Relation between Two Ordinals
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
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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.
$\blacksquare$
