2 typo

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.

1

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 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.