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 $g$.

Corollary

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.

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.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.8$