Size of largest subset $F$ of $\mathcal{p}(X)$ such that any two subsets in $F$ intersect no trivially.

Question

Help with the following Putnam problem: let $S$ be a finite set, and suppose that a collection $\mathcal{F}$ of subsets of $S$ has the property that any two members of $\mathcal{F}$ have at least one element in common, but cannot be extended(while keeping this property). Prove that $\mathcal{F}$ contains exactly half of the subsets of $S$.

Answer 1

Hint: For any $A\subseteq S$, let $S_A=\{A,S\backslash A\}$. Show that $\mathcal F$ contains at most one element of $S_A$ for each $A\subseteq S$. Then, prove by contradiction that $\mathcal F$ contains one element of $S_A$ for each $A\subseteq S$.

Answer 2

First note that if $A \in \mathcal{F}$ and $B$ is a subset of $S$ with $A \subseteq B$ then $\mathcal{F} \cup \{B\}$ also has the intersection property.

If $A \notin \mathcal{F}$ and $\mathcal{F}$ is maximal, this means that $\mathcal{F} \cup \{A\}$ must have two disjoint sets, one of which must be $A$. Otherwise it would contradict the maximality of $\mathcal{F}$.

So there exists $B\in \mathcal{F}$ such that $B \cap A = \emptyset$, and so $B \subseteq S\setminus A$. By the first remark, $\mathcal{F} \cup \{S\setminus A\}$ also obeys the intersection property and by maximality $\mathcal{F} \cup \{S\setminus A\} = \mathcal{F}$, or equivalently, $S\setminus A \in \mathcal{F}$.

So if $\mathcal{F}$ is a maximal intersecting family of $S$, for every $A \subseteq S$ exactly one of the sets $A$ or $S\setminus A$ is in $\mathcal{F}$. (we cannot have both as these sets in the family as they're disjoint).

This implies what you're asked to show.