Relation between Two Ordinals

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Theorem

Let $S$ and $T$ be ordinals.

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


Corollary

Let $S$ and $T$ be ordinals.

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: Conjunction of Negations:

$\left({\neg \left({S \subseteq T}\right)}\right) \land \left({\neg \left({T \subseteq S}\right)}\right)$

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

Now from Intersection is Subset, we have $S \cap T \subseteq S$ and $S \cap T \subseteq T$.

Also, by Intersection with Subset is Subset, we have $S \cap T \ne S$ and $S \cap T \ne T$.

That is, $S \cap T \subsetneq S$ and $S \cap T \subsetneq T$.


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

So by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, we have:

$S \cap T \in S$
$S \cap T \in T$

But then $S \cap T \in S \cap T$, which contradicts Ordinal is not Element of Itself.

$\blacksquare$