The Empty Set, Finite and Infinite Sets, Equal Sets

There is no natural number less than 0, because natural numbers are non-negative in the usual school definition. Hence no value of x satisfies the condition in option C, so that set has no elements and equals ∅. In contrast, {0}, {1}, and {∅} each contain one element; particularly, {∅} is not the empty set.

Order is not relevant in set notation. For example, {1, 2, 3} and {3, 1, 2} contain exactly the same elements, so they represent the same set. Equality of sets depends on membership, not on the sequence in which elements are displayed. Therefore, the two sets are equal and option C is correct.

A prime number is a natural number greater than 1 with exactly two positive divisors: 1 and itself. Checking numbers below 10 gives 2, 3, 5, and 7 as primes. The number 1 is not prime, and 4, 6, 8, and 9 are composite because they have more than two divisors. Therefore A = {2, 3, 5, 7}, so option B is correct.

The even natural numbers strictly less than 7 are 2, 4, and 6. Thus the set-builder description gives B = {2, 4, 6}. Set equality depends on having exactly the same elements, not on how the elements are obtained or ordered. Since A already contains precisely these three members, A and B are equal. Therefore option A is correct; A is not empty and B is finite, not infinite.

The condition requires an integer strictly greater than 0 and strictly less than 1. There is no integer between 0 and 1; numbers such as 0.5 are not integers. Consequently, no integer satisfies the condition, so A contains no elements and is the empty set ∅. Hence option A is correct.

A finite set has a limited, definite number of elements, so its elements can be counted and the counting ends at a fixed number. A finite set may contain no elements, one element, or many elements; therefore, it is not necessarily empty. The description of an endless collection belongs to an infinite set. Thus option C is correct.

An infinite set has more elements than can be completely counted to a final number; its counting does not terminate. Examples include the natural numbers {1, 2, 3, ...} and the integers. An infinite set can contain many elements, but it is not empty and does not have a fixed finite size such as five. Therefore, option B is correct.

Two sets are equal only when every element of the first set is also an element of the second set, and vice versa. Here, A contains p but B does not contain p. Similarly, B contains s but A does not contain s. Although q and r are common to both sets, the different elements p and s show that A and B are unequal sets. Therefore, option B is correct.

A week has exactly seven days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. The set described in the question therefore contains seven distinct elements. Since its elements can be completely counted and the counting ends at seven, this is a finite set, not an infinite or empty set.

The multiples of 5 include 5, 10, 15, 20, 25, and so on. For every multiple listed, another larger multiple can be obtained by adding 5, so the list never terminates. Therefore, the set contains infinitely many elements and is an infinite set. It is not empty, finite, or a singleton. Hence, option C is correct.

In the standard school classification of the English alphabet, the vowels are a, e, i, o, and u. These are five distinct letters, and a set counts each distinct element only once. Therefore, the cardinality, or number of elements, of A is 5. The other letters are consonants in this context.

A set is non-empty if at least one object satisfies its defining condition. The number 2 is an even prime number because it has exactly two positive divisors, 1 and 2. Thus option C contains the element 2. There is no natural number less than 1 under the usual school convention, no integer strictly between 2 and 3, and no month with 40 days.

In set notation, repetition of an element does not create a new element, and the order in which elements are written does not matter. Thus, B = {1, 2, 2} is the same set as {1, 2}, because the repeated 2 is counted only once. Also, C = {2, 1} has exactly the same elements as A, only in a different order. Therefore, A = B = C, so option C is correct.

The empty set is the set that contains no elements. Its standard symbols are ∅ and, alternatively, { }. The symbol {0} does not mean empty because it contains the element 0, and {1} contains the element 1. The symbol ℕ denotes the set of natural numbers, which is not empty. Therefore, option A is correct.

The two-digit natural numbers begin at 10 and end at 99. Thus the set can be listed completely as 10, 11, 12, through 99. It has 90 elements, since 99 − 10 + 1 = 90. Because the list has a definite beginning and ending, the set is finite. It is not empty or infinite.

The integers are ..., −3, −2, −1, 0, 1, 2, 3, .... They extend without end in both the negative and positive directions. For any integer, another larger integer can be found, and another smaller integer can also be found. Consequently, the set of all integers has infinitely many elements and is an infinite set.

A factor of 6 is a positive integer that divides 6 exactly, leaving remainder zero. The factor pairs are 1 × 6 and 2 × 3, so the positive factors are 1, 2, 3, and 6. Hence A = {1, 2, 3, 6}, making option A correct. Option B omits 1, option C omits 2 and 3, and option D incorrectly includes 4, which does not divide 6 exactly.

Natural numbers less than 2 are considered from the counting sequence 1, 2, 3, and so on. Among them, only 1 is less than 2, and 1 is odd. Therefore, the set contains exactly one element, written as A = {1}. It is a singleton set, not an empty set, because one number satisfies the stated condition.

Two sets are equal when they contain exactly the same elements; the order of listing does not matter, and repeated elements are written only once. Options A and D contain 1, 2, and 3 in a different order. Option B also represents {1, 2, 3}, because repetition does not create a new element. Option C contains the additional element 4, so it is not equal.

Taking square numbers as positive natural-number squares, list the squares below 20: 1² = 1, 2² = 4, 3² = 9, and 4² = 16. The next square is 5² = 25, which is not less than 20. Therefore A = {1, 4, 9, 16}, containing four distinct elements. It is finite, so option B is correct; 3 misses one square, 5 includes an excluded value, and the set is not infinite.