Ordering is Equivalent to Subset Relation

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

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

where:
 * $\struct {\mathbb S, \subseteq}$ 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.

Specifically:

Let
 * $\mathbb S := \set {a^\preceq: a \in S}$

where $a^\preceq$ is the lower closure of $a$.

That is:
 * $a^\preceq := \set {b \in S: b \preceq a}$

Let the mapping $\phi: S \to \mathbb S$ be defined as:
 * $\map \phi a = a^\preceq$

Then $\phi$ is an order isomorphism from $\struct {S, \preceq}$ to $\struct {\mathbb S, \subseteq}$.