# Set Theory | Operations, Unions, Intersections

In this post of set theory, we will discuss the basic operations that we can operate of sets. Just like we do mathematical operations like addition, subtraction, multiplication etc. on numbers, we can perform these functions on the sets.

But remember, set operations has its own rules and we have to study them before getting into the problem solving. We have tried to cover the theory in as simple manner as possible, hopefully after reading this post you will have fair understanding of the concepts.

## Operation on sets

We will start with the basic concepts and then advance towards the complex ones. I have tried to explain the concept with the help of examples, i hope you will enjoy the theory.

### Union of sets

In this operation we basically merge the elements of the set.

Let there are two sets A & B, and you have been asked to form union of both
A = { 1, 2, 3, 4 }
B = { 3, 4, 5, 6}
The union of A & B (A U B) is = { 1, 2, 3, 4, 5, 6}

The union of A & B contains:
a. All the elements of set A
b. All the elements of set B
c. Common elements of A & B

The union of A & B can be represented in the form of following venn diagram

Symbol of Union
Th symbol to represent union is ” U” and the union between two sets can be technically defined as follows:
A ∪ B = { x : x ∈A or x ∈B }
The union of A & B is expressed in the form of symbol x, where all the elements either belong to set A or set B

Example 01
Let A = { 2, 4, 6, 8}
& B = { 6, 8, 10, 12}
Find A U B

Solution
A U B = { 2, 4, 6, 8, 10, 12}

Example 02
Let X = { Ram, Geeta, Akbar}
& Y = { Geeta, Preet, Karan}
Find X U Y

Solution
X U Y = {Ram, Geeta, Akbar, Preet, Karan}

#### Properties of Union

1. Commutative Law
The law states that even if we swap the set in the operation of union, the end result will be the same
i.e. A U B = B U A

Example 01
Let A = { 1, 2, 3, 4}
& B = ( 3, 4, 5, 6}
You will see that for ( A U B ) and ( B U A) operation, the end result will be same
i.e A U B = B U A = { 1, 2, 3, 4, 5, 6}

2. Associative Property of Union
The property states that you can do union calculation of set forming different groups with the same end result
i.e. (A U B) U C= A U (B U C)

Example 01
Let A={1, 2, 3, 4}
B = { 3, 4, 5, 6}
& C = { 6, 7, 8}

When you do he analysis you will find that the result of ( A U B ) U C = A U (B U C) is the same
(A U B) U C = A U (B U C) = {1, 2, 3, 4, 5, 6, 7, 8}

3. Union with Null
If you perform union with the empty set, the final result will be the same
A ∪ φ = A

Example
A = { 1, 2, 3}
φ = { }
Then A U φ = { 1, 2, 3}

4. Idempotent Law
Applying union operation to same set will lead to same set
A U A = A
If you apply union to the same set A, then the final result after operation will be the set A

5. Law of union
Union of set with universal set will result in universal set as final result
(Universal Set) U A = Universal Set

### Intersection of sets

When you perform intersection of two sets, you are basically taking common elements between the two sets.

Example
A = { 1, 2, 3, 4}
B = { 3, 4, 5, 6}
The intersection of set A & B is:
A ∩ B = { 3, 4 }

The {3, 4} is the answer as they are the common element present in both sets.

The symbol for intersection is “∩”
Technically we write intersection set as A ∩ B = {x : x ∈ A and x ∈ B}

#### Properties of Set Intersection

1. Commutative Law of intersection
A B = B A
The law states that even if we swap the location of set in the operation, the end result will be the same.

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

A ∩ B = 3
B ∩ A = 3

Hence A∩B = B∩A

2. Associative law of intersection of set
( A ∩ B ) ∩ C = A ∩ ( B ∩ C )
In the intersection calculation of multiple set, if we change the order of calculation of intersection the result will be the same.

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

A ∩ B = {3, 4}
( A ∩ B ) ∩ C = {3}

B ∩ C = {3}
A ∩ ( B ∩ C ) = {3}

Hence you can see that ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

3. Intersection with Null
φ ∩ A = φ
Intersection of any set with empty set will result in empty set

4. Intersection with union
U ∩ A = A
Intersection of any set with union will result in same set

5. Idempotent Law
Intersection of set with itself will result in same set
A ∩ A = A

6. Distributive Law of Sets
A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )

Here we are just opening up the brackets as we used to do in basic addition or multiplication operation
a ( b + c ) = ab + ac

Please remember these properties as it would help us to solve worksheets questions.

### Difference of Sets

As we do subtraction of numbers in algebra, we can do the same operations in sets also.
If we subtract one set from another we will remove the common elements present in each set.

Example 01
A = { 1, 2, 3, 4, 5}
B = { 4, 6, 9 }
Find (A – B) and (B-A)

Solution
(A – B) = {1, 2, 3, 5}
Here we remove common element of B present in A

(B – A) = {6, 9}
Here we remove common element of A present in B

Example 02
V = {a, e, i, o, u}
B = {a, i, k, u}
Find (V – B) and (B – V)

Solution
(V-B) = { e, o}
Here we remove common element of B from set V

(B – V) = {k}
Here we remove all the common element of V from set B

### Complement of Sets

Let U be the universal set containing all the element ( as shown by the rectangle box above)
And A is a set containing some element (shown by white circle)

Then the complement of set A is the leftover when we remove elements of set A from Universal set ( here shown by blue color part of the rectangle)

The complement of any set A is denoted by A′

Example 01
Let U = {1, 2, 3, 4, 5, 6}
A = {2, 3}
B = {3, 4, 5}
Find A′, B′ , A′ ∩ B′

Solution
A′ = {1, 4, 5, 6}
We have removed element of A from set U

B′ = {1, 2, 6}
We have removed element of B from set U

A′ ∩ B′ = {1, 6}
We have taken common element from A′ & B′

#### Properties of complement of sets

1. Complement laws:
(i) A ∪ A′ = U
Union of set & its complement will result in universal set

Example 01
Let U = { 11, 12, 13, 14, 15}
& A = {11, 13}
Then the complement of A will be A′ = { 12, 14, 15}

Now let us do the union of A & A′ = { 11, 13} U {12, 14, 15} ==> {11, 12, 13, 14, 15}
You can see that union of A & A′ is equal to the universal set

2. A ∩ A′ = φ
Intersection of set with its complement will lead to Empty set

3. De Morgan’s law

(i) (A ∪ B)´ = A′ ∩ B′
The complement of A U B is equal to intersection of complement of individual sets

(ii) (A ∩ B)′ = A′ ∪ B′
The complement of A ∩ B is equal to the union of complement of individual sets

Both these laws are extremely important as it help us to solve lot off questions from the exercise. From all the laws we have till now discussed, you have to remember this one as a priority.

4. Double Complement
Double complement of any set will result in same set
(A′)′ = A

### Questions

(01) If A = { 3, 5, 7, 9, 11 },
B = {7, 9, 11, 13},
C = {11, 13, 15} and
D = {15, 17};
Find:
(i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D (iv) A ∩ C (v) A ∩ (B ∪ C)

Solution
(i) A ∩ B = { 7, 9, 11}
We will take elements which are common between both A & B

(ii) B ∩ C = { 11, 13}
we will take elements which are common between both B & C

(iii) A ∩ C ∩ D
=> (A ∩ C) ∩ D => {11} ∩ {15, 17} => ( \phi )

(iv) A ∩ C = {11}
11 is the only element which is common between A & C

(v) A ∩ (B ∪ C)
(B ∪ C) => {7, 9, 11, 13, 15}
A ∩ (B ∪ C) = {7, 9, 11}

(02) Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }
(ii) { a, e, i, o, u } and { c, d, e, f }

Solution
Two sets are disjoint when there is no common element between them

(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }
There are two sets given
A = { 1, 2, 3, 4}
B = { 4, 5, 6}
These sets are not disjoint because there is common element 4 between them

(ii) { a, e, i, o, u } and { c, d, e, f }
you can see in both sets element “e” is common
Hence the sets are not disjoint

(03) Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}
Find A′, B′ , A′ ∩ B′, A ∪ B

Solution
(i) A′ = { 1, 4, 5, 6}
(ii) B′ = { 1, 2, 6}
(iii) A′ ∩ B′ = {1, 6}
(iv) A ∪ B = { 2, 3, 4, 5}

(04) Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}
(ii) { x : x is an odd natural number }
(iii) {x : x is a positive multiple of 3}

Solution
Universal Set (U) = ( x : 0 < x < infinity)
This Universal set is the collection of natural numbers

(i) A = {x : x is an even natural number}
This set is the collection of even numbers.
If we take complement of set A (A′), we will be left with odd numbers
A′ = {x : x is an odd natural number)

(ii) A = { x : x is an odd natural number }
This set is the collection of odd natural number
If we take complement of set A (A′), we will be left with even numbers
A′ = {x : x is an even natural number)

(iii) A= {x : x is a positive multiple of 3}
If we take complement of set A (A′), we will get numbers which are non multiple of 3
A′ = {x : x is natural number & is not multiple of 3}