If $X$ is a non-empty set and ${\cal A}, {\cal B}$ are covers, then we say that ${\cal A} \leq_{\text{fin}} {\cal B}$ if for all $A\in {\cal A}$ there is $B\in{\cal B}$ such that $A\subseteq B$ and we say that ${\cal A}$ *refines* ${\cal B}$.

Let $G=(V,E)$ be a simple, undirected graph. A *clique decomposition* is a set ${\cal C} \subseteq {\cal P}(V)$ such that

- $\emptyset \notin {\cal C}$,
- $C\in {\cal C}$ and $x\neq y \in C$ imply that $\{x,y\}\in E$ (that is every member of ${\cal C}$ is a clique),
- $\bigcup {\cal C} = V$, and
- $e=\{x,y\} \in E \implies$ there is $C\in{\cal C}$ such that $x,y\in C$.

The collection of all edges plus the isolated points is a clique decomposition, and it is easily seen to be minimal with respect to refinement amongst all clique decompositions. Does the collection of clique decompositions also have maximal elements with respect to refinement? (The answer is yes for finite graphs, but I don't know about infinite graphs.)