Complement of a Set and Its Properties

The closed interval A = [−1,4] includes both endpoints −1 and 4 as well as every real number between them. Consequently, its complement in ℝ must exclude both endpoints and contain only numbers less than −1 or greater than 4. Therefore Aᶜ = (−∞,−1) ∪ (4,∞), so option A is correct.

A set and its complement partition the universal set: they contain no common elements and together contain every element of U. Since Aᶜ = {2,5,8}, obtain A by removing these three elements from U = {1,2,3,4,5,6,7,8,9}. The remaining elements are {1,3,4,6,7,9}. Thus option B is correct. Option A is the given complement, not A itself.

The complement of A is defined relative to a specified universal set U. It consists of exactly those elements that belong to U but do not belong to A. Therefore, in set-builder notation, Aᶜ = {x : x ∈ U and x ∉ A}. Option B states both required conditions and is correct.

The complement Aᶜ contains exactly those elements of the universal set U that are not in A. Therefore, an element cannot belong to both A and Aᶜ at the same time. The two sets are disjoint, so their intersection contains no element: A ∩ Aᶜ = ∅. Hence option D is correct; A, Aᶜ, and U represent other set expressions, not this intersection.

For every element of the universal set U, exactly one of two possibilities holds: it belongs to A, or it does not belong to A and therefore belongs to Aᶜ. Thus A together with Aᶜ covers every element of U. By the complement law, A ∪ Aᶜ = U. Option D is correct; the empty set would describe A ∩ Aᶜ, not their union.

The complement Bᶜ is taken with respect to U, so Bᶜ = U − B = {1,2,3,8,9,10}. Now intersect this set with A = {1,2,3,4,5}. The common elements are only 1, 2, and 3. Therefore, A ∩ Bᶜ = {1,2,3}, which is option A. The elements 4 and 5 are excluded because they belong to B.

The complement Aᶜ contains all elements of the universal set U that are not members of A. Since A = {p,r,t}, removing these elements from U leaves Aᶜ = {q,s}. Among the given choices, q belongs to this complement, whereas p, r, and t are already elements of A. Hence option C is the only correct answer.

By definition, the complement Aᶜ consists of the elements of the universal set U that are not in A. Therefore, the statement x ∈ Aᶜ means both x ∈ U and x ∉ A. Option A contradicts the definition, option C places x outside the universal set, and option D would require x to belong to both A and its complement, which is impossible.

The multiples of 3 that lie in U are 3, 6, 9, and 12, so A = {3,6,9,12}. The complement contains every element of U that is not a multiple of 3. Removing these four multiples from U gives Aᶜ = {1,2,4,5,7,8,10,11}. Thus option B is correct; options A and D list only multiples of 3.

The prime numbers in U are 2, 3, 5, and 7, because each has exactly two positive divisors. Thus A = {2,3,5,7}. The complement contains the remaining elements of U, namely {1,4,6,8,9}. Notice that 1 is neither prime nor composite, so it belongs to the complement. Therefore option B is correct.

With respect to U, Aᶜ = {4,5,6} and Bᶜ = {1,2,6}. Taking their union gives every element appearing in either complement: {1,2,4,5,6}. This also follows from De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ, because A ∩ B = {3} and its complement in U is {1,2,4,5,6}. Hence option A is correct.

Calculate each complement within U: Aᶜ = {3,5,6} and Bᶜ = {1,3,5}. Their common elements are 3 and 5, so Aᶜ ∩ Bᶜ = {3,5}. De Morgan’s law gives the same result because Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, A ∪ B = {1,2,4,6}, and its complement is {3,5}. Therefore option A is correct.

The complement is defined by Aᶜ = U − A. If Aᶜ is empty, there is no element of U outside A. Therefore A must contain every element of U, which means A = U. Option A would make the complement equal to U, not empty. Also, A ∩ U = A, so option D would be empty only if A itself were empty. Hence option B is the unique correct answer.

Since Aᶜ = U − A, the complement can equal the entire universal set only when A contains no elements. Thus A must be the empty set, ∅. Equivalently, the complement of ∅ is U. Option A would give an empty complement, while A ∪ U is always U, so option D cannot be true for a nonempty universal set. Therefore option B is correct.

The universal set has n(U) = 8 elements, while the complement has n(Aᶜ) = 3 elements. Since A and Aᶜ partition U, n(A) + n(Aᶜ) = n(U). Therefore n(A) = 8 − 3 = 5. Equivalently, A consists of the remaining elements {2,3,5,6,8}, which also confirms that its cardinality is 5. Hence option B is correct.

The complement Aᶜ contains every element of the universal set U that is not present in A. Removing 1, 2, 3, and 4 from U gives Aᶜ = {5,6,7,8,9,10}. Since Aᶜ is a subset of B, every one of these six elements must occur in B. B may contain additional elements, but it cannot omit any of them; therefore option B is the required minimum set.

The difference U \ A consists of the elements that belong to U but do not belong to A. From U = {1,2,3,4,5,6}, remove 2, 3, and 5. The remaining set is {1,4,6}. By the definition of a complement relative to U, this remaining set is Aᶜ. Therefore U \ A = Aᶜ, so option B is correct. The actual elements are {1,4,6}.

The complement Aᶜ contains all elements of U that are not in A, so Aᶜ = {2,3,5,6}. Taking the union with A gives {2,3,5,6} ∪ {1,4,7} = U. Thus Aᶜ ∪ A contains every element of the universal set, and its cardinality is |U| = 7. This is the standard identity A ∪ Aᶜ = U, so option C is correct.

Relative to U, the complement of A is Aᶜ = {1,3,5,7}. No element can belong to both A and its complement: A contains {2,4,6}, while Aᶜ contains {1,3,5,7}. Therefore their intersection is Aᶜ ∩ A = ∅. The empty set has cardinality zero, so the number of elements is 0 and option A is correct.

The complement of A is formed by selecting from U every element that is not in A. Since U contains 1 through 8 and A contains 1, 2, 5, and 8, the elements left outside A are {3,4,6,7}. This set is exactly C. Hence C = Aᶜ relative to U, making option A correct. The other options either equate C with A, all of U, or the empty set.