Successor Set of Ordinal is Ordinal

Theorem
Let $S$ be an ordinal.

Then its successor set $S\,^{+} = S \cup \left\{{S}\right\}$ is also an ordinal.

Proof
Since $S$ is transitive, it follows by Successor Set of Transitive Set is Transitive that $S\,^{+}$ is transitive.

We now have to show that $S\,^{+}$ is strictly well-ordered by the epsilon restriction $\Epsilon \! \restriction_{S^+}$.

So suppose that a subset $A \subseteq S\,^{+}$ is non-empty.

Then:

Let us refer to the above equation by the symbol $\left({1}\right)$.

We need to show that $A$ has a smallest element.

We first consider the case where $A \cap S$ is empty.

By equation $\left({1}\right)$, it follows that $A \cap \left\{{S}\right\}$ is non-empty (because $A$ is non-empty).

Therefore, $S \in A$. That is, $\left\{{S}\right\} \subseteq A$.

By Union with Empty Set and Intersection with Subset is Subset, equation $\left({1}\right)$ implies that $A \subseteq \left\{{S}\right\}$.

Therefore, $A = \left\{{S}\right\}$ by the definition of set equality.

So $S$ is the smallest element of $A$.

We now consider the case where $A \cap S$ is non-empty.

By Intersection Subset, $A \cap S \subseteq S$; by the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap S$.

Let $y \in A$.

If $y \in S$, then $y \in A \cap S$; therefore, by the definition of the smallest element, either $x \in y$ or $x = y$.

Otherwise, $y = S$, and so $x \in S = y$.

That is, $x$ is the smallest element of $A$.