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  "body": "\n## Proof\n\nWe translate matchings into transversals of the family of neighbourhoods.\n\n1. Let $G$ be a finite [[bipartite graph|07DVHJ8N0A]] with parts $X$ and $Y$. For $x \\in X$, let $N(x) \\subseteq Y$ be its [[neighbourhood|01T3TMGTTE]]. For $W \\subseteq X$, the set $N(W)$ of the statement equals $\\bigcup_{x \\in W} N(x)$.\n2. Consider the finite family $(N(x))_{x \\in X}$ of finite sets, indexed by $X$.\n3. Suppose $M$ is a [[matching|0Z87BHAFTB]] that covers every vertex of $X$. For $x \\in X$, let $f(x)$ be the other end of the edge of $M$ at $x$. This edge is unique, because no two edges of $M$ share an end.\n4. Each edge joins $X$ to $Y$, so $f(x) \\in N(x)$. If $f(x) = f(x')$ with $x \\ne x'$, two edges of $M$ would share the end $f(x)$. So $f$ is injective, and it is a transversal of the family.\n5. Conversely, let $f$ be a transversal of the family. Let $M = \\set{x f(x) : x \\in X}$. These are edges, since $f(x) \\in N(x)$.\n6. Two distinct edges $x f(x)$ and $x' f(x')$ of $M$ have $x \\ne x'$ and $f(x) \\ne f(x')$. Ends in $X$ and ends in $Y$ cannot coincide, as the parts are disjoint. So $M$ is a matching, and it covers $X$.\n7. By Steps 3 to 6, $G$ has a matching covering $X$ if and only if the family $(N(x))_{x \\in X}$ has a transversal.\n8. By [[Hall's marriage theorem|08WCB90WND]], applied after numbering the elements of $X$, this holds if and only if $\\abs{\\bigcup_{x \\in W} N(x)} \\ge \\abs{W}$ for every $W \\subseteq X$.\n9. By Step 1, this condition says $\\abs{N(W)} \\ge \\abs{W}$ for every $W \\subseteq X$.\n\n## Idea\n\nA matching that covers $X$ is the same thing as a choice of a distinct neighbour for each vertex of $X$. That is a transversal of the neighbourhoods, so the marriage theorem applies directly.\n",
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