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.

General Result

Let $\mathbb S$ be a set of sets.

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.

Class Theoretical Result

Let $C$ be a class.

Then the subset relation $\subseteq$ is an ordering on $C$.

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.

$\Box$

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.

$\Box$

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.

$\Box$

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

$\blacksquare$

Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $2 \ \text {(b)}$