The Empty Set, Finite and Infinite Sets, Equal Sets

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.

Numbers divisible by both 2 and 3 are multiples of 6. The positive multiples of 6 that are less than 20 are 6, 12, and 18. There are only three members because the condition x < 20 gives an upper bound. Hence Z is a finite set, even though the unrestricted set of multiples of 6 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. Neither value is an integer, because √2 is irrational. The condition x ∈ ℤ restricts us to integers only, so no value satisfies both requirements. Therefore A₁ contains no elements and is the empty set, A₁ = ∅.

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.

Sets are equal when they contain exactly the same elements; the order of listing does not matter, and repeating an element does not create a new element. Options A, B, and D therefore describe equal sets. In option C, the first set contains 0 while the second contains 6, so the sets are not equal.

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.