Difference between revisions of "Inductive set"
From Maths
m (Definition) |
(See also) |
||
| Line 7: | Line 7: | ||
#* {{M|S(x)}} denotes the [[successor set]] of {{M|X}} | #* {{M|S(x)}} denotes the [[successor set]] of {{M|X}} | ||
{{Caveat|Note that this certainly describes the natural numbers as we require {{M|\emptyset\in I}}, so they're in there. The problem is that rule 2 seems to require that for every element {{M|n}} that {{M|n\cup\{n\} }} is in there too.}} - this seems to be intended{{rSTTJ}} | {{Caveat|Note that this certainly describes the natural numbers as we require {{M|\emptyset\in I}}, so they're in there. The problem is that rule 2 seems to require that for every element {{M|n}} that {{M|n\cup\{n\} }} is in there too.}} - this seems to be intended{{rSTTJ}} | ||
| + | ==See also== | ||
| + | * [[The natural numbers]] | ||
| + | * [[The axiom of infinity]] - positing that an inductive set exists. | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Set Theory}} | {{Definition|Set Theory}} | ||
Latest revision as of 15:56, 3 February 2017
Stub grade: A*
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Needs another reference
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.
