论文标题
随机版本的Sperner定理的尖锐阈值
A sharp threshold for a random version of Sperner's Theorem
论文作者
论文摘要
布尔晶格$ \ MATHCAL {p}（n）$由$ [n] = \ {1，\ dots，n \} $部分订购的所有子集组成。 Sperner的定理指出，布尔晶格的最大抗小节是由中间层给出的：所有尺寸$ \ lfloor {n/2} \ rfloor $的集合的集合，或者，如果$ n $是奇怪的，则所有大小$ \ lceil \ lceil \ lceil \ lceil {n/2} \ rceil $的收集。给定$ p $，独立选择$ [n] $的每个子集，概率$ p $。我们表明，对于每一个常数$ p> 3/4 $，这些子集中最大的敌人也由中层给出，概率趋于$ 1 $ as $ n $倾向于无穷大。这款$ 3/4 $是最好的，我们还表征了每个常数$ p> 1/2 $的最大抗小选。我们的证明是基于Sapozhenko图形容器方法的一些新变化。
The Boolean lattice $\mathcal{P}(n)$ consists of all subsets of $[n] = \{1,\dots, n\}$ partially ordered under the containment relation. Sperner's Theorem states that the largest antichain of the Boolean lattice is given by a middle layer: the collection of all sets of size $\lfloor{n/2}\rfloor$, or also, if $n$ is odd, the collection of all sets of size $\lceil{n/2}\rceil$. Given $p$, choose each subset of $[n]$ with probability $p$ independently. We show that for every constant $p>3/4$, the largest antichain among these subsets is also given by a middle layer, with probability tending to $1$ as $n$ tends to infinity. This $3/4$ is best possible, and we also characterize the largest antichains for every constant $p>1/2$. Our proof is based on some new variations of Sapozhenko's graph container method.
