Subset Relation on Power Set is Partial Ordering

Theorem
Let $$S$$ be a set.

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

Let $$\left({\mathcal{P} \left({S}\right); \subseteq}\right)$$ be the relational structure defined on $$\mathcal{P} \left({S}\right)$$ by the relation $$\subseteq$$.

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

The ordering $$\subseteq$$ is partial iff $$S$$ is neither empty nor a singleton; otherwise it is total.

Proof
From Subset Relation on Set of Subsets is an Ordering, we have that $$\subseteq$$ is an ordering on any set of subsets of a given set.

Suppose $$S$$ is neither a singleton nor the empty set.

Then $$\exists a, b \in S$$ such that $$a \ne b$$.

Then $$\left\{{a}\right\} \in \mathcal{P} \left({S}\right)$$ and $$\left\{{b}\right\} \in \mathcal{P} \left({S}\right)$$.

However, $$\left\{{a}\right\} \not \subseteq \left\{{b}\right\}$$ and $$\left\{{b}\right\} \not \subseteq \left\{{a}\right\}$$.

So by definition, $$\subseteq$$ is a partial ordering.

Now suppose $$S = \varnothing$$.

Then $$\mathcal{P} \left({S}\right) = \left\{{\varnothing}\right\}$$ and, by Empty Set Subset of All, $$\varnothing \subseteq \varnothing$$.

Hence, trivially, $$\subseteq$$ is a total ordering on $$\mathcal{P} \left({S}\right)$$.

Now suppose $$S$$ is a singleton: let $$S = \left\{{a}\right\}$$.

Then $$\mathcal{P} \left({S}\right) = \left\{{\varnothing, \left\{{a}\right\}}\right\}$$.

So there are only two elements of $$\mathcal{P} \left({S}\right)$$, and we see that $$\varnothing \subseteq \left\{{a}\right\}$$ from Empty Set Subset of All.

So, trivially again, $$\subseteq$$ is a total ordering on $$\mathcal{P} \left({S}\right)$$.