The Empty Set, Finite and Infinite Sets, Equal Sets

For every real number x, x² is at least 0. Consequently, x² + 4 is at least 4, so it can never equal 0. Equivalently, the equation would require x² = −4, which has no real solution. Since F₁ contains real numbers satisfying the condition and there are none, F₁ = ∅, the empty set.

Option A is the set of positive integers {1, 2, 3, …}. It has infinitely many elements because there is no largest positive integer, but it is not the set of all integers because it excludes zero and every negative integer. Option B is finite, option C is finite, and option D equals {−1, 1}, which is also finite. Hence A is the unique answer.

Taking n = 1, 2, 3, 4, and so on produces K₁ = {1, 8, 27, 64, 125, …}. Natural numbers continue without an ending value, and each distinct natural number gives a distinct cube, so the list of elements never terminates. Therefore K₁ is infinite. It is neither empty nor a singleton, and it is not finite because no final cube exists.

Since x belongs to the natural numbers and x² < 10, test the natural numbers in order: 1² = 1, 2² = 4, and 3² = 9, all of which are less than 10. However, 4² = 16, so 4 is excluded. Therefore L₁ = {1, 2, 3}, which is exactly M₁. Two sets are equal when they contain precisely the same elements, regardless of how those elements are written or listed.

For natural numbers, evaluate x² ≤ 10. The values x = 1, 2, and 3 work because their squares are 1, 4, and 9. The next natural number, 4, does not work because 4² = 16, which is greater than 10. Thus N₁ = {1, 2, 3}. This is exactly O₁, so the correct statement is N₁ = O₁. The inclusive sign ≤ does not add 4 because no natural-number square equals 10.

Solving the equation 2x + 1 = 0 gives 2x = −1 and hence x = −1/2. Although this is a solution over the real or rational numbers, it is not an integer. The definition of Q₁ specifically restricts x to ℤ, the set of integers. Therefore no integer satisfies the condition, and Q₁ contains no elements. Hence Q₁ is the empty set, written as ∅.

Solving 2x + 1 = 0 gives x = −1/2. The number −1/2 is rational because it can be expressed as a ratio of two integers with a nonzero denominator. It therefore satisfies the stated domain x ∈ ℚ. Since the equation has exactly one solution, R₁ = {−1/2} and contains one element. A set with exactly one element is called a singleton set.

A natural number leaves remainder 2 on division by 5 precisely when it has the form 5k + 2, where k is a nonnegative integer (with the usual convention that natural numbers include positive values). The resulting members begin 2, 7, 12, 17, 22, and continue indefinitely by adding 5. There is no upper bound on k, so the set has infinitely many elements and is therefore infinite.

By the division algorithm, when a number is divided by a positive divisor 6, the remainder must be one of 0, 1, 2, 3, 4, or 5. A remainder must always be strictly less than the divisor. Therefore remainder 6 is impossible for every natural number. No x satisfies the defining condition, so U₁ has no elements and is the empty set, denoted by ∅.

The integer solutions of x² = 25 are x = 5 and x = −5, so X₁ = {5, −5}. In set notation, repeating an element does not create a new element; duplicates are ignored. Therefore Y₁ = {5, −5, 5} represents the same set as {5, −5}. Both sets contain exactly the two elements 5 and −5, so X₁ = Y₁. Statement A is correct.

A two-digit natural number ranges from 10 through 99, inclusive. The condition x > 99 asks for a two-digit number strictly greater than the largest possible two-digit number, 99. No such natural number exists. Consequently, the defining conditions cannot be satisfied simultaneously, and Z₁ contains no elements. Therefore Z₁ is the empty set, written as ∅; it is not the set containing 99 because the inequality is strict.

The condition 10 ≤ x < 100 selects the natural numbers 10, 11, 12, ..., 99. There is a definite first element, 10, and a definite last element, 99. Therefore, the set contains only 90 elements and is finite. A set may have many elements and still be finite if its total number of elements is limited.

The integers satisfying −5 < x < 5 are −4, −3, −2, −1, 0, 1, 2, 3, and 4. Among these, the odd integers are −3, −1, 1, and 3. Their squares are 9, 1, 1, and 9 respectively. None has x² > 9; the values −3 and 3 only give x² = 9, which is not greater than 9. Therefore, C is the empty set and has 0 elements.

Because x is restricted to real numbers, x² is always non-negative. Consequently, x² + 4 is always at least 4 and can never equal zero. Thus the equation has no real solution, so A contains no elements and A = ∅. Since B is also given as the empty set, A and B are equal. Option A is therefore correct.

The natural numbers are discrete: consecutive natural numbers have no natural number between them. Since 5 and 6 are consecutive, there is no natural number x satisfying the strict inequalities 5 < x < 6. The endpoints are excluded as well. Therefore A has no elements, so it is the empty set, written as ∅.

In a set, an element is either present or absent; writing it more than once does not increase the set’s membership. Thus A simplifies to {1, 2, 3}. The order of elements is also irrelevant in set notation, so B = {3, 2, 1} represents the same set {1, 2, 3}. Hence A and B contain exactly the same elements and are equal.

There is no integer whose square is 2. The integers nearest to a possible square root are 1 and 2, but 1² = 1 and 2² = 4; in fact, √2 is irrational and is not an integer. Therefore no integer satisfies the defining condition, so A = ∅. The empty set has cardinality zero and is finite.

A natural number is a factor of 24 if it divides 24 exactly. The positive factors are 1, 2, 3, 4, 6, 8, 12, and 24. This list is precisely the roster description of B. Since two sets are equal when they contain exactly the same elements, A = B. The set is finite, not infinite, and 24 is correctly included.

The symbol ∅ denotes a set with no elements, whereas {∅} denotes a set whose only element is the empty set itself. Therefore, {∅} contains exactly one element and is a singleton set. It is not equal to ∅, because ∅ has zero elements while {∅} has one element. This distinction is fundamental in set notation.

The set contains all prime natural numbers not exceeding 100. There are only finitely many natural numbers from 1 through 100, so the set can have only finitely many elements. It is also non-empty because numbers such as 2, 3, 5, 7, and 11 are prime and satisfy the condition. Therefore, A is finite and non-empty.

Because x must be an integer strictly greater than −3 and strictly less than 3, the boundary values −3 and 3 are excluded. The integers satisfying the condition are −2, −1, 0, 1, and 2. Thus, A has five distinct elements, so its cardinality is n(A) = 5. The notation n(A) means the number of elements in set A.

A set records membership, not the number of times an element appears in the original word. Although LEVEL has five positions, its distinct letters are only L, E, and V. Therefore, A = {L, E, V}. This is exactly the set B. Repeated letters are written only once in a set, so A and B are equal and A has three elements, not five.

The natural-number multiples of 3 include 3, 6, 9, 12, 15, and so on. For every multiple 3k, where k is a natural number, another larger multiple 3(k+1) can be found. Therefore, the sequence never ends and there is no greatest element. Hence A is an infinite set, not merely the singleton {3}.

To determine A, solve x² = 9 over the real numbers. Taking square roots gives x = 3 or x = -3, because both 3² and (-3)² equal 9. Therefore A = {-3, 3}. This is exactly the same collection of elements as B, so A = B. The negative root must not be omitted when solving a square equation.

For every integer x, the square x² is non-negative: it is either positive when x ≠ 0 or equal to 0 when x = 0. Therefore no integer can satisfy x² < 0. The set A has no elements, so A is the empty set. The cardinality of the empty set is n(A) = 0, making option A correct.