Complement of a Set

Introduction

The complement of a set is the set where all the elements of the universal set are present except the set itself. To give you more clarity, let’s say A is a set for all Italian food, which is a subset of a universal set that includes food from all cuisines. Hence, the complement of set A is all the other cuisines except the Italian one. In this article, we will go through the complement of a set in detail.

Definition of a Complement of a set

If the Universal Set (U) is having a subset A, then the complement of the subset A, which is represented as A′, includes everything in the universal set except the elements of A. This means that A′ = U – A. This is because the complement of a set is nothing but the Universal Set (U) minus the subset A.

Complement of a Set Examples

Let the Universal Set, U, be the set which contains all even numbers less than and equal to 10

U = {2,4,6,8,10}

Let A be the subset of U, which are all even numbers less and equal to 4 in the Universal set

A = {2,4}

Now let’s find the complement of the set A.

A′ = U – A

Hence, A’ includes all the elements present in the universal set that are not present in A.

A′ = {2,4,6,8,10} – {2,4}

A′ = {6,8,10}

Representation of the Complement of a Set in a Venn Diagram

This diagram represents that A′ is clearly everything but the subset A itself in the Universal Set (U).

Some properties of the complement of sets

(i) Law of complements: 

(a) A ∪ A′ = U 

This means that the union of A and A′ is A+A′ = Universal Set 

Example: 

U = {1,2,3,4,5,6}

A = {2,4,6}

A′ = {1,3,5}

A ∪ A′ = {2,4,6} + {1,3,5} = U

(b) A ∩ A′ = φ

This means that the intersection of A and A′ is a null set as they have nothing in common.

Example:

U = {1,2,3,4,5,6}

A = {2,4,6}

A′ = {1,3,5}

Since there is nothing in common,  A ∩ A′ = φ

(ii) De Morgan’s law 

(a) (A ∪ B)′ = A′ ∩ B′ 

It shows that the complement of the union of two sets, A and B, is equal to the intersection of the complement of those two sets separately.

Example:

U = {1,2,3,4,5,6}

A = {2,4,6}

A′ = {1,3,5}

B = {1,2,3}

B′ = {4,5,6}

(A ∪ B)′ is nothing but {5}, where as A′ = {1,3,5} and B′ = {4,5,6} which means the intersection is also {5}, hence proved

(b) (A ∩ B)′ = A′ ∪ B′ 

It tells us that the complement of the intersection of two sets is equal to the complement of the sets and its union.

Example:

Taking same values from the previous example, 

(A ∩ B)′ = { 1,3,4,5,6 }, where as A′ = { 1,3,5 } and B′ = { 4,5,6 } and its union is { 1,3,4,5,6 }, hence proved

(iii) (A′ )′ = A 

The complement of a set is nothing but the set itself.

(iv) U′ = φ and φ′ = U 

The complement of a Universal Set is a null set or an empty set. This is because the Universal Set contains all the elements and hence the complement of that has nothing in it, hence becoming the null set.

Some Problems and Solutions of Complement of a Set

Problems

1. If A = { 1, 2, 3, 4} and U = { 1, 2, 3, 4, 5, 6, 7, 8} then find A complement ( A’).

1. Let A = {3, 5, 7}, B = {2, 3, 4, 6} and C = {2, 3, 4, 5, 6, 7, 8}

(i) Verify (A ∩ B)’ = A’ ∪ B’

(ii) Verify (A ∪ B)’ = A’ ∩ B’

1. Let the universal set U have all the letters of the English alphabet. What is the complement of the empty set?

1. U
2. {a,b,c,d}
3. ϴ
4. ϴ – U

Solutions:

1.  A = { 1, 2, 3, 4} and Universal set = U = { 1, 2, 3, 4, 5, 6, 7, 8}

Complement of set A contains the elements present in the universal set but not in set A. Elements are 5, 6, 7, 8. ∴ A complement = A’ = { 5, 6, 7, 8}.

1. (i) (A ∩ B)’ = A’ ∪ B’

L.H.S. = (A ∩ B)’

A ∩ B = {3}

(A ∩ B)’ = {2, 4, 5, 6, 7, 8}     ……………….. (1)

R.H.S. = A’ ∪ B’

A’ = {5, 7, 8}

B’ = {2, 4, 6}

A’∪B’ = {2, 4, 5, 6, 7, 8}     ……………….. (2)

From (1) and (2), we conclude that;

(A ∩ B)’ = (A’ ∪ B’)

(ii) (A ∪ B)’ = A’ ∩ B’

L.H.S. = (A ∪ B)’

A∪B = {2, 3, 4, 5, 6, 7}

(A ∪ B)’ = {8}     ……………….. (1)

R.H.S. = A’ ∩ B’

A’ = {2, 4, 6, 8}

B’ = {5, 7, 8}

A’ ∩ B’ = {8}     ……………….. (2)

From (1) and (2), we conclude that;

(A ∪ B)’ = A’ ∩ B’

Conclusion

The understanding of the complement of a set will set up basics for a student in the chapter of Sets and this will set him further up for the Class 11 chapter of sets. The problems will help them test their understanding and the basic concept of the complement of a set