The Empty Set, Finite and Infinite Sets, Equal Sets

Solving x² = 2 over the real numbers gives x = √2 and x = −√2. Neither value is an integer, because √2 is irrational. Since the domain in the set-builder notation is ℤ, neither real solution is allowed. Thus no integer satisfies the condition, so the set is empty and option A 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.

The empty set, written ∅, has no elements, so its cardinality is 0. In contrast, {∅} is a set whose single element is the empty set itself; therefore, its cardinality is 1 and it is a singleton. The braces create a new set and do not leave the set empty. Hence option C is correct; A and B confuse an element with a set, while D is false.

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.

The empty set has no elements. The statement ∅ ⊆ A means that every element of ∅ is also an element of A. Because there are no elements in ∅, there is no counterexample to this requirement; therefore the statement is true for every set A. In contrast, A ⊆ ∅ is true only when A itself is empty, and ∅ = A is not always true. Hence option A is correct.

The order of elements does not matter in a set. Two sets are equal when they contain exactly the same elements. Both given sets contain 1, 2, and 3, even though the elements are written in different orders. Therefore, {1, 2, 3} = {3, 2, 1}; they are equal sets, and option A is correct.

The positive natural-number divisors of 9 are 1, 3, and 9. Therefore the set described by A is \(A=\{1,3,9\}\), which is exactly the given set B. Since two sets are equal when they contain precisely the same elements, A and B are equal. They are not disjoint, empty, or in a proper-subset relationship. The order in which elements are listed would not affect equality, although here the order is already the same.

A set is unchanged when the order of its elements is rearranged. Therefore, {2, {3}, 1} contains exactly the same three elements as A: 1, 2, and the set {3}; hence it equals A. Notice that 3 and {3} are different objects, so option B is not equal to A. Options C and D also change the nesting or the elements.

Using the standard school convention N = {1, 2, 3, ...}, the natural numbers not exceeding 4 are 1, 2, 3, and 4. Hence A = {1, 2, 3, 4}. Set B contains exactly these same elements, merely written in reverse order. The order of elements does not matter in a set, so A = B. Option B is false for this reason; option C is false because equal sets are not proper subsets of one another; and option D is false because B has four elements.