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 a priori $\O \in S$.
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, a priori:
- $\map g x \in S$
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.
$\blacksquare$
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: Exercise $4.2$