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During research, I hit a set theoretic claim that I could neither proof nor disproof.

Let $S_1,S_2,S_3$ be three set systems over the same finite universe $U$ such that

  1. $S_1,S_2,S_3$ are closed w.r.t. $\cap$ and $\cup$ and

  2. for each pair $u,v \in U$, there are two disjoint sets $X \in S_i, Y \in S_j$, $i,j \in \{1,2,3\}$, $i \neq j$, such that $u \in X$ and $v \in Y$.

Then, there is a partition of $U$ into at most three sets, each from a different set system.

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  • $\begingroup$ Is $U$ finite? Countable? $\endgroup$ – kodlu Apr 28 '15 at 1:45
  • $\begingroup$ This is not set theory but set systems. $\endgroup$ – domotorp Apr 28 '15 at 9:12
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They don't necessarily imply a partition, here is a counterexample. Each set in each $S_i$ will have an even size, while $U$ will have an odd number of elements, this already guarantees that a partition is impossible. $U$ is the nine vertices of a cycle of length $9$, whose edges are colored with $3$ colors, in an alternating way. With the union operation each color class generates a set system, these will be the $S_i$. In more detail, the generating edges are 01, 34, 67 for $S_1$, 12, 45, 78 for $S_2$ and 23, 56, 80 for $S_3$. It is easy to check that your separating condition holds.

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