Subset Relation is Ordering/General Result

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

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 relation $\subseteq$.

Then $\struct {\mathbb S, \subseteq}$ is an ordered set.


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:


\(\ds \forall T \in \mathbb S: \, \) \(\ds T\) \(\subseteq\) \(\ds T\) Set is Subset of Itself

So $\subseteq$ is reflexive.



\(\ds \forall S_1, S_2 \in \mathbb S: \, \) \(\ds S_1 \subseteq S_2 \land S_2 \subseteq S_1\) \(\iff\) \(\ds S_1 = S_2\) Definition 2 of Set Equality

So $\subseteq$ is antisymmetric.



\(\ds \forall S_1, S_2, S_3 \in \mathbb S: \, \) \(\ds S_1 \subseteq S_2 \land S_2 \subseteq S_3\) \(\implies\) \(\ds S_1 \subseteq S_3\) Subset Relation is Transitive

That is, $\subseteq$ is transitive.


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