Given a subset A of the universal set π, the complement of A, denoted Aβ 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 Aβ, is π-A, or equivalently,
Aβ={x|xβA}.
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
Δ={x:xβπ and xβA}. 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] If A is 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 and xβBΜ}
={1,3,5,7,9}=B
It 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] The complement of a set S is the collection of objects in the universal set that are not in S. The complement is written Sβ. In curly brace notation
Sβ={xπxβπ)β§(xβS)}
or more compactly as
Sβ={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
Sβ={x:Β¬(xβS)}
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 Kβ={4,5} while the complement of L={1,3,5} is Lβ={2,4}.
(iii) Let the universal set be the letters {a,e,i,o,u,y}. Then {y}β={a,e,i,o,u}.
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:xβπ and x is not a divisor of 42}. We see that 2βπ but 2βO, because 2 is divisor of 42. Similarly, 3βπ but 3βO, and 7βπ but 7βO. 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:xβπ and xβO}. 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,
Δ={x:xβπ and xβA}. 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|xβπ and xβΓ} 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,
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 PββͺQβͺR.
β Solution:
