Difference between revisions of "Set subtraction"

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(Created page with "==Definition== Given two sets, {{M|A}} and {{M|B}} we define ''set subtraction'' as follows: * {{M|1=A-B=\{x\in A\vert x\notin B\} }} ==Other names== * Relative complement **...")
 
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Given two sets, {{M|A}} and {{M|B}} we define ''set subtraction'' as follows:
 
Given two sets, {{M|A}} and {{M|B}} we define ''set subtraction'' as follows:
 
* {{M|1=A-B=\{x\in A\vert x\notin B\} }}
 
* {{M|1=A-B=\{x\in A\vert x\notin B\} }}
 +
===Equivalent definitions===
 +
{{Begin Inline Theorem}}
 +
* {{M|1=A-B=(A^c\cup B)^c}}
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{{Begin Inline Proof}}
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{{Todo|Be bothered to do this}}
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{{End Proof}}{{End Theorem}}
 
==Other names==
 
==Other names==
 
* Relative complement
 
* Relative complement
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* {{M|A\setminus B}}
 
* {{M|A\setminus B}}
  
==Expressions that are equal to set subtraction==
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==Trivial expressions for set subtraction==
 
{{Begin Inline Theorem}}
 
{{Begin Inline Theorem}}
* {{M|1=A-B=(A^c\cup B)^c}}
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'''Claim:''' {{M|1=(A-B)-C=A-(B\cup C)}}
 
{{Begin Inline Proof}}
 
{{Begin Inline Proof}}
{{Todo|Be bothered to do this}}
+
'''Proof:'''
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* 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}}
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** But: {{M|1=(A^c)^c=A}} so:
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*** {{M|1=(A-B)-C=(A^c\cup B\cup C)^c=(A^c\cup(B\cup C))^c=A-(B\cup C)}}
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{{Todo|Make this proof neat}}
 
{{End Proof}}{{End Theorem}}
 
{{End Proof}}{{End Theorem}}
 +
 
==See also==
 
==See also==
 
* [[Set complement]]
 
* [[Set complement]]

Revision as of 15:35, 9 December 2015

Definition

Given two sets, A and B we define set subtraction as follows:

  • AB={xA|xB}

Equivalent definitions

[Expand]

  • AB=(AcB)c

Other names

  • Relative complement
    • This comes from the fact that the complement of a subset of X, A is just XA

Notations

Other notations include:

  • AB

Trivial expressions for set subtraction

[Expand]

Claim: (AB)C=A(BC)


See also

References



TODO: Find references