Disjoint refinement of a set of sets Say I have some set $F$ of sets. This is obviously a cover of $\cup F$. Is there a general algorithm that makes "this" cover disjoint? That is, a set $F'$ of pairwise disjoint sets with the properties $\cup F = \cup F'$ and $(\forall A \in F')(\exists B \in F)(A \subset B)$?
This question covers the countable case.

Ok, let's do it by Zorn's lemma, and see whether this answer has appeared by the time I finish typing it...
Let $Z$ be the set of ordered pairs $(S,f)$ such that $S\subset F$, $f:S\to\bigcup F$, $f(x)\subset x$ for every $x\in S$, the sets $f(x)$ for $x\in S$ are pairwise disjoint, and $\bigcup_{x\in S}f(x)=\bigcup S$.
Order $Z$ by extension: $(S,f)\le(S',f')$ if $S\subset S'$ and $f'|_S=f'$.
Zorn's lemma says $Z$ has a maximal element $(S,f)$. An argument like the construction in the other proof shows that $S=F$.
