Set of Subsets of Element of Minimally Inductive Class is Finite

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

Let $x \in M$.

Let $S$ be the set of all $y \in M$ such that $y \subseteq x$.

Then $S$ is finite.

Proof
The proof proceeds by general induction.

For all $x \in M$, let $\map P x$ be the proposition:
 * $\set {y \in M: y \subseteq x}$ is finite.

Basis for the Induction
We have that:
 * $\set {y \in M: y \subseteq \O} = \O$

From Cardinality of Empty Set, $\O$ is finite.

Thus $\map P \O$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true.

So this is the induction hypothesis:
 * $\set {y \in M: y \subseteq x}$ is finite.

from which it is to be shown that:
 * $\set {y \in M: y \subseteq \map g x}$ is finite.

Induction Step
This is the induction step:

Let $S = \set {y \in M: y \subseteq x}$.

By the induction hypothesis, $S$ is finite.

Consider the set:
 * $S' = \set {y \in M: y \subseteq \map g x}$

By definition of progressing mapping:


 * $x \subseteq \map g x$

Hence it follows that:
 * $S \subseteq S'$

Let $y \in S' \setminus S$.

Then:
 * $y \notin S$

and so:
 * $y \nsubseteq x$

From Minimally Inductive Class under Progressing Mapping induces Nest:
 * $\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

As it is not the case that $y \subseteq x$, it must be that:
 * $\map g x \subseteq y$

From sandwich principle:
 * $x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$

As $x \ne y$ (or then $y \subseteq x$), it must be the case that:
 * $y = \map g x$

So:
 * $S' \setminus S = \map g x$

and so is finite, containing $1$ element.

We have that:
 * $S' = S \cup \paren {S' \setminus S}$

and it follows from Union of Finite Sets is Finite that $S'$ is finite.

So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction.

Therefore:
 * $\forall x \in M: \set {y \in M: y \subseteq x}$ is finite.