View source for Set subtraction

{{Stub page|Cleanup and further expansion}}
{{Requires references}}
==Definition==
Given two sets, {{M|A}} and {{M|B}} we define ''set subtraction'' ({{AKA}}: ''relative complement''{{rMTH}}) as follows:
* {{M|1=A-B=\{x\in A\vert x\notin B\} }}
===Alternative forms===
{{Begin Inline Theorem}}
* {{M|1=A-B=(A^c\cup B)^c}}
{{Begin Inline Proof}}
{{Requires proof|Be bothered to do this}}
{{End Proof}}{{End Theorem}}
==Terminology==
* '''Relative complement'''<ref name="MTH"/>
** This comes from the idea of a [[complement]] of a subset of {{M|X}}, say {{M|A}} being just {{M|X-A}}, so if we have {{M|A,B\in\mathcal{P}(X)}} then {{M|A-B}} can be thought of as the complement of {{M|B}} if you consider it relative (to be in) {{M|A}}.
==Notations==
Other notations include:
* {{M|A\setminus B}}

==Trivial expressions for set subtraction==
{{Begin Inline Theorem}}
'''Claim:''' {{M|1=(A-B)-C=A-(B\cup C)}}
{{Begin Inline Proof}}
'''Proof:'''
* Note that {{M|1=A-B=(A^c\cup B)^c}} so {{M|1=(A-B)-C = ((A-B)^c\cup B)^c =(((A^c\cup B)^c)^c\cup C)^c}}
** But: {{M|1=(A^c)^c=A}} so:
*** {{M|1=(A-B)-C=(A^c\cup B\cup C)^c=(A^c\cup(B\cup C))^c=A-(B\cup C)}}
{{Todo|Make this proof neat}}
{{End Proof}}{{End Theorem}}

==See also==
* [[Set complement]]
==References==
<references/>
{{Set operations navbox|plain}}
{{Definition|Set Theory}}
{{Theorem Of|Set Theory}}
[[Category:Set operations]]