The Empty Set, Finite and Infinite Sets, Equal Sets

Taking N as the natural numbers 1, 2, 3, and so on, the condition x ≤ 50 restricts J to J = {1, 2, 3, ..., 50}. This set has exactly 50 elements, so its cardinality is |J| = 50. Because the list has a fixed final element and cannot continue beyond 50, J is finite. It is not empty or infinite, while “equal set” is not a classification that answers the question.

The natural numbers satisfying x ≥ 50 are 50, 51, 52, 53, and so on. There is no greatest natural number, so more elements can always be added to the list. Consequently, K has infinitely many elements and is an infinite set, not a finite or singleton set.

For every real number x, x² ≥ 0. Consequently, x² + 1 ≥ 1, so the expression can never equal 0. Equivalently, solving the equation gives x² = −1, which has no real solution because the square of a real number cannot be negative. Therefore no real number belongs to L, and L is the empty set. The alternatives involving one, two, or infinitely many elements would require real solutions that do not exist.

The set-builder description of N selects every natural number less than 5. Thus N = {1, 2, 3, 4}, assuming natural numbers begin with 1. This is exactly the roster form of M. Two sets are equal when they contain precisely the same elements, regardless of how those elements are written.

The square of every natural number is non-negative: for every x ∈ N, x² ≥ 0. Hence the inequality x² < 0 cannot be satisfied by any natural number. There is therefore no element that belongs to Q. By definition, a set with no elements is the empty set, so Q = ∅. The choices Q = N, {0}, or {1} are impossible because each would contain elements whose squares are not negative.

Two-digit natural numbers start at 10 and end at 99, so S = {10, 11, 12, ..., 99}. The number of elements is 99 − 10 + 1 = 90. Since the lower and upper bounds are fixed, the list has a limited number of members and is therefore finite. It is not empty because such numbers exist, not infinite because the list stops at 99, and “equal set” does not describe its cardinality.

The set contains natural numbers strictly greater than 3 and strictly less than 4. The only integers near this interval are 3 and 4, but 3 is not greater than 3 and 4 is not less than 4. There is no natural number between consecutive integers 3 and 4. Therefore, the set has no elements and is the empty set, written as ∅.

Because x is an integer and both inequalities are inclusive, list every integer from −2 through 2: U = {−2, −1, 0, 1, 2}. This gives five distinct elements, so |U| = 5. The count can also be found by upper bound minus lower bound plus one: 2 − (−2) + 1 = 5. Excluding an endpoint would produce four, but the ≤ signs include both endpoints.

The positive integers divisible by 7 are 7, 14, 21, 28, 35, and so on. For every positive multiple 7n, where n is a positive integer, there is another larger multiple 7(n + 1). Thus the list never ends, so V is an infinite set. It is neither empty nor a singleton.

The natural-number factors of 12 are 1, 2, 3, 4, 6, and 12. Hence W = {1, 2, 3, 4, 6, 12}, which contains exactly six elements. A fixed positive integer has only finitely many positive factors because every factor can be paired with another factor whose product is 12. Therefore, W is a finite set.

The natural-number multiples of 12 are 12, 24, 36, 48, and so on. In general, every number of the form 12n, where n is a positive natural number, belongs to X. Since there is no largest natural number, the sequence of multiples continues indefinitely. Therefore, X has infinitely many elements and is an infinite set.

A natural number divisible by both 2 and 3 is divisible by their least common multiple, 6. The members therefore begin 6, 12, 18, 24, 30, and continue as 6n for every positive natural number n. Since there is no upper bound, infinitely many such numbers exist, so Y is infinite.

The symbol ∅, also written as { }, denotes the empty set or null set. By definition, it contains no elements, so its cardinality is zero: |∅| = 0. It must not be confused with {0}, which contains the single element 0 and therefore has cardinality one. Thus option B is correct.

The symbol ∅ denotes the empty set, which contains no elements, so its cardinality is 0. The notation {0} denotes a set whose only element is the number 0, so its cardinality is 1. The braces are important: ∅ has no member, while {0} contains one member. Therefore, statement B is correct and the two sets are not equal.

Solving x² = 2 over the real numbers gives x = √2 and x = −√2. Therefore, B₁ = {−√2, √2}, which contains exactly two distinct elements. A set with a fixed, countable number of elements is finite. Thus B₁ is a two-element finite set, not an empty or infinite set.

The odd natural numbers less than 30 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29. This list contains 15 elements and stops at 29 because the condition requires the number to be less than 30. A set with a limited, countable number of elements is called a finite set. Therefore, C₁ is finite, not empty or infinite, and it is not equal to the complete set of natural numbers.

The odd natural numbers greater than 30 are 31, 33, 35, 37, 39, and so on. After every odd number, another larger odd natural number can be found, so the sequence does not terminate. There is no greatest odd natural number and no upper limit in the definition of D₁. Hence D₁ has infinitely many elements and is an infinite set. It is not empty, finite, or a singleton.

The two inequalities 5 ≤ x and x ≤ 5 together force x to be exactly 5. Therefore, E₁ = {5}, which contains precisely one element. A set containing exactly one element is called a singleton set. It is also finite, but the most specific answer among the choices is singleton set. It cannot be empty because 5 satisfies the condition, and it cannot be infinite because there is only one permitted value.

The condition requires the same natural number x to be less than 15 and greater than 15 simultaneously. No number can satisfy both strict inequalities: a number below 15 cannot also be above 15, and 15 itself is neither less than nor greater than 15. Thus there is no element in F₁, so F₁ = ∅. A set with no elements is called the empty or null set.

The natural-number factors of 9 are 1, 3, and 9. All of these factors are odd, so none of them satisfies the additional condition of being even. Because no natural number satisfies both conditions simultaneously, G₁ contains no elements and therefore equals the empty set, ∅.

The positive natural-number factors of 16 are 1, 2, 4, 8, and 16. Among them, the even factors are 2, 4, 8, and 16, so H₁ = {2, 4, 8, 16}. This set has exactly four elements. Because the factors are selected from a fixed number and the list ends, H₁ is a finite set. It is not empty, infinite, or equal to all natural numbers.

A particular class contains a limited number of students, and each student is assigned one roll number. Therefore, the set of roll numbers has the same limited number of elements as the class. It is non-empty when students are enrolled and is generally a finite set, not an infinite or real-number set.

A circle is a continuous geometric curve, not a polygon made of a limited number of points. Between any two distinct points on the circle, there are further points, and this process never ends. Thus the collection of all points on a circle contains infinitely many elements and is an infinite set.

Every horizontal line has an equation of the form y = c, where c is a real number. Since there are infinitely many possible real values of c, there are infinitely many distinct lines parallel to the x-axis. The x-axis itself is one such line, and lines such as y = 1, y = 2, and y = -3 are other examples. Therefore, the set has infinitely many elements and is an infinite set.

The condition requires x to be a natural number and also to satisfy x ≤ 0. Under the definition given in the question, the natural numbers are 1, 2, 3, and so on; none of them is zero or negative. Consequently, no element satisfies both conditions simultaneously. Therefore I₁ contains no elements and is the empty set, written as ∅.