Subset Relation is Ordering

Theorem
Let $S$ be a set.

Let $\powerset S$ be the power set of $S$.

Let $\mathbb S \subseteq \powerset S$ be any subset of $\powerset S$, that is, an arbitrary set of subsets of $S$.

Then $\subseteq$ is an ordering on $\mathbb S$.

In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the subset relation $\subseteq$.

Then $\struct {\mathbb S, \subseteq}$ is an ordered set.

Proof
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.

So, checking in turn each of the criteria for an ordering:

Reflexivity
From Subset Relation is Reflexive:


 * $\forall T \in \mathbb S: T \subseteq T$

So $\subseteq$ is reflexive.

Antisymmetry
From Subset Relation is Antisymmetric:
 * $\forall S_1, S_2 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_1 \iff S_1 = S_2$

So $\subseteq$ is antisymmetric.

Transitivity
From Subset Relation is Transitive:


 * $\forall S_1, S_2, S_3 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_3 \implies S_1 \subseteq S_3$

That is, $\subseteq$ is transitive.

So we have shown that $\subseteq$ is an ordering on $\mathbb S$.