论文标题
塔尔伯特定理的简单证明用于相交的分离集
A simple proof of Talbot's theorem for intersecting separated sets
论文作者
论文摘要
一个子集$ a $ a $ of $ [n] = \ {1，\ dots，n \} $是$ k $ - 分配的，如果在圆圈上考虑$ [n] $的元素时，在$ a $的任何两个元素之间，至少有$ k $ elements $ k $ elements $ [n] $不在$ a $中。如果每两个集合中的每两个集合中的每两个集合{a} $相交，则$ \ mathcal {a} $正在相交。我们简短而简单地证明了塔尔伯特（Talbot，2003）的显着结果，并指出，如果$ n \ geq（k + 1）r $和$ \ mathcal {a} $是$ k $ r $ r $ r $ - element子集的$ r $ - element子集的$ [n] $的相交家族，则\ leq \ binom {n -kr -1} {r -1} $。这是最好的。
A subset $A$ of $[n] = \{1, \dots, n\}$ is $k$-separated if, when the elements of $[n]$ are considered on a circle, between any two elements of $A$ there are at least $k$ elements of $[n]$ that are not in $A$. A family $\mathcal{A}$ of sets is intersecting if every two sets in $\mathcal{A}$ intersect. We give a short and simple proof of a remarkable result of Talbot (2003), stating that if $n \geq (k + 1)r$ and $\mathcal{A}$ is an intersecting family of $k$-separated $r$-element subsets of $[n]$, then $|\mathcal{A}| \leq \binom{n - kr - 1}{r - 1}$. This bound is best possible.
