${{\mathcal C}}$ has $2^n$ ST-sets. However, as the following technical results show, ${{\mathcal C}}$ will have at most $n$ *maximal* ST-sets, and they will partition $\mathcal{X}$ i.e. they are mutually disjoint and entirely cover $\mathcal{X}$. Several of the proofs have been deferred to the appendix.

\[lem:stcompute\] Let $\mathcal{C}$ be a set of clusters on ${{\mathcal X}}$ and let $S_1 \neq S_2$ be two ST-sets of $\mathcal{C}$. If $S_1 \cap S_2 \neq \emptyset$ then $S_1 \cup S_2$ is an ST-set.

Deferred to the appendix.

\[cor:maxstdisjoint\] Let $\mathcal{C}$ be a set of clusters on ${{\mathcal X}}$ and let $S_1 \neq S_2$ be two maximal ST-sets of $\mathcal{C}$. Then $S_1 \cap S_2 = \emptyset$.

\[cor:mostn\] Let $\mathcal{C}$ be a set of clusters on ${{\mathcal X}}$. Then there are at most $n$ maximal ST-sets with respect to ${{\mathcal C}}$, they are uniquely defined and they partition ${{\mathcal X}}$.

The disjointness of maximal ST-sets guarantees that there at most $n$ of them. Consider the set ${{\mathcal S}}$ of all (necessarily disjoint) maximal ST-sets of ${{\mathcal X}}$. Suppose the maximal ST-sets in ${{\mathcal S}}$ do not entirely cover ${{\mathcal X}}$. Then there is some $x \in {{\mathcal X}}$ which is disjoint from all maximal ST-sets in 