Difference between revisions of "Intersections"
		
		
		
		
		
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== Definition ==  | == Definition ==  | ||
| − | [[File:Intersections.png|thumb|The Venn diagram displays two sets <math>X</math> and <math>Y</math> with the intersection <math>X\cap Y</math> shaded]]  | + | [[File:Intersections.png|thumb|The Venn diagram displays two sets <math style="vertical-align: 0px">X</math> and <math style="vertical-align: 0px">Y</math> with the intersection <math style="vertical-align: -1px">X\cap Y</math> shaded]]  | 
| − | Let <math>X</math> and <math>Y</math> be subsets of some universal set <math>U</math>. The '''''intersection of <math>X</math> and <math>Y</math>''''', written <math>X\cap Y</math>, is the set of all <math>x</math> in <math>U</math> which are in both of the sets <math>X</math> and <math>Y</math>.<br />  | + | Let <math style="vertical-align: 0px">X</math> and <math style="vertical-align: 0px">Y</math> be subsets of some universal set  <math style="vertical-align: 0px">U</math>. The '''''intersection of <math style="vertical-align: 0px">X</math> and <math style="vertical-align: 0px">Y</math>''''', written <math style="vertical-align: -1px">X\cap Y</math>, is the set of all <math style="vertical-align: 0px">x</math> in  <math style="vertical-align: 0px">U</math> which are in both of the sets <math style="vertical-align: 0px">X</math> and <math style="vertical-align: 0px">Y</math>.<br />  | 
| − | Symbolically, <math>X\cap Y=\lbrace x\in U : x\in X \text{ and } x\in Y\rbrace</math>.  | + | Symbolically, <math style="vertical-align: -5px">X\cap Y=\lbrace x\in U : x\in X \text{ and } x\in Y\rbrace</math>.  | 
== Examples ==  | == Examples ==  | ||
Latest revision as of 11:47, 1 July 2015
Definition
Let  and  be subsets of some universal set  . The intersection of  and , written , is the set of all  in   which are in both of the sets  and .
Symbolically, .
Examples
Example 1
Determine the intersection of the sets and .
Solution. By definition, we wish to find the set of all elements which are in both of the sets. The only such element is . Thus, our solution is .
Example 2
Prove that for any sets and , .
Proof. Let . That is, and . In particular, since we have that .