The Empty Set, Finite and Infinite Sets, Equal Sets

Set A contains the elements 5, 10, and 15. Set B contains exactly the same three elements; only their written order is different. In set theory, order does not affect membership, and repeated elements would also not create new elements. Since every element of A is in B and every element of B is in A, A and B are equal. Hence option B is correct.

Using the usual school definition N = {1, 2, 3, ...}, the smallest natural number is 1. The condition x < 1 therefore cannot be satisfied by any natural number. Since no element meets both conditions, the set is empty and its correct roster form is A = ∅. Zero is excluded because it is not included in this convention for N.

The positive odd numbers less than 8 are obtained by checking 1, 2, 3, 4, 5, 6, and 7: they are 1, 3, 5, and 7. Thus Q = {1, 3, 5, 7}, which is exactly the set P. Since sets are equal when they contain the same elements, P = Q. Q is finite, not infinite.

A set is empty when no element satisfies its defining condition. In option A, x² = 0 gives x = 0; depending on the convention for natural numbers, this may be considered empty if ℕ starts at 1. Options B and D are clearly empty because no integer lies strictly between 0 and 1 and no natural number is negative. In option C, x + 2 = 5 gives x = 3, and 3 is a natural number. Therefore, option C contains one element and is not empty.

There are only 40 students, so at most 40 birth-date entries can be collected. Some students may share the same birth date, which would reduce the number of distinct elements, but it can never make the set infinite. The set contains a limited number of dates and is therefore finite. Hence, option C, finite set, is correct.

The empty set ∅ and the number 0 are different mathematical objects. The symbol ∅ denotes a set having no elements, whereas 0 is a number. In A = {∅, 0}, both objects are placed inside the outer braces as separate members. Thus A contains two distinct elements, so n(A) = 2. The correct answer is option C.

The condition for B allows n = 1, 2, and 3 only. Squaring these values gives x = 1² = 1, x = 2² = 4, and x = 3² = 9. Therefore B = {1, 4, 9}, which has exactly the same elements as A. Sets are equal when they contain the same elements, regardless of how they are represented. Hence A = B.

To find the members of A, solve x² = 1 over the integers. Factoring gives (x − 1)(x + 1) = 0, so x = 1 or x = −1. Both solutions are integers and both satisfy the condition. Therefore A = {−1, 1}, which contains two distinct elements. Thus the correct answer is option C.

The set requires natural numbers that are even and strictly less than 10. Under the usual school convention ℕ = {1, 2, 3, ...}, the even natural numbers below 10 are 2, 4, 6, and 8. Zero is excluded by this convention, odd numbers do not satisfy the condition, and 10 is not less than 10. Therefore option B is correct.

The symbol ∅ represents the empty set, so it has zero elements. The notation {∅} is different: the outer braces create a set whose only member is the empty set itself. Thus {∅} is a singleton set and has exactly one element. The empty set and the singleton containing it are not equal. Therefore, option C is correct.

A standard week has seven named days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. Each name is a distinct element of the set A. Since the list is fixed and contains exactly seven different names, A is a finite set with cardinality 7. Therefore, option C is the correct answer.

The natural-number factors of 13 are 1 and 13 because 13 is prime. Thus A = {1, 13}. This set has exactly two elements, so it is neither empty nor infinite. The term unequal set is not a classification of a set by its number of elements and is irrelevant here. Therefore, A is a finite set and option B is correct.

The only natural number less than 2 is 1. However, 1 is not a prime number because a prime number must have exactly two distinct positive divisors, namely 1 and itself. Therefore, no natural number satisfies both conditions simultaneously. A set with no element is called the empty set, so A = ∅.

The natural numbers greater than 50 are 51, 52, 53, 54, and so on. There is no greatest natural number, so this list never ends. Therefore, set A contains infinitely many elements and is an infinite set. The condition gives only a lower bound, not a final upper bound.

A set records whether an element is present, not how many times it is written. Thus, the repeated 3 in B is counted only once, so B is effectively {3, 6, 9}. Both A and B contain exactly the same elements. The order of elements and repetition do not affect set equality; therefore, A = B.

The governing concept is the cardinality of a finite set. The positive multiples of 10 below 100 are obtained as 10n with n = 1, 2, ..., 9: 10, 20, 30, 40, 50, 60, 70, 80, and 90. Since “less than 100” excludes 100, there are 9 distinct elements. Therefore option B is correct; 8 omits one multiple, 10 includes 100, and the set is not infinite.

For every integer x, the square x² is non-negative: it is either zero or positive. It can never equal −4. Although the equation has complex solutions, the set is restricted to integers, so those solutions are not allowed. Hence no integer satisfies the condition and A is the empty set.

For option A, substitute the natural-number values n = 1, 2, and 3 into x = n³. The resulting values are 1³ = 1, 2³ = 8, and 3³ = 27. Hence option A produces exactly the elements of the given set. Equality of sets depends on having precisely the same elements, regardless of their order.

The integers are ..., −3, −2, −1, 0, 1, 2, 3, ... . They extend without end in the negative direction and also without end in the positive direction. Consequently, there is no final integer and the set cannot be listed completely. Therefore, the set of all integers is infinite.

The positive natural-number factors of 9 are 1, 3, and 9. All three are odd, so none satisfies the additional condition of being even. Because both conditions must hold simultaneously, no element qualifies. Therefore A has no elements and is the empty set, written as ∅.

The governing concept is the cardinality of a finite set. A standard die has six faces, conventionally marked with the distinct numbers 1, 2, 3, 4, 5, and 6. Thus the set of numbers written on it is A = {1, 2, 3, 4, 5, 6}, whose cardinality is 6. Option C is correct. The values 4 and 5 undercount the faces, while “infinitely many” is impossible for a finite die.

The natural numbers less than 6 are 1, 2, 3, 4, and 5, so A = {1, 2, 3, 4, 5}. Set B contains exactly the same five elements, merely written in reverse order. In set theory, order does not matter; equality depends only on having the same elements. Hence A and B are equal sets.

The condition requires natural numbers strictly greater than 10 and strictly less than 15. Therefore, 10 and 15 are excluded because the inequalities are strict. The members are 11, 12, 13 and 14. Since there are four distinct members, the set is finite and its cardinality is 4. Hence option B is correct.

The empty set, written as ∅, contains no elements, so its cardinality is 0. A set is finite when it has a fixed, countable number of elements, and zero is a finite number. Thus the empty set is both finite and empty. In contrast, {0} has one element, while the natural-number sequence and the integers are infinite. Therefore option B is correct.

The two-digit natural numbers are 10, 11, 12, ..., 99. This collection has a definite first element and a definite last element, so its members can be counted completely. In fact, it contains 99 − 10 + 1 = 90 elements. Therefore, A is a finite set, even though it has many elements. It is neither empty, nor infinite, nor a singleton.