The Empty Set, Finite and Infinite Sets, Equal Sets

For option A, substitute the natural-number values n = 1, 2, and 3 into x = n³. The resulting values are 1³ = 1, 2³ = 8, and 3³ = 27. Hence option A produces exactly the elements of the given set. Equality of sets depends on having precisely the same elements, regardless of their order.

The integers are ..., −3, −2, −1, 0, 1, 2, 3, ... . They extend without end in the negative direction and also without end in the positive direction. Consequently, there is no final integer and the set cannot be listed completely. Therefore, the set of all integers is infinite.

The positive natural-number factors of 9 are 1, 3, and 9. All three are odd, so none satisfies the additional condition of being even. Because both conditions must hold simultaneously, no element qualifies. Therefore A has no elements and is the empty set, written as ∅.

The governing concept is the cardinality of a finite set. A standard die has six faces, conventionally marked with the distinct numbers 1, 2, 3, 4, 5, and 6. Thus the set of numbers written on it is A = {1, 2, 3, 4, 5, 6}, whose cardinality is 6. Option C is correct. The values 4 and 5 undercount the faces, while “infinitely many” is impossible for a finite die.

The natural numbers less than 6 are 1, 2, 3, 4, and 5, so A = {1, 2, 3, 4, 5}. Set B contains exactly the same five elements, merely written in reverse order. In set theory, order does not matter; equality depends only on having the same elements. Hence A and B are equal sets.

The condition requires natural numbers strictly greater than 10 and strictly less than 15. Therefore, 10 and 15 are excluded because the inequalities are strict. The members are 11, 12, 13 and 14. Since there are four distinct members, the set is finite and its cardinality is 4. Hence option B is correct.

The empty set, written as ∅, contains no elements, so its cardinality is 0. A set is finite when it has a fixed, countable number of elements, and zero is a finite number. Thus the empty set is both finite and empty. In contrast, {0} has one element, while the natural-number sequence and the integers are infinite. Therefore option B is correct.

The two-digit natural numbers are 10, 11, 12, ..., 99. This collection has a definite first element and a definite last element, so its members can be counted completely. In fact, it contains 99 − 10 + 1 = 90 elements. Therefore, A is a finite set, even though it has many elements. It is neither empty, nor infinite, nor a singleton.

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 = ∅.