g \in \FF$.

Let $F$ be a finite subset of $I$.

Then $g \restriction F$ is obviously in $\FF$.

Suppose instead that $g$ is a partial function from $I$ to $Y$.

Let the restriction of $g$ to each finite subset of $\map {\operatorname{Dom}} g$ be in $\FF$.

Then $g$ is one-to-one, as follows:

Let $p, q \in \map {\operatorname{Dom} } g$ with $p \ne q$.

Then $\set {p, q}$ is a finite subset of $\map {\operatorname{Dom}} g$.

Thus:
 * $g \restriction \set {p, q} \in \FF$

Thus:
 * $\map g p \ne \map g q$

Let $\tuple {k, y} \in g$.

Then:
 * $\tuple {k, y} \in g \restriction \set k$

So:
 * $y \in S_k$

Thus:
 * $g \in \FF$

So we see that $\FF$ has finite character.

Let $F$ be a finite subset of $I$.

By the finite case, there is an element of $\FF$ with domain $F$.

By the Cowen-Engeler Lemma, $\FF$ has an element $\phi$ whose domain is $I$.

That is, $\phi$ is a mapping from $I$ to $Y$.

But a one-to-one mapping is an injection, so $\phi$ is an injection from $X$ to $Y$.

By the definition of $\FF$:
 * $\forall k \in I: \map \phi k \in S_k$