Let \(n > k > 1\) be integers, \([n] = \{1, \ldots, n\}\) the standard \(n\)-element set and \({[n]\choose k}\) the collection of all its \(k\)-subsets. The families \(\mathcal F_0, \ldots, \mathcal F_s \subset {[n]\choose k}\) are said to be cross-union if \(F_0 \cup \cdots \cup F_s \neq [n]\) for all choices of \(F_i \in \mathcal F_i\). It is known [13] that for \(n \leq k(s + 1)\) the geometric mean of \(|\mathcal F_i|\) is at most \({n - 1\choose k}\). We conjecture that the same is true for the arithmetic mean for the range \(ks < n < k(s + 1)\), \(s > s_0(k)\) (Conjecture 8.1) and prove this in several cases. The proof for the case \(n = ks + 2\) relies on a novel approach, a combination of shifting and Katona’s cyclic permutation method.

令\（n> k> 1 \）为整数，\（[n] = \ {1，\ ldots，n \} \）为标准\（n \）元素集，而\（{[n] \ choose k} \）所有\（k \）-子集的集合。家庭\（\ mathcal F_0，\ ldots，\ mathcal F_S \子集{[N] \选择k} \）被说成是交联合如果\（F_0 \杯\ cdots \杯F_S \ NEQ [N] \ ）的\（F_i \ in \ mathcal F_i \）中的所有选择。已知[13]对于\（n \ leq k（s + 1）\），\（| \ mathcal F_i | \）的几何平均值最多为 \（{n -1 \ choose k} \）。我们猜想范围的算术平均值也是如此 \（ks <n <k（s + 1）\），\（s> s_0（k）\）（猜想8.1）并在几种情况下证明这一点。\（n = ks + 2 \）的情况的证明依赖于一种新颖的方法，即位移法和卡托纳循环置换法的结合。