# Definition:Inductive Class

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## 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 $x^+ = x \cup \set x$ |

### Inductive Set

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 $x^+ = x \cup \set x$ |

## General Definition

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 $g$: | \(\ds \forall x:\) | \(\ds \paren {x \in A \implies \map g x \in A} \) |

## Also see

- Results about
**inductive classes**can be found**here**.

## Sources

- 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 $2.1$