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LECTURE # 7
Sets Continued
UNION:
Let A and B be subsets of a universal set U. The union of sets A and B is the set of all elements in U that belong to A or to B or to both, and is denoted A ∪ B. Symbolically: A ∪ B = {x ∈U | x ∈A or x ∈ B} EMAMPLE: Let U = {a, b, c, d, e, f, g} A = {a, c, e, g}, B = {d, e, f, g} Then A ∪ B = {x ∈U | x ∈A or x ∈ B} ={a, c, d, e, f, g} VENN DIAGRAM FOR UNION:
A
B
U
A ∪ B is shaded
REMARK:
- A ∪ B = B ∪ A that is union is commutative you can prove this very easily only by using definition.
- A ⊆ A ∪ B and B ⊆ A ∪ B The above remark of subset is easily seen by the definition of union. MEMBERSHIP TABLE FOR UNION:
A B A ∪ B
REMARK:
This membership table is similar to the truth table for logical connective, disjunction (∨).
INTERSECTION:
Let A and B subsets of a universal set U. The intersection of sets A and B is the set of all elements in U that belong to both A and B and is denoted A ∩ B. Symbolically: A ∩ B = {x ∈U | x ∈ A and x ∈B} EXMAPLE: Let U = {a, b, c, d, e, f, g} A = {a, c, e, g}, B = {d, e, f, g} Then A ∩ B = {e, g}
VENN DIAGRAM FOR INTERSECTION:
U
A
B
∩ B is shaded EMARK:
A
R
- A ∩ B = B ∩ A
- A ∩ B ⊆ A ∩ B
- If A ∩ B = φ, re called disjoint sets. EMBERSHIP TABLE FOR INTERSECT N:
and A ⊆ B then A & B a M IO
REMARK:
This membership table is similar to the truth table for logical connective, onjunc on (∧). E:
c ti DIFFERENC difference of “A and B” (or relative omple ent of in A) i belong to A but not to B, and is enoted – B or A \ B.
A – B = {x ∈U | x ∈ A and x ∈B} EXAMPLE:
Let A and B be subsets of a universal set U. The c m B s the set of all elements in U that d A Symbolically:
Let U = {a, b, c, d, e, f, g} A = {a, c, e, g}, B = {d, e, f, g} Then A – B = {a, c} VENN DIAGRAM FOR SET DIFFERENCE:
U
A-B is shaded
A B
REMARK:
- A – B ≠ B – A that is Set difference is not commutative.
- A – B ⊆ A
- A – B, A ∩ B and B A are utually disjoint sets. MEMBERSHIP TABLE FOR SET DIFFERENCE:
A B A ∩ B
U = {1, 2, 3, …, 10},
(2) Y ∪ Z = , 6
X – Z = {1, 2, 3, 4, 5} – {2, 7}
(5) X – Z = {6, 7, 8, 9, 10} – {1, 3, 4, 5, 6, 8, 9, 10}
OTE Xc - Zc
X =
Y = {y | y = 2 x, x ∈X} = {2, 4, 6, 8, 10} Z = {z | z2 – 9 z + 14 = 0} = {2, 7} (1) X ∩ Y = {1, 2, 3, 4, 5} ∩ {2, 4 = {2, 4} {2, 4 , 8, 10} ∪ {2, 7} 4, 6, , 8, 10} (3) = {1, 3, 4, 5} (4) Yc^ = U – Y = {1, 2, 3, …, 10 = {1, 3, 5, 7, 9 c c = {7} (6 (X – Z)c = U – (X – Z) = {1, 2, 3, …, 10} – {1, 3, 4, 5} = {2, 6, 7, 8, 9, 10}
N (X – Z)c ≠ EXERCISE: Given the following universal set U and its two subsets P and Q, where U = {x | x ∈ Z,0 ≤ x ≤ 10} P = {x | x is a prime number} Q = {x | x2 < 70} (i) Draw a Venn diagram for the above (ii) List the elements in Pc ∩ Q SOLUTION: First we write the sets in Tabular form. U = {x | x ∈Z, 0 ≤ x ≤ 10} Since it is the set of integers that are greater then or equal 0 and less or equal to 10. So we have U= {0, 1, 2, 3, …, 10} number} s between 0 and 10. Remember Prime numbers are those
x uare less or equal to 70.
e write th m.
P = {x | x is a prime It is the set of prime number numbers which have only two distinct divisors. P = {2, 3, 5, 7} Q = { | x2 < 70} The set Q contains the elements between 0 and 10 which has their sq Q= {0, 1, 2, 3, 4, 5, 6, 7, 8} Thus w e sets in Tabular for VENN DIAGRAM:
U
Q
0
2,3,5,7^ P
0,1,4,6, 9, (i) Pc^ ∩ Q =?
P c^ = U – P = {0, 1, 2, 3, …, 10}- {2, 3, 5, 7} = {0, 1, 4, 6, 8, 9, 10}
and , 1 6, 7, 8} = {0, 1, 4, 6, 8}
P c^ ∩ Q = {0, 1, 4, 6, 8, 9, 10} ∩ {0 , 2, 3, 4, 5,
EXERCISE: Let U = {1, 2, 3, 4, 5}, C = {1, 3} and A and B are no n empty sets. F ind A in each of the following: i) A ∪ B = U, A ∩ B = φ and B = {1}
(i 3}, A ∪ B = {2, 3, 4} and B ∪ C = {1,2,3} (iv) A and B are disjoint, B and C are disjoint, and the union of A and B is the set {1, 2}. (i) A ∪ B = U, A ∩ B = φ and B = {1} SOLUTION
(ii) A ⊂ B and A ∪ B = {4, 5} ii) A ∩ B = {
Since A ∪ B = U = {1, 2, 3, 4, 5} and A ∩ B = φ, Therefore A = Bc^ = {1}c^ = {2, 3, 4, 5}
(i) A ⊂ B and A ∪ B = {4, 5} also C = {1, 3}
SOLUTION When A ⊂ B, then A ∪ B = B = {4, 5} Also A being a proper subset of B implies A = {4} or A = {5}
B = {2, 3, 4}and B ∪ C = {1,2,3} = {1, 3}
SOLUTION
(iii) A ∩ B = {3}, A ∪ Also C
A U
4 3 B
C 1
Since we have 3 in the intersection of A and B as well as in C so we place 3 in common Venn diagram. Now since 1 is in the union of B and C it means that 1 may be in C or may be in B, but 1can t be in B because if 1 is in the B then it must be in A ∪ B but 1 is not there, thus we place 1 in the part of C which is not shared by e is the reason for 4 and we place it in the set which is not shared by any in B, 2 cannot be in A because A ∩ B = {3}, and is not in C. So A = {3, 4} and B = {2, 3}
(i) A ∩ B = φ , B ∩ C = φ , A ∪ B = {1, 2}. Also C = {1, 3} SOLUTION
part shared by the three sets in the no
any other set. Sam other set. Now 2 will be
U
A
B
C
(A – B) ∩ C is shaded
(iii) (^) (A ∩ Bc) ∪ Cc (^) is shad ed.
U
8
1
2
3
4
5 6
7
A
B
C
PROVING SET IDENTITIES BY VENN DIAGRAMS:
Prove the following using Venn Diagrams: (i) A – (A – B) = A ∩ B (ii) (A ∩ B)c^ = A c^ ∪ B c (iii) A – B = A ∩ B c
SOLUTION (i) A - (A – B) = A ∩ B
U A B
1 2 3
4
(a)
A – B is shaded
A = { 1, 2 } B = { 2, 3 } A – B ={ 1 }
(b)
A – (A – B) is shaded
U
A B
A = { 1, 2 }
A – B = { 1 }
A – (A – B) = { 2 }
A B^ U 1 2 3
A ∩ B is shaded
(c)
4
A= { 1, 2 } B = { 2, 3 } A ∩ B = {2}
RESULT: A – (A – B) = A ∩ B
SOLUTION (ii) (A ∩ B)c^ = A c^ ∪ B c
U A B
1
2 3
A ∩ B
(a)
4
(b)
(A ∩ B)c
U
A^ B
U
A B
(a)
A – B is shaded.
(b)
A B U
(c)
A
B is shaded
U
A ∩ c
B c^ is shaded.
say
RESULT: A – B = PROVING SET ID ERSHIP TABLE:
A B
From diagrams (a) and (b) we can
A ∩ Bc ENTITIES BY MEMB
A – B = A ∩ B
Prove the following using Membership Table: (i) A – (A – B) = A ∩ B (ii) (A ∩ B)c^ = A c^ ∪ B c (iii)^ c
SOLUTION (i) A – (A – B) = A ∩ B
A B A-B A-(A-B) A∩B
Since the last two colu mns of the above table are sa me hence the corresponding set expressions are same. That is A – (A – B) = A ∩ B
SOLUTION (ii)
(A ∩ B)c^ = A c^ ∪ B c
A B (^) A ∩ B (A ∩ B) c^ A c^ B c^ A c^ ∪ B c 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0 0 1 1 1 1
Sinc e e above table are same hence the corre t is (A ∩ B)c^ = A c^ ∪ B c
SOLUTION (iii)
e th fourth and last columns of th sponding set expressions are same. Tha
A B A – B Bc^ A ∩ Bc 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 0 0 1 0
