Sets

A collection is a set only if it is a well defined collection of objects.

(i) All states of India — This is a well defined collection. It is a set.

(ii) Four cities of India having more than one lakh population — There are many cities having more than one lakh population, so it is not clear which four cities are being referred to. This is not well defined. It is not a set.

(iii) All tall students of your school — The word 'tall' is not well defined as it varies from person to person. It is not a set.

(iv) Four colours of a rainbow — A rainbow has seven colours, so it is not clear which four colours are being referred to. This is not well defined. It is not a set.

(v) All beautiful flowers — The word 'beautiful' is not well defined as it varies from person to person. It is not a set.

(vi) All clever people of Lucknow — The word 'clever' is not well defined as it varies from person to person. It is not a set.

(vii) Last three days of a week — This is a well defined collection. It is a set.

(viii) All months of a year having atleast 30 days — This is a well defined collection. It is a set.

Hence, (i), (vii) and (viii) are sets; whereas (ii), (iii), (iv), (v) and (vi) are not sets because these collections are not well defined.

A = {vowels of English alphabet} = {a, e, i, o, u}

(i) c ∈ A — False, because c is not a vowel, so c ∉ A.

(ii) {a} ∈ A — False, because {a} is a set and not an element.

(iii) a, i, u ∈ A — True, because a, i and u are all vowels.

(iv) {a, u} ∉ A — True, because {a, u} is a set and not an element, so it does not belong to A.

(v) {a, i, u} ∈ A — False, because {a, i, u} is a set and not an element.

(vi) a, b ∈ A — False, because b is not a vowel, so b ∉ A. (Of course, a ∈ A.)

Hence, statements (iii) and (iv) are true.

(i) {a, b, c, d, e, f}

The set of first six letters of the English alphabet.

(ii) {2, 3, 5, 7, 11, 13, 17, 19}

{prime numbers less than 20}

(iii) {Friday, Saturday, Sunday}

{last three days of a week}

(iv) {April, August, October}

The names April, August and October all begin with a vowel.

{months of a year whose name begin with a vowel}

(i) The set of even whole numbers which lie between 10 and 50

The even whole numbers between 10 and 50 are 12, 14, 16, ..., 48

Tabular form : {12, 14, 16, ..., 48}

Set builder form : {x : x = 2n, n ∈ N and 5 < n < 25}

(ii) {months of year having more than 30 days}

The months having more than 30 days (i.e. 31 days) are January, March, May, July, August, October and December.

Tabular form : {January, March, May, July, August, October, December}

Set builder form : {x | x is a month of a year having more than 30 days}

(iii) The set of single digit whole numbers which are perfect square

The single digit whole numbers are 0, 1, 2, ..., 9. Out of these, the perfect squares are 0, 1, 4 and 9.

Tabular form : {0, 1, 4, 9}

Set builder form : {x | x is a perfect square one digit number}

(iv) The set of factors of 36

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Tabular form : {1, 2, 3, 4, 6, 9, 12, 18, 36}

Set builder form : {x | x is a factor of 36}

(i) {x | x = 4n, n ∈ W and n < 5}

Here n ∈ W and n < 5, so n = 0, 1, 2, 3, 4

When n = 0, x = 4 × 0 = 0

When n = 1, x = 4 × 1 = 4

When n = 2, x = 4 × 2 = 8

When n = 3, x = 4 × 3 = 12

When n = 4, x = 4 × 4 = 16

Roster form : {0, 4, 8, 12, 16}

Description form : {whole numbers which are divisible by 4 and less than 20}

(ii) {x : x = n², n ∈ N and n < 8}

Here n ∈ N and n < 8, so n = 1, 2, 3, 4, 5, 6, 7

When n = 1, x = 1² = 1

When n = 2, x = 2² = 4

When n = 3, x = 3² = 9

When n = 4, x = 4² = 16

When n = 5, x = 5² = 25

When n = 6, x = 6² = 36

When n = 7, x = 7² = 49

Roster form : {1, 4, 9, 16, 25, 36, 49}

Description form : {squares of first seven natural numbers}

(iii) {y : y = 2x - 1, x ∈ W and x < 5}

Here x ∈ W and x < 5, so x = 0, 1, 2, 3, 4

When x = 0, y = 2 × 0 - 1 = -1

When x = 1, y = 2 × 1 - 1 = 1

When x = 2, y = 2 × 2 - 1 = 3

When x = 3, y = 2 × 3 - 1 = 5

When x = 4, y = 2 × 4 - 1 = 7

Roster form : {-1, 1, 3, 5, 7}

Description form : {odd integers which lie between -2 and 8}

(iv) {x : x is a letter in word ULTIMATUM}

The letters of the word ULTIMATUM are U, L, T, I, M, A, T, U, M. Writing each letter only once, we get U, L, T, I, M and A.

Roster form : {U, L, T, I, M, A}

Description form : {letters in the word ULTIMATUM}

(i) {x | x ∈ N, 5 ≤ x < 10}

Here x ∈ N and 5 ≤ x < 10, so x = 5, 6, 7, 8, 9

{5, 6, 7, 8, 9}

(ii) {x | x = 6p, p ∈ I and -2 ≤ p ≤ 2}

Here p ∈ I and -2 ≤ p ≤ 2, so p = -2, -1, 0, 1, 2

When p = -2, x = 6 × (-2) = -12

When p = -1, x = 6 × (-1) = -6

When p = 0, x = 6 × 0 = 0

When p = 1, x = 6 × 1 = 6

When p = 2, x = 6 × 2 = 12

{-12, -6, 0, 6, 12}

(iii) {x | x = n² - 1, n ∈ N and n < 5}

Here n ∈ N and n < 5, so n = 1, 2, 3, 4

When n = 1, x = 1² - 1 = 0

When n = 2, x = 2² - 1 = 3

When n = 3, x = 3² - 1 = 8

When n = 4, x = 4² - 1 = 15

{0, 3, 8, 15}

(iv) {x | x - 1 = 0}

x - 1 = 0

⇒ x = 1

{1}

(v) {x | x is a consonant in word NOTATION}

The letters of the word NOTATION are N, O, T, A, T, I, O, N. The consonants among these are N, T, T, N. Writing each consonant only once, we get N and T.

{N, T}

(vi) {x | x is a digit in the numeral 11056771}

The digits in the numeral 11056771 are 1, 1, 0, 5, 6, 7, 7, 1. Writing each digit only once, we get 1, 0, 5, 6 and 7.

{1, 0, 5, 6, 7}

(i) {1, 3, 5, 7, ..., 29}

These are odd natural numbers less than 30.

{x : x is an odd natural number and x < 30}

(ii) {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

These are prime numbers less than 30.

{x | x is a prime number and x < 30}

(iii) {1, 4, 9, 16, 25, ...}

These are squares of natural numbers, i.e. 1², 2², 3², 4², 5², ...

{x | x = n², n ∈ N}

(iv) $\Big\lbrace\frac{1}{5}, \frac{1}{6}, \frac{1}{7}, ..., \frac{1}{20}\Big\rbrace$

Each element is of the form $\frac{1}{n}$ where n takes the values from 5 to 20.

Hence, $\Big\lbrace x \mid x = \frac{1}{n}, n ∈ N \text{ and } 5 ≤ n ≤ 20\Big\rbrace$

(v) {-16, -8, 0, 8, 16, 24, 32, 40}

These are multiples of 8, i.e. 8 × (-2), 8 × (-1), 8 × 0, 8 × 1, 8 × 2, 8 × 3, 8 × 4, 8 × 5

{x | x = 8p, p ∈ I and -2 ≤ p ≤ 5}

(vi) {January, June, July}

The names January, June and July all begin with the letter 'J'.

{x : x is a month of a year whose name begins with letter 'J'}

The letters of the word COMPETITION are C, O, M, P, E, T, I, T, I, O, N. The vowels among these are O, E, I, I, O. Writing each vowel only once, we get O, E and I.

(i) Description form : {vowels in the word COMPETITION}

(ii) Set builder form : {x : x is a vowel in the word COMPETITION}

(iii) Roster form : {O, E, I}

(i) {all colours of a rainbow}

A rainbow has 7 colours, i.e. violet, indigo, blue, green, yellow, orange and red.

Finite set; cardinal number = 7

(ii) {x | x is a prime number between 7 and 11}

There is no prime number between 7 and 11 (8, 9 and 10 are not prime).

Empty set

(iii) {multiples of 5}

The multiples of 5 are 5, 10, 15, 20, ... which never end.

Infinite set

(iv) {all straight lines drawn in a plane}

An unlimited number of straight lines can be drawn in a plane.

Infinite set

(v) {x | x is a digit in the numeral 550131527}

The digits in the numeral are 5, 5, 0, 1, 3, 1, 5, 2, 7. Writing each digit only once, we get 5, 0, 1, 3, 2 and 7, which are 6 in number.

Finite set; cardinal number = 6

(vi) {x | x is a letter in word SUFFICIENT}

The letters of the word SUFFICIENT are S, U, F, F, I, C, I, E, N, T. Writing each letter only once, we get S, U, F, I, C, E, N and T, which are 8 in number.

Finite set; cardinal number = 8

(vii) {x | x = 4n, n ∈ I and x < 10}

Here n ∈ I (integers), so n = ..., -2, -1, 0, 1, 2 (as 4 × 2 = 8 < 10).

Then x = ..., -8, -4, 0, 4, 8 which never ends on the negative side.

Infinite set

(viii) {x | x ∈ N, x is a prime factor of 180}

180 = 2² × 3² × 5

The prime factors of 180 are 2, 3 and 5, which are 3 in number.

Finite set; cardinal number = 3

(ix) {x : x is a vowel in the word WHY}

There is no vowel in the word WHY.

Empty set

(x) {x : x = 5n, n ∈ W and x < 60}

Here x = 5n and x < 60, so n = 0, 1, 2, ..., 11 (as 5 × 11 = 55 < 60).

Then x = 0, 5, 10, ..., 55 which are 12 in number.

Finite set; cardinal number = 12

(i) {vowels of English alphabet} = {a, e, i, o, u} and {e, a, u, i, o}

Both sets have the same elements.

They describe the same set.

(ii) {a, b, d} and {d, a, b, b} = {d, a, b}

Both sets have the same elements.

They describe the same set.

(iii) {letters of PUPPET} = {P, U, E, T} and {E, T, P, U}

Both sets have the same elements.

They describe the same set.

(iv) {1, 2, 3} and {2, 3, 4}

1 is in the first set but not in the second, and 4 is in the second set but not in the first.

They do not describe the same set.

(v) {1, 2, 3, 4, 5} and {x | x ∈ N, x ≤ 5} = {1, 2, 3, 4, 5}

Both sets have the same elements.

They describe the same set.

Hence, (i), (ii), (iii) and (v) describe the same sets.

Let us write each set in roster form.

A = {0, 1, 2, 3}

B = {x : x² < 10, x ∈ W} = {0, 1, 2, 3} (since 0² = 0, 1² = 1, 2² = 4, 3² = 9 are all less than 10)

C = {letters of word FOLLOW} = {F, O, L, W}

D = {days of a week} = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

E = {x | x ∈ W, x < 4} = {0, 1, 2, 3}

F = {letters of word FLOW} = {F, L, O, W}

G = {Monday, Tuesday, ..., Sunday} = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

H = {letters of word WOLF} = {W, O, L, F}

Comparing the sets :

A, B and E are equal; C, F and H are equal; D and G are equal.

Two sets are equivalent if they have the same number of elements. Let us find the cardinal number of each set.

A = {colours of a rainbow} → 7 elements

B = {letters of word GOOD} = {G, O, D} → 3 elements

C = {x : x is a digit in the numeral 371011489} = {3, 7, 1, 0, 4, 8, 9} → 7 elements

D = {letters of word TOM} = {T, O, M} → 3 elements

E = {x : x ∈ I, x² < 10} = {-3, -2, -1, 0, 1, 2, 3} → 7 elements

F = {months of a year} → 12 elements

G = {days of a week} → 7 elements

H = {x | x = 3n, n ∈ W and n < 12} = {0, 3, 6, ..., 33} → 12 elements

I = {all even numbers between 1 and 53} = {2, 4, 6, ..., 52} → 26 elements

J = {all letters of English alphabet} → 26 elements

Comparing the cardinal numbers :

A, C, E and G are equivalent (these have 7 elements); B ↔ D; F ↔ H; I ↔ J.

(i) A = {1, 2, 3}, B = {2, 3, 3, 3, 1, 3} = {1, 2, 3}

Every element of A is in B and every element of B is in A.

A ⊂ B, B ⊂ A. In fact, A = B.

(ii) A = {2, 4, 6, ...}, B = {all natural numbers} = {1, 2, 3, 4, ...}

Every even natural number is a natural number, so A ⊂ B. But 1 ∈ B and 1 ∉ A, so B ⊄ A.

A ⊂ B but B ⊄ A

(iii) A = {x | x ∈ I, x² < 20} = {-4, -3, -2, -1, 0, 1, 2, 3, 4} (since 4² = 16 < 20), B = {0, 1, 2, 3, 4}

Every element of B is in A, so B ⊂ A. But -1 ∈ A and -1 ∉ B, so A ⊄ B.

B ⊂ A but A ⊄ B

(iv) A = {letters of KING} = {K, I, N, G}, B = {letters of QUEEN} = {Q, U, E, N}

Not every element of A is in B, and not every element of B is in A.

Neither A ⊂ B nor B ⊂ A

A = {letters of CLOUD} = {C, L, O, U, D}

B = {letters of KOLKATA} = {K, O, L, A, T}

(i) A ⊂ B

C, U and D are in A but not in B, so A is not a subset of B.

False

(ii) B ⊂ A

K, A and T are in B but not in A, so B is not a subset of A.

False

(iii) A ↔ B

n(A) = 5 and n(B) = 5, so both sets have the same number of elements.

True

(i) $\phi$

The only subset of the empty set is the empty set itself.

$\phi$

(ii) {3, 5}

The subsets are :

$\phi$, {3}, {5}, {3, 5}

(iii) {2, 4, 6}

The subsets are :

$\phi$, {2}, {4}, {6}, {2, 4}, {4, 6}, {2, 6}, {2, 4, 6}

A = {x : x = 2n, n < 5}

(i) $\xi$ = N

When the universal set is the set of all natural numbers, n = 1, 2, 3, 4

x = 2n gives 2, 4, 6, 8

A = {2, 4, 6, 8}

(ii) $\xi$ = W

When the universal set is the set of all whole numbers, n = 0, 1, 2, 3, 4

x = 2n gives 0, 2, 4, 6, 8

A = {0, 2, 4, 6, 8}

(iii) $\xi$ = I

When the universal set is the set of all integers, n = ..., -2, -1, 0, 1, 2, 3, 4

x = 2n gives ..., -4, -2, 0, 2, 4, 6, 8

A = {..., -4, -2, 0, 2, 4, 6, 8}

(i) If x is not a member of the set A, then symbolically we write it as x ∉ A.

(ii) Each element of a set is listed once and only once, repetitions are removed.

(iii) A set that contains a limited number of different elements is called a finite set.

(iv) Two finite sets are called equivalent if and only if they have equal number of elements.

(i) A collection of books is a set.

A general collection of books does not specify which books, so it is not a well defined collection.

False

(ii) If X = {letters of the word PRINCIPAL}, then cardinal number of set X is 9.

X = {letters of PRINCIPAL} = {P, R, I, N, C, A, L}, so n(X) = 7, not 9.

False

(iii) If P = {letters of the word AHMEDABAD}, then n(P) = 6.

P = {letters of AHMEDABAD} = {A, H, M, E, D, B}, so n(P) = 6.

True

(iv) If A is any set, then A ⊂ A.

Every set is a subset of itself.

True

(v) Empty set is a subset of every set.

The empty set is a subset of every set.

True

(vi) If set A = {0}, then n(A) = 0.

A = {0} has one element, namely 0, so n(A) = 1, not 0.

False

(vii) If A and B are two sets such that A ↔ B, then A = B.

Equivalent sets need not be equal; they only have the same number of elements.

False

A set with a limited (countable) number of distinct elements is called a finite set.

Hence, option 1 is the correct option.

The symbol ↔ is used between two equivalent sets.

Hence, option 3 is the correct option.

The empty set has no elements and is denoted by { } (or by $\phi$). Note that {$\phi$} and {0} are not empty as each has one element.

Hence, option 2 is the correct option.

The odd prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19. (2 is prime but it is even.)

So A = {3, 5, 7, 11, 13, 17, 19} and n(A) = 7.

Hence, option 2 is the correct option.

A = {x | x is a positive multiple of 3 less than 20} = {3, 6, 9, 12, 15, 18}, so n(A) = 6

B = {x | x is a prime number less than 20} = {2, 3, 5, 7, 11, 13, 17, 19}, so n(B) = 8

n(A) + n(B) = 6 + 8 = 14

Hence, option 4 is the correct option.

A = {x | x ∈ N, x² > 100} = {11, 12, 13, ...}, which is an infinite set, so n(A) is not defined and is certainly not 9. So Statement I is false.

The cardinal number of an infinite set is not defined. So Statement II is true.

Hence, option 2 is the correct option.

A = {x | x is a colour in the rainbow} has 7 elements, so n(A) = 7.

B = {x | x is a vowel in the English alphabet} = {a, e, i, o, u}, so n(B) = 5.

Since 5 < 7, n(B) < n(A) is true. So Statement I is true.

A singleton set has exactly one element, so its cardinal number is 1. So Statement II is true.

Hence, option 3 is the correct option.

A = {x | x ∈ N, x² ≤ 81} = {1, 2, 3, 4, 5, 6, 7, 8, 9} (since 9² = 81)

B = {x | x ∈ N, 3x ≤ 27} = {1, 2, 3, 4, 5, 6, 7, 8, 9} (since 3 × 9 = 27)

Both sets have exactly the same elements, so A = B. So Statement I is true.

If two sets are equal, they have the same number of elements, so they are also equivalent, i.e. A ↔ B. So Statement II is true.

Hence, option 3 is the correct option.

(i) The set of even integers which lie between -6 and 10

The even integers between -6 and 10 are -4, -2, 0, 2, 4, 6 and 8.

Tabular form : {-4, -2, 0, 2, 4, 6, 8}

Set builder form : {x : x = 2n, n ∈ I and -3 < n < 5}

(ii) The set of two digit numbers which are perfect square

The two digit perfect squares are 16, 25, 36, 49, 64 and 81 (i.e. 4², 5², 6², 7², 8², 9²).

Tabular form : {16, 25, 36, 49, 64, 81}

Set builder form : {x : x = n², n ∈ N and 4 ≤ n ≤ 9}

(iii) {factors of 42}

The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Tabular form : {1, 2, 3, 6, 7, 14, 21, 42}

Set builder form : {x : x is a factor of 42}

(i) {x : x = 5n, n ∈ I and -3 < n ≤ 3}

Here n = -2, -1, 0, 1, 2, 3

When n = -2, x = 5 × (-2) = -10

When n = -1, x = 5 × (-1) = -5

When n = 0, x = 5 × 0 = 0

When n = 1, x = 5 × 1 = 5

When n = 2, x = 5 × 2 = 10

When n = 3, x = 5 × 3 = 15

{-10, -5, 0, 5, 10, 15}

(ii) {x : x = n², n ∈ W and n < 5}

Here n = 0, 1, 2, 3, 4

When n = 0, x = 0² = 0

When n = 1, x = 1² = 1

When n = 2, x = 2² = 4

When n = 3, x = 3² = 9

When n = 4, x = 4² = 16

{0, 1, 4, 9, 16}

(iii) {x : x = n² - 2, n ∈ W and n < 4}

Here n = 0, 1, 2, 3

When n = 0, x = 0² - 2 = -2

When n = 1, x = 1² - 2 = -1

When n = 2, x = 2² - 2 = 2

When n = 3, x = 3² - 2 = 7

{-2, -1, 2, 7}

(i) {-14, -7, 0, 7, 14, 21, 28}

These are multiples of 7, i.e. 7 × (-2), 7 × (-1), 7 × 0, 7 × 1, 7 × 2, 7 × 3, 7 × 4

{x | x = 7n, n ∈ I and -2 ≤ n ≤ 4}

(ii) {1, 2, 3, 6, 9, 18}

These are the factors of 18.

{x | x ∈ N, x is a factor of 18}

(i) The set of even prime numbers

The only even prime number is 2, so the set is {2}.

Finite set; cardinal number = 1

(ii) {multiples of 9}

The multiples of 9 are 9, 18, 27, 36, ... which never end.

Infinite set

(iii) {x : x is a prime factor of 84}

84 = 2² × 3 × 7

The prime factors of 84 are 2, 3 and 7, which are 3 in number.

Finite set; cardinal number = 3

(iv) {x : 2x + 5 = 1, x ∈ N}

2x + 5 = 1

⇒ 2x = 1 - 5

⇒ 2x = -4

⇒ x = -2

Since -2 is not a natural number, there is no such x.

Empty set

(v) {x : x is a month of a year having less than 30 days}

Only February has less than 30 days (28 or 29 days), so the set is {February}.

Finite set; cardinal number = 1

(vi) {x | x is a month of a leap year having 28 days}

In a leap year, February has 29 days. So no month of a leap year has 28 days.

Empty set

(i) A = {1, 3, 5}, B = {Red, Blue, Green}

n(A) = 3 and n(B) = 3, so the sets are equivalent. But their elements are different, so the sets are not equal.

A ↔ B; A ≠ B

(ii) A = {prime factors of 70}, B = {prime factors of 60}

70 = 2 × 5 × 7, so A = {2, 5, 7} and n(A) = 3

60 = 2² × 3 × 5, so B = {2, 3, 5} and n(B) = 3

n(A) = n(B), so the sets are equivalent. But their elements are different (7 is in A but not in B), so the sets are not equal.

A ↔ B; A ≠ B

(iii) A = {even natural numbers less than 10}, B = {odd natural numbers less than 10}

A = {2, 4, 6, 8}, so n(A) = 4

B = {1, 3, 5, 7, 9}, so n(B) = 5

Since n(A) ≠ n(B), the sets are not equivalent.

A is not equivalent to B

A = {x | x ∈ N, x < 40 and x is a multiple of 6} = {6, 12, 18, 24, 30, 36}, so n(A) = 6

B = {x | x ∈ W, x ≤ 40 and x is a multiple of 8} = {0, 8, 16, 24, 32, 40}, so n(B) = 6

C = {x | x is a factor of 28} = {1, 2, 4, 7, 14, 28}, so n(C) = 6

(i) A ↔ B

n(A) = 6 = n(B), so A and B are equivalent.

True

(ii) B ↔ C

n(B) = 6 = n(C), so B and C are equivalent.

True

(iii) A ↔ C

n(A) = 6 = n(C), so A and C are equivalent.

True