User:Dfeuer/Subclass is Order-Like

Theorem

Let $A$, $B$, and $C$ be classes.

Then:

$A \subseteq A$

If $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$

If $A \subseteq B$ and $B \subseteq A$ then $A = B$.

Proof

By the law of identity, $\forall x: (x \in A \implies x \in A)$.

Thus by the definition of subclass, $A \subseteq A$.

Suppose that $A \subseteq B$ and $B \subseteq C$.

Let $x \in A$.

Then by the definition of subclass, $x \in B$.

Thus by the definition of subclass, $x \in C$.

As this holds for all $x \in A$, $A \subseteq C$.

Suppose that $A \subseteq B$ and $B \subseteq A$.

Then $\forall x: x \in A \iff x \in B$.

Thus by the axiom of extensionality, $A = B$.

$\blacksquare$

Remark

Although these properties look a lot like the properties that define an ordering, there is no "subclass relation", and hence no "subclass ordering". Specifically, a relation is a class of ordered pairs of sets; each component of an ordered pair must be a set. However, there is a subset relation on the universal class.