Let $\mathcal F\subset 2^{[n]}$ be a family in which any three sets have non-empty intersection and any two sets have at least $38$ elements in common. The nearly best possible bound $|\mathcal F|\le 2^{n-2}$ is proved. We believe that $38$ can be replaced by $3$ and provide a simple-looking conjecture that would imply this.

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不兼容的交点属性

令 $\mathcal F\subset 2^{[n]}$ 是一个族，其中任何三个集合都具有非空交集，并且任何两个集合至少有 $38$ 个共同元素。证明了几乎最好的边界 $|\mathcal F|\le 2^{n-2}$。我们相信 $38$ 可以替换为 $3$ 并提供一个简单的猜想来暗示这一点。