**Universal set and Complements**

Given a subset *A* of the universal set π, the **complement** of *A*, denoted *A*^{β} or Δ (read *A* with macron) or *A*‘, consists of all elements of π which are __not__ elements in *A*.

[__Definition__] Let π be the universal set. The **complement** of a set *A*, denoted Δ or *A*^{β}, is π-*A*, or equivalently,

*A*

^{β}={

*x*|

*x*β

*A*}.

Complement of a set. Let π be the universal set and *A* is a subset of π. Then the complement of *A* is the set of all elements of π which are not the elements of *A*. Symbolically, we write

*x*:

*x*βπ and

*x*β

*A*}. Also Δ=π-

*A*

Memorize these complement symbols Μ , ‘,

^{β}, they have the same meaning. So, donβt debate with anyone what the differences among them are

β² Example 1. Let π={1,2,3,4,5,6,7,8,9,10} and *B*={1,3,5,7,9}. Find BΜ.

β Solution: We note that 2,4,6,8,10 are the only elements of π which do not belong to *B*. Hence BΜ={2,4,6,8,10}.

[

Note] IfAis a subset of the universal set π, then its complement Δ is also a subset of π.Again in Example 1 above, we have BΜ={2,4,6,8,10}

Hence (BΜ)’={

x:xβπ andxβBΜ}

={1,3,5,7,9}= BIt is clear from the definition of the complement that for any subset of the universal set π, we have (Δ)’=

A

β² Ex2. If π={a,b,c,d,e,f,g,h}, then find the complement of the following sets.

(i) *C*={a,b,c} (ii) *D*={d,e,f,g} (iii) *E*={a,c,e,g} (iv) *F*={f,g,h}

β Solution: We have,π={a,b,c,d,e,f,g,h}

(i) Complement of *C* is CΜ. CΜ=π-*C*={a,b,c,d,e,f,g,h}-{a,b,c}={d,e,f,g,h} (ii) Complement of *D* is DΜ. DΜ=π-*D*={a,b,c,d,e,f,g,h}-{d,e,f,g}={a,b,c,h} (iii) Complement of *E* is Δ. Δ=π-*E*={a,b,c,d,e,f,g,h}-{a,c,e,g}={b,d,f,h} (iv) Complement of *F* is FΜ. FΜ=π-*F*={a,b,c,d,e,f,g,h}-{f,g,h}={a,b,c,d,e}

β² Ex3. If π={a,b,c,d,e,f,g,h},find the complements of the following sets:

(i) *G*={a,b,c}

(ii) *H*={d,e,f,g}

(iii) *I*={a,c,e,g}

(iv) *J*={f,g,h,a}

β Solution:

(i) *G*={a,b,c} β αΈ ={d,e,f,g,h}

(ii) *H*={d,e,f,g} β HΜ={a,b,c,h}

(iii) *I*={a,c,e,g} β Δͺ={b,d,f,h}

(iv) *J*={f,g,h,a} β JΜ={b,c,d,e}

[

Definition] Thecomplementof a setSis the collection of objects in the universal set that are not in S. The complement is writtenS^{β}. In curly brace notation

S^{β}={xπxβπ)β§(xβS)}

or more compactly as

S^{β}={x:xβS}

however it should be apparent that the complement of a set always depends on which universal set is chosen.

There is also a Boolean symbol associated with the complementation operation: the not operation. The notation for not is Β¬. There is not much savings in space as the definition of complement becomes

*S*

^{β}={

*x*:Β¬(

*x*β

*S*)}

β² Example 4: **Set Complements**

(i) Let the universal set be the integers. Then the complement of the even integers is the odd integers.

(ii) Let the universal set be {1,2,3,4,5},then the complement of *K*={1,2,3} is *K*^{β}={4,5} while the complement of *L*={1,3,5} is *L*^{β}={2,4}.

(iii) Let the universal set be the letters {a,e,i,o,u,y}. Then {y}^{β}={a,e,i,o,u}.

β² Ex5. Let

*M*={a,b,e,f},

*N*={b,d,e,g,h.

Find each of the following sets.

(a)

*M*‘

Set

*M*‘ contains all the elements of set π that are not in set

*M*. Since set

*M*contains the elements a, b, e, and f, these elements will be disqualified from belonging to set

*M*‘, and consequently set

*M*‘ will contain c, d, g, and h, or

*M*‘={c, d, g, h}.

(b)

*N*‘

Set

*N*‘ contains all the elements of π that are not in set

*N*, so

*N*‘={a,c,f}.

**Complement of a Set**

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

[__Definition__]] Let π be the universal set and *A* is a subset of π. Then the complement of *A* is the set of all elements of π which are not the elements of *A*. Symbolically, we write Δ to denote the complement of *A* with respect to π. Thus,

*x*:

*x*βπ and

*x*β

*A*}. Obviously Δ=π-

*A*.

We note that the complement of a set

*A*can be looked upon, alternatively, as the

difference between a universal set π and the set

*A*.

Consider the complement of the universal set, π’. The set π’ is found by selecting all the elements of π that do not belong to π. There are no such elements, so there can be no elements in set π’. This means that for any universal set π, π’=Γ.

Now consider the complement of the empty set, Γ’. Since Γ’={*x*|*x*βπ and *x*βΓ} and set Γ contains no elements, every member of the universal set π satisfies this definition. Therefore for any universal set π, Γ’=π.

**Properties of the empty set and complements**

The empty set has the following properties: For any set *A*,

*A*=

*A*, Γβ©

*A*=Γ and Γβ

*A*.

Complements have the following properties:

*A*β©Δ=Γ, (Δ)

^{β}=

*A*and

*A*βͺΔ=π

**Some Properties of Complement Sets**

1. Complement laws: (i) *A*βͺΔ=π (ii) *A*β©Δ=Γ

2. De Morgan’s law: (i) (*A*βͺ*B*)=Δβ©BΜ (ii) (*A*β©*B*)=ΔβͺBΜ

3. Law of double complementation: (Δ)’=*A*

4. Laws of empty set and universal set π=Γ and Γ=π.

These laws can be verified by using Venn diagrams.

β² Ex6. If *P*={1,2,3,4}, *Q*={2,4,6,8} and *R*={3,4,5,6} are subsets of the universal set π={1,2,3,β¦,10}, list the elements of the set *P*^{β}βͺ*Q*βͺ*R*.

β Solution:

*P*

^{β}={5,6,7,8,9,10},

*Q*βͺ

*R*={2,3,4,5,6,8},

*P*

^{β}βͺ(

*Q*βͺ

*R*)={2,3,4,5,6,7,8,9,10}.