Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 109 / 3

Discard ground points lying in no set. For every point $x$ that lies in at least two members, formEvery pair of indices lies in exactly one $L_x$, and no $L_x$ equals $[n]$ because the total intersection is empty. Thus the $L_x$ form a finite linear space on the $n$ indices. The number of its lines through index $i$ is at most $a_i$, with equality unless $A_i$ contains private points.

The De Bruijn--Erdos pair-covering inequality says that if $r_i$ is the number of lines through point $i$ in a nontrivial finite linear space on $n$ points, thenApplying it here givesMoreover, a pair of ground points can lie in at most one $A_i$, since two different members meet in only one point. HenceCombining the inequalities gives $\left(\genfrac{}{}{0ex}{}{m}{2}\right)\ge \left(\genfrac{}{}{0ex}{}{n}{2}\right)$, and therefore $m\ge n$.

No example exists for $n=2$, because the unique point in $A_1\cap A_2$ would also lie in the total intersection; the one-member case is excluded by nontriviality. For every $n\ge 3$, take ground set $[n]$ and define the near-pencilTwo small sets meet in $n$, each small set meets $A_0$ in its other point, and the total intersection is empty. Thus the required values are exactly $n\ge 3$.

Yes. For every prime power $q\ge 2$, a finite projective plane of order $q$ haspoints and the same number of lines; every two lines meet in exactly one point, and the intersection of all lines is empty. Its lines therefore form one required family on $n$ points. The near-pencil from part (ii) is another. They are non-isomorphic because every projective-plane line has size $q+1$, whereas the near-pencil has a member of size $n-1$. There are infinitely many prime powers, so infinitely many such $n$.