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  "body": "\n## Proof\n\nWe prove a slightly more general form, for families indexed by any finite set, by strong induction on the number of sets.\n\n*Setup.*\n\n1. Let $I$ be a finite set and $(S_i)_{i \\in I}$ a family of finite sets. For $J \\subseteq I$, write $S(J) = \\bigcup_{i \\in J} S_i$. A transversal is an injective function $f$ on $I$ with $f(i) \\in S_i$ for all $i \\in I$.\n2. Say that the family satisfies Hall's condition if $\\abs{S(J)} \\ge \\abs{J}$ for all $J \\subseteq I$. For $I = \\set{1, \\dots, n}$ this is the condition of the statement.\n\n*The condition is necessary.*\n\n3. Let $f$ be a transversal and $J \\subseteq I$. Then $f(J) \\subseteq S(J)$, since $f(i) \\in S_i$ for each $i \\in J$.\n4. As $f$ is injective, $\\abs{f(J)} = \\abs{J}$. So $\\abs{S(J)} \\ge \\abs{J}$.\n\n*The condition is sufficient.*\n\n5. We prove by [[strong induction|0KMVFSHFGZ]] on $\\abs{I}$ that a family with Hall's condition has a transversal. If $I = \\varnothing$, the empty function is a transversal.\n6. Let $\\abs{I} = n \\ge 1$, and suppose the claim holds for all index sets with fewer than $n$ elements. Call $J \\subseteq I$ *critical* if $J \\ne \\varnothing$, $J \\ne I$ and $\\abs{S(J)} = \\abs{J}$.\n\n*Case 1: no subset of $I$ is critical.*\n\n7. Then $\\abs{S(J)} \\ge \\abs{J} + 1$ for every $J$ with $\\varnothing \\ne J \\subsetneq I$.\n8. Fix $i_0 \\in I$. By Hall's condition for $J = \\set{i_0}$, the set $S_{i_0}$ is not empty. Choose $x \\in S_{i_0}$.\n9. Let $I' = I \\setminus \\set{i_0}$ and $T_i = S_i \\setminus \\set{x}$ for $i \\in I'$. Write $T(J) = \\bigcup_{i \\in J} T_i$.\n10. The family $(T_i)_{i \\in I'}$ satisfies Hall's condition. For $J = \\varnothing$ this is clear. For $\\varnothing \\ne J \\subseteq I'$, the set $J$ is a proper subset of $I$, so $\\abs{T(J)} \\ge \\abs{S(J)} - 1 \\ge \\abs{J}$ by Step 7.\n11. Since $\\abs{I'} = n - 1$, the induction hypothesis gives a transversal $g$ of $(T_i)_{i \\in I'}$. Its values differ from $x$, since $x \\notin T_i$.\n12. Define $f = g$ on $I'$ and $f(i_0) = x$. Then $f(i) \\in S_i$ for all $i$, and $f$ is injective. So $f$ is a transversal.\n\n*Case 2: some $J_0 \\subseteq I$ is critical.*\n\n13. Let $A = S(J_0)$, so $\\abs{A} = \\abs{J_0}$. The subfamily $(S_i)_{i \\in J_0}$ satisfies Hall's condition, because subsets of $J_0$ are subsets of $I$.\n14. Since $\\abs{J_0} \u003c n$, the induction hypothesis gives a transversal $g$ of $(S_i)_{i \\in J_0}$. All its values lie in $A$.\n15. Let $K_0 = I \\setminus J_0$ and $U_i = S_i \\setminus A$ for $i \\in K_0$. Write $U(K) = \\bigcup_{i \\in K} U_i$.\n16. Let $K \\subseteq K_0$. Then $S(K \\cup J_0) = S(K) \\cup A = U(K) \\cup A$, and $U(K)$ is disjoint from $A$. So $\\abs{S(K \\cup J_0)} = \\abs{U(K)} + \\abs{A}$.\n17. Since $K$ and $J_0$ are disjoint, Hall's condition gives $\\abs{U(K)} + \\abs{A} \\ge \\abs{K} + \\abs{J_0}$. As $\\abs{A} = \\abs{J_0}$, we get $\\abs{U(K)} \\ge \\abs{K}$.\n18. So $(U_i)_{i \\in K_0}$ satisfies Hall's condition. Since $J_0 \\ne \\varnothing$, we have $\\abs{K_0} \u003c n$, and the induction hypothesis gives a transversal $h$ of it. All its values lie outside $A$.\n19. Define $f = g$ on $J_0$ and $f = h$ on $K_0$. Then $f(i) \\in S_i$ for all $i \\in I$.\n20. The map $f$ is injective: $g$ and $h$ are injective, values of $g$ lie in $A$, and values of $h$ lie outside $A$. So $f$ is a transversal.\n\n*Conclusion.*\n\n21. By induction, every finite family with Hall's condition has a transversal. With Steps 3 and 4, a family $(S_1, \\dots, S_n)$ has a transversal if and only if $\\abs{\\bigcup_{i \\in I} S_i} \\ge \\abs{I}$ for every $I \\subseteq \\set{1, \\dots, n}$.\n\n## Idea\n\nIf every proper subfamily has room to spare, choose any representative for one set and delete it from the others; the condition survives. If some subfamily $J_0$ is tight, its union $A$ must be used up exactly by $J_0$. Solve $J_0$ inside $A$ and the rest outside $A$, and both smaller problems still satisfy the condition.\n",
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