Inductive Set under Mapping has Minimally Inductive Subset

Theorem
Let $A$ be an inductive class under a mapping $g$.

Let $A$ be a set.

Then there exists some subset $S$ of $A$ such that $S$ is minimally inductive under $g$.

Proof
Let $S$ be the set defined as:
 * $S = \set {s: \exists t \in A: t = \map g s} \cup \set \O$

By definition, we have that:
 * $\O \in S$
 * $s \in S \implies \map g s \in S$

So, by definition, $S$ is inductive under $g$.

Because $A$ is inductive under $g$:
 * $\O \in A$

and so by definition of subset:
 * $S \subseteq A$

It remains to be shown that $S$ is minimally inductive under $g$.

Let $T \subseteq S$ such that $T$ is inductive under $g$.

It will be demonstrated by the Principle of General Induction that $T = S$.

That is:
 * $\forall x \in T: x \in S$

Let $P: T \to \set {\T, \F}$ be the propositional function defined as:
 * $\forall x \in T: \map P x \iff x \in S$

First it is noted that as $T$ is inductive under $g$:
 * $\O \in T$

But $\O \in S$.

Hence we have that:
 * $\map P \O = \T$

This is the basis for the induction.

Now suppose $x \in T$ such that $\map P x = \T$.

That is:
 * $x \in S$

This is the induction hypothesis.

The induction step is to show that $\map P {\map g x} = \T$.

That is:
 * $\map g x \in S$

Indeed, as $T$ is inductive under $g$:
 * $x \in T \implies \map g x \in T$

But again, :
 * $\map g x \in S$

Hence we have shown that:
 * $\map P x = \T \implies \map P {\map g x} = \T$

Hence it follows by the Principle of General Induction:
 * $\forall x \in T: \map P x = \T$

That is:
 * $\forall x \in T: x \in S$

That is:
 * $T = S$

Thus a subset of $S$ which is inductive under $g$ is necessarily equal to $S$.

Hence $S$ has no proper subset which is inductive under $g$.

Hence, by definition, $S$ is minimally inductive under $g$.

The result follows.