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 is 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 $\mathcal P \left({S}\right)$.

$\blacksquare$

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 14$: Order
• A.N. Kolmogorov and S.V. Fomin‎: Introductory Real Analysis (1968): $\S 3.1$: Example $3$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.5$