Sets

A set is a well-defined collection of objects, meaning there is no ambiguity about whether an object belongs to the collection or not.

(i) Yes, it is a set. The months of a year beginning with 'J' are January, June, and July — clearly and unambiguously defined. So this is a set: {January, June, July}.

(ii) No, it is not a set. The term 'most talented' is subjective and varies from person to person. There is no definite criterion, so the collection is not well-defined.

(iii) No, it is not a set. The term 'best-cricket batsmen' is subjective. Different selectors may choose different players, so the collection is not well-defined.

(iv) Yes, it is a set. The collection of all boys in your class is well-defined — for any boy, it can be determined whether he belongs to your class or not.

(v) Yes, it is a set. The natural numbers less than 100 are precisely 1, 2, 3, …, 99. This is a well-defined collection.

(vi) Yes, it is a set. The novels written by Munshi Prem Chand are well-defined — one can verify whether a given novel was written by him or not.

(vii) Yes, it is a set. Even integers are well-defined: …, −4, −2, 0, 2, 4, … There is no ambiguity.

(viii) Yes, it is a set. The questions in this chapter are fixed and well-defined.

(ix) No, it is not a set. The term 'most dangerous' is subjective and not well-defined. Different people may have different opinions.

Given: $A = \{1, 2, 3, 4, 5, 6\}$

We check whether each number belongs to A or not:

(i) $5 \in A$ — since 5 is an element of A.

(ii) $8 \notin A$ — since 8 is not an element of A.

(iii) $0 \notin A$ — since 0 is not an element of A.

(iv) $4 \in A$ — since 4 is an element of A.

(v) $2 \in A$ — since 2 is an element of A.

(vi) $10 \notin A$ — since 10 is not an element of A.

(i) The integers satisfying -3 \leq x < 7 are: $-3, -2, -1, 0, 1, 2, 3, 4, 5, 6$.
$$A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$$

(ii) The natural numbers less than 6 are: 1, 2, 3, 4, 5.
$$B = \{1, 2, 3, 4, 5\}$$

(iii) Two-digit natural numbers whose digits sum to 8:
- 17 (1+7=8), 26 (2+6=8), 35 (3+5=8), 44 (4+4=8), 53 (5+3=8), 62 (6+2=8), 71 (7+1=8), 80 (8+0=8)
$$C = \{17, 26, 35, 44, 53, 62, 71, 80\}$$

(iv) $60 = 2 \times 2 \times 3 \times 5$. The prime divisors of 60 are 2, 3, and 5.
$$D = \{2, 3, 5\}$$

(v) The word TRIGONOMETRY has letters: T, R, I, G, O, N, O, M, E, T, R, Y. Removing repetitions:
$$E = \{T, R, I, G, O, N, M, E, Y\}$$

(vi) The word BETTER has letters: B, E, T, T, E, R. Removing repetitions:
$$F = \{B, E, T, R\}$$

(i) The elements 3, 6, 9, 12 are multiples of 3 not exceeding 12.
$$\{3, 6, 9, 12\} = \{x : x = 3n,\ n \in \mathbf{N},\ 1 \leq n \leq 4\}$$

(ii) The elements 2, 4, 8, 16, 32 are powers of 2: $2^1, 2^2, 2^3, 2^4, 2^5$.
$$\{2, 4, 8, 16, 32\} = \{x : x = 2^n,\ n \in \mathbf{N},\ 1 \leq n \leq 5\}$$

(iii) The elements 5, 25, 125, 625 are powers of 5: $5^1, 5^2, 5^3, 5^4$.
$$\{5, 25, 125, 625\} = \{x : x = 5^n,\ n \in \mathbf{N},\ 1 \leq n \leq 4\}$$

(iv) The elements 2, 4, 6, … are all positive even integers.
$$\{2, 4, 6, \ldots\} = \{x : x \text{ is an even natural number}\}$$
or equivalently $\{x : x = 2n,\ n \in \mathbf{N}\}$.

(v) The elements 1, 4, 9, …, 100 are perfect squares of natural numbers from 1 to 10.
$$\{1, 4, 9, \ldots, 100\} = \{x : x = n^2,\ n \in \mathbf{N},\ 1 \leq n \leq 10\}$$

(i) Odd natural numbers are 1, 3, 5, 7, 9, …
$$A = \{1, 3, 5, 7, 9, \ldots\}$$
This is an infinite set.

(ii) We need integers $x$ such that -\frac{1}{2} < x < \frac{9}{2}, i.e., -0.5 < x < 4.5.
The integers in this range are: 0, 1, 2, 3, 4.
$$B = \{0, 1, 2, 3, 4\}$$

(iii) We need integers $x$ such that $x^2 \leq 4$, i.e., $-2 \leq x \leq 2$.
The integers are: $-2, -1, 0, 1, 2$.
$$C = \{-2, -1, 0, 1, 2\}$$

(iv) The letters in the word LOYAL are L, O, Y, A, L. Removing repetition:
$$D = \{L, O, Y, A\}$$

(v) Months not having 31 days: February (28/29 days), April (30), June (30), September (30), November (30).
$$E = \{\text{February, April, June, September, November}\}$$

(vi) Consonants in the English alphabet that precede k (i.e., come before k):
The letters before k are: a, b, c, d, e, f, g, h, i, j. Among these, the consonants are b, c, d, f, g, h, j.
$$F = \{b, c, d, f, g, h, j\}$$

Concept: Match each roster set with its corresponding set-builder description.

(i) $\{1, 2, 3, 6\}$: These are all the natural number divisors of 6 (since $6 = 1 \times 6 = 2 \times 3$). This matches (c).

(ii) $\{2, 3\}$: These are the prime numbers that are also divisors of 6. This matches (a).

(iii) $\{M, A, T, H, E, I, C, S\}$: The word MATHEMATICS has letters M, A, T, H, E, M, A, T, I, C, S. Removing repetitions gives $\{M, A, T, H, E, I, C, S\}$. This matches (d).

(iv) $\{1, 3, 5, 7, 9\}$: These are the odd natural numbers less than 10. This matches (b).

Summary: (i) → (c), (ii) → (a), (iii) → (d), (iv) → (b).

A null (empty) set is a set that contains no elements.

(i) Null set. No odd natural number is divisible by 2 (odd and even are mutually exclusive). So this set has no elements: it is a null set.

(ii) Not a null set. The number 2 is an even prime number. So this set = {2}, which is non-empty.

(iii) Null set. There is no natural number that is simultaneously less than 5 and greater than 7. So this set has no elements.

(iv) Null set. Two parallel lines never intersect, so they have no common point. This set is empty.

(i) Finite. There are exactly 12 months in a year.

(ii) Infinite. The set of all natural numbers $\{1, 2, 3, \ldots\}$ has no last element; it is infinite.

(iii) Finite. The set contains exactly 100 elements.

(iv) Infinite. The positive integers greater than 100 are 101, 102, 103, … — there is no end, so the set is infinite.

(v) Finite. There are finitely many prime numbers less than 99 (in fact, 24 such primes).

(i) Infinite. For every real number $c$, the line $y = c$ is parallel to the $x$-axis. Since there are infinitely many real numbers, there are infinitely many such lines.

(ii) Finite. The English alphabet has exactly 26 letters.

(iii) Infinite. Multiples of 5 are 5, 10, 15, 20, … — this list never ends, so the set is infinite.

(iv) Finite. Although the number is very large, the number of animals living on earth at any given time is a definite (finite) number.

(v) Infinite. Infinitely many circles can pass through the origin (they can have any centre and the radius is determined by the centre's distance from the origin). So this set is infinite.

Two sets are equal if and only if they have exactly the same elements.

(i) $A = \{a, b, c, d\}$ and $B = \{d, c, b, a\}$. Both sets contain exactly the same elements a, b, c, d (order does not matter in sets). Therefore, $A = B$.

(ii) $A = \{4, 8, 12, 16\}$ and $B = \{8, 4, 16, 18\}$. The element 12 is in A but not in B, and 18 is in B but not in A. Therefore, $A \neq B$.

(iii) $B = \{x : x \text{ is a positive even integer and } x \leq 10\} = \{2, 4, 6, 8, 10\}$. This is the same as A. Therefore, $A = B$.

(iv) $A = \{10, 20, 30, 40, \ldots\}$ (multiples of 10) and $B = \{10, 15, 20, 25, 30, \ldots\}$ (multiples of 5). The element 15 is in B but not in A (since 15 is not a multiple of 10). Therefore, $A \neq B$.

(i) Solve $x^2 + 5x + 6 = 0$:
$$x^2 + 5x + 6 = (x+2)(x+3) = 0 \implies x = -2 \text{ or } x = -3$$
So $B = \{-2, -3\}$.

But $A = \{2, 3\}$. Since $2 \neq -2$ and $3 \neq -3$, we have $A \neq B$. The sets are not equal.

(ii) Letters in FOLLOW: F, O, L, L, O, W → removing repetitions: $A = \{F, O, L, W\}$.
Letters in WOLF: W, O, L, F → $B = \{W, O, L, F\}$.

Both sets contain exactly the same elements F, O, L, W. Therefore, $A = B$. The sets are equal.

We compare each set element by element:

- $A = \{2, 4, 8, 12\}$
- $B = \{1, 2, 3, 4\}$
- $C = \{4, 8, 12, 14\}$
- $D = \{3, 1, 4, 2\} = \{1, 2, 3, 4\}$
- $E = \{-1, 1\}$
- $F = \{0, a\}$
- $G = \{1, -1\} = \{-1, 1\}$
- $H = \{0, 1\}$

Comparing:
- $B = \{1, 2, 3, 4\}$ and $D = \{1, 2, 3, 4\}$ → $B = D$.
- $E = \{-1, 1\}$ and $G = \{-1, 1\}$ → $E = G$.
- No other pairs are equal (A ≠ C since 8, 12 ∈ A but 14 ∉ A; F and H differ from all others).

Equal sets: $B = D$ and $E = G$.

$A \subset B$ means every element of A is also in B.

(i) Every element of $\{2,3,4\}$ is in $\{1,2,3,4,5\}$.
$$\{2,3,4\} \subset \{1,2,3,4,5\}$$

(ii) $a \in \{a,b,c\}$ but $a \notin \{b,c,d\}$.
$$\{a,b,c\} \not\subset \{b,c,d\}$$

(iii) Every Class XI student of your school is also a student of your school.
$$\{x: x \text{ is a student of Class XI}\} \subset \{x: x \text{ is a student of your school}\}$$

(iv) A circle in the plane need not have radius 1 unit. So not every circle belongs to the second set.
$$\{x: x \text{ is a circle in the plane}\} \not\subset \{x: x \text{ is a circle with radius 1 unit}\}$$

(v) A triangle is not a rectangle, so no triangle belongs to the set of rectangles.
$$\{x: x \text{ is a triangle}\} \not\subset \{x: x \text{ is a rectangle}\}$$

(vi) Every equilateral triangle is a triangle.
$$\{x: x \text{ is an equilateral triangle}\} \subset \{x: x \text{ is a triangle}\}$$

(vii) Every even natural number is an integer.
$$\{x: x \text{ is an even natural number}\} \subset \{x: x \text{ is an integer}\}$$

(i) $\{a,b\} \not\subset \{b,c,a\}$: Both $a$ and $b$ are in $\{b,c,a\}$. So $\{a,b\} \subset \{b,c,a\}$, which means the statement $\{a,b\} \not\subset \{b,c,a\}$ is False.

(ii) $\{a,e\} \subset \{x: x \text{ is a vowel}\}$: The vowels are a, e, i, o, u. Both $a$ and $e$ are vowels. So the statement is True.

(iii) $\{1,2,3\} \subset \{1,3,5\}$: The element $2 \in \{1,2,3\}$ but $2 \notin \{1,3,5\}$. So the statement is False.

(iv) $\{a\} \subset \{a,b,c\}$: The only element of $\{a\}$ is $a$, and $a \in \{a,b,c\}$. So the statement is True.

(v) $\{a\} \in \{a,b,c\}$: The elements of $\{a,b,c\}$ are $a$, $b$, $c$ — these are individual letters, not sets. The set $\{a\}$ is not an element of $\{a,b,c\}$. So the statement is False.

(vi) Even natural numbers less than 6: $\{2, 4\}$. Natural numbers that divide 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Both 2 and 4 divide 36. So $\{2,4\} \subset \{1,2,3,4,6,9,12,18,36\}$. The statement is True.

Note: $A = \{1, 2, \{3,4\}, 5\}$ has four elements: $1$, $2$, $\{3,4\}$ (a set), and $5$.

(i) $\{3,4\} \subset A$: For this to be true, both 3 and 4 must be elements of A. But 3 and 4 are not elements of A (only $\{3,4\}$ as a whole is). Incorrect. The correct statement is $\{3,4\} \in A$.

(ii) $\{3,4\} \in A$: The set $\{3,4\}$ is indeed one of the elements of A. Correct.

(iii) $\{\{3,4\}\} \subset A$: The only element of $\{\{3,4\}\}$ is $\{3,4\}$, and $\{3,4\} \in A$. So $\{\{3,4\}\} \subset A$. Correct.

(iv) $1 \in A$: 1 is an element of A. Correct.

(v) $1 \subset A$: The symbol $\subset$ is used between sets. 1 is an element (not a set), so writing $1 \subset A$ is incorrect. Incorrect. The correct statement is $1 \in A$.

(vi) $\{1,2,5\} \subset A$: Elements 1, 2, 5 all belong to A. Correct.

(vii) $\{1,2,3\} \subset A$: The element 3 is not in A (only $\{3,4\}$ is). Incorrect.

(ix) $\phi \in A$: The empty set $\phi$ is not listed as an element of A. Incorrect.

(x) $\phi \subset A$: The empty set is a subset of every set. Correct.

(xi) $\{\phi\} \subset A$: For this, $\phi$ must be an element of A. But $\phi \notin A$. Incorrect.

A set with $n$ elements has $2^n$ subsets.

(i) $\{a\}$ has 1 element, so $2^1 = 2$ subsets:
$$\phi,\ \{a\}$$

(ii) $\{a,b\}$ has 2 elements, so $2^2 = 4$ subsets:
$$\phi,\ \{a\},\ \{b\},\ \{a,b\}$$

(iii) $\{1,2,3\}$ has 3 elements, so $2^3 = 8$ subsets:
$$\phi,\ \{1\},\ \{2\},\ \{3\},\ \{1,2\},\ \{1,3\},\ \{2,3\},\ \{1,2,3\}$$

(iv) $\phi$ has 0 elements, so $2^0 = 1$ subset:
$$\phi$$
(The empty set is the only subset of itself.)

Recall: (a,b) = \{x: a < x < b\}, $[a,b] = \{x: a \leq x \leq b\}$, (a,b] = \{x: a < x \leq b\}, [a,b) = \{x: a \leq x < b\}.

(i) -4 < x \leq 6 means $x$ is greater than $-4$ (not included) and at most 6 (included):
$$(-4, 6]$$

(ii) -12 < x < -10 means both endpoints are excluded:
$$(-12, -10)$$

(iii) 0 \leq x < 7 means 0 is included and 7 is excluded:
$$[0, 7)$$

(iv) $3 \leq x \leq 4$ means both endpoints are included:
$$[3, 4]$$

(i) $(-3, 0)$ means all real numbers strictly between $-3$ and $0$:
\{x : x \in \mathbb{R},\ -3 < x < 0\}

(ii) $[6, 12]$ means all real numbers from 6 to 12, both included:
$$\{x : x \in \mathbb{R},\ 6 \leq x \leq 12\}$$

(iii) $(6, 12]$ means 6 is excluded and 12 is included:
\{x : x \in \mathbb{R},\ 6 < x \leq 12\}

(iv) $[-23, 5)$ means $-23$ is included and 5 is excluded:
\{x : x \in \mathbb{R},\ -23 \leq x < 5\}

A universal set must contain all elements of the set under consideration.

(i) For the set of right triangles, a suitable universal set is:
$$U = \{x : x \text{ is a triangle in a plane}\}$$
or more broadly, the set of all polygons in a plane.

(ii) For the set of isosceles triangles, a suitable universal set is:
$$U = \{x : x \text{ is a triangle in a plane}\}$$
or more broadly, the set of all polygons in a plane.

In both cases, the set of all triangles in a plane serves as a natural universal set.

A universal set U must contain all elements of A, B, and C as subsets, i.e., every element of A, B, C must be in U.

$A = \{1,3,5\}$, $B = \{2,4,6\}$, $C = \{0,2,4,6,8\}$.

All elements needed: 0, 1, 2, 3, 4, 5, 6, 8.

(i) $\{0,1,2,3,4,5,6\}$: The element $8 \in C$ but $8 \notin \{0,1,2,3,4,5,6\}$. So this cannot be a universal set.

(ii) $\phi$: The empty set contains no elements, so A, B, C cannot be subsets of it. Cannot be a universal set.

(iii) $\{0,1,2,3,4,5,6,7,8,9,10\}$: Contains 0,1,2,3,4,5,6,8 — all required elements. $A \subset U$, $B \subset U$, $C \subset U$. Can be a universal set. ✓

(iv) $\{1,2,3,4,5,6,7,8\}$: The element $0 \in C$ but $0 \notin \{1,2,3,4,5,6,7,8\}$. Cannot be a universal set.

Answer: Only (iii) can be considered as a universal set for A, B and C.

$A \cup B = \{x : x \in A \text{ or } x \in B\}$

(i) $X \cup Y = \{1, 2, 3, 5\}$

(ii) $A \cup B = \{a, b, c, e, i, o, u\}$

(iii) $A = \{3, 6, 9, 12, \ldots\}$, $B = \{1, 2, 3, 4, 5\}$.
$$A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, 15, \ldots\}$$
More precisely: $A \cup B = \{x : x = 1, 2, 4, 5 \text{ or } x \text{ is a multiple of 3}\}$.

(iv) $A = \{2, 3, 4, 5, 6\}$, $B = \{7, 8, 9\}$.
$$A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9\}$$

(v) $A \cup \phi = \{1, 2, 3\}$ (union with empty set gives the set itself).

Given: $A = \{a, b\}$, $B = \{a, b, c\}$.

Is $A \subset B$? Every element of A (i.e., $a$ and $b$) is also in B. Therefore, $A \subset B$. Yes.

$A \cup B$: Since $A \subset B$, every element of A is already in B.
$$A \cup B = \{a, b, c\} = B$$

Given: $A \subset B$, meaning every element of A is also in B.

Concept: $A \cup B$ is the set of all elements in A or B (or both).

Since every element of A is already in B, combining A and B gives no new elements beyond B.

$$A \cup B = B$$

Given: $A = \{1,2,3,4\}$, $B = \{3,4,5,6\}$, $C = \{5,6,7,8\}$, $D = \{7,8,9,10\}$.

(i) $A \cup B = \{1, 2, 3, 4, 5, 6\}$

(ii) $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(iii) $B \cup C = \{3, 4, 5, 6, 7, 8\}$

(iv) $B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$

(v) $A \cup B \cup C = (A \cup B) \cup C = \{1,2,3,4,5,6\} \cup \{5,6,7,8\} = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(vi) $A \cup B \cup D = \{1,2,3,4,5,6\} \cup \{7,8,9,10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

(vii) $B \cup C \cup D = \{3,4,5,6,7,8\} \cup \{7,8,9,10\} = \{3, 4, 5, 6, 7, 8, 9, 10\}$

$A \cap B = \{x : x \in A \text{ and } x \in B\}$

(i) Common elements of $\{1,3,5\}$ and $\{1,2,3\}$: 1 and 3.
$$X \cap Y = \{1, 3\}$$

(ii) Common elements of $\{a,e,i,o,u\}$ and $\{a,b,c\}$: only $a$.
$$A \cap B = \{a\}$$

(iii) $A = \{3,6,9,12,\ldots\}$, $B = \{1,2,3,4,5\}$. Common element: 3.
$$A \cap B = \{3\}$$

(iv) $A = \{2,3,4,5,6\}$, $B = \{7,8,9\}$. No common elements.
$$A \cap B = \phi$$

(v) $A \cap \phi = \phi$ (intersection with empty set is always empty).

Given: $A = \{3,5,7,9,11\}$, $B = \{7,9,11,13\}$, $C = \{11,13,15\}$, $D = \{15,17\}$.

(i) $A \cap B$: Common elements of A and B: 7, 9, 11.
$$A \cap B = \{7, 9, 11\}$$

(ii) $B \cap C$: Common elements of B and C: 11, 13.
$$B \cap C = \{11, 13\}$$

(iii) $A \cap C \cap D$: $A \cap C = \{11\}$; $\{11\} \cap D = \phi$ (11 ∉ D).
$$A \cap C \cap D = \phi$$

(iv) $A \cap C$: Common elements: 11.
$$A \cap C = \{11\}$$

(v) $B \cap D$: B = {7,9,11,13}, D = {15,17}. No common elements.
$$B \cap D = \phi$$

(vi) $A \cap (B \cup C)$: First, $B \cup C = \{7,9,11,13,15\}$. Then $A \cap \{7,9,11,13,15\} = \{7,9,11\}$.
$$A \cap (B \cup C) = \{7, 9, 11\}$$

(vii) $A \cap D$: A = {3,5,7,9,11}, D = {15,17}. No common elements.
$$A \cap D = \phi$$

(viii) $A \cap (B \cup D)$: $B \cup D = \{7,9,11,13,15,17\}$. $A \cap \{7,9,11,13,15,17\} = \{7,9,11\}$.
$$A \cap (B \cup D) = \{7, 9, 11\}$$

(ix) $(A \cap B) \cap (B \cup C)$: $A \cap B = \{7,9,11\}$; $B \cup C = \{7,9,11,13,15\}$. Intersection: $\{7,9,11\}$.
$$(A \cap B) \cap (B \cup C) = \{7, 9, 11\}$$

(x) $(A \cup D) \cap (B \cup C)$: $A \cup D = \{3,5,7,9,11,15,17\}$; $B \cup C = \{7,9,11,13,15\}$. Common elements: 7, 9, 11, 15.
$$(A \cup D) \cap (B \cup C) = \{7, 9, 11, 15\}$$

$A = \mathbf{N}$ (all natural numbers), $B$ = even natural numbers, $C$ = odd natural numbers, $D$ = prime numbers.

(i) $A \cap B$: Every even natural number is a natural number, so $B \subset A$.
$$A \cap B = B = \{x : x \text{ is an even natural number}\}$$

(ii) $A \cap C$: Every odd natural number is a natural number, so $C \subset A$.
$$A \cap C = C = \{x : x \text{ is an odd natural number}\}$$

(iii) $A \cap D$: Every prime number is a natural number, so $D \subset A$.
$$A \cap D = D = \{x : x \text{ is a prime number}\}$$

(iv) $B \cap C$: A number cannot be both even and odd simultaneously.
$$B \cap C = \phi$$

(v) $B \cap D$: Even prime numbers. The only even prime is 2.
$$B \cap D = \{2\}$$

(vi) $C \cap D$: Odd prime numbers. All primes except 2 are odd.
$$C \cap D = \{x : x \text{ is an odd prime number}\} = \{3, 5, 7, 11, 13, \ldots\}$$

Two sets are disjoint if their intersection is the empty set.

(i) Second set $= \{4, 5, 6\}$. The element 4 is in both $\{1,2,3,4\}$ and $\{4,5,6\}$. So their intersection $= \{4\} \neq \phi$. Not disjoint.

(ii) The element $e$ is in both $\{a,e,i,o,u\}$ and $\{c,d,e,f\}$. Intersection $= \{e\} \neq \phi$. Not disjoint.

(iii) An integer cannot be both even and odd. So the intersection of even integers and odd integers is $\phi$. Disjoint.

$A - B = \{x : x \in A \text{ and } x \notin B\}$

Given: $A = \{3,6,9,12,15,18,21\}$, $B = \{4,8,12,16,20\}$, $C = \{2,4,6,8,10,12,14,16\}$, $D = \{5,10,15,20\}$.

(i) $A - B$: Elements in A not in B. 12 ∈ B, so remove it.
$$A - B = \{3, 6, 9, 15, 18, 21\}$$

(ii) $A - C$: Elements in A not in C. 6 ∈ C, 12 ∈ C, so remove them.
$$A - C = \{3, 9, 15, 18, 21\}$$

(iii) $A - D$: Elements in A not in D. 15 ∈ D, so remove it.
$$A - D = \{3, 6, 9, 12, 18, 21\}$$

(iv) $B - A$: Elements in B not in A. 12 ∈ A, so remove it.
$$B - A = \{4, 8, 16, 20\}$$

(v) $C - A$: Elements in C not in A. 6 ∈ A, 12 ∈ A, so remove them.
$$C - A = \{2, 4, 8, 10, 14, 16\}$$

(vi) $D - A$: Elements in D not in A. 15 ∈ A, so remove it.
$$D - A = \{5, 10, 20\}$$

(vii) $B - C$: Elements in B not in C. 4 ∈ C, 8 ∈ C, 12 ∈ C, 16 ∈ C; only 20 ∉ C.
$$B - C = \{20\}$$

(viii) $B - D$: Elements in B not in D. 20 ∈ D, so remove it.
$$B - D = \{4, 8, 12, 16\}$$

(ix) $C - B$: Elements in C not in B. 4 ∈ B, 8 ∈ B, 12 ∈ B, 16 ∈ B; remove these.
$$C - B = \{2, 6, 10, 14\}$$

(x) $D - B$: Elements in D not in B. 20 ∈ B, so remove it.
$$D - B = \{5, 10, 15\}$$

(xi) $C - D$: Elements in C not in D. 10 ∈ D, so remove it.
$$C - D = \{2, 4, 6, 8, 12, 14, 16\}$$

(xii) $D - C$: Elements in D not in C. 10 ∈ C, so remove it.
$$D - C = \{5, 15, 20\}$$