Subset Relation is Ordering

Theorem
Let $$S$$ be a set.

Let $$\mathcal{P} \left({S}\right)$$ be the power set of $$S$$.

Let $$\mathbb{S} \subseteq \mathcal{P} \left({S}\right)$$ be any subset of $$\mathcal{P} \left({S}\right)$$, that is, an arbitrary set of subsets of $$S$$.

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

In other words, let $$\left({\mathbb{S}; \subseteq}\right)$$ be the relational structure defined on $$\mathbb{S}$$ by the relation $$\subseteq$$.

Then $$\left({\mathbb{S}; \subseteq}\right)$$ is a poset.

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
$$

So $$\subseteq$$ is reflexive.

Antisymmetry
$$

So $$\subseteq$$ is antisymmetric.

Transitivity
$$

So $$\subseteq$$ is transitive.

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