2.  Union Symbol ∪ If A and B are sets, their union is equal to all elements in both A & B A = {1,2,3,4} B = {2,4,5,6,7,8} A ∪ B = {1,2,3,4,5,6,7,8}

3.  Venn Diagram Square with U The U indicates the universal set. The universal set is all objects in a particular discussion.

4.  The intersection ∩ is the set of all elements that belong to both A & B. A ={1,2,3,4} B = {2,4,5,6,7,8} A & B have the numbers 2 and 4 in common A∩B = {2,4}

5.  Disjoint sets are sets with no common elements.  A = {1,2,3}  B = {4,5,6}  A ∩ B = {}

6.  Compliment If A and B are 2 sets, the compliment of B, with respect to A, is the set of all elements that belong to A but not to B. Compliment of B = A – B denoted B Compliment of A = B – A denoted A

7.  The universal set includes A The compliment of A is denoted A and is indicated by the red shaded area. The complement of A with regard to the universal set.

8.  Symmetric Difference ⊕ S = {a,b,c,d} T= {a,c,e,f,g} S⊕T = {b,d,e,f,g} What is in S that is not in T What is in T that is not in S Formula S⊕T = (S-T) ∪ (T-S)

9.  Cardinality of a finite set is the number of distinct elements in a set. The cardinality of a finite Set A is denoted |A|. A = {a,b,c} |A| = 3 There are 3 distinct elements in set A  A finite set has a limited size.  An infinite set has no size limitation.

10. Be familiar with the properties on pages 8 & 9  Cummutative  Associative  Distributive  Idempotent  Properties of the compliment  Properties of a universal set  Properties of the empty set

11.  The Addition Principle 1. Disjoint |A ∪ B| = |A| + |B| 2. Find A ∪ B Cardinality of A = 10 Cardinality of B = 7 A ∩B = 4 A ∪ B counts the intersection twice, therefore, subtract A ∩B. | A ∪ B |=|A|+|B|-| A ∩B | | A ∪ B |=10 + 7 – 4 | A ∪ B |= 13

12.  The formula for three sets |A∪B∪C|=|A|+|B|+|C|-|A∩B|-|B∩C|-|A∩C|+|A∩B∩C| People were asked what mode of transportation they used: Bus, Train, Automobile. Can select any or all. |B| = 30, |T| = 35, |A| = 100 |B ∩ T| = 15, |B ∩ A| = 15, |T ∩ A| = 20, |B ∩T ∩A| = 5 |B∪T∪A|=|B|+|T|+|A|-|B∩T|-|B∩A|-|T∩A|+|B∩T∩A| |B∪T∪A|=30+35+100-15 -15 - 20 +5 |B∪T∪A|=120