Subset Relation is Ordering
Contents |
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 a poset.
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
| \(\displaystyle \) | \(\displaystyle \forall T \in \mathbb S:\) | \(\displaystyle \) | \(\displaystyle T\) | \(\subseteq\) | \(\displaystyle T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Subset of Itself |
So $\subseteq$ is reflexive.
$\Box$
Antisymmetry
| \(\displaystyle \) | \(\displaystyle \forall S_1, S_2 \in \mathbb S:\) | \(\displaystyle \) | \(\displaystyle S_1 \subseteq S_2 \land S_2 \subseteq S_1\) | \(\iff\) | \(\displaystyle S_1 = S_2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equality of Sets |
So $\subseteq$ is antisymmetric.
$\Box$
Transitivity
| \(\displaystyle \) | \(\displaystyle \forall S_1, S_2, S_3 \in \mathbb S:\) | \(\displaystyle \) | \(\displaystyle S_1 \subseteq S_2 \land S_2 \subseteq S_3\) | \(\implies\) | \(\displaystyle S_1 \subseteq S_3\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Subsets Transitive |
So $\subseteq$ is transitive.
$\Box$
So we have shown that $\subseteq$ is an ordering on $\mathbb S$.
$\blacksquare$
Sources
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1$: Theorem $1.1$
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 7$: Example $7.1$
- Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.5$
