The Empty Set, Finite and Infinite Sets, Equal Sets

For an integer x, the inequality |x| < 3 is equivalent to -3 < x < 3. The integers strictly between -3 and 3 are -2, -1, 0, 1, and 2. Thus J₁ = {-2, -1, 0, 1, 2}, which contains five distinct elements. The endpoints -3 and 3 are excluded because the inequality is strict, so the correct answer is 5.

The absolute value |x| represents the distance of x from zero, so it can equal zero only when x = 0. Since 0 is an integer, it belongs to the stated domain. Hence K₁ = {0}. A set containing exactly one distinct element is called a singleton set. There are not two solutions, because |0| = 0 has only the single solution x = 0.

The absolute value of every real number, and therefore of every integer, is always non-negative. It can be zero or positive, but it can never equal -2. Thus there is no integer x satisfying |x| = -2. A set defined by a condition with no possible solution contains no elements, so L₁ is the empty set, denoted by ∅.

The positive perfect squares are obtained by squaring natural numbers: 1² = 1, 2² = 4, 3² = 9, and 4² = 16. The next square, 5² = 25, is not less than 20 and must be excluded. Therefore the complete set of perfect-square natural numbers below 20 is M₁ = {1, 4, 9, 16}.

The perfect-square natural numbers begin as 1, 4, 9, 16, 25, and continue as 36, 49, 64, and so on. For every natural number n, the number n² is a perfect square, and there is no largest natural number. Consequently, new perfect squares continue without end, so N₁ contains infinitely many elements and is an infinite set.

The only even prime number is 2. Every other even natural number is divisible by 2 and therefore has at least the divisors 1, 2, and itself, so it is not prime. Hence O₁ = {2}. Because this set contains exactly one element, it is a singleton set. It is neither empty nor infinite, and it certainly does not contain three elements.

Two sets are equal when they contain exactly the same elements, regardless of the order in which those elements are written. P₁ and Q₁ both contain 1, 2, and 3, so P₁ = Q₁. However, R₁ contains 4 instead of 3, so it does not have the same elements as P₁. Therefore, P₁ = Q₁ but P₁ ≠ R₁. None of these sets is empty because each contains three elements.

The natural-number factors of 8 are 1, 2, 4, and 8. The additional condition requires the factor x to be greater than 8, but none of these factors satisfies x > 8; 8 itself is equal to 8, not greater than it. Therefore no natural number meets both conditions, so A has no elements and A = ∅.

Since x = 2n and n ranges over the natural numbers, C contains even natural numbers such as 2, 4, 6, 8, and so on. There is no greatest natural number, so n can always be increased to produce another even member of C. Consequently, the listing never terminates and C is an infinite set. This conclusion is unchanged whether 0 is included in ℕ.

Two sets are equal precisely when they have the same elements, meaning every element of the first set belongs to the second and every element of the second belongs to the first. The order of listing is irrelevant because sets are unordered collections. For example, {1, 2, 3} and {3, 1, 2} are equal, whereas {1, 2} and {1, 3} are not equal even though both have two elements.

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.

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 ∅.

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.

Subtracting 4 from both sides of x² + 4 = 4 gives x² = 0. The only integer, and indeed the only real number, whose square is zero is x = 0. Therefore E₁ contains exactly one element: E₁ = {0}. A set with precisely one element is called a singleton set, so option B is correct; there are not two roots because +0 and −0 are the same number.