Ordering is Equivalent to Subset Relation/Proof 2

Theorem
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 $\struct {\mathbb S, \subseteq}$.

Proof
From Subset Relation is Ordering, we have that $\struct {\mathbb S, \subseteq}$ 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 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 $\struct {S, \preceq}$ and $\struct {\mathbb S, \subseteq}$.