Difference between revisions of "Unions"

From Grad Wiki
Jump to navigation Jump to search
(Created page with "== Definition == Let <math>X</math> and <math>Y</math> be subsets of some universal set <math>U</math>. The '''''union of <math>X</math> and <math>Y</math>''''', written <math...")
 
 
Line 1: Line 1:
 
== Definition ==
 
== Definition ==
 +
[[File:Union.png|thumb|The Venn diagram displays two sets <math>X</math> and <math>Y</math> with the union <math>X\cup Y</math> shaded]]
 
Let <math>X</math> and <math>Y</math> be subsets of some universal set <math>U</math>. The '''''union of <math>X</math> and <math>Y</math>''''', written <math>X\cup Y</math>, is the set of all <math>x</math> in <math>U</math> which are in at least one of the sets <math>X</math> or <math>Y</math>.<br />
 
Let <math>X</math> and <math>Y</math> be subsets of some universal set <math>U</math>. The '''''union of <math>X</math> and <math>Y</math>''''', written <math>X\cup Y</math>, is the set of all <math>x</math> in <math>U</math> which are in at least one of the sets <math>X</math> or <math>Y</math>.<br />
 
Symbolically, <math>X\cup Y=\lbrace x\in U : x\in X \text{ or } x\in Y\rbrace</math>.
 
Symbolically, <math>X\cup Y=\lbrace x\in U : x\in X \text{ or } x\in Y\rbrace</math>.

Latest revision as of 13:05, 29 June 2015

Definition

The Venn diagram displays two sets and with the union shaded

Let and be subsets of some universal set . The union of and , written , is the set of all in which are in at least one of the sets or .
Symbolically, .

Examples

Example 1

Determine the union of the sets and .

Solution By definition, we wish to find the set of all elements which are in at least one of the two sets. Thus, we will collect all unique elements into a set as long as they appear once or more between the sets. Our solution is .

Example 2

Prove that for any sets and , .

Proof Let . We wish to show that , so this means showing that such that or . Since , we have that and by definition of we know that . Thus the “or” statement is true and hence . This shows that .