Strengthening the union-closed sets conjecture

Christopher Bouchard

Published 2023-10-03Version 1

A family of sets $\mathcal{F}$ is union-closed if $X,Y \in \mathcal{F} \implies X \cup Y \in \mathcal{F}$. Let $\mathcal{A} \neq \{\emptyset\}$ be a finite union-closed family of sets. The union-closed sets conjecture, also called Frankl's conjecture, states that there exists an element of $\bigcup_{A \in \mathcal{A}} A $ that appears in at least $\frac{|\mathcal{A}|}{2}$ members of $\mathcal{A}$. Let $\mathcal{A}_B=\{A\in\mathcal{A}|A \cap B = B\}$ and $\mathcal{A}_{\bar{B}}=\{A\in\mathcal{A}|A \cap B = \emptyset\}$ where $B \subseteq \bigcup_{A \in \mathcal{A}}A$. Further, let ${S \choose k}$ be the set of all $k$-element subsets of a set $S$, and $[n]=\{1,2,\cdots,n\}=\bigcup_{A \in \mathcal{A}}A$. The union-closed sets conjecture can then be stated as $\exists B \in {[n] \choose 1}$ $|\mathcal{A}_{B}| \geq |\mathcal{A}_{\bar{B}}|$. With this notation, we introduce the stronger conjecture that $\forall x \in [n]$ $\exists B \in {[n] \choose n+1-x}$ $|\mathcal{A}_B| \geq |\mathcal{A}_{\bar{B}}|$, and we prove the new conjecture for $x \in [\lceil \frac{n}{3} \rceil + 1]$, where $\lceil \frac{n}{3} \rceil$ is the smallest integer greater than or equal to $\frac{n}{3}$. Other related conjectures are investigated.