The Empty Set, Finite and Infinite Sets, Equal Sets

The notation Y = {0} contains one element, namely the number 0. A set with exactly one element is called a singleton set. It is not empty because the element 0 is present, and it is not infinite because no other elements are listed or implied. The set {0} must also be distinguished from the empty set ∅, which has no elements.

A set is empty when no element satisfies its defining condition. Under the usual school convention, natural numbers are 1, 2, 3, …, so no natural number is less than 1. Therefore, C contains no elements and is the empty set, written as ∅. The other sets are non-empty because months, two-digit prime numbers, and vowels all exist.

A finite set has a definite, countable number of elements. Set A contains exactly four distinct elements: 2, 4, 6, and 8. Since its cardinality is 4, it is a finite set. It is not empty because it has elements, and it is not infinite because the list ends after 8. A universal set cannot be identified without a stated context.

Two sets are equal when they contain exactly the same elements, regardless of the order in which those elements are written. Both A and B contain 1, 2, and 3, and neither contains any additional element. Therefore, A = B. The order matters in an ordered tuple, but it does not matter in ordinary set notation. Both sets are also finite and non-empty.

The ellipsis symbol (…) indicates that the sequence continues without an ending. The natural numbers 1, 2, 3, 4, and so on have no greatest member, so their collection cannot be completely counted to a final element. Therefore, N is an infinite set. It is neither empty nor a singleton, and it is not finite because the list does not terminate.

There is no integer strictly between two consecutive integers, 5 and 6. The numbers between them are real numbers such as 5.5, but none of those is an integer. Hence the set A has no elements and is the empty set. Its cardinality, written as n(A) or |A|, is therefore 0. The endpoints are not between 5 and 6.

A singleton set is a set containing exactly one element. B = {0} contains one element, namely 0, so it is a singleton and it is not empty. The symbol ∅ in option A denotes a set with no elements. C has two elements and D has three elements, so neither is a singleton. The braces around 0 must not be confused with the empty-set symbol.

In a set, each element is counted only once, even if it is written repeatedly. Therefore, B = {a, c, b, b} is the same as {a, b, c}. Since A and B contain exactly the same elements, their order does not matter and A = B. Also, B has only three distinct elements, not four.

The positive even numbers less than 10 are 2, 4, 6, and 8. Thus P = {2, 4, 6, 8}. This set has exactly four elements, so all its elements can be counted and listed. A set with a fixed, limited number of elements is called a finite set; therefore, option B is correct.

The empty set, written as ∅ or {}, is defined as the set containing no elements. It must not be confused with {0}, which is a set containing one element, namely 0. The empty set is finite because its number of elements is zero. Hence the statement that it has no elements is the only correct choice.

Solving x² = 4 gives x = 2 or x = −2. The condition that x must be positive eliminates −2, leaving only x = 2. Thus A = {2}. A set containing exactly one element is called a singleton set. It is finite, but the more specific and correct classification among the options is singleton set.

The English vowels are a, e, i, o, and u. The descriptive, or set-builder, form used for A identifies exactly these five letters, while B lists the same elements in roster form. Sets are equal when they contain the same elements, and the order in which elements are written is irrelevant. Therefore, A = B.

The prime numbers less than 12 are 2, 3, 5, 7, and 11. Thus A = {2, 3, 5, 7, 11}, which contains five distinct elements. The number 1 is not prime because it has only one positive divisor, and 12 is excluded because the condition says less than 12. Therefore, the answer is 5.

A is empty because 2 is even, not odd. B is empty because a square cannot have three sides. Under the usual school convention, natural numbers are 1, 2, 3, and so on; therefore, the natural numbers less than 3 are 1 and 2. D is empty because no natural number is less than 0. Hence C is non-empty.

Two sets are equal only when every element of one set is also an element of the other set, with no extra element in either set. Here, 7 belongs to B but does not belong to A. Therefore, the sets do not contain exactly the same elements, so A ≠ B. Both sets are finite, not empty or infinite.

The positive multiples of 5 are 5, 10, 15, 20, and so on. The condition x < 20 excludes 20 and every larger multiple. Also, 0 is not positive, so it cannot be included. The remaining elements are exactly 5, 10, and 15. Therefore, A = {5, 10, 15}, which is option A. This is a finite set with three elements.

Set equality depends on the elements present, not on the order in which they are written. Both sets in option B contain exactly p and q, so they are equal. In option A, 3 and 4 differ; in option C, {0} has one element while ∅ has none; and in option D, B has the additional element 6.

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.