# Relation between Two Ordinals/Corollary

## Corollary to Relation between Two Ordinals

Let $S$ and $T$ be ordinals.

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

## Proof 1

By Ordinal Membership is Trichotomy, either $S \in T$ or $T \in S$.

By definition, every element of an ordinal is an initial segment.

Hence the result.

$\blacksquare$

## Proof 2

If either $S \subset T$ or $T \subset S$ then we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.

Aiming for a contradiction, suppose $S \not \subset T$ and $T \not \subset S$.

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

By Intersection of Two Ordinals is Ordinal, $S \cap T$ is an ordinal.

So by Ordinal Subset of Ordinal is Initial Segment, we have:

- $S \cap T = S_a$ for some $a \in S$
- $S \cap T = S_b$ for some $b \in T$

Then:

- $a = S_a = S \cap T = T_b = b$

But $a \in S, b \in T$.

Thus $a = b = S \cap T$.

But $S \cap T = S_a$, so:

- $x \in S \cap T \implies x \subset a$

In particular, this means $a \subset a$, which is a contradiction.

So either $S \subset T$ or $T \subset S$, and again we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.

$\blacksquare$

## Proof 3

We have that $S \ne T$

Therefore, from Relation between Two Ordinals either $S \subset T$ or $T \subset S$.

By Ordering on Ordinal is Subset Relation or Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, either $S \in T$ or $T \in S$.

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

$\blacksquare$