Hall's Marriage Theorem/Finite Set

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

Suppose that for each subset $F$ of $S$, $|F| \le \left| \bigcup F \right|$.

Then $S$ has an injective choice function.

Proof
If one or more elements of $S$ are infinite, then those elements can be handled trivially.

Suppose, then that each element of $S$ is finite.

The proof proceeds by induction on the number of elements in $S$.

If $|S| = 0$, then $\varnothing$ is an injective choice function for $S$.

Suppose the theorem holds for $|S| = n$.

Let $|S| = n + 1$ and let $S$ satisfy the premises.

Then $S ≠ \varnothing$.

Suppose for the sake of contradiction that $S$ has no injective choice function.

Let $T \in S$. Note that $1 = |\{T\}| \le |\bigcup \{T\}| = |T|$

If $|T| = 1$ for each $T \in S$, then $S$ only has one choice function, and it is injective.

Suppose instead that there is a $U \in S$ such that $|U| \ge 2$.

Let $Q$ be the set of injective choice functions on $S \setminus \{ T \}$.

By the inductive hypothesis, $Q$ has an element.

Since we assume $S$ has no injective choice function, $x \in g[S \setminus T]$ for each $g \in Q$.

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