Subset Relation is Ordering/General Result

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Theorem

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.


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