# Complement of a Set (Set theory Part 8 NCERT)

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Complement of a Set

Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42. Thus, A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these three prime numbers, i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by A’. So we have A’ = {2, 3, 7}. Thus, we see that
A’ = {x : x ∈ U and x ∉ A }. This leads to the following definition.

Definition 8 Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A. Symbolically, we write A′ to denote the complement of A with respect to U. Thus,
A’ = {x : x ∈ U and x ∉ A }. Obviously A′ = U – A
We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A.

Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.
Solution We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A.
Hence A′ = { 2, 4, 6, 8,10 }.

Example 21 Let U be universal set of all the students of Class XI of a coeducational school and A be the set of all girls in Class XI. Find A′.
Solution Since A is the set of all girls, A’ is clearly the set of all boys in the class.

Note:- If A is a subset of the universal set U, then its complement A′ is also a subset of U.
Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 }
Hence (A′ )′ = {x : x ∈ U and x ∉ A′ } = {1, 3, 5, 7, 9} = A
It is clear from the definition of the complement that for any subset of the universal set U,
we have ( A′ )′ = A

Now, we want to find the results for ( A ∪ B )′ and A′ ∩ B′ in the following example.

Example 22 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′.
Solution Clearly A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′= { 1, 6 }
Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B )′ = { 1, 6 }
(A ∪ B)′ = { 1, 6 } = A′ ∩ B′
It can be shown that the above result is true in general. If A and B are any two subsets of the universal set U, then
( A ∪ B)′ = A′ ∩ B′. Similarly, ( A ∩ B )′ = A′ ∪ B′. These two results are stated in words as follows :

The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgan’s laws
These are named after the mathematician De Morgan.
The complement A′ of a set A can be represented by a Venn diagram as shown in Fig 1.10.
The shaded portion represents the complement of the set A. Some Properties of Complement Sets
1. Complement laws: (i) A ∪ A′ = U (ii) A ∩ A′ = φ
2. De Morgan’s law: (i) (A ∪ B)´ = A′ ∩ B′ (ii) (A ∩ B )′ = A′ ∪ B′
3. Law of double complementation : (A′ )′ = A
4. Laws of empty set and universal set φ′ = U and U′ = φ.

These laws can be verified by using Venn diagrams.

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