CBSE Class 11 Mathematics Sets Notes Set 01

Important Notes and Formulas

Set

A set is a collection of well-defined and well distinguished objects of our perception or thought.

Notations

The sets are usually denoted by capital letters A, B, C, etc. and the members or elements of the set are denoted by lower-case letters a, b, c, etc. If x is a member of the set A, we write x ∈ A (read as 'x belongs to A') and if x is not a member of the set A, we write x ∉ A (read as 'x does not belong to A'). If x and y both belong to A, we write x, y ∈ A.

Representation of a Set

Usually, sets are represented in the following two ways:
(i) Roster form or Tabular form
(ii) Set Builder form or Rule Method

Roster Form: In this form, we list all the members of the set within braces (curly brackets) and separate these by commas. For example, the set A of all odd natural numbers less than 10 in the Roster form is written as:
A = {1, 3, 5, 7, 9}
Note:
(i) In roster form, every element of the set is listed only once.
(ii) The order in which the elements are listed is immaterial. For example, each of the following sets denotes the same set {1, 2, 3}, {3, 2, 1}, {1, 3, 2}.

Set-Builder Form: In this form, we write a variable (say x) representing any member of the set followed by a property satisfied by each member of the set.
For example, the set A of all prime numbers less than 10 in the set-builder form is written as
A = {x | x is a prime number less than 10}
The symbol '|' stands for the words 'such that'. Sometimes, we use the symbol ':' in place of the symbol '|'.

Types of Sets

(1) Empty Set or Null Set: A set which has no element is called the null set or empty set. It is denoted by the symbol φ.
For example, each of the following is a null set:
(a) The set of all real numbers whose square is −1.
(b) The set of all rational numbers whose square is 2.
(c) The set of all those integers that are both even and odd.
A set consisting of at least one element is called a non-empty set.

(2) Singleton Set: A set having only one element is called singleton set.
For example, {0} is a singleton set, whose only member is 0.

(3) Finite and Infinite Set: A set which has finite number of elements is called a finite set. Otherwise, it is called an infinite set.
For example, the set of all days in a week is a finite set whereas the set of all integers, denoted by {…, −2, −1, 0, 1, 2, …} or {x | x is an integer}, is an infinite set.
The empty set φ, which has no element, is regarded as a finite set.

(4) Cardinal Number: The number of elements in a finite set A is represented by n(A), known as the cardinal number of A.

(5) Equal Sets: Two sets A and B are said to be equal, written as A = B, if every element of A is in B and every element of B is in A.

(6) Equivalent Sets: Two finite sets A and B are said to be equivalent, if n(A) = n(B). Clearly, equal sets are equivalent but equivalent sets need not be equal.
For example, the sets A = {4, 5, 3, 2} and B = {1, 6, 8, 9} are equivalent but are not equal.

Subset

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 and we write A ⊂ B or B ⊃ A (read as 'A is contained in B' or 'B contains A'). B is called superset of A.
Note:
(i) Every set is a subset and a superset of itself.
(ii) If A is not a subset of B, we write A ⊄ B.
(iii) The empty set is a subset of every set.
For example, if A = {3, 4}, then φ, {3}, {4}, {3, 4} are subsets of A. Also {3} ⊂ {3, 4} and {2, 3} ⊄ {3, 4}.

Union of Two Sets

The union of two sets A and B, written as A ∪ B (read as 'A union B'), is the set consisting of all the elements which are either in A or in B or in both. Thus,
A ∪ B = {x : x ∈ A or x ∈ B}
Clearly, x ∈ A ∪ B ⇒ x ∈ A or x ∈ B, and x ∉ A ∪ B ⇒ x ∉ A and x ∉ B.

For example, if A = {a, b, c, d} and B = {c, d, e, f}, then A ∪ B = {a, b, c, d, e, f}.

Intersection of Two Sets

The intersection of two sets A and B, written as A ∩ B (read as 'A intersection B'), is the set consisting of all the common elements of A and B. Thus,
A ∩ B = {x : x ∈ A and x ∈ B}
Clearly, x ∈ A ∩ B ⇒ x ∈ A and x ∈ B, and x ∉ A ∩ B ⇒ x ∉ A or x ∉ B.

For example, if A = {a, b, c, d} and B = {c, d, e, f}, then A ∩ B = {c, d}.

Disjoint Sets

Two sets A and B are said to be disjoint, if A ∩ B = φ, i.e. A and B have no element in common.

For example, if A = {1, 3, 5} and B = {2, 4, 6}, then A ∩ B = φ, so A and B are disjoint sets.

Difference of Two Sets

If A and B are two sets, then their difference A − B is defined as:
A − B = {x : x ∈ A and x ∉ B}.
Similarly, B − A = {x : x ∈ B and x ∉ A}.

For example, if A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9} then A − B = {2, 4} and B − A = {7, 9}.
Important Results
(a) A − B ≠ B − A
(b) The sets A − B, B − A and A ∩ B are disjoint sets
(c) A − B ⊆ A and B − A ⊆ B
(d) A − φ = A and A − A = φ

Symmetric Difference of Two Sets

The symmetric difference of two sets A and B, denoted by A Δ B, is defined as
A Δ B = (A − B) ∪ (B − A).
For example, if A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9} then
A Δ B = (A − B) ∪ (B − A) = {2, 4} ∪ {7, 9} = {2, 4, 7, 9}.

Complement of a Set

If U is a universal set and A is a subset of U, then the complement of A is the set which contains those elements of U, which are not contained in A and is denoted by A′ or A^c. Thus,
A^c = {x : x ∈ U and x ∉ A}
For example, if U = {1, 2, 3, 4, …} and A = {2, 4, 6, 8, …}, then A^c = {1, 3, 5, 7, …}.
Important Results
(a) U^c = φ    (b) φ^c = U    (c) A ∪ A^c = U    (d) A ∩ A^c = φ

Algebra of Sets

1. For any set A, we have (a) A ∪ A = A    (b) A ∩ A = A
2. For any set A, we have (c) A ∪ φ = A    (d) A ∩ φ = φ    (e) A ∪ U = U    (f) A ∩ U = A
3. For any two sets A and B, we have (g) A ∪ B = B ∪ A    (h) A ∩ B = B ∩ A
4. For any three sets A, B and C, we have
(i) A ∪ (B ∪ C) = (A ∪ B) ∪ C
(j) A ∩ (B ∩ C) = (A ∩ B) ∩ C
5. For any three sets A, B and C, we have
(k) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(l) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
6. If A is any set, we have (A^c)^c = A.
7. De Morgan's Laws: For any three sets A, B and C, we have
(m) (A ∪ B)^c = A^c ∩ B^c
(n) (A ∩ B)^c = A^c ∪ B^c
(o) A − (B ∪ C) = (A − B) ∩ (A − C)
(p) A − (B ∩ C) = (A − B) ∪ (A − C)

Important Results on Operations on Sets

(i) A ⊆ A ∪ B, B ⊆ A ∪ B, A ∩ B ⊆ A, A ∩ B ⊆ B
(ii) A − B = A ∩ B^c
(iii) (A − B) ∪ B = A ∪ B
(iv) (A − B) ∩ B = φ
(v) A ⊆ B ⇔ B^c ⊆ A^c
(vi) A − B = B^c − A^c
(vii) (A ∪ B) ∩ (A ∪ B^c) = A
(viii) A ∪ B = (A − B) ∪ (B − A) ∪ (A ∩ B)
(ix) A − (A − B) = A ∩ B
(x) A − B = B − A ⇔ A = B
(xi) A ∪ B = A ∩ B ⇔ A = B
(xii) A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

Solved Examples

Example 1. Write the set of all positive integers whose cube is odd.
Answer:
The elements of the required set are not even.
[∵ Cube of an even integer is also an even integer]
Moreover, the cube of a positive odd integer is a positive odd integer.
⇒ The elements of the required set are all positive odd integers.
Hence, the required set, in the set builder form, is: {2k + 1 : k ≥ 0, k ∈ Z}.

Example 2. Write the set \( \left\{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8} \right\} \) in the set builder form.
Answer:
In each element of the given set the denominator is one more than the numerator.
Also the numerators are from 1 to 7.
Hence the set builder form of the given set is: \( \left\{ x : x = \frac{n}{n + 1},\ n \in N \text{ and } 1 \le n \le 7 \right\} \).

Example 3. Write the set {x : x is a positive integer and x² < 30} in the roster form.
Answer:
The positive integers whose squares are less than 30 are: 1, 2, 3, 4, 5.
Hence the given set, in roster form, is {1, 2, 3, 4, 5}.

Example 4. Write the set {0, 1, 4, 9, 16, …} in set builder form.
Answer:
The elements of the given set are squares of integers: 0, ±1, ±2, ±3, ±4, …
Hence the given set, in set builder form, is {x² : x ∈ Z}.

Example 5. State which of the following sets are finite and which are infinite
(i) A = {x : x ∈ N and x² − 3x + 2 = 0}
(ii) B = {x : x ∈ N and x² = 9}
(iii) C = {x : x ∈ N and x is even}
(iv) D = {x : x ∈ N and 2x − 3 = 0}
Answer:
(i) A = {1, 2}. [∵ x² − 3x + 2 = 0 ⇒ (x − 1)(x − 2) = 0 ⇒ x = 1, 2]
Hence A is finite.
(ii) B = {3}. [∵ x² = 9 ⇒ x = ±3. But −3 ∉ N]
Hence B is finite.
(iii) C = {2, 4, 6, …}
Hence C is infinite.
(iv) D = φ. [∵ 2x − 3 = 0 ⇒ \( x = \frac{3}{2} \notin N \)]
Hence D is finite.

Example 6. Which of the following are empty (null) sets?
(i) Set of odd natural numbers divisible by 2
(ii) {x : 3 < x < 4, x ∈ N}
(iii) {x : x² = 25 and x is an odd integer}
(iv) {x : x² − 2 = 0 and x is rational}
(v) {x : x is common point of any two parallel lines}
Answer:
(i) Since there is no odd natural number which is divisible by 2, it is an empty set.
(ii) Since there is no natural number between 3 and 4, it is an empty set.
(iii) Now x² = 25 ⇒ x = ±5, both are odd. ∴ The set {−5, 5} is non-empty.
(iv) Since there is no rational number whose square is 2, the given set is an empty set.
(v) Since any two parallel lines have no common point, the given set is an empty set.
Hence the sets in (i), (ii), (iv) and (v) are empty sets.

Example 7. Find the pairs of equal sets from the following sets, if any, giving reasons:
A = {0}, B = {x : x > 15 and x < 5},
C = {x : x − 5 = 0}, D = {x : x² = 25},
E = {x : x is a positive integral root of the equation x² − 2x − 15 = 0}.
Answer:
Here we have,
A = {0}
B = φ [∵ There is no number which is greater than 15 and less than 5]
C = {5} [∵ x − 5 = 0 ⇒ x = 5]
D = {−5, 5} [∵ x² = 25 ⇒ x = ±5]
and E = {5} [∵ x² − 2x − 15 = 0 ⇒ (x − 5)(x + 3) = 0 ⇒ x = 5, −3. Out of these two, 5 is positive integral]
Clearly C = E.

Example 8. Are the following pairs of sets equal? Give reasons.
(i) A = {1, 2}, B = {x : x is a solution of x² + 3x + 2 = 0}
(ii) A = {x : x is a letter in the word FOLLOW}, B = {y : y is a letter in the word WOLF}.
Answer:
(i) A = {1, 2}, B = {−2, −1}
[∵ x² + 3x + 2 = 0 ⇒ (x + 2)(x + 1) = 0 ⇒ x = −2, −1]
Clearly A ≠ B.
(ii) A = {F, O, L, L, O, W} = {F, O, L, W}
B = {W, O, L, F} = {F, O, L, W}.
Clearly A = B.

Example 9. Let A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7}, C = {6, 7, 8, 9} and D = {7, 8, 9, 10}. Find:
(a) (i) A ∪ B    (ii) B ∪ D    (iii) A ∪ B ∪ C    (iv) B ∪ C ∪ D
(b) (i) A ∩ B    (ii) B ∩ D    (iii) A ∩ B ∩ C
Answer:
(a) (i) A ∪ B = {1, 2, 3, 4, 5} ∪ {3, 4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7}.
(ii) B ∪ D = {3, 4, 5, 6, 7} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}.
(iii) A ∪ B ∪ C = {1, 2, 3, 4, 5} ∪ {3, 4, 5, 6, 7} ∪ {6, 7, 8, 9} = {1, 2, 3, 4, 5, 6, 7} ∪ {6, 7, 8, 9} = {1, 2, 3, 4, 5, 6, 7, 8, 9}.
(iv) B ∪ C ∪ D = {3, 4, 5, 6, 7} ∪ {6, 7, 8, 9} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}.
(b) (i) A ∩ B = {1, 2, 3, 4, 5} ∩ {3, 4, 5, 6, 7} = {3, 4, 5}.
(ii) B ∩ D = {3, 4, 5, 6, 7} ∩ {7, 8, 9, 10} = {7}.
(iii) A ∩ B ∩ C = {1, 2, 3, 4, 5} ∩ {3, 4, 5, 6, 7} ∩ {6, 7, 8, 9} = {3, 4, 5} ∩ {6, 7, 8, 9} = φ.

Example 10. If A₁ = {2, 3, 4, 5}, A₂ = {3, 4, 5, 6}, A₃ = {4, 5, 6, 7}, find \( \bigcup A_i \) and \( \bigcap A_i \), where i ∈ {1, 2, 3}.
Answer:
(i) \( \bigcup A_i \) = A₁ ∪ A₂ ∪ A₃ = {2, 3, 4, 5} ∪ {3, 4, 5, 6} ∪ {4, 5, 6, 7}
= {2, 3, 4, 5} ∪ {3, 4, 5, 6, 7} = {2, 3, 4, 5, 6, 7}.
(ii) \( \bigcap A_i \) = A₁ ∩ A₂ ∩ A₃ = {2, 3, 4, 5} ∩ {3, 4, 5, 6} ∩ {4, 5, 6, 7}
= {2, 3, 4, 5} ∩ {4, 5, 6} = {4, 5}.

Example 11. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}. Find:
(i) A^c    (ii) B^c    (iii) (A^c)^c    (iv) (A ∪ B)^c
Answer:
(i) A^c = Set of those elements of U which are not in A = {5, 6, 7, 8, 9}.
(ii) B^c = Set of those elements of U which are not in B = {1, 3, 5, 7, 9}.
(iii) (A^c)^c = Set of those elements of U which are not in A^c = {1, 2, 3, 4} = A.
(iv) A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}.
∴ (A ∪ B)^c = Set of those elements of U which are not in (A ∪ B) = {5, 7, 9}.

Example 12. If U = {x : x is a letter in English alphabet}, A = {x : x is a vowel in English alphabet}. Find A^c and (A^c)^c.
Answer:
(i) Since A = {x : x is a vowel in English alphabet},
∴ A^c is the set of those elements of U which are not vowels = {x : x is a consonant in English alphabet}.
(ii) (A^c)^c is the set of those elements of U which are not consonants = {x : x is a vowel in English alphabet} = A.
Hence (A^c)^c = A.

Example 13. Let A = {1, 2, 3, 4, 5, 6}, B = {3, 4, 5, 6, 7, 8}. Find (A − B) ∪ (B − A).
Answer:
We have, A = {1, 2, 3, 4, 5, 6} and B = {3, 4, 5, 6, 7, 8}.
∴ A − B = {1, 2} and B − A = {7, 8}
∴ (A − B) ∪ (B − A) = {1, 2} ∪ {7, 8} = {1, 2, 7, 8}.

Example 14. If A = {1, 2, 3}, B = {4, 5, 6} and C = {7, 8, 9}, verify that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Answer:
We have, A = {1, 2, 3}, B = {4, 5, 6} and C = {7, 8, 9}.
∴ A ∪ B = {1, 2, 3} ∪ {4, 5, 6} = {1, 2, 3, 4, 5, 6} …(1)
A ∪ C = {1, 2, 3} ∪ {7, 8, 9} = {1, 2, 3, 7, 8, 9} …(2)
and B ∩ C = {4, 5, 6} ∩ {7, 8, 9} = φ …(3)
Now A ∪ (B ∩ C) = {1, 2, 3} ∪ φ = {1, 2, 3} …(4)
and (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} ∩ {1, 2, 3, 7, 8, 9} = {1, 2, 3} …(5)
From (4) and (5), A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C), which verifies the result.

Example 15. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)^c = A^c ∩ B^c    (ii) (A ∩ B)^c = A^c ∪ B^c.
Answer:
We have, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}.
(i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B)^c = {1, 9} …(1)
Also A^c = {1, 3, 5, 7, 9} and B^c = {1, 4, 6, 8, 9}
∴ A^c ∩ B^c = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9} …(2)
From (1) and (2), (A ∪ B)^c = A^c ∩ B^c, which verifies the result.
(ii) A ∩ B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}
∴ (A ∩ B)^c = {1, 3, 4, 5, 6, 7, 8, 9} …(3)
and A^c ∪ B^c = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9} …(4)
From (3) and (4), (A ∩ B)^c = A^c ∪ B^c, which verifies the result.

Example 16. If A and B are any two sets, prove using Venn diagrams
(i) A − B = A ∩ B^c    (ii) (A − B) ∪ B = A ∪ B.
Answer:
(i) The first diagram shows A − B shaded. In the second diagram, A ∩ B^c is the region common to A and to the region outside B (B^c), shown shaded.

The shaded regions in both diagrams are the same. Hence A − B = A ∩ B^c.
(ii) The first diagram shows A ∪ B shaded. In the second diagram, the region A − B together with the region B is shaded.

The combination of all the shaded regions represents the union in both cases. Hence (A − B) ∪ B = A ∪ B.

Example 17. Prove that: A ∩ (B − C) = (A ∩ B) − (A ∩ C)
Answer:
Let x be an arbitrary element of A ∩ (B − C). Then x ∈ A ∩ (B − C)
⇒ x ∈ A and x ∈ (B − C)
⇒ x ∈ A and (x ∈ B and x ∉ C)
⇒ (x ∈ A and x ∈ B) and (x ∈ A and x ∉ C)
⇒ x ∈ (A ∩ B) and x ∉ (A ∩ C)
⇒ x ∈ {(A ∩ B) − (A ∩ C)}
∴ A ∩ (B − C) ⊆ (A ∩ B) − (A ∩ C) …(1)
Let y be an arbitrary element of (A ∩ B) − (A ∩ C). Then y ∈ (A ∩ B) − (A ∩ C)
⇒ y ∈ (A ∩ B) and y ∉ (A ∩ C)
⇒ (y ∈ A and y ∈ B) and (y ∈ A and y ∉ C)
⇒ y ∈ A and (y ∈ B and y ∉ C)
⇒ y ∈ A and y ∈ (B − C)
⇒ y ∈ A ∩ (B − C)
∴ (A ∩ B) − (A ∩ C) ⊆ A ∩ (B − C) …(2)
Combining (1) and (2), A ∩ (B − C) = (A ∩ B) − (A ∩ C).

Example 18. Prove the following: A ⊂ B ⇔ B^c ⊂ A^c
Answer:
Let x ∈ B^c, where x is arbitrary.
Now x ∈ B^c ⇒ x ∉ B ⇒ x ∉ A [∵ A ⊂ B] ⇒ x ∈ A^c
∴ B^c ⊂ A^c …(1)
Conversely: Let x ∈ A, where x is arbitrary.
Now x ∈ A ⇒ x ∉ A^c ⇒ x ∉ B^c [∵ B^c ⊂ A^c] ⇒ x ∈ B
∴ A ⊂ B …(2)
Combining (1) and (2), A ⊂ B ⇔ B^c ⊂ A^c.

Example 19. Prove the following: A − B = A − (A ∩ B), where U is the universal set.
Answer:
Let x ∈ (A − B), where x is arbitrary.
Now x ∈ (A − B)
⇔ x ∈ A and x ∉ B
⇔ (x ∈ A and x ∈ A) and x ∉ B [Note this step]
⇔ x ∈ A and (x ∈ A and x ∉ B) [Associative Law]
⇔ x ∈ A and x ∉ (A ∩ B)
⇔ x ∈ A − (A ∩ B)
Hence A − B = A − (A ∩ B).