Let $latex \mathfrak{A}$ is a complete lattice. Let $latex a\in\mathfrak{A}$.

I will call *weak partitioning* of $latex a$ a set $latex S\in\mathscr{P}\mathfrak{A}\setminus\{0\}$ such that

$latex \bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall x\in S: x\cap^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}(S\setminus\{x\}) = 0$.

I will call *strong partitioning* of $latex a$ a set $latex S\in\mathscr{P}\mathfrak{A}\setminus\{0\}$ such that

$latex \bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall A,B\in\mathscr{P}S: (A\cap B=0\Rightarrow \bigcup{}^{\mathfrak{A}}A\cap{}^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}B = 0)$.

**Question:** Do exist complete lattices for which weak partitioning and strong partitioning are not the same?

If this conjecture (that it is the same for arbitrary complete lattices) is indeed false for arbitrary complete lattices, we must find the cases when it is true. (I strongly suspect that it is true for distributive complete lattices.)

## 3 thoughts on “Partitioning of a lattice element: a conjecture”