Minimally Inductive Class under Progressing Mapping is Well-Ordered under Subset Relation/Proof 1

Theorem

Let $M$ be a class which is minimally inductive under a progressing mapping $g$.

Let $x$ be a fixed point of $g$.

Then $M$ is well-ordered under the subset relation.

Proof

According to hypothesis, let $M$ be minimally inductive under $g$.

By Minimally Inductive Class under Progressing Mapping induces Nest:

$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

By Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element:

if $x$ is a fixed point of $g$, then $x$ is the greatest element of $M$.

Thus the conditions for Closed Class under Progressing Mapping Lemma hold.

In order to show that $M$ is well-ordered under the subset relation, it is to be shown that every non-empty subclass $A$ of $M$ has a smallest element.

Let $A$ be an arbitrary non-empty subclass of $N$.

Let $L$ be the class of all elements $y$ of $N$ such that $y$ is a proper subset of all elements of $A$.

If $\O \in A$, then $\O$ is the smallest element of $A$.

So, consider the case where $\O \notin A$.

Then $\O$ is a proper subset of every element of $A$.

Thus $\O \in L$ and so $L$ is non-empty.

By definition, $L$ is bounded by all elements of $A$.

Since $A$ is non-empty, $L$ is bounded by at least one element of $A$.

Hence by Non-Empty Bounded Subset of Minimally Inductive Class under Progressing Mapping has Greatest Element:

$L$ has a greatest element $x$.

Therefore, by Closed Class under Progressing Mapping Lemma, $\map g x$ is the smallest element of $A$.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.16$