Element of Ordinal is Ordinal

Theorem
Let $n$ be an ordinal.

Let $m \in n$.

Then $m$ is also an ordinal.

Proof
By the definition of ordinal, $n$ is transitive.

Thus $m \subseteq n$.

By the definition of ordinal, $n$ is strictly well-ordered by the epsilon relation, $\Epsilon$.

Since a subset of a strictly well-ordered set is strictly well-ordered, $m$ is strictly well-ordered by $\Epsilon$.

Let $x \in m$.

Let $y \in x$.

Since $n$ is an ordinal, it is transitive, so $x \in n$.

Thus since $y \in x$, $y \in n$.

Since $n$ is strictly well-ordered by $\Epsilon$, it is strictly totally ordered by $\Epsilon$.

Thus:


 * $y \in m$ or $m \in y$.

But if $m \in y$, then $y \in x \in m \in y$, so $y \in y$, contradicting the fact that $\Epsilon$ is a strict ordering, and hence antireflexive on $n$.

Thus $y \in m$.

As this holds for all such $x$ and $y$, $m$ is transitive.

As $m$ is transitive and is well-ordered by $\Epsilon$, it is by definition an ordinal.