=== $(2)$ 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. {{qed|lemma}} === $(1)$ 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 $\SS = \sequence {S_k}_{k \in I}$ be an indexed family of finite sets that satisfies the '''marriage condition'''. Let $e \in I$. Let $y \in S_e$. Let $\SS_y = \sequence {S_k \setminus \set y}_{k \in I \setminus \set e}$. 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 $\displaystyle g: I \setminus \set e \to \bigcup \SS_y$ 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. {{qed|lemma}} === Case $(b)$ === Suppose $\SS_y$ does not satisfy the marriage condition. Then there exists $P \subseteq I \setminus \set e \subsetneqq I$ such that: :$\displaystyle \card P > \card {\bigcup_{i \mathop \in P} \paren {S_i \setminus \set y} }$ Note that $P$ cannot be empty. {{begin-eqn}} {{eqn | l = \bigcup_{i \mathop \in P} \paren {S_i \setminus \set y}       | r = \paren {\bigcup_{i \mathop \in P} S_i} \setminus \set y       | c =  }} {{eqn | r = Y_P \setminus \set y       | c =  }} {{eqn | ll= \leadsto       | l = \card P       | o = >       | r = \card {Y_P \setminus \set y}       | c =  }} {{end-eqn}} 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 $\TT = \sequence {S_k \setminus Y_P}_{k \in I \setminus P}$ 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: {{begin-eqn}} {{eqn | l = \card P + \card Q       | r = \card {P \cup Q}       | c =  }} {{eqn | o = \le       | r = \card {Y_{P \cup Q} }       | c =  }} {{eqn | r = \card {Y_P} + \card {Y_Q \setminus Y_P}       | c =  }} {{eqn | ll= \leadsto       | l = \card Q       | o = \le       | r = \card {Y_Q \setminus Y_P}       | c = as $\card P = \card {Y_P}$ }} {{end-eqn}} 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. {{qed}} {{namedfor|Philip Hall}} Category:Set Theory q3tlw4ywqsne7lz69saystli1adfpbm