Set which is Superinductive under Progressing Mapping has Fixed Point

Theorem
Let $S$ be a set.

Let $g: S \to S$ be a progressing mapping on $S$.

Let $S$ be superinductive under $g$.

Then there exists $x \in S$ such that $x$ is a fixed point of $g$.

Proof
Let us assume the hypothesis.

Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.

From:
 * Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
 * Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions

it follows by definition that $M$ is minimally superinductive under $g$.

Hence by definition $M$ is a $g$-Tower.

By Intersection of Non-Empty Class is Set, $M$ is a set.

Thus by Union of $g$-Tower is Greatest Element and Unique Fixed Point:
 * $\ds \bigcup M$ is a fixed point of $g$
 * $\ds \bigcup M \in M$

Because $M \subseteq S$ it follows that:
 * $\ds \bigcup M \in S$

and so $\ds \bigcup M$ is the $x$ that is being proved to exist.