# Inductive set

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## Contents

## Definition

Let [ilmath]I[/ilmath] be a set. We call [ilmath]I[/ilmath] an *inductive set* if^{[1]} both of the following properties hold:

- [ilmath]\emptyset\in I[/ilmath] - often written [ilmath]0\in I[/ilmath] as [ilmath]0[/ilmath] is represented by the [ilmath]\emptyset[/ilmath] - and
- [ilmath]\forall n[n\in I\implies[/ilmath][ilmath]S(n)[/ilmath][ilmath]\in I][/ilmath] - often written as "if [ilmath]n\in I[/ilmath] then [ilmath](n+1)\in I[/ilmath]"
- [ilmath]S(x)[/ilmath] denotes the successor set of [ilmath]X[/ilmath]

**Caveat:**Note that this certainly describes the natural numbers as we require [ilmath]\emptyset\in I[/ilmath], so they're in there. The problem is that rule 2 seems to require that for every element [ilmath]n[/ilmath] that [ilmath]n\cup\{n\} [/ilmath] is in there too. - this seems to be intended^{[2]}

## See also

- The natural numbers
- The axiom of infinity - positing that an inductive set exists.