The Empty Set, Finite and Infinite Sets, Equal Sets

The natural-number factors of 25 are 1, 5, and 25. All three factors are odd, so none satisfies the additional condition that x must be even. Therefore, H has no elements and H = ∅. A set with no elements is called the empty set. The two conditions must be applied simultaneously, not separately.

The positive multiples of 4 that are less than 30 are 4, 8, 12, 16, 20, 24, and 28. The next multiple, 32, is not less than 30, so it must not be included. Therefore, the set is I = {4, 8, 12, 16, 20, 24, 28}, which contains 7 distinct elements. Hence, the cardinality of I is 7, so option B is correct. Specifying positive natural numbers avoids the convention-related issue of whether 0 is included in ℕ.

Since n can be any positive natural number, substituting n = 1, 2, 3, 4, ... gives J = {4, 7, 10, 13, ...}. There is no greatest permitted value of n, so new elements continue to appear without end. The set therefore has infinitely many elements and is an infinite set, not a finite or singleton set.

Because n is a natural number and n ≤ 5, its possible values are only 1, 2, 3, 4, and 5. The corresponding values of x = 3n + 1 are 4, 7, 10, 13, and 16. Thus K = {4, 7, 10, 13, 16}, which has five elements and is therefore finite. The upper bound on n is decisive.

The condition x ≤ 20 restricts x to the finite collection of natural numbers from 1 through 20. Excluding multiples of 3 removes 3, 6, 9, 12, 15, and 18, but it cannot create infinitely many elements. In fact, L = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}, so L is finite.

Natural numbers that are not divisible by 2 are the odd natural numbers: 1, 3, 5, 7, 9, and so on. For every odd natural number, another larger odd natural number exists, so the list never ends. There is no upper bound on x. Consequently, M contains infinitely many elements and is an infinite set.

The condition -2 ≤ x < 3 includes the integer -2 because the left inequality is inclusive, and excludes 3 because the right inequality is strict. The integers in this interval are -2, -1, 0, 1, and 2. This is exactly the roster given for O. Since two sets are equal when they have precisely the same elements, N = O.

The positive natural-number factors of 36 are obtained by checking which natural numbers divide 36 exactly: 1, 2, 3, 4, 6, 9, 12, 18, and 36. This list contains every element written in Q and no other elements. Therefore P and Q have exactly the same members, so P = Q. The number 8 is not a factor because 36 is not divisible by 8.

The positive natural-number multiples of 36 are 36, 72, 108, 144, and so on, so R continues indefinitely and is infinite. In contrast, Q is explicitly listed and contains only nine elements, so it is finite. The two sets are not equal: Q lists factors of 36, whereas R contains multiples of 36. Hence option B is correct.

A prime number has exactly two positive divisors: 1 and itself. The only even prime number is 2; every other even natural number is divisible by 2 and therefore has at least one additional divisor, so it is composite. Since 2 is less than 50, it satisfies both conditions. Thus S = {2}, a singleton set containing one element.

The only natural number with exactly one positive factor is 1, because its only positive divisor is 1. However, the definition of T also requires x > 1, which excludes 1. Every natural number greater than 1 has at least two positive factors, namely 1 and the number itself. Therefore, no value satisfies both conditions, so T is the empty set, written as ∅.

The condition defining U is x² = x. Rearranging gives x² − x = 0, which factors as x(x − 1) = 0. Therefore, x = 0 or x = 1. Both values are integers, so both satisfy the stated domain condition x ∈ ℤ. Hence U contains exactly the two elements 0 and 1, and U = {0, 1}.

Factor the defining equation: x² = x gives x² − x = 0, so x(x − 1) = 0. The algebraic solutions are x = 0 and x = 1. Here ℕ is taken as {1, 2, 3, ...}, so 0 is not in the stated domain. The only permitted element is 1, and therefore V = {1}. If zero were included in the convention for ℕ, option A would apply, but not under the convention used here.

For integers, x² < 4 means |x| < 2, or equivalently −2 < x < 2. The integers strictly between −2 and 2 are −1, 0, and 1. Thus Y = {−1, 0, 1}, which contains three elements. The endpoints −2 and 2 are excluded because the inequality is strict, using < rather than ≤.

The inequality x² ≤ 4 is equivalent to |x| ≤ 2, so −2 ≤ x ≤ 2. The integers in this closed interval are −2, −1, 0, 1, and 2. Therefore Z = {−2, −1, 0, 1, 2}, which has five elements. Unlike a strict inequality, ≤ includes both boundary values −2 and 2.

The set {0} contains the element 0, so it has one element and is a singleton set, not an empty set. The empty set is written as ∅ or {}, and it contains no elements. Statement A is true because the empty set is finite with cardinality zero. Statement C correctly describes equal sets, and statement D expresses the unending nature of an infinite set.

A number that is a multiple of both 10 and 15 must be a multiple of their least common multiple. Since lcm(10, 15) = 30, the set is A₁ = {30, 60, 90, 120, ...}. There is no upper bound on x, so new common multiples can always be found by continuing the sequence. Therefore, A₁ is an infinite set.

The least common multiple of 10 and 15 is 30, so every common multiple is a multiple of 30. The positive multiples of 30 that are less than 100 are 30, 60, and 90. The next multiple, 120, is not less than 100. Thus B₁ = {30, 60, 90}, and its cardinality is 3, so option B is correct.

The positive factors of 12 are 1, 2, 3, 4, 6, and 12. The positive factors of 18 are 1, 2, 3, 6, 9, and 18. The elements common to both lists are therefore 1, 2, 3, and 6, so C₁ = {1, 2, 3, 6}. Because a fixed natural number has only finitely many factors, this is a finite set.

The common natural-number factors of 12 and 18 are 1, 2, 3, and 6. The additional condition requires x to be greater than 6, but none of these common factors satisfies that inequality. Hence no element belongs to D₁, and the correct description is D₁ = ∅, the empty set. It is not {6}, because 6 is not greater than 6.