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Source. Theorem 2, p. 132, of P. Frankl, "On families of finite sets no two of which intersect in a singleton," Bull. Austral. Math. Soc. 17 (1977), no. 1, 125-134, doi:10.1017/S0004972700025521. Pages are the journal's own, as on the source card.

Statement

Theorem 2 (p. 132). Let $\mathcal F$ be an $(n,\{0,2,3,\ldots,k-1\},k)$-system, that is, a family of $k$-subsets of an $n$-set $X$ no two different members of which meet in exactly one element, with $k\ge4$. Suppose $n>n_0(k)+2\binom{n_0(k)}{k}$, where $n_0(k)$ is the bound of Theorem 1. Then either there are two different elements $x,y$ with $\mathcal F$ equal to the family of all $k$-subsets of $X$ containing $\{x,y\}$, or $\lvert\mathcal F\rvert<\binom{n-2}{k-2}$.

In particular $\lvert\mathcal F\rvert\le\binom{n-2}{k-2}$ in this range, which is the main theorem (p. 125), and the family of all $k$-sets through a fixed pair is the only family attaining the bound.

Read depth. Claims checked: the statement was read clause by clause on the print and the proof read through.

Proof pointer

Page 133, by contradiction. If $\lvert\mathcal F\rvert=\binom{n-2}{k-2}+d$ with $d\ge0$ and $\mathcal F$ is not of the first kind, Theorem 1 yields a point or a pair whose deletion leaves a system on fewer points that exceeds the corresponding bound by at least $d+1$. Repeating until at most $n_0(k)$ points remain leaves more than $\binom{n_0(k)}{k}$ $k$-subsets of a set of at most $n_0(k)$ points, which is impossible.

Bears on

• Problem 702: proves the problem's corrected Statement, with the explicit range $n>n_0(k)+2\binom{n_0(k)}{k}$ in terms of the threshold of Theorem 1, and adds that the family of all $k$-sets through a fixed pair is the only extremal family. The paper says nothing about smaller $n$.