Problem Statement
Let $\mathcal{F}$ be a family of sets closed under taking subsets (i.e. if $B\subseteq A\in\mathcal{F}$ then $B\in \mathcal{F}$). There exists some element $x$ such that whenever $\mathcal{F}'\subseteq \mathcal{F}$ is an intersecting subfamily we have\[\lvert \mathcal{F}'\rvert \leq \lvert \{ A\in \mathcal{F} : x\in A\}\rvert.\]
Categories:
Combinatorics Intersecting Family
Progress
A problem of Chvátal [Ch74], who proved it replacing the closed under subsets condition with the (stronger) condition that, assuming all sets in $\mathcal{F}$ are subsets of $\{1,\ldots,n\}$, whenever $A\in \mathcal{F}$ and there is an injection $f:B\to A$ such that $x\leq f(x)$ for all $x\in B$, then $B\in \mathcal{F}$.Sterboul [St74] proved this when, letting $\mathcal{G}$ be the maximal sets (under inclusion) in $\mathcal{F}$, all sets in $\mathcal{G}$ have the same size, $\lvert A\cap B\rvert\leq 1$ for all $A\neq B\in \mathcal{G}$, and at least two sets in $\mathcal{G}$ have non-empty intersection.
Frankl and Kupavskii [FrKu23] have proved this when $\mathcal{F}$ has covering number $2$.
Borg [Bo11] has proposed a weighted generalisation of this conjecture, which he proves under certain additional assumptions.
Source: erdosproblems.com/701 | Last verified: January 16, 2026