Definition:Inductive Class

Definition

Let $A$ be a class.

Then $A$ is inductive if and only if:

$(1)$	$:$	$A$ contains the empty set:		$\ds \quad \O \in A $
$(2)$	$:$	$A$ is closed under the successor mapping:	$\ds \forall x:$	$\ds \paren {x \in A \implies x^+ \in A} $	where $x^+$ is the successor of $x$
					That is, where $x^+ = x \cup \set x$

The same definition can be applied when $A$ is a set:

Let $S$ be a set of sets.

Then $S$ is inductive if and only if:

$(1)$	$:$	$S$ contains the empty set:		$\ds \quad \O \in S $
$(2)$	$:$	$S$ is closed under the successor mapping:	$\ds \forall x:$	$\ds \paren {x \in S \implies x^+ \in S} $	where $x^+$ is the successor of $x$
					That is, where $x^+ = x \cup \set x$

Let $A$ be a class.

Let $g: A \to A$ be a mapping on $A$.

Then $A$ is inductive under $g$ if and only if:

$(1)$	$:$		$A$ contains the empty set:		$\ds \quad \O \in A $
$(2)$	$:$		$A$ is closed under $g$:	$\ds \forall x:$	$\ds \paren {x \in A \implies \map g x \in A} $

2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers: Definition $2.1$

\((1)\)	$:$		$A$ contains the empty set:		\(\ds \quad \O \in A \)
\((2)\)	$:$		$A$ is closed under $g$:	\(\ds \forall x:\)	\(\ds \paren {x \in A \implies \map g x \in A} \)