Ordinal Subset is Well-Ordered

Theorem

Let $S$ be a class.

Let every element of $S$ be an ordinal.

Then $\struct {S, \in}$ is a strict well-ordering.

Proof

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

By definition of subset, $S \subseteq \On$.

But by Class of All Ordinals is Ordinal $\On$ is an ordinal.

Therefore $\On$ is well-ordered by $\in$.

This means that $S$ is also well-ordered by $\in$.

$\blacksquare$