Ordering is Equivalent to Subset Relation

Theorem
Let $$\left({S, \preceq}\right)$$ be a poset.

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.

Hence the subset relation is, up to order isomorphism, the only ordering there is on a given set of subsets.

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

For each $$a \in S$$:
 * Let $$S_a = \left({b \in S: b \preceq a}\right\}$$;
 * Let $$T = \left({S_a: a \in S}\right\}$$.

Let us define a mapping $$\phi: S \to T$$ as:
 * $$\phi \left({a}\right) = S_a$$.

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

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

Now suppose $$S_x, S_y \in T: S_x = S_y$$.

Then $$\left({b \in S: b \preceq x}\right\} = \left({b \in S: b \preceq y}\right\}$$.

We have that $$x \in S_x = S_y$$ and $$y \in S_y = S_x$$ 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, $$a_1 \in S_{a_2}$$.

Let $$a_3 \in S_{a_1}$$.

Then by definition, $$a_3 \preceq a_1$$.

As an ordering is transitive, it follows that $$a_3 \preceq a_2$$ and so $$a_3 \in S_{a_2}$$.

So by definition of a subset, $$S_{a_1} \subseteq S_{a_2}$$.

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