Principle of General Induction/Minimally Closed Class
Theorem
Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $b \in M$ such that $M$ is minimally closed under $g$ with respect to $b$.
Let $P: M \to \set {\T, \F}$ be a propositional function on $M$.
Suppose that:
- $(1): \quad \map P b = \T$
- $(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$
Then:
- $\forall x \in M: \map P x = \T$
Proof
We are given that $M$ is minimally closed under $g$ with respect to $b$.
That is, $M$ is closed under $g$ with the extra property that $M$ has no proper class containing $b$ which is also closed under $g$.
Let $P$ be a propositional function on $M$ which has the properties specified:
- $(1): \quad \map P b = \T$
- $(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$
Let $S \subseteq M$ be the subclass of $M$ defined as:
- $S = \set {x \in M: \map P x}$
We have:
\(\ds \map P b\) | \(=\) | \(\ds \T\) | ||||||||||||
\(\ds b\) | \(\in\) | \(\ds M\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(\in\) | \(\ds S\) |
By definition, the class $S$ of all elements of $M$ such that $\map P x = \T$ is closed under $g$.
But because $M$ is minimally closed under $g$ with respect to $b$, if $b \in S$ then $S$ contains all elements of $M$.
That is:
- $S = M$
and so:
- $\forall x \in M: \map P x = \T$
as we were to show.
$\blacksquare$
Terminology
Basis for the Induction
The step that shows that the propositional function $P$ holds for the distinguished $b \in M$ is called the basis for the induction.
Induction Hypothesis
The assumption made that $\map P x$ is true for some $x \in M$ is called the induction hypothesis.
Induction Step
The step which shows that $\map P x = \T \implies \map P {\map g x} = \T$ is called the induction step.