Hall's marriage theorem states that a finite bipartite graph with parts has a matching in a graph saturating exactly when for every . Necessity follows because the matched neighbours of are distinct.
For sufficiency use induction on . The empty case is immediate. If a nonempty proper subset has , the induced graph on satisfies Hall's condition and has a saturating matching in a graph by induction. The remaining graph on also satisfies it: for ,Apply induction again and unite the two matchings in a graph. If there is no such tight proper subset, every nonempty proper has . Choose a vertex and a neighbour , which exists by Hall's condition. Delete both. Every remaining subset loses at most one neighbour and still satisfies Hall's condition. Induction supplies a matching in a graph of the remainder; add . For this last step directly proves the assertion.
Let be the two parts of the bipartite graph. Counting edges gives , hence . For , the edges incident to all end in , whose vertices have total capacity . Thus . Hall's marriage theorem gives a perfect matching, and no matching in a graph can have more than edges. The matching number is .
Take a maximal matching in a graph with edges and let be its endpoints. Maximality makes its complement an independent set, since an edge wholly outside could be added. Every vertex outside therefore sends all its edges into , giving crossing edges. Every vertex in already uses at least one incident edge inside , its matched edge, so it sends at most outside. HenceThe maximum matching in a graph is at least this large. For , take any disjoint union of triangles. It is 2-regular, has vertices and matching number , since each triangle contributes exactly one edge. As varies this gives infinitely many equality examples.
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