# Ordering is Equivalent to Subset Relation/Proof 2

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

## Proof

First a lemma:

### Lemma

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

Then:

- $\forall a_1, a_2 \in S: \paren {a_1 \preceq a_2 \implies {a_1}^\preceq \subseteq {a_2}^\preceq}$

where ${a_1}^\preceq$ denotes the lower closure of $a_1$.

$\Box$

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

$\blacksquare$