Set which is Superinductive under Progressing Mapping has Fixed Point/Corollary

Corollary to Set which is Superinductive under Progressing Mapping has Fixed Point
Let $S$ be a non-empty set of sets.

Let $g: S \to S$ be a progressing mapping on $S$ such that:
 * $(1): \quad S$ is closed under $g$
 * $(2): \quad S$ is closed under chain unions.

Let $b \in S$.

Then there exists $x \in S$ such that:
 * $b \subseteq x$

and:
 * $\map g x = x$

Proof
Let us assume the hypothesis.


 * Case $b = \O$

This case reduces to Set which is Superinductive under Progressing Mapping has Fixed Point.

Hence the result follows for $b = \O$.


 * Case $b \ne \O$

Let $S_b$ be the set of all elements $x$ of $S$ such that $b \subseteq x$, together with $\O$:
 * $S_b = \set {x \in S: b \subseteq x} \cup \set \O$

It is clear by inspection that $S_b$ is closed under chain unions.

Let us define the mapping $g': S_b \to S_b$ as follows:
 * $\forall x \in S_b: \map {g'} x = \begin{cases} b & : x = \O \\ \map g x & : x \ne \O \end{cases}$

Then:
 * $g'$ is a progressing mapping on $S_b$
 * $S_b$ is closed under $g'$
 * $\O \in S_b$

So $S_b$ is superinductive under $g'$.

Hence by Set which is Superinductive under Progressing Mapping has Fixed Point:
 * $\exists x \in S_b: x = \map {g'} x$

For such an $x$ we have that $x \ne \O$ because $\map {g'} \O = b$.

Hence:
 * $b \subseteq x$

Also because $x \ne \O$:
 * $\map {g'} x = \map g x$

and so:
 * $\map g x = x$

Hence:
 * $b \subseteq x$

and:
 * $\map g x = x$

and the proof is complete.