For a subfamily ${F}\subseteq 2^{[n]}$ of the Boolean lattice, consider the graph $G_{F}$ on ${F}$ based on the pairwise inclusion relations among its members. Given a positive integer $t$, how large can ${F}$ be before $G_{F}$ must contain some component of order greater than $t$? For $t=1$, this question was answered exactly almost a century ago by Sperner: the size of a middle layer of the Boolean lattice. For $t=2^n$, this question is trivial. We are interested in what happens between these two extremes. For $t=2^{g}$ with $g=g(n)$ being any integer function that satisfies $g(n)=o(n/\log n)$ as $n\to\infty$, we give an asymptotically sharp answer to the above question: not much larger than the size of a middle layer. This constitutes a nontrivial generalisation of Sperner's theorem. We do so by a reduction to a Tur\'an-type problem for rainbow cycles in properly edge-coloured graphs. Among other results, we also give a sharp answer to the question, how large can ${F}$ be before $G_{F}$ must be connected?

翻译：对于布尔格$2^{[n]}$的子族${F}\subseteq 2^{[n]}$，考虑基于其成员间两两包含关系所定义的图$G_{F}$。给定正整数$t$，在$G_{F}$必然包含阶数大于$t$的连通分支之前，${F}$的规模最大能达到多少？当$t=1$时，该问题已由Sperner在近一个世纪前给出精确解答：即布尔格中间层的规模。当$t=2^n$时，该问题平凡无趣。我们关注的是这两个极端情形之间的过渡状态。对于$t=2^{g}$（其中$g=g(n)$是满足$n\to\infty$时$g(n)=o(n/\log n)$的任意整数函数），我们给出了上述问题的渐近紧确界：其规模不会显著超过中间层的规模。这构成了Sperner定理的一个非平凡推广。我们的证明方法是通过归约到恰当边着色图中彩虹圈的Turán型问题。除上述结果外，我们还对“在$G_{F}$必须连通之前${F}$的最大规模”这一问题给出了紧确解答。