The Complement-Pairing Bound for Two-Intersecting Families

Let $F$ be a family of $2n$-element subsets of a $4n$-element set in which any two members meet in at least two points. We give the short proof, by pairing each $2n$-set with its complement, that $|F| \\le \\tfrac12 \\binom{4n}{2n}$. The argument uses nothing beyond the fact that a $2n$-subset of a $4n$-set and its complement are themselves complementary $2n$-subsets, so the family cannot contain both without violating 2-intersection. The sharp constant --- lower than this half-bound by a term of order $\\binom{2n}{n}^2$ --- belongs to the general theory of $t$-intersecting families and is stated here as the established result it is, not reproved.