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}} | ||
| + | {{Begin Inline Proof}} | ||
| + | {{Todo|Be bothered to do this}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
==Other names== | ==Other names== | ||
* Relative complement | * Relative complement | ||
| Line 9: | Line 15: | ||
* {{M|A\setminus B}} | * {{M|A\setminus B}} | ||
| − | == | + | ==Trivial expressions for set subtraction== |
{{Begin Inline Theorem}} | {{Begin Inline Theorem}} | ||
| − | + | '''Claim:''' {{M|1=(A-B)-C=A-(B\cup C)}} | |
{{Begin Inline Proof}} | {{Begin Inline Proof}} | ||
| − | {{Todo| | + | '''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}} | {{End Proof}}{{End Theorem}} | ||
| + | |||
==See also== | ==See also== | ||
* [[Set complement]] | * [[Set complement]] | ||
Revision as of 15:35, 9 December 2015
Contents
Definition
Given two sets, [ilmath]A[/ilmath] and [ilmath]B[/ilmath] we define set subtraction as follows:
- [ilmath]A-B=\{x\in A\vert x\notin B\}[/ilmath]
Equivalent definitions
- [ilmath]A-B=(A^c\cup B)^c[/ilmath]
TODO: Be bothered to do this
Other names
- Relative complement
- This comes from the fact that the complement of a subset of [ilmath]X[/ilmath], [ilmath]A[/ilmath] is just [ilmath]X-A[/ilmath]
Notations
Other notations include:
- [ilmath]A\setminus B[/ilmath]
Trivial expressions for set subtraction
Claim: [ilmath](A-B)-C=A-(B\cup C)[/ilmath]
Proof:
- Note that [ilmath]A-B=(A^c\cup B)^c[/ilmath] so [ilmath](A-B)-C = ((A-B)^c\cup B)^c =(((A^c\cup B)^c)^c\cup C)^c[/ilmath]
- But: [ilmath](A^c)^c=A[/ilmath] so:
- [ilmath](A-B)-C=(A^c\cup B\cup C)^c=(A^c\cup(B\cup C))^c=A-(B\cup C)[/ilmath]
- But: [ilmath](A^c)^c=A[/ilmath] so:
TODO: Make this proof neat
See also
References
TODO: Find references
