CBSE Class 11 Mathematics Sets Notes Set 02

Important Notes and Formulas

Sets and Their Representation

A set is a well-defined class or collection of objects. By a well-defined collection we mean that there exists a rule with the help of which it is possible to tell whether a given object belongs or does not belong to the given collection. The objects in sets may be anything: numbers, people, mountains, rivers etc. The objects constituting the set are called elements or members of the set.
Sets are usually represented by capital letters A, B, C, X, Y, Z etc. and their elements by small letters a, b, c, x, y, z etc. If a is an element of a set A, we say that “a belongs to A” and write a ∈ A; if b is not an element of A, we write b ∉ A.

(1) Description method: A description about the set is made and it is enclosed in curly brackets { }. For example, the set of composite numbers less than 30 is written as {Composite numbers less than 30}.

(2) Roster method or Listing method: In this method a set is described by listing its elements, separated by commas, within braces { }. The set of vowels of the English alphabet may be described as {a, e, i, o, u}. The set of even natural numbers can be described as {2, 4, 6, …}. Here the dots stand for ‘and so on’.
Note: The order in which the elements are written in a set makes no difference. Thus {a, e, i, o, u} and {e, a, i, o, u} denote the same set. Also, the repetition of an element has no effect; for example, {1, 2, 3, 2} is the same set as {1, 2, 3}. So each element of a set is listed once and only once.

(3) Set-builder method or Rule method: In this method a set is described by a characterising property P(x) of its elements x. In such a case the set is described by {x : P(x) holds} or {x | P(x) holds}, which is read as ‘the set of all x such that P(x) holds’. The symbol ‘|’ or ‘:’ is read as ‘such that’.
The set E of all even natural numbers can be written as
E = {x | x is a natural number and x = 2n for n ∈ N}
or E = {x | x ∈ N, x = 2n, n ∈ N}
or E = {x ∈ N | x = 2n, n ∈ N}
The set A = {0, 1, 4, 9, 16, …} can be written as A = {x² | x ∈ Z}.
Similarly, the set L of all integers greater than 5 and less than 9 is L = {6, 7, 8} in roster form and L = {x : x is an integer greater than 5 and less than 9} in set-builder form.

Some important symbols:

Types of Sets

(1) Null set or Empty set: The set which contains no element at all is called the null set. This set is sometimes also called the ‘empty set’ or the ‘void set’. It is denoted by the symbol φ or { }. A set which has at least one element is called a non-empty set.
Let A = {x : x² + 1 = 0 and x is real}. Since there is no real number which satisfies the equation x² + 1 = 0, the set A is the empty set.
Note: If A and B are any two empty sets, then x ∈ A iff x ∈ B is satisfied because there is no element x in either A or B to which the condition may be applied. Thus A = B. Hence there is only one empty set and we denote it by φ. Therefore, the article ‘the’ is used before empty set.

(2) Singleton set: A set consisting of a single element is called a singleton set. The set {5} is a singleton set.

(3) Finite set: A set is called a finite set if it is either the void set or its elements can be listed (counted, labelled) by the natural numbers 1, 2, 3, … and the process of listing terminates at a certain natural number n (say). Since the empty set has no element, it is regarded as a finite set.
Cardinal number of a finite set: The number n in the above definition, i.e. the number of distinct elements in a finite set A, is called the cardinal number or order of A and is denoted by n(A) or O(A). As the empty set has no elements, n(φ) = 0.

(4) Infinite set: A set whose elements cannot be listed by the natural numbers 1, 2, 3, …, n, for any natural number n is called an infinite set, i.e. a set having an unlimited number of different elements.

(5) Equivalent sets: Two finite sets A and B are equivalent if their cardinal numbers are the same, i.e. n(A) = n(B). We write it as A ↔ B.
Example: A = {1, 3, 5, 7} and B = {10, 12, 14, 16} are equivalent sets [∵ O(A) = O(B) = 4].

(6) Equal sets: Two sets A and B are said to be equal iff every element of A is an element of B and also every element of B is an element of A. We write “A = B” if the sets A and B are equal and “A ≠ B” if they are not equal. Symbolically, A = B if x ∈ A ⇔ x ∈ B. This statement is also known as the axiom of extension.
Example: If A = {2, 3, 5, 6} and B = {6, 5, 3, 2}, then A = B, because each element of A is an element of B and vice-versa. Similarly, X = {letters of the word STONE} = {S, T, O, N, E} and Y = {letters of the word NOTES} = {N, O, T, E, S} are equal sets.
Note: Equal sets are always equivalent, but equivalent sets need not be equal.

(7) Universal set: A set containing all the elements which occur in a given discussion is called a universal set and is denoted by U (sometimes ξ). Thus, in any particular discussion, no element can exist outside the universal set. It should be noted that the universal set is not unique; it may differ from problem to problem.

Subsets

Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B. We write A ⊆ B, which is read as “A is a subset of B” or “A is contained in B”. In this case B is called a superset of A, written B ⊇ A.
Thus, A ⊆ B ⇔ (a ∈ A ⇒ a ∈ B).
If there exists at least one element in A which is not an element of B, then A is not a subset of B, written A ⊄ B.

Proper subset: If A ⊆ B and A ≠ B, i.e. every element of A is in B and there is at least one element of B which is not in A, then A is called a proper subset of B and we write A ⊂ B. A subset which is not a proper subset is called an improper subset.

Important points:
(a) Every set is a subset of itself.
(b) The empty set is a subset of every set.
(c) Every set has only one improper subset and that is the set itself. If A and B are equal sets, they are improper subsets of each other.
(d) The empty set is a proper subset of every set except itself.
(e) Every set is a subset of the universal set.
(f) If X ⊆ Y and Y ⊆ X, then X = Y.
Example: Let A = {1, 2}. Then the subsets of A are φ, {1}, {2}, {1, 2}; of these φ, {1} and {2} are proper subsets and {1, 2} is the improper subset.

Venn Diagrams

The combination of rectangles and circles used to represent sets is called a Venn–Euler diagram or simply a Venn diagram. The universal set U is represented by the points within a rectangle and its subsets are represented by points in closed curves (usually circles) within the rectangle.

If a set A is a subset of a set B, the circle representing A is drawn inside the circle representing B. If A and B are not equal but have some common elements, we draw two intersecting circles. Two disjoint sets are represented by two non-intersecting circles.
Points marked inside a closed figure can show the elements of the set. For example, the set of letters of the word AMERICA is {A, M, E, R, I, C}:

In the diagram below, A ⊆ B (and since A ≠ B, A ⊂ B), C ⊄ B and C ⊄ A, and U is the universal set.

Operations on Sets

(1) Union of sets: Let A and B be two sets. The union of A and B is the set of all elements which are in A or in B (common elements taken only once). It is denoted by A ∪ B, read as “A union B”.
Symbolically, A ∪ B = {x : x ∈ A or x ∈ B}.
Here “or” is used in the standard mathematical sense: when we say x ∈ A or x ∈ B, we do not exclude the possibility that x is a member of both A and B.
For example, if X = {2, 4, 6, 8, 10} and Y = {4, 8, 12}, then X ∪ Y = {2, 4, 6, 8, 10, 12}.
If A₁, A₂, …, A_n is a finite family of sets, their union is written A₁ ∪ A₂ ∪ … ∪ A_n.
Properties: (i) A ∪ B = B ∪ A   (ii) A ∪ φ = A   (iii) A ∪ A = A   (iv) (A ∪ B) ∪ C = A ∪ (B ∪ C) (associative law)   (v) U ∪ A = U (law of U)

(2) Intersection of sets: The intersection of A and B is the set of all those elements that belong to both A and B. It is denoted by A ∩ B (read as “A intersection B”).
Thus, A ∩ B = {x : x ∈ A and x ∈ B}. Clearly, x ∈ A ∩ B ⇔ x ∈ A and x ∈ B. Evidently A ∩ B ⊆ A and A ∩ B ⊆ B.
For example, if X = {a, e, i, o, u} and Y = {a, b, c, d, e}, then X ∩ Y = {a, e}.
If A₁, A₂, …, A_n is a finite family of sets, their intersection is written A₁ ∩ A₂ ∩ … ∩ A_n.
Properties: (i) A ∩ B = B ∩ A   (ii) φ ∩ A = φ   (iii) A ∩ A = A   (iv) (A ∩ B) ∩ C = A ∩ (B ∩ C) (associative law)   (v) U ∩ A = A (law of U)

(3) Disjoint sets: Two sets A and B are said to be disjoint if A ∩ B = φ, i.e. they have no element in common. If A ∩ B ≠ φ, i.e. A and B have at least one common element, they are called overlapping (or joint) sets. Example: the sets {1, 2} and {3, 4} are disjoint.

(4) Difference of sets: The difference of A and B, written A − B, is the set of all those elements of A which do not belong to B.
Thus, A − B = {x : x ∈ A and x ∉ B} or A − B = {x ∈ A : x ∉ B}. Clearly, x ∈ A − B ⇔ x ∈ A and x ∉ B.
Similarly, B − A = {x ∈ B : x ∉ A}.
Example: For A = {1, 2, 3} and B = {3, 4, 5}, A − B = {1, 2} and B − A = {4, 5}. As another example, R − Q is the set of all irrational numbers.

(5) Complement of a set: Let U be the universal set and A ⊆ U. The complement of A with respect to U is denoted by A′ or A^c or U − A and is the set of all those elements of U which are not in A.
Thus, A′ = {x ∈ U : x ∉ A}. Clearly, x ∈ A′ ⇔ x ∉ A.
Example: If U = {1, 2, …, 10} and A = {1, 3, 5, 7, 9}, then A′ = {2, 4, 6, 8, 10}.
Properties of a set and its complement: (a) A ∩ A′ = φ   (b) A ∪ A′ = U   (c) U′ = φ   (d) φ′ = U   (e) (A′)′ = A

Venn Diagrams for Union and Intersection

When A and B are overlapping sets, the shaded portions show A ∪ B and A ∩ B respectively:

When A and B are disjoint, A ∪ B is the whole of both circles, while A ∩ B = φ (nothing is shaded):

When B is contained in A, A ∪ B = A and A ∩ B = B:

For three sets A, B and C, the shaded portions show A ∪ B ∪ C and A ∩ B ∩ C respectively:

Laws of Algebra of Sets

(1) Idempotent laws: For any set A, (i) A ∪ A = A   (ii) A ∩ A = A

(2) Identity laws: For any set A, (i) A ∪ φ = A   (ii) A ∩ U = A
i.e. φ and U are the identity elements for union and intersection respectively.

(3) Commutative laws: For any two sets A and B, (i) A ∪ B = B ∪ A   (ii) A ∩ B = B ∩ A
i.e. union and intersection are commutative. But A − B ≠ B − A in general, i.e. difference is not commutative.

(4) Associative laws: If A, B and C are any three sets, then (i) (A ∪ B) ∪ C = A ∪ (B ∪ C)   (ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C
i.e. union and intersection are associative. But (A − B) − C ≠ A − (B − C) in general.

(5) Distributive laws: If A, B and C are any three sets, then
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)   (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
i.e. union and intersection are distributive over intersection and union respectively.

(6) De Morgan’s laws: If A and B are any two sets, then
(i) (A ∪ B)′ = A′ ∩ B′   (ii) (A ∩ B)′ = A′ ∪ B′
(iii) A − (B ∪ C) = (A − B) ∩ (A − C)   (iv) A − (B ∩ C) = (A − B) ∪ (A − C)

Some useful results: If A, B and C are any sets, then
(i) A − B = A ∩ B′   (ii) B − A = B ∩ A′
(iii) A − B = A ⇔ A ∩ B = φ   (iv) (A − B) ∪ B = A ∪ B
(v) (A − B) ∩ B = φ   (vi) A ⊆ B ⇔ B′ ⊆ A′
(vii) (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B)
(viii) A ∩ (B − C) = (A ∩ B) − (A ∩ C)

Solved Examples

Example 1. The set of intelligent students in a class is [AMU 1998]
(a) A null set
(b) A singleton set
(c) A finite set
(d) Not a well defined collection
Answer:
Ans. (d) Not a well defined collection
Since intelligence is not defined for students in a class, i.e. it is not a well defined collection.

Example 2. Which of the following is the empty set [Karnataka CET 1990]
(a) {x : x is a real number and x² − 1 = 0}
(b) {x : x is a real number and x² + 1 = 0}
(c) {x : x is a real number and x² − 9 = 0}
(d) {x : x is a real number and x² = x + 2}
Answer:
Ans. (b) {x : x is a real number and x² + 1 = 0}
Since x² + 1 = 0 gives x² = −1 ⇒ x = ±i
∴ x is not real, but x is real (given)
∴ No value of x is possible.

Example 3. The set A = {x : x ∈ R, x² = 16 and 2x = 6} equals [Karnataka CET 1995]
(a) φ
(b) {14, 3, 4}
(c) {3}
(d) {4}
Answer:
Ans. (a) φ
x² = 16 ⇒ x = ±4
2x = 6 ⇒ x = 3
There is no value of x which satisfies both the above equations. Thus, A = φ.

Example 4. If X = {8ⁿ − 7n − 1 : n ∈ N} and Y = {49(n − 1) : n ∈ N}, then
(a) X ⊆ Y
(b) Y ⊆ X
(c) X = Y
(d) None of these
Answer:
Ans. (a) X ⊆ Y
Since 8ⁿ − 7n − 1 = (7 + 1)ⁿ − 7n − 1 = 7ⁿ + ⁿC₁7ⁿ⁻¹ + ⁿC₂7ⁿ⁻² + … + ⁿC_n−17 + ⁿC_n − 7n − 1
= ⁿC₂7² + ⁿC₃7³ + … + ⁿC_n7ⁿ   (∵ ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n−1)
= 49[ⁿC₂ + ⁿC₃(7) + … + ⁿC_n7ⁿ⁻²]
∴ 8ⁿ − 7n − 1 is a multiple of 49 for n ≥ 2.
For n = 1, 8ⁿ − 7n − 1 = 8 − 7 − 1 = 0; for n = 2, 8ⁿ − 7n − 1 = 64 − 14 − 1 = 49.
∴ 8ⁿ − 7n − 1 is a multiple of 49 for all n ∈ N.
∴ X contains elements which are multiples of 49, and Y contains all multiples of 49 (including 0). ∴ X ⊆ Y.

Example 5. Given the sets A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then A ∪ (B ∩ C) is [MNR 1988; Kurukshetra CEE 1996]
(a) {3}
(b) {1, 2, 3, 4}
(c) {1, 2, 4, 5}
(d) {1, 2, 3, 4, 5, 6}
Answer:
Ans. (b) {1, 2, 3, 4}
B ∩ C = {4}, ∴ A ∪ (B ∩ C) = {1, 2, 3, 4}.

Example 6. If A ⊆ B, then A ∪ B is equal to
(a) A
(b) B ∩ A
(c) B
(d) None of these
Answer:
Ans. (c) B
Since A ⊆ B ⇒ A ∪ B = B.

Example 7. If A and B are any two sets, then A ∪ (A ∩ B) is equal to
(a) A
(b) B
(c) A^c
(d) B^c
Answer:
Ans. (a) A
A ∩ B ⊆ A. Hence A ∪ (A ∩ B) = A.

Example 8. If A and B are two given sets, then A ∩ (A ∩ B)^c is equal to [AMU 1998; Kurukshetra CEE 1999]
(a) A
(b) B
(c) φ
(d) A ∩ B^c
Answer:
Ans. (d) A ∩ B^c
A ∩ (A ∩ B)^c = A ∩ (A^c ∪ B^c) = (A ∩ A^c) ∪ (A ∩ B^c) = φ ∪ (A ∩ B^c) = A ∩ B^c.

Example 9. If N_a = {an : n ∈ N}, then N₃ ∩ N₄ =
(a) N₇
(b) N₁₂
(c) N₃
(d) N₄
Answer:
Ans. (b) N₁₂
N₃ ∩ N₄ = {3, 6, 9, 12, 15, …} ∩ {4, 8, 12, 16, 20, …} = {12, 24, 36, …} = N₁₂
Trick: N₃ ∩ N₄ = N₁₂ [∵ 3 and 4 are relatively prime numbers]

Example 10. If aN = {ax : x ∈ N} and bN ∩ cN = dN, where b, c ∈ N are relatively prime, then
(a) d = bc
(b) c = bd
(c) b = cd
(d) None of these
Answer:
Ans. (a) d = bc
bN = the set of positive integral multiples of b, cN = the set of positive integral multiples of c.
∴ bN ∩ cN = the set of positive integral multiples of bc = bcN [∵ b, c are relatively prime]
∴ d = bc.

Example 11. If the sets A and B are defined as A = {(x, y) : y = \( \frac{1}{x} \), 0 ≠ x ∈ R}, B = {(x, y) : y = −x, x ∈ R}, then
(a) A ∩ B = A
(b) A ∩ B = B
(c) A ∩ B = φ
(d) None of these
Answer:
Ans. (c) A ∩ B = φ
The curves y = \( \frac{1}{x} \) and y = −x would meet when −x = \( \frac{1}{x} \) ⇒ x² = −1, which does not give any real value of x.
Hence A ∩ B = φ.

Example 12. Let A = {x : x ∈ R, |x| < 1}, B = {x : x ∈ R, |x − 1| ≥ 1} and A ∪ B = R − D, then the set D is
(a) {x : 1 < x ≤ 2}
(b) {x : 1 ≤ x < 2}
(c) {x : 1 ≤ x ≤ 2}
(d) None of these
Answer:
Ans. (b) {x : 1 ≤ x < 2}
A = {x : x ∈ R, −1 < x < 1}
B = {x : x ∈ R : x − 1 ≤ −1 or x − 1 ≥ 1} = {x : x ∈ R : x ≤ 0 or x ≥ 2}
∴ A ∪ B = {x : x < 1 or x ≥ 2} = R − D, where D = {x : x ∈ R, 1 ≤ x < 2}.

Example 13. If X = {4ⁿ − 3n − 1 : n ∈ N} and Y = {9(n − 1) : n ∈ N}, then X ∪ Y is equal to [Karnataka CET 1997]
(a) X
(b) Y
(c) N
(d) None of these
Answer:
Ans. (b) Y
Since 4ⁿ − 3n − 1 = (3 + 1)ⁿ − 3n − 1 = 3ⁿ + ⁿC₁3ⁿ⁻¹ + ⁿC₂3ⁿ⁻² + … + ⁿC_n−13 + ⁿC_n − 3n − 1
= ⁿC₂3² + ⁿC₃3³ + … + ⁿC_n3ⁿ = 9[ⁿC₂ + ⁿC₃(3) + … + ⁿC_n3ⁿ⁻²]
∴ 4ⁿ − 3n − 1 is a multiple of 9 for n ≥ 2.
For n = 1, 4ⁿ − 3n − 1 = 4 − 3 − 1 = 0; for n = 2, 4ⁿ − 3n − 1 = 16 − 6 − 1 = 9.
∴ 4ⁿ − 3n − 1 is a multiple of 9 for all n ∈ N.
∴ X contains elements which are multiples of 9, and Y contains all multiples of 9 (including 0).
∴ X ⊆ Y, ∴ X ∪ Y = Y.

Example 14. If A = {(x, y) : x² + y² = 25} and B = {(x, y) : x² + 9y² = 144}, then A ∩ B contains [AMU 1996; Pb. CET 2002]
(a) One point
(b) Three points
(c) Two points
(d) Four points
Answer:
Ans. (d) Four points
A = set of all points (x, y) on the circle x² + y² = 5².
B = set of all points on the ellipse \( \frac{x^2}{144} + \frac{y^2}{16} = 1 \), i.e. \( \frac{x^2}{12^2} + \frac{y^2}{4^2} = 1 \).
The circle (radius 5) cuts the ellipse (semi-axes 12 and 4) in one point in each quadrant, as shown below.

Clearly, A ∩ B consists of four points.

Example 15. If A, B and C are any three sets, then A − (B ∪ C) is equal to
(a) (A − B) ∪ (A − C)
(b) (A − B) ∩ (A − C)
(c) (A − B) ∪ C
(d) (A − B) ∩ C
Answer:
Ans. (b) (A − B) ∩ (A − C)
It is De Morgan’s law: A − (B ∪ C) = (A − B) ∩ (A − C).

Example 16. If A = {x : x is a multiple of 3} and B = {x : x is a multiple of 5}, then A − B is (Ā means complement of A) [AMU 1998]
(a) Ā ∩ B
(b) A ∩ B̄
(c) Ā ∩ B̄
(d) \( \overline{A \cap B} \)
Answer:
Ans. (b) A ∩ B̄
A − B = A ∩ B^c = A ∩ B̄.

Example 17. If A, B and C are non-empty sets, then (A − B) ∪ (B − A) equals [AMU 1992, 1998; DCE 1998]
(a) (A ∪ B) − B
(b) A − (A ∩ B)
(c) (A ∪ B) − (A ∩ B)
(d) (A ∩ B) ∪ (A ∪ B)
Answer:
Ans. (c) (A ∪ B) − (A ∩ B)
From the Venn diagram, (A − B) ∪ (B − A) is the part of A ∪ B outside the common region A ∩ B.

∴ (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).