The Empty Set, Finite and Infinite Sets, Equal Sets

The empty set, denoted by ∅, contains no elements, so its cardinality is 0. A set with a definite, limited number of elements is finite; therefore a set with zero elements is also finite. The sets in options C and D continue without end and are infinite, while {0} is finite but not empty. Hence option A is correct.

A positive divisor of 6 is a positive integer that divides 6 exactly, with zero remainder. The factor pairs 1 × 6 and 2 × 3 show that the positive divisors are 1, 2, 3, and 6. No other positive integer divides 6 without a remainder. Thus the roster form is A = {1, 2, 3, 6}, so option A is correct. Options B and C omit valid divisors, while D incorrectly treats the set as empty.

An infinite set has endlessly many distinct elements, whereas a finite set has a fixed number of elements. The odd natural numbers and the integers continue without end, and the positive multiples of 5 are also unlimited. Natural numbers less than 10 are only {1, 2, 3, 4, 5, 6, 7, 8, 9}, so this is a finite set. Therefore, option C is correct.

The positive even numbers less than 8 are 2, 4, and 6. Thus the condition defining B produces B = {2, 4, 6}, which contains exactly the same elements as A. Two sets are equal when they have precisely the same elements, regardless of how they are written. Therefore A = B and option A is correct.

By definition, a triangle is a polygon with exactly three sides. An object cannot simultaneously be a triangle and have four sides, so no object satisfies the condition defining A. A set with no elements is called the empty set and is represented by ∅. Hence A is empty, and option A is correct.

A perfect-square natural number is a natural number that can be written as the square of a natural number. The natural numbers less than 4 are 1, 2, and 3. Of these, only 1 is a perfect square because 1 = 1². Although 4 = 2², it is not included because the condition is strictly less than 4. Therefore A = {1}, so option A is correct; the other options either include excluded numbers or wrongly make the set empty.

The notation n(A) denotes the cardinality of set A, meaning the number of distinct elements in it. The empty set has 0 elements, {5} has 1 element, {5, 7} has 2 distinct elements, and {1, 2, 3} has 3 elements. Therefore, the set satisfying n(A) = 2 is {5, 7}, which is option C.

In the usual weekly order, Sunday is followed immediately by Monday. There is no named day between these two consecutive days. Consequently, no element satisfies the condition defining A, so A has zero elements and is the empty set. Therefore option A is correct; it is not a singleton or an infinite set.

The odd natural numbers in the inclusive interval from 1 through 10 are numbers that are not divisible by 2: 1, 3, 5, 7, and 9. The number 10 is even, and 11 lies outside the stated interval. Hence the roster form of the set is A = {1, 3, 5, 7, 9}, so option A is correct.

In a set, each element is listed only once; repetition does not create a new element. Therefore, {b, a, a} has the same distinct elements as {a, b}, namely a and b. The order of elements also does not matter in a set. Hence both sets contain exactly the same elements and are equal. Option A is correct.

Assuming natural numbers begin with 1, the members of A are 1, 2, 3, ..., 99. This list contains exactly 99 elements and has a final member, so it cannot continue indefinitely. A set with a fixed, countable number of elements is called a finite set. Therefore, A is finite.

The integers strictly between 14 and 16 contain only 15. However, 15 is not prime because it has divisors 1, 3, 5, and 15; in particular, it is divisible by 3 and 5. Thus no number satisfies both conditions in the set-builder description. Therefore A has no elements and A = ∅.

A set records distinct elements only once, so repeated writing does not increase its membership. The set {0,0,0} therefore contains just one distinct element, namely 0. It is exactly the same set as {0}. The empty set has no elements, and {1} contains 1 rather than 0, so neither of those is equal to {0}.

A standard year has twelve distinct months: January, February, March, April, May, June, July, August, September, October, November, and December. Each month is one element of the set, and the list is fixed and complete. Therefore, the set contains 12 elements and is a finite set, not an infinite one.

A prime number has exactly two positive divisors: 1 and the number itself. The only even number with this property is 2; every other even number is divisible by 2 and has additional divisors. Thus the set of even prime numbers contains exactly one element, 2, and is therefore A = {2}.

The elements of A are a, b, and c. In B, the repeated a is not counted as a separate element, so the distinct elements are also a, b, and c. Set equality depends on membership, not on order or repetition. Since A and B have exactly the same members, they are equal sets.

Whole numbers are 0, 1, 2, 3, 4, and so on. There is no whole number strictly between two consecutive whole numbers, 3 and 4. Because no value satisfies the condition defining A, the set has no members. A set with no elements is called the empty set, so A = ∅.

Natural numbers less than 5 are 1, 2, 3, and 4, so B = {1, 2, 3, 4}. This is exactly the same collection as A. Two sets are equal when they contain the same elements; the way a set is written, such as roster form or set-builder form, does not affect equality. Therefore, A = B, and option A is correct.

A set contains distinct objects, and its cardinality is the number of distinct elements. The letters appearing in MATH are M, A, T, and H. Since all four letters are different, A = {M, A, T, H} and n(A) = 4. Thus the set is finite and option B is correct. Option A undercounts the letters, option C overcounts them, and option D incorrectly treats this word-based set as infinite.

The defining property of a set is that repetition does not create additional elements. In A = {1, 1, 2, 2, 3}, the repeated entries are written more than once, but each value is counted only once. Therefore the set is equivalent to A = {1, 2, 3}, whose cardinality is 3. Option C is correct; 5 and 4 count repetitions, while 2 omits one distinct value.