Abstract

Let [n]:={1,2,…,n}$[n]:=\left\{1,2,\dots ,n\right\}$$[n]:=\left\{1,2,\dots ,n\right\}$ and let a k-set denote a set of cardinality k. A family of sets is union-closed (UC) if the union of any two sets in the family is also in the family. Frankl’s conjecture states that for any nonempty UC family F⊆2[n]$\mathcal{F}\subseteq 2^{[n]}$ such that $\mathcal{F}\ne \{\varnothing \}$, there exists an element $i\in [n]$ that is contained in at least half the sets of $\mathcal{F}$, where $2^{[n]}$ denotes the power set on $[n]$. The 3-sets conjecture of Morris states that the smallest number of distinct 3-sets (whose union is an n-set) that ensure Frankl’s conjecture is satisfied (in an element of the n-set) for any UC family that contains them is $\lfloor n/2\rfloor +1$ for all $n\ge 4$. For an UC family $\mathcal{A}\subseteq 2^{[n]}$, Poonen’s theorem characterizes the existence of weights on $[n]$ which ensure all UC families that contain $\mathcal{A}$ satisfy Frankl’s conjecture, however the determination of such weights for specific $\mathcal{A}$ is nontrivial even for small n. We classify families of 3-sets on $n\le 9$ using a polyhedral interpretation of Poonen’s theorem and exact rational integer programming. This yields a proof of the 3-sets conjecture.

摘要

让[ n ] : = { 1 , 2 , … , n }$[n]:=\left\{1,2,\dots \dots ,n\right\}$$[n]:=\left\{1,2,\dots ,n\right\}$并令k集表示基数k的集合。如果一个集合族中任意两个集合的并集也在该族中，则该集合族是并闭的 (UC)。弗兰克尔猜想指出，对于任何非空 UC 族F⊆2[ ] _$\mathcal{F}\subseteq 2^{[n]}$这样$\mathcal{F}\ne \{\varnothing \}$，存在一个元素$我\epsilon [n]$至少包含在一半的集合中$\mathcal{F}$， 在哪里$2^{[n]}$表示电源已开启$[n]$。Morris 的 3 集猜想指出，对于包含它们的任何 UC 族，确保满足 Frankl 猜想（在n集的元素中）的不同 3 集（其并集是n集）的最小数量是$\lfloor n/2\rfloor +1$对全部$n\ge 4$。对于 UC 家庭$\mathcal{A}\subseteq 2^{[n]}$, Poonen 定理描述了权重的存在性$[n]$这确保所有 UC 系列包含$\mathcal{A}$满足弗兰克尔猜想，但是这种权重的确定要针对具体情况$\mathcal{A}$即使对于小n来说也是不平凡的。我们将 3 组的家庭分类为$n\le 9$使用 Poonen 定理的多面体解释和精确有理整数规划。这产生了三集猜想的证明。