Relation between Two Ordinals

Theorem
Let $S$ and $T$ be ordinals.

Then either $S \subseteq T$ or $T \subseteq S$.

Corollary
If $S \ne T$, then either $S$ is an initial segment of $T$, or vice versa.

Proof
Aiming for a contradiction, suppose that the claim is false.

That is, by De Morgan's laws:
 * $\left({\neg \left({S \subseteq T}\right)}\right) \land \left({\neg \left({T \subseteq S}\right)}\right)$

where $\neg$ denotes logical not and $\land$ denotes logical and.

Now from Intersection Subset, we have $S \cap T \subset S$ and $S \cap T \subset T$; equality does not hold by Intersection with Subset is Subset.

Note that by Intersection of Two Ordinals is Ordinal‎, $S \cap T$ is an ordinal.

So by Ordering on an Ordinal is Subset Relation or Ordinal Proper Subset Membership, we have:
 * $S \cap T \in S$
 * $S \cap T \in T$

But then $S \cap T \in S \cap T$, which contradicts the definition that the $\in$-relation is the strict well-ordering on (the ordinal) $S \cap T$.

Alternatively, one could again use Ordering on an Ordinal is Subset Relation to conclude that $S \cap T \subset S \cap T$, also a contradiction.

Proof of Corollary
If $S \ne T$, then the main theorem implies that either $S \subset T$ or $T \subset S$.

By Ordering on an Ordinal is Subset Relation or Ordinal Proper Subset Membership, either $S \in T$ or $T \in S$.

By definition, every element of an ordinal is an initial segment; hence the result.