1.11 Complement of a Set

Complement of a set

Let \(\mathrm{U}\) be the universal set and A a subset of \(\mathrm{U}\). Then the complement of \(\mathrm{A}\) is the set of all elements of \(\mathrm{U}\) which are not the elements of \(\mathrm{A}\). Symbolically, we write
\(
\mathrm{A}^{\prime}=\{x: x \in \mathrm{U} \text { and } x \notin \mathrm{A}\} . \mathrm{Also} \mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}
\)

De Morgan’s laws

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 named after the mathematician De Morgan.

The complement \(\mathrm{A}^{\prime}\) of a set \(\mathrm{A}\) can be represented by a Venn diagram as shown in Fig 1.8. The shaded portion represents the complement of the set \(\mathrm{A}\).

Some Properties of Complement Sets

1. Complement laws:
(i) \(\mathrm{A} \cup \mathrm{A}^{\prime}=\mathrm{U}\)
(ii) \(\mathrm{A} \cap \mathrm{A}^{\prime}=\phi\)

2. De Morgan’s law:
(i) \((\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\)
(ii) \((\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\)

3. Law of double complementation: \(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\)

4. Laws of empty set and universal set \(\phi^{\prime}=\mathrm{U}\) and \(\mathrm{U}^{\prime}=\phi\).

Example 1:
Let \(\mathrm{U}=\{1,2,3,4,5,6,7,8,9,10\}\) and \(\mathrm{A}=\{1,3,5,7,9\}\). Find \(\mathrm{A}^{\prime}\).

Solution: We note that \(2,4,6,8,10\) are the only elements of \(\mathrm{U}\) which do not belong to A. Hence \(\mathrm{A}^{\prime}=\{2,4,6,8,10\}\).

Example 2:

Let \(\mathrm{U}=\{1,2,3,4,5,6\}, \mathrm{A}=\{2,3\}\) and \(\mathrm{B}=\{3,4,5\}\). Find \(\mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}, \mathrm{A} \cup \mathrm{B}\) and hence show that \((\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\).

Solution: Clearly \(\mathrm{A}^{\prime}=\{1,4,5,6\}, \mathrm{B}^{\prime}=\{1,2,6\}\). Hence \(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}=\{1,6\}\)
Also \(\mathrm{A} \cup \mathrm{B}=\{2,3,4,5\}\), so that \((\mathrm{A} \cup \mathrm{B})^{\prime}=\{1,6\}\)
\((\mathrm{A} \cup \mathrm{B})^{\prime}=\{1,6\}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\)
It can be shown that the above result is true in general. If \(\mathrm{A}\) and \(\mathrm{B}\) are any two subsets of the universal set \(\mathrm{U}\), then
\((\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\). Similarly, \((\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\).