Element of Ordinal is Ordinal/Proof 1

Proof
Let $\On$ denote the class of all ordinals.

From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.

By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.

We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
 * $0 := \O$

Vacuously, every element of $\O$ is an ordinal.

Let $\alpha \in \On$ be an ordinal.

Let us assume that every element of $\alpha$ is an ordinal.

Let $\alpha^+$ denote the successor to $\alpha$.

Because $\On$ is superinductive:
 * $\alpha \in \On \implies \alpha^+ \in \On$

By definition of successor mapping:
 * $\alpha^+ = \alpha \cup \set \alpha$

Then every element of $\alpha^+$ is either an element of $\alpha$ or $\alpha$ itself.

We have that:
 * $\alpha$ is an ordinal
 * every element of $\alpha$ is an ordinal.

That is, $\alpha^+$ is an ordinal.

Let $C$ be a chain of ordinals such that:
 * each element $\alpha$ of $C$ has the property that each element of $\alpha$ is itself an ordinal.

Let $\beta \in \bigcup C$.

Then $\beta \in \alpha$ for some $\alpha \in C$.

Hence, by definition of $C$, $\beta$ is an ordinal which has the property that each element of $\beta$ is itself an ordinal.

It follows by the Principle of Superinduction that every element of an ordinal is an ordinal.

Let $\alpha \in \On$ be an arbitrary ordinal.

Let $\beta \subseteq \alpha$.

From Class of All Ordinals is Well-Ordered by Subset Relation, $\beta$ is an ordinal

Hence from the above:
 * $\beta \in \alpha$

As $\alpha$ is arbitrary:
 * $\forall \alpha \in \On: \beta \subseteq \alpha \implies \beta \in \alpha$

proving that $\On$ is transitive.