论文标题
关于$ t $的尺寸 - $ k $ - 无链家庭
On the sizes of $t$-intersecting $k$-chain-free families
论文作者
论文摘要
设置系统$ \ MATHCAL {F} $是$ t $ - \ textit {Intersecting}，如果其每对元素的相交的大小至少具有至少$ t $。设置系统$ \ MATHCAL {F} $是$ K $ - \ textit {sperner}，如果它不包含长度$ k+1 $的链条。 我们的主要结果是：假设$ k $和$ t $是固定的正整数，其中$ n+t $甚至使用$ t \ le n $，而$ n $也足够大。如果$ \ Mathcal {f} \ subseteq 2^{[n]} $是$ t $ - 缩放$ k $ -sperner家族，则$ | | \ nathcal {f} | $的大小的大小$ k $ layers的大小，尺寸$（n+t）/2，\ ldots $ k $ layers lays $（n+t）$ n+ldots $ n+k-1（n n+k-k-k-k）。 这是最好的。 $ n+t $奇数时的情况保持开放。
A set system $\mathcal{F}$ is $t$-\textit{intersecting}, if the size of the intersection of every pair of its elements has size at least $t$. A set system $\mathcal{F}$ is $k$-\textit{Sperner}, if it does not contain a chain of length $k+1$. Our main result is the following: Suppose that $k$ and $t$ are fixed positive integers, where $n+t$ is even with $t\le n$ and $n$ is large enough. If $\mathcal{F}\subseteq 2^{[n]}$ is a $t$-intersecting $k$-Sperner family, then $|\mathcal{F}|$ has size at most the size of the sum of $k$ layers, of sizes $(n+t)/2,\ldots, (n+t)/2+k-1$. This bound is best possible. The case when $n+t$ is odd remains open.
