The Empty Set, Finite and Infinite Sets, Equal Sets

Solve the inequality: 2x + 1 < 20 gives 2x < 19, so x < 9.5. Since x belongs to the natural numbers and, in the usual school convention, ℕ = {1, 2, 3, ...}, the possible values are 1 through 9. Therefore, the set is {1, 2, 3, 4, 5, 6, 7, 8, 9} and has 9 elements. Hence, option B is correct.

The set contains all rational numbers strictly between 0 and 1. It includes 1/2, 1/3, 1/4, 2/5, and infinitely many other rational numbers. In fact, for every positive integer n, the number 1/(n+1) is rational and lies between 0 and 1, giving infinitely many distinct members. Therefore, the set is infinite, so option D is correct.

The equation x² = 4 has solutions x = 2 and x = −2 over the integers. However, the condition specifies x ∈ ℕ, so −2 is excluded. The only natural-number solution is x = 2; therefore A = {2}. Since B is also {2}, both sets contain exactly the same element and are equal. Option A is correct.

Set equality requires the two sets to contain exactly the same elements. A contains the numbers 1, 2, and 3, while B contains the symbols a, b, and c. Thus the sets are not equal. However, each set has three elements, so they have the same cardinality and are equivalent in size. Therefore, option C is correct.

For every integer x, x² is non-negative, so x² + 1 is at least 1 and can never equal 0. Thus the equation x² + 1 = 0 has no integer solution, which means B contains no elements. Therefore B is the empty set, just like A. Since two sets are equal when they have exactly the same elements, A = B. Hence, option A is correct.

The empty set, written as ∅ or {}, is the set containing no elements. Its cardinality is therefore zero: n(∅) = 0. This does not mean that 0 is an element of the empty set; the number 0 describes how many elements it has. The empty set is also finite because its number of elements is limited. Hence option B is correct.

Factor the equation: x² − 2x + 1 = (x − 1)². Setting this equal to zero gives (x − 1)² = 0, so x = 1 is the only real solution. Therefore, the set is {1}, which contains exactly one element. A set with exactly one element is called a singleton set. Hence, option B is the correct answer.

Taking n = 1, 2, 3, … produces the elements 2, 4, 6, 8, … . For every natural number n, another even number 2n can be produced, so the list never terminates. Repetition does not occur for different natural values of n. Hence the set contains infinitely many elements and is an infinite set. Option C is correct.

The condition requires x to be a natural number strictly greater than 5 and strictly less than 6. There is no integer, and therefore no natural number, between two consecutive integers 5 and 6. No value satisfies the condition, so the set has no elements. Hence it is the empty set, and option A is correct.

Because x is an integer and both inequalities are strict, −3 and 3 are excluded. The integers strictly between them are −2, −1, 0, 1, and 2. Counting this list gives five distinct elements. Therefore, the cardinality of the set is 5, so option B is the correct answer.

The empty set contains no elements and is denoted by ∅ or by a pair of empty braces, {}. In contrast, {0} contains the element 0, {1} contains the element 1, and {∅} contains the empty set as one element. Therefore only option C represents a set with no elements.

The natural-number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. This list contains eight elements and is complete because a positive integer has only finitely many positive divisors. Therefore, the set is non-empty and finite, not infinite or a singleton. Option B is correct.

The governing concept is counting the positive natural-number divisors of a finite number. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Each divides 36 exactly, and the list is complete; alternatively, 36 = 2² × 3², so the number of positive divisors is (2 + 1)(2 + 1) = 9. Hence option B is correct. The set is finite, so option D cannot be true.

The absolute value |x| represents the distance of x from zero and is always non-negative for every real number. Therefore, it can never equal −1. No real number satisfies the condition, so the set has no elements and is the empty set. Hence option A is correct.

To determine B, solve the defining condition x² = 0. The only solution is x = 0, and 0 belongs to the integers, so B = {0}. Thus B has one element, whereas A = ∅ has no elements. A singleton cannot equal the empty set, so A ≠ B. Option B is correct; C and D misclassify B, and A ignores the different contents.

The positive natural-number factors of 12 are 1, 2, 3, 4, 6, and 12. These are exactly the elements listed in B, with no missing or additional element. Since two sets are equal when they contain precisely the same elements, A = B. The order in which elements are written does not affect equality.

The only natural number that divides 1 is 1 itself. However, 1 is not prime because a prime number must have exactly two distinct positive divisors, namely 1 and itself. Therefore, no natural number satisfies both conditions simultaneously. The set has no elements and is consequently the empty set.

For every natural number n, the set contains the ordered pair (n, n + 1). Taking n = 1, 2, 3, 4, and so on produces (1,2), (2,3), (3,4), (4,5), and infinitely many further pairs. Different n values give different first coordinates, so no pairs repeat. Hence the set is infinite.

To determine A, solve x³ = x. Rearranging gives x³ − x = 0, so x(x² − 1) = 0 and therefore x(x − 1)(x + 1) = 0. Hence x = 0, 1, or −1. All these values are integers, so A = {-1, 0, 1}. Since B contains exactly the same elements, A and B are equal sets. Therefore, option A is correct.

The open interval (0, 1) contains every real number strictly between 0 and 1. There are infinitely many such numbers; for example, 1/2, 1/3, 1/4, and infinitely many others lie in the interval. Hence (0, 1) is an infinite subset of R. Each other option is finite: the singleton has one element, the two-element set has two, and the empty set has none.

A finite set with n distinct elements has 2ⁿ subsets. Each element has two independent choices in forming a subset: it is either included or excluded. Here A = {1, 3, 5} has n = 3 elements, so the number of subsets is 2³ = 2 × 2 × 2 = 8. These include the empty set, three one-element subsets, three two-element subsets, and the full set.

A closed interval contains both of its finite endpoints. Square brackets show inclusion, so [4,9] represents all real numbers x for which 4≤x≤9, including both 4 and 9. Option A excludes both endpoints and is open. Options C and D include one endpoint but exclude the other, so they are half-open intervals rather than closed intervals.

The empty set has no elements. For a set to fail to be a subset of another set, it must contain at least one element that is missing from the other set. Since the empty set has no element that can violate this condition, it is a subset of every set, including itself. It is different from {0}, which has one element.

To find the members of A, solve the equation x² = 4 over the real numbers. Taking square roots gives x = 2 or x = −2. Therefore, both values belong to the solution set, so A = {−2, 2}. The braces indicate a set containing two separate numbers; they do not represent every number in an interval. Hence option B is correct.

The interval [−5, −1) includes −5 because the left square bracket means the left endpoint is included. It contains every number greater than −5 and less than −1. The right round bracket means −1 itself is excluded. Thus −3 and −4.5 are inside the interval, −5 is included, and −1 is the only number that is not in A. Therefore option C is correct.