The Empty Set, Finite and Infinite Sets, Equal Sets

The greatest three-digit number is 999. Every natural number from 1000 onward is greater than every three-digit number, so A = {1000, 1001, 1002, ...}. This list has no final member and continues without bound. Therefore A has infinitely many elements, making option C correct.

A factor divides a number exactly, with no remainder. The natural-number factors of 4 are 1, 2, and 4, while the natural-number factors of 6 are 1, 2, 3, and 6. The elements common to both lists are 1 and 2. Hence A = {1, 2}, so option A is correct. The set has two elements and is therefore finite.

The natural numbers divisible by 11 are 11, 22, 33, 44, and so on. In general, every number of the form 11n, where n is a positive natural number, belongs to A. Since n can increase without limit, there is no greatest member and infinitely many multiples exist. Hence option C is correct.

The symbol Z denotes all integers. We need integers strictly greater than −2 and strictly less than 2. The integers satisfying both inequalities are −1, 0, and 1. The endpoints −2 and 2 are excluded because the inequalities are strict, not inclusive. Therefore A = {−1, 0, 1}, which is option B and is a finite set with three elements.

Every integer, including every natural number, is either even or odd, but no integer can be both even and odd. An even number is divisible by 2, whereas an odd number leaves remainder 1 when divided by 2; these conditions cannot hold simultaneously. Thus no natural number satisfies the defining condition, so A has no elements and A = ∅. Option C is correct.

A set counts distinct objects only once, even when an object is repeated in the description. The letters of KAMAL are K, A, M, A and L. Although A appears twice, it is included only once in the set. Thus A = {K, A, M, L}, which has four elements. Option B is correct.

Under the stated convention, the natural numbers are 1, 2, 3, and so on; zero and negative numbers are not included. Every natural number is therefore greater than 0 and fails the condition x ≤ 0. No element satisfies the rule, so A = ∅, the empty set. Hence option A is correct.

The numbers whose square is 16 are 4 and −4, because 4² = 16 and (−4)² = 16. However, the condition specifies that x belongs to ℕ, the set of natural numbers. Under the usual school convention, −4 is not a natural number, so it must be excluded. Therefore the only member of A is 4, and A = {4}.

The natural numbers that are less than or equal to 4 are 1, 2, 3, and 4, assuming ℕ = {1, 2, 3, ...}. Thus B can be written in roster form as {1, 2, 3, 4}. Set A contains exactly the same elements, and the order in which elements are written does not matter. Therefore A = B, so they are equal sets.

Every positive factor of 18 is at most 18. Its natural-number factors are 1, 2, 3, 6, 9, and 18; none is greater than 18. Consequently, no natural number can satisfy both conditions “x is a factor of 18” and “x > 18.” A set with no element satisfying its defining condition is the empty set, so A = ∅.

Sets contain distinct elements only, so repetition is ignored. In the set {1, 0, 0}, the second occurrence of 0 does not create a new element. After removing the repetition, the set is {1, 0}, which has exactly the same elements as {0, 1}. Since order also does not matter in a set, option A represents a set equal to {0, 1}.

The smallest prime number is 2. Therefore, there is no prime number smaller than 2. The defining condition for A asks for a natural number that is simultaneously prime and less than 2, but no such number exists. Whenever no element satisfies the rule defining a set, the set has no members and is called the empty set. Hence A = ∅.

A calendar year has exactly twelve months: January, February, March, April, May, June, July, August, September, October, November, and December. These names form a set with a fixed and limited number of distinct elements. Because its number of elements is 12, not unlimited, A is a finite set. Therefore option B is correct; it is neither empty nor infinite.

The natural-number factors of 6 are 1, 2, 3, and 6, so A = {1, 2, 3, 6}. In B, the number 6 appears twice, but repeated entries are counted only once in a set. Thus B also represents {1, 2, 3, 6}. Since A and B contain exactly the same elements, A = B. Therefore option A is correct.

Solving x² + 1 = 0 gives x² = −1. For every natural number x, x² is non-negative, so it can never equal −1. Hence no natural number satisfies the defining equation. A set whose defining condition has no solution is the empty set. Therefore A = ∅, and the correct statement is that A is an empty set, represented by option A.

The months having 31 days are January, March, May, July, August, October, and December. There are exactly seven such months, and this list can be completely counted and has a final element. A set with a fixed, countable number of elements is called a finite set. Therefore, B is finite, not empty or infinite.

For B, we list the natural numbers that are multiples of 3 and less than 15. They are 3, 6, 9, and 12; 15 is excluded because the condition is x < 15. Thus B = {3, 6, 9, 12}, which contains exactly the same elements as A. Since two sets are equal when they have precisely the same elements, A = B. Therefore, option A is correct.

For every natural number n, the expression x = 2n gives an even number. If N begins with 1, the set contains 2, 4, 6, 8, and so on; if a convention includes 0, 0 is also included, without changing the conclusion. Since natural numbers continue indefinitely, there is no last generated element. Therefore C is infinite.

The prime natural numbers less than 10 are 2, 3, 5, and 7. Therefore, the set described in option A is exactly {2, 3, 5, 7}. Number 1 is not prime, so option B has an extra element. Option C contains even numbers rather than the required primes, and option D contains 11 while omitting 2. Hence only option A represents a set equal to the given set.

The prime factorisation of 24 is 24 = 2 × 2 × 2 × 3 = 2³ × 3. The prime factors are therefore 2 and 3. A set records membership, not repeated occurrences, so 2 is written only once even though it appears three times in the factorisation. Neither 1 nor 24 is a prime factor, and 8 is not prime. Hence A = {2, 3}, making option A correct.

List the positive factors of each number. The factors of 18 are 1, 2, 3, 6, 9, and 18, while the factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The elements appearing in both lists are 1, 2, 3, and 6. Therefore, P is the set of common factors, so P = {1, 2, 3, 6}; option A is correct.

Because x is an integer, list every integer from −3 up to but not including 2. The condition −3 ≤ x includes −3, and x < 2 excludes 2. Thus A = {−3, −2, −1, 0, 1}. Counting these elements gives 5. The set is finite, not infinite, so option B is the only correct answer.

A prime number has exactly two distinct positive divisors, 1 and itself. Among natural numbers, 2 is the only number that is both even and prime. Every other even natural number greater than 2 is divisible by 2 and another number, so it is composite. Hence the set contains only one element and is A = {2}, making option B correct.

The set of natural-number factors of 12 is {1, 2, 3, 4, 6, 12}. It has six elements, so it is finite, and it has several elements, so it is not empty. Option A is empty because no integer has square −9. Options B and D are infinite. Therefore, only option C satisfies both required conditions.

The natural numbers less than 40 that are divisible by 5 are 5, 10, 15, 20, 25, 30, and 35. The number 40 is excluded because the condition is x < 40, not x ≤ 40. There are seven listed elements, so the set is finite and its cardinality is 7. Therefore, option A is correct.