Ordering is Equivalent to Subset Relation/Proof 1

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

Then let $T$ be defined as:
 * $T := \set {a^\prec: a \in S}$

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

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

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

Now suppose $x^\prec, y^\prec \in T: x^\prec = y^\prec$.

Then:
 * $\set {b \in S: b \preceq x} = \set {b \in S: b \preceq y}$

We have that:
 * $x \in x^\prec = y^\prec$ and $y \in y^\prec = x^\prec$

which means:
 * $x \preceq y$ and $y \preceq x$

So as an ordering is antisymmetric, we have $x = y$ and so $\phi$ is injective.

Hence by definition, $\phi$ is a bijection.

Now let $a_1 \preceq a_2$.

Then by definition:
 * $a_1 \in {a_2}^\prec$

Let $a_3 \in {a_1}^\prec$.

Then by definition:
 * $a_3 \preceq a_1$

As an ordering is transitive, it follows that:
 * $a_3 \preceq a_2$

and so:
 * $a_3 \in {a_2}^\prec$

So by definition of a subset:
 * ${a_1}^\prec \subseteq {a_2}^\prec$

Therefore, $\phi$ is order-preserving

Conversely, suppose that ${a_1}^\prec \subseteq {a_2}^\prec$.

Then, since $a_1 \in {a_1}^\prec$, also $a_1 \in {a_2}^\prec$ by definition of subset.

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

Hence it is seen that $\phi^{-1}$ is also order-preserving.

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