Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 109 / 2 / Solution

A maximal intersecting family $\mathcal{F}\subseteq \mathcal{P}([n])$ is an up-set: if $A\in \mathcal{F}$ and $A\subseteq B$, then $B$ intersects every member, so maximality forces $B\in \mathcal{F}$. For every complementary pair $\{A,A^c\}$, at most one member lies in $\mathcal{F}$. If neither did, maximality would give $B\in \mathcal{F}$ disjoint from $A$; then $B\subseteq A^c$, and the up-set property would put $A^c$ in $\mathcal{F}$, a contradiction. Exactly one set from each complementary pair occurs, so

For $n\ge 4$, three pairwise nonisomorphic examples are:

Their smallest members have different sizes except at $n=4$, where the inclusion graphs of the minimal two-sets are respectively a triangle and a three-edge star. Thus they are nonisomorphic.

The Erdős-Ko-Rado theorem says that if $n\ge 2r$ and $\mathcal{F}\subseteq [n{]}^{(r)}$ is intersecting, thenFor the Kruskal-Katona theorem proof, the family of complements ${\mathcal{F}}^c\subseteq [n{]}^{(n-r)}$ is disjoint from the $(n-r)$th upper shadow of $\mathcal{F}$: otherwise one member of $\mathcal{F}$ would be contained in the complement of another. Kruskal–Katona says that when $|\mathcal{F}|=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this upper shadow has at least $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ members, with a strict corresponding inequality above that threshold. Since these two disjoint families lie in one level of size $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, the EKR bound follows.

For the averaging proof, place $[n]$ in a cyclic order. Among the $n$ cyclic intervals of length $r$, an intersecting family contains at most $r$: after fixing one interval, the possible intersecting intervals can be paired by their first separating endpoint. Double-counting pairs consisting of a cyclic order and a member of $\mathcal{F}$ that appears as an interval giveswhich is the same bound. This is the Katona circle method.

The minimum size $f_r(n)$ does not tend to infinity. Fix a core $S\subseteq [n]$ of size $2r-1$ and takeAny two members intersect. If an $r$-set $A$ is not contained in $S$, then $|A\cap S|\le r-1$, so the remaining at least $r$ points of $S$ contain a member of $\mathcal{F}$ disjoint from $A$. Thus $\mathcal{F}$ is maximal intersecting in $[n{]}^{(r)}$, andindependently of $n$.