Subset Relation is Ordering/General Result
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
- 1963: George F. Simmons: Introduction to Topology and Modern Analysis ... (previous) ... (next): $\S 1$: Sets and Set Inclusion
- 1967: Garrett Birkhoff: Lattice Theory (3rd ed.) ... (next): $\S \text I.1$
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1.5$: Relations