The Empty Set, Finite and Infinite Sets, Equal Sets

The even natural numbers are 2, 4, 6, 8, 10, and so on. For every even natural number, a larger even natural number can be found by adding 2. Therefore, the list never terminates and the set has infinitely many elements. Hence, E is an infinite set, not a finite or singleton set.

The condition requires an integer x to satisfy 2 < x < 3. However, consecutive integers 2 and 3 have no integer strictly between them. The endpoints are excluded because the inequalities are strict. Therefore, no value of x satisfies the condition, so the set contains no elements and is the empty set: A = ∅. Hence, option D is correct.

The notation A = {∅} means that the empty set itself is enclosed in braces and is being treated as one element of A. Thus, A has exactly one element, namely ∅, and its cardinality is |A| = 1. This must be distinguished from A = ∅, which has no elements. Therefore, option B is correct.

Solving x² = 9 gives x = 3 or x = −3. Both 3 and −3 are odd integers, so neither satisfies the additional condition that x must be even. Since no number satisfies both conditions simultaneously, A has no elements. Therefore A is the empty set, represented by ∅.

A week has exactly seven days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. This collection has a fixed and countable number of distinct elements. Since its elements can be completely listed and the list ends after seven items, A is a finite set, not an infinite or empty set.

To compare the sets, write B in roster form. The prime numbers less than 10 are 2, 3, 5, and 7, so B = {2, 3, 5, 7}. This is exactly the roster form of A. Since two sets are equal when they contain precisely the same elements, A = B. Therefore, option A is correct.

In the usual school convention, natural numbers begin with 1. The one-digit natural numbers are therefore 1 through 9. Zero is not included under this convention, and 10 is a two-digit number. Hence the correct roster form is {1, 2, 3, 4, 5, 6, 7, 8, 9}, shown in option B.

The natural numbers greater than 7 are 8, 9, 10, 11, and so on. The phrase “and so on” continues without an ending because there is no upper bound. For every member, another larger natural number can be found. Consequently, A has infinitely many elements and is an infinite set.

A set records membership, not the number of times an element is written. Repeated occurrences of 2 and 3 are counted only once. Thus the given set is equivalent to {1, 2, 3}. It has three distinct elements, so its cardinality, denoted by n(A), is 3. Therefore, option A is correct.

The counting numbers from 1 through 5 are 1, 2, 3, 4, and 5. Thus, A has exactly five distinct elements. A set with a fixed and limited number of elements is called a finite set. It is not empty because it has elements, not infinite because its elements can be completely counted, and not an equal set because equality requires comparison with another set.

An empty set is a set that contains no elements. Its standard symbol is ∅, and it may also be written as { }. The symbol 0 represents the number zero, not a set. The set {0} contains one element, namely zero, so it is not empty. Similarly, 1 is a number rather than the notation for an empty set.

The names of the months in a year are January through December, giving exactly twelve distinct elements. Since the number of month names is fixed and limited, A is a finite set. It is not empty because month names exist, not infinite because there are only twelve of them, and not a singleton because a singleton has exactly one element.

Evaluate the rule defining B for each allowed value of x: 1² = 1, 2² = 4, and 3² = 9. Therefore, B = {1, 4, 9}. Set equality depends on having exactly the same elements, regardless of how the sets are represented or ordered. Since A has precisely these elements, A = B.

The empty set ∅ has no elements. There is no natural number less than zero, so the condition in option C cannot be satisfied by any natural number. Consequently, that set contains no elements and is equal to ∅. In contrast, {0}, {1}, and {∅} each contain one element; notably, {∅} contains the empty set as an element and is not itself empty.

The positive multiples of 3 that are less than 15 are 3, 6, 9, and 12. The number 15 is excluded because the condition says less than 15, not less than or equal to 15. Therefore, A = {3, 6, 9, 12}, which has four elements. The set is finite because this list ends after a limited number of values.

Two sets are equal precisely when they contain the same elements. The order in which elements are written does not matter, and repeated listing of an element does not create a new element. Equal sets need not be empty; for example, {1, 2} and {2, 1} are equal. Hence, option B is correct.

A prime number has exactly two positive divisors: 1 and itself. Among all even numbers, only 2 has this property; every other even number is divisible by 2 and another integer, so it is not prime. Hence the set of even prime numbers is A = {2}. It contains one element, so it is also a singleton and a finite set.

The natural numbers greater than 50 are 51, 52, 53, 54, and so on. There is no greatest natural number, so new members can always be found beyond any listed value. The set therefore continues without end and has infinitely many elements. Hence, the correct answer is option D, infinite.

Sets are equal when they have exactly the same distinct elements; the order does not matter, and repeated elements are counted only once. Thus options A and D contain the same elements as A, while option B also represents {1, 2, 3} because 1 is repeated. Option C has the additional element 4, so it is not equal.

Solving x + 1 = 1 gives x = 0. Under the usual school convention used here, natural numbers are 1, 2, 3, ...; therefore 0 is not a natural number. No value satisfies both conditions simultaneously, so A contains no element and is the empty set, written as ∅. Thus option C is correct.