Subset Relation is Ordering

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Theorem

Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.


Let $\mathbb S \subseteq \mathcal P \left({S}\right)$ be any subset of $\mathcal P \left({S}\right)$, that is, an arbitrary set of subsets of $S$.

Then $\subseteq$ is an ordering on $\mathbb S$.


In other words, let $\left({\mathbb S, \subseteq}\right)$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$.

Then $\left({\mathbb S, \subseteq}\right)$ is an ordered set.


General Result

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


Then $\subseteq$ is an ordering on $\mathbb S$.


In other words, let $\left({\mathbb S, \subseteq}\right)$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$.

Then $\left({\mathbb S, \subseteq}\right)$ 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

\(\displaystyle \forall T \in \mathbb S: \ \ \) \(\displaystyle T\) \(\subseteq\) \(\displaystyle T\) $\quad$ Set is Subset of Itself $\quad$

So $\subseteq$ is reflexive.

$\Box$


Antisymmetry

\(\displaystyle \forall S_1, S_2 \in \mathbb S: \ \ \) \(\displaystyle S_1 \subseteq S_2 \land S_2 \subseteq S_1\) \(\iff\) \(\displaystyle S_1 = S_2\) $\quad$ Definition of Set Equality $\quad$

So $\subseteq$ is antisymmetric.

$\Box$


Transitivity

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

That is, $\subseteq$ is transitive.

$\Box$


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

$\blacksquare$


Sources