Minimal Element of an Ordinal

Theorem
The minimal element of any nonempty ordinal is the empty set.

That is, if $S$ is a nonempty ordinal, $\bigcap S = \O$

Proof
Let $S$ be an ordinal.

Let the minimal element of $S$ be $s_0$.

This exists by dint of an ordinal being a woset.

From Ordering on Ordinal is Subset Relation, $S$ is well-ordered by $\subseteq$.

So, by definition of an ordinal:
 * $s_0 = \set {s \in S: s \subset s_0}$

But as $s_0$ is minimal, there are no elements of $S$ which are a subset of it.

So:
 * $\set {s \in S: s \subset s_0} = \O$

Hence the result.