Ordering is Equivalent to Subset Relation

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Then there exists a set $\mathbb S$ of subsets of $S$ such that:
 * $\left({S, \preceq}\right) \cong \left({\mathbb S, \subseteq}\right)$

where:
 * $\left({\mathbb S, \subseteq}\right)$ is the relational structure consisting of $\mathbb S$ and the subset relation
 * $\cong$ denotes order isomorphism.

Hence any ordering on a set can be modelled uniquely by a set of subsets of that set under the subset relation.