Complement of a Set and Its Properties

By definition, an element belongs to Aᶜ exactly when it belongs to U but does not belong to A. An element in A∩Aᶜ would therefore have to satisfy both x∈A and x∉A at the same time. This is logically impossible, so no element can be common to A and Aᶜ. Consequently, A∩Aᶜ=∅ for every subset A of U, regardless of whether A is empty or equal to U.

The complement of A relative to U contains every element of U that is not in A. Removing 1, 2, and 3 from U gives Aᶜ={4,5,6,7,8}. This is also exactly U\A and the set described by {x∈U | x∉A}. Option A, {4,5,6,7}, omits 8, which belongs to U and is not in A; therefore it is not the complement.

De Morgan’s law states that the complement of a union is the intersection of the complements: (A∪B)ᶜ=Aᶜ∩Bᶜ. An element is outside A∪B precisely when it is outside A and outside B, so it must belong to both Aᶜ and Bᶜ. The other statements either keep the wrong operation or incorrectly place complements, so they are not valid identities in general.

Because \(B=A^c\), set \(B\) contains exactly those elements of the universal set that are not in \(A\). Here, \(B=\{2,4,6,8,10\}\). Combining \(A\) and \(B\) includes every element from 1 through 10, so \(A\cup B=U\). This is the complement law: a set and its complement together cover the whole universal set.

The complement of \(A\) in \(U\) is \(B=\{1,3,5,7,9\}\). The sets \(A\) and \(B\) have no common elements: every element of \(U\) belongs to exactly one of them. Therefore their intersection is empty, \(A\cap B=A\cap A^c=\varnothing\). This is the standard complement identity and not merely a result of the particular numbers used.

The union combines all elements of the two sets: \(A\cup B=\{1,2,3,4,5,6,7,8\}=U\). The complement of a set consists of elements of U that are outside that set. Since the union already equals all of U, there are no elements left outside it. Consequently, \((A\cup B)^c=U^c=\varnothing\), making option A correct.

For any set \(A\) defined inside a universal set \(U\), its complement \(A^c\) contains precisely the elements of U that are not in A. Thus no element can belong to both sets, so \(A\cap A^c=\varnothing\). At the same time, every element of U belongs to A or to its complement, so \(A\cup A^c=U\). Therefore option A states both correct properties.

The elements of A are the positive multiples of 6 not exceeding 60: 6, 12, 18, …, 60. Their number is 60 ÷ 6 = 10. Since U has 60 elements, the complement Aᶜ contains all elements of U that are not divisible by 6. Hence n(Aᶜ) = n(U) − n(A) = 60 − 10 = 50. Therefore, option A is correct; option B counts A itself.

By definition, the complement of a set A contains all elements of U that are not in A. If A is the empty set, it contains no elements, so every element of U lies outside A. Therefore, ∅ᶜ = U. The complement of U is instead ∅, and a singleton or infinite set generally has a complement that is neither necessarily empty nor the whole universal set. Hence option A is the unique correct answer.

The integers from −6 through 6 form the universal set U, so n(U) = 13 because there are 6 negative integers, zero, and 6 positive integers. Solving x² = 16 gives x = −4 or x = 4, both of which belong to U. Thus A = {−4,4} and n(A) = 2. Therefore, n(Aᶜ) = n(U) − n(A) = 13 − 2 = 11. Option A is correct.

The complement has one-third as many elements as the universal set, so n(Aᶜ) = (1/3) × 96 = 32. A set and its complement partition the universal set into two non-overlapping parts. Therefore, n(A) + n(Aᶜ) = n(U), and n(A) = 96 − 32 = 64. Thus, option A is correct; 32 is the size of Aᶜ, not A.

The universal set U contains all integers from 1 through 36, so n(U) = 36. The perfect squares in this range are 1, 4, 9, 16, 25, and 36; hence n(A) = 6. The complement contains every element of U that is not a perfect square. Therefore, n(Aᶜ) = n(U) − n(A) = 36 − 6 = 30, so option A is correct.

The equation Bᶜ = A means that B must be the complement of A in the given universal set U. Remove the elements 2, 4, and 6 from U = {1, 2, 3, 4, 5, 6, 7}; the remaining elements are 1, 3, 5, and 7. Hence B = Aᶜ = {1, 3, 5, 7}, making option A correct. Option B is A itself, not its complement.

The universal set contains a, b, c, d, e, and g. Set A contains a, d, and g, while the remaining elements b, c, and e form B. Thus B contains exactly the elements of U that are not in A, so B = U \ A = Aᶜ. The sets are disjoint and their union is U; therefore option A is correct.

A set and its complement are disjoint, and their union is the universal set. Hence the elements of A and Aᶜ can be counted separately and added: n(U) = n(A ∪ Aᶜ) = n(A) + n(Aᶜ) = 27 + 33 = 60. The subtraction 33 − 27 = 6 is irrelevant here, so option A is correct.

The double-complement law states that taking the complement twice returns the original set, so (Aᶜ)ᶜ = A. The required expression therefore becomes A ∩ {3, 6, 9}. Since A = {1, 2, 3, 4, 5, 6}, the common elements are 3 and 6; 9 is not in A. Thus the intersection is {3, 6}, so option A is correct.

The complement A^c consists precisely of the elements of the universal set U that are not in A. Therefore, no element can belong to both A and A^c, which gives A∩A^c=∅. At the same time, every element of U belongs to either A or A^c, so A∪A^c=U. Hence option A is the only universally true statement.

The digits in the universal set are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, so n(U) = 10. The prime digits are 2, 3, 5, and 7, giving n(A) = 4. The complement Aᶜ contains all digits in U that are not prime: {0, 1, 4, 6, 8, 9}. Therefore, n(Aᶜ) = n(U) − n(A) = 10 − 4 = 6. Hence, option A is correct.

The complement Aᶜ consists of all elements of the universal set U that are not in A. Every element of U is therefore either in A or in Aᶜ, so their union is U: A ∪ Aᶜ = U. They have no common element, which means A ∩ Aᶜ = ∅. Thus options B, C, and D are not generally true.

For every element x of U, the law of excluded middle says that either x ∈ A or x ∉ A. If x ∈ A, it is included in A; if x ∉ A, the definition of the complement places it in Aᶜ. Thus every element of U belongs to A or Aᶜ, and both sets are subsets of U. Consequently, A ∪ Aᶜ = U.

Because \(B=A^c\), set \(B\) contains every element of the universal set that is not in \(A\). Here, \(B=\{1,4,6,8,9,10,12\}\). Together, \(A\) and \(B\) contain every element from 1 through 12, so \(A\cup B=U\). This illustrates the identity \(A\cup A^c=U\).

The complement \(B=A^c\) contains exactly those elements of the universal set \(U\) that are not members of \(A\). Therefore, no element can belong to both \(A\) and \(B\) at the same time. Hence, \(A\cap B=A\cap A^c=\varnothing\). The correct answer is option A. This is a standard complement identity, and it is true for every set relative to its universal set.

The union combines every element that belongs to either set. Here, \(A\cup B=\{1,2,3,4,5,6,7,8\}=U\), because the two sets together contain all elements of the universal set. Therefore, \((A\cup B)^c=U^c\). The complement of the universal set relative to itself contains no elements, so \(U^c=\varnothing\). Thus, option A is the only correct answer.

Set \(A\) contains only odd numbers from the universal set, whereas \(B\) contains only even numbers. Hence they have no common element and \(A\cap B=\varnothing\). The complement of the empty set relative to \(U\) is the whole universal set, so \((A\cap B)^c=\varnothing^c=U\).

The complement \(A^c\) consists of every element of the universal set \(U\) that is not an element of \(A\). Removing 2, 4, and 6 from \(U=\{1,2,3,4,5,6\}\) leaves \(\{1,3,5\}\). Therefore, option A is correct. The complement always depends on the stated universal set.