<p>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.</p>

CC BY-SA 3.0
2015-04-28T09:42:50.343Z
2015-04-28T09:52:43.093Z
2015-04-28T09:52:43.093Z
domotorp
<p>During research, I hit a set theoretic claim that I could neither proof nor disproof.</p>
<blockquote>
<p>Let <span class="math-container">$S_1,S_2,S_3$</span> be three set systems over the same finite universe <span class="math-container">$U$</span> such that</p>
<ol>
<li><p><span class="math-container">$S_1,S_2,S_3$</span> are closed w.r.t. <span class="math-container">$\cap$</span> and <span class="math-container">$\cup$</span> and</p>
</li>
<li><p>for each pair <span class="math-container">$u,v \in U$</span>, there are two disjoint sets <span class="math-container">$X \in S_i, Y \in S_j$</span>, <span class="math-container">$i,j \in \{1,2,3\}$</span>, <span class="math-container">$i \neq j$</span>, such that <span class="math-container">$u \in X$</span> and <span class="math-container">$v \in Y$</span>.</p>
</li>
</ol>
<p>Then, there is a partition of <span class="math-container">$U$</span> into at most three sets, each from a different set system.</p>
</blockquote>

CC BY-SA 3.0
2015-04-28T01:45:14.890Z
Is $U$ finite? Countable?
kodlu
CC BY-SA 3.0
2015-04-28T09:12:31.493Z
This is not set theory but set systems.
domotorp
<set-system>
Do these set systems imply a partition?
Oliver Witt
# Do these set systems imply a partition?

(Asked by: username_0 on 2015-04-27)

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.

**Tags:** set-system
> **Comments:**
> **username_1**: Is $U$ finite? Countable?  
> **username_2**: This is not set theory but set systems.  

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## 1 answers

### Answer by username_2 on 2015-04-28. Score: 5 | Accepted answer

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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