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 $2 \ \text {(b)}$