Union of Sets

Introduction

A set is a well-defined collection of objects represented as {A, B, C, D, X, Y, Z} etc. The elements of sets are separated by commas and enclosed within brackets { }. Subsets define the concepts of relations and functions. We will always get a new number after performing mathematical operations of addition, subtraction and multiplication on a pair of numbers. Similarly, we will get a new set after performing set operations on a pair of sets. There are different set operations along with different properties. Union of sets is similar to operations of addition. The intersection of sets contains common elements of both sets.

Union of sets

Union of sets is one of the set operations similar to arithmetic operations. On taking common elements a single time, the union of sets A and B contains all the elements of A and B. Symbolically, the union of sets is represented as ∪(union). A ∪ B means A union B.

A ∪ B = {x: x ∈A or x ∈B}

if x is present in both sets A and B,

Example: If A = {1, 3, 5, 7} and B = {5, 7, 9, 11}. Find A ∪ B.

Solution: A ∪ B = {1, 3, 5, 7, 9, 11}. The common elements 5 and 7 are written once.

Properties of operation of union of sets

Commutative law 

A ∪ B = B ∪ A

Example: P = {a, b, c, d}, Q = {1, 2, 3, 4}

P ∪ Q = {a, b, c, d} U {1, 2, 3, 4}

P ∪ Q = {a, b, c, d, 1, 2, 3, 4}

Q ∪ P = {1, 2, 3, 4} U {a, b, c, d} 

Q ∪ P = {1, 2, 3, 4, a, b, c, d} 

Associative law

(A ∪ B) ∪ C = A ∪ (B ∪ C)

Example: A = {2, 3, 4}, B = {2, 5, 6}, C = {1, 6, 9}

(A ∪ B) = {2, 3, 4} U {2, 5, 6} = {2, 3, 4, 5, 6}

(A ∪ B) ∪ C = {2, 3, 4, 5, 6} U {1, 6, 9} = {1, 2, 3, 4, 5, 6, 9}

(B ∪ C) = {2, 5, 6} ∪ {1, 6, 9} = {1, 2, 5, 6, 9}

A ∪ (B ∪ C) = {2, 3, 4} ∪ {1, 2, 5, 6, 9} = {1, 2, 3, 4, 5, 6, 9}

Law of the identity element 

A ∪ φ = A (φ is an empty set)

Example: A = {a, b, c}

Thus, A ∪ ∅ = {a, b, c} ∪ {} = {a, b, c}

Idempotent law

A ∪ A = A

Example: A = {1, 3, 5, 7}

Thus, A ∪ A = {1, 3, 5, 7} ∪ {1, 3, 5, 7} = {1, 3, 5, 7} = A

Law of U

U ∪ A = U(U is the universal set)

Example: A = {a, e} and U = {a, b, c, d, e}

then A ∪ U = {a, e} ∪ {a, b, c, d, e} = {a, b, c, d, e} = U

Intersection of sets

The intersection of sets A and B forms a set that contains all the common elements in both sets. Symbolically, the intersection of sets is represented as ∩ (intersection).

Represent it as A ∩ B = {x: x ∈ A and x ∈ B}

Example:

Let X = {Abhor, Anika, Akash} be the set of students of class XI who are in the school cricket team.

Let Y = {Anika, Deep, Harsh} be the set of students from Class XI who are on the school football team. Find X ∩ Y.

Solution: Anika is the only element common in both sets, so X ∩ Y = {Anika}.

If A and B are two sets with A ∩ B = φ, they are called disjoint sets. It means they have no common elements between them.

Example:

A = {2, 4, 6} and B = {1, 3, 5}. Since there is no common element in A and B, they are disjoint sets.

Properties of set operations of the intersection of sets

Commutative law

A ∩ B = B ∩ A

Associative law

(A ∩ B) ∩ C = A ∩ (B ∩ C)

Law of φ and U

φ ∩ A = φ (φ is an empty set)

U ∩ A = A(U is the universal set)

Idempotent law

A ∩ A = A

Distributive law

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Practical problems on union and intersection of two sets

Let A and B be the finite sets. If A ∩ B = φ, then n (A ∪ B) = n (A) + n (B)

It indicates that the elements of A ∪ B are present either in A or B.

If A and B are finite sets,

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

So,

n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)

     = n (A – B) + n (A ∩ B) + n (B – A) + n (A ∩ B) – n (A ∩ B)

     = n (A) + n (B) – n (A ∩ B)

If A, B and C are finite sets, then

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B∩C) – n (A∩C) + n (A ∩ B∩C)

We know from 2, that

n (A ∪ B ∪ C) = n (A) + n (BUC) – n [A ∩ (BUC)]

= n (A) + n (B) + n (C) – n (B∩C) – n [A ∩ (BUC)]

We know that A ∩ (BUC) = (A ∩ B) ∪ (A∩C), so

 n [A ∩ (BUC)] = n (A ∩ B) + n (A∩C) – n [(A ∩ B) ∩ (A∩C)] 

= n (A ∩ B) + n (A∩C) – n (A ∩ B∩C)

So,

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (BC) – n (A-C) + n (A ∩ BC)

Example: If X and Y are two sets and X ∪ Y has 40 elements. X has 18 elements and Y has 22 elements. How many elements does X ∩ Y have?

Solution: n (X ∪ Y) = 40, n (X) = 28, n (Y) = 22, n (X ∩ Y) =?

According to the formula, n (X ∪ Y) = n (X) + n (Y) – n (X ∩ Y)

We find that,

n (X ∩ Y) = n (X) + n (Y) – n (X ∪ Y)

                  = 28 + 22 – 40

                  = 10

Answer: There are 10 elements in X ∩ Y

Conclusion

A set is a well-defined collection of objects. The number of the sets is 2n (n=number of elements in the set). The union of two sets A and B forms a set C containing all the elements present in A and B. The intersection of two sets is the set of all the elements which belong to both A and B.