Ordering is Equivalent to Subset Relation/Proof 2

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.

Specifically:

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

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

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

Let the mapping $\phi: S \to \mathbb S$ be defined as:
 * $\phi \left({a}\right) = a^\preceq$

Then $\phi$ is an order isomorphism from $\left({S, \preceq}\right)$ to $\left({\mathbb S, \subseteq}\right)$.

Proof
From Subset Relation is Ordering, we have that $\left({\mathbb S, \subseteq}\right)$ is an ordered set.

We are to show that $\phi$ is an order isomorphism.

$\phi$ is clearly surjective, as every $a^\preceq$ is defined from some $a \in S$.

By the Ordering Equivalent to Subset Relation: Lemma, $\phi$ is order-preserving.

Suppose that ${a_1}^\preceq \subseteq {a_2}^\preceq$.

We have that:
 * $a_1 \in {a_1}^\preceq$

Thus by definition of subset:
 * $a_1 \in {a_2}^\preceq$

By definition of ${a_2}^\preceq$:
 * $a_1 \preceq a_2$

Thus $\phi$ is also order-reflecting.

Thus it follows that $\phi$ is an order isomorphism between $\left({S, \preceq}\right)$ and $\left({\mathbb S, \subseteq}\right)$.