Let $U$ is a complete lattice with least element 0.


*Weak partitioning* is a collection $S$ of nonempty subsets of $U$ such that $\forall x\in S: x\cap\bigcup(S\setminus\{x\})=0$.


*Strong partitioning* is a collection $S$ of nonempty subsets of $U$ such that $\forall A,B\in PS:(A\cap B=\emptyset \Rightarrow \bigcup A\cap\bigcup B=0)$.


Easy to show that every strong partitioning is weak partitioning.


Is weak and strong partitioning the same?


If not, under which additional conditions these are the same?



Weak partitioning vs. strong partitioning
If the lattice $U$ satisfies the meet distributive law
$$x \wedge \bigvee\_{i \in I} y\_i = \bigvee\_{i \in I} x \wedge y\_i$$
where $(y\_i)\_{i \in I}$ is an arbitrary collection of elements of $U$, then "weak partitioning" implies "strong partitioning." More precisely, you only need the above to hold when the right hand side is $0$.


An example of a complete lattice where weak and strong partitioning are inequivalent is the lattice $U$ consisting of all closed subsets of $\{1,\frac12,\frac13,\ldots,0\}$ (as a subspace of $\mathbb{R}$) and the collection $S = \{\{\frac1n\}: n \geq 1\}$. The weak-partitioning property is easily verified since the points $\frac1n$ are isolated. The strong partitioning property fails for the two sets $A = \{\{\frac1{2n}\}: n \geq 1\}$ and $B = \{\{\frac1{2n+1}\} : n \geq 0\}$, for example, since $\bigvee A = \overline{\bigcup A}$ and $\bigvee B = \overline{\bigcup B}$ both contain the point $0$.


PS: In your formulation of weak and strong partitioning, I interpret $S$ as a collection of *nonzero elements* of $U$, since "nonempty subsets" doesn't make much sense in context.


