TITLE: On pairwise non-disjointness of a sequence of finite sets
QUESTION [5 upvotes]: Let $\{A_n\}_{n=1}^\infty $ be a sequence of finite sets such that no two sets of the sequence are pairwise disjoint ; then must there exist a finite set $F$ such that no two sets in the sequence $\{ A_n \cap F\}_{n=1}^\infty $ are pairwise disjoint ?

REPLY [2 votes]: Counterexample. For $n\in\mathbb N$ define
$$A_n=\{x\in\mathbb N:x\le n\ \text{ and }\ x\equiv n\!\!\!\!\!\pmod2\}\cup\{n+1\}.$$
The sets $A_1,A_2,A_3,\dots$ are pairwise intersecting: for $m\lt n,$ if $m$ and $n$ have the same parity, then $m\in A_m\cap A_n;$ otherwise, $m+1\in A_m\cap A_n;$ in either case, $A_m\cap A_n\ne\emptyset.$
If $F\subseteq\mathbb N$ is such that the sets $A_1\cap F,A_2\cap F,A_3\cap F,\dots$ are pairwise intersecting, then we must have $F=\mathbb N,$ since $A_n\cap A_{n+1}=\{n+1\}$ and $A_1\cap A_3=\{1\}.$