Complement of a Set and Its Properties

The power set P(A) is the set whose elements are all subsets of A. Since 1 and 2 both belong to A, the set {1, 2} is a subset of A and therefore is an element of P(A). The number 4 is not in A, so {4} and {1, 4} are not subsets. Also, 4 itself is not a subset of A in this context.

The complement A' contains exactly those elements of the universal set U that are not members of A. Removing 2 and 4 from U gives A' = U \ A = {1, 2, 3, 4, 5} \ {2, 4} = {1, 3, 5}. Option A is the original set A, option C is U itself, and option D is incorrect because some elements remain after removing A.

The complement of B is defined relative to the universal set U and contains every element of U that does not belong to B. Thus B′ = U − B. Removing a and d from U = {a, b, c, d} leaves b and c, so B′ = {b, c}. Option A is B itself, option C is the entire universal set, and option D incorrectly retains d, which belongs to B.

The complement of A is defined with respect to U: A′ = U − A = {x ∈ U : x ∉ A}. Every element of A′ therefore comes from the universal set U, so A′ ⊆ U. It is not generally a subset of A, because A and A′ are disjoint. P(A) is the collection of subsets of A, not the natural containing set for A′, and the complement need not be empty. Thus option B is correct.

The complement A′ consists of elements of U that are not in A. Since A is equal to the entire universal set U, every element of U already belongs to A, so there are no elements left outside A. Using the rule A′ = U − A, we obtain A′ = U − U = ∅. Hence option C is correct. The set {0} cannot be the answer because 0 is not in U.

The complement \(A'\) contains all elements of the universal set \(U\) that are not present in \(A\). From \(U=\{0,1,2,3,4\}\), remove the elements of \(A=\{0,2,4\}\). The remaining elements are \(A'=U\setminus A=\{1,3\}\), so \(A'\) has 2 elements. The same result follows from \(n(A')=n(U)-n(A)=5-3=2\). Hence option B is correct.

The complement E′ consists of all elements in the universal set U that are not members of E. Starting with U = {1, 2, 3, 4, 5, 6}, remove the elements 2, 4, and 6 that belong to E. The remaining elements are 1, 3, and 5, so E′ = {1, 3, 5}. Therefore, option A is correct.

The complement A′ is defined relative to the universal set U. It contains every element of U that is not an element of A, so A′ = U \ A. Starting with U = {a, b, c, d} and removing a and c, which belong to A, leaves b and d. Hence A′ = {b, d}, making option B the only correct answer.

A set and its complement partition the universal set into two disjoint parts. Therefore, A is obtained by removing A′ from U: A = U \ A′. Removing 4 and 5 from {1, 2, 3, 4, 5} leaves {1, 2, 3}. Thus option A is correct. Notice that option B gives the complement itself, not the original set.

The complement of A relative to the universal set U contains every element of U that is not an element of A. Starting with U = {1, 2, 3, 4, 5, 6, 7}, remove 1, 3, 5, and 7 because these belong to A. The elements left are 2, 4, and 6. Therefore, A′ = U − A = {2, 4, 6}. Option B is A itself, not its complement, while options C and D do not contain exactly all the required elements.

The complement A′ contains every element of the universal set U that is not present in A. If A is the empty set, it contains no elements to remove from U, so all elements of U remain in the complement. Therefore, ∅′ = U. If A = U, the complement would instead be ∅, not U.

The complement A' is defined relative to the universal set U: A' = U \ A. It contains every element of U that is not an element of A. Starting with U = {red, blue, green} and removing red, which is the only element of A, leaves {blue, green}. Therefore A' = {blue, green}, so option B is correct. The other options either give A itself, an incomplete set, or an empty set.

Since B = A', we need the complement of A in the universal set U. The complement contains elements of U that are not in A. From U = {1,2,3,4}, remove the elements 1 and 4 belonging to A. The remaining elements are 2 and 3, so B = A' = {2,3}. Option A is A itself, while options C and D contain one element of A and therefore cannot be the complete complement. Thus option B is correct.

The complement A' consists of all elements that belong to the universal set U but do not belong to A. Removing {3,6,9} from U = {1,2,3,4,5,6,7,8,9} gives A' = {1,2,4,5,7,8}. Among the given choices, 8 is present in this complement, whereas 3, 6, and 9 are members of A and therefore cannot belong to A'. Hence option C is the only correct answer.

The complement of \(A\), written \(A'\), contains exactly those elements of the universal set \(U\) that are not in \(A\). Starting with \(U=\{a,e,i,o,u\}\), remove \(a\), \(i\), and \(u\), because these belong to \(A\). The elements left are \(e\) and \(o\). Hence \(A'=U-A=\{e,o\}\).

A universal set specifies the complete collection of objects under consideration in a particular problem. The complement of a set \(A\) is defined relative to this collection: \(A'=U-A\). Therefore, changing \(U\) can change the complement even when \(A\) remains the same. The universal set is not used only for empty sets, equal sets, or ordered pairs, so option A is correct.

The complement \(A'\) consists of every element in the universal set \(U\) that is absent from \(A\). Begin with \(U=\{1,2,3,4,5,6,7\}\) and remove 2, 5, and 7, the elements of \(A\). The remaining elements are 1, 3, 4, and 6. Hence \(A'=U-A=\{1,3,4,6\}\).

The complement B′ contains all elements of the universal set U that are not elements of B. Removing b and e from U gives B′ = {a, c, d}. Therefore, b does not belong to B′ because it already belongs to B. The other listed elements, a, c, and d, are in U but not in B, so they belong to B′.

The complement of A, written as A′, contains every element of the universal set U that is not present in A. From U = {0,1,2,3,4,5} and A = {0,1,5}, we get A′ = {2,3,4}. This set has three elements, so n(A′) = 3 and option B is correct. Equivalently, n(A′) = n(U) − n(A) = 6 − 3 = 3. Option D gives the size of U, not its complement.

The complement of G is taken relative to the universal set U. Therefore, G′ contains every student in the class who is not in G, meaning every student who does not wear glasses. It does not represent teachers or only absent students, because those groups are not defined by the given universal set and condition.

For every element of the universal set U, exactly one of two possibilities holds: it is in A, or it is not in A and therefore belongs to A′. Consequently, combining A and A′ includes every element of U and no element outside U. Hence, A ∪ A′ = U. This is one of the fundamental complement identities.

The complement A′ contains elements of U that are not in A. Therefore, no element can belong to both A and A′ at the same time. Their common part, or intersection, contains no elements and is the empty set. Hence, A ∩ A′ = ∅. This identity is paired with A ∪ A′ = U and applies for complements taken relative to the same universal set U.

The complement A′ contains exactly those elements of the universal set U that are not members of A. Starting with U = {1, 2, 3, 4, 5, 6, 8}, remove 2, 4, 6, and 8, because these belong to A. The elements left are 1, 3, and 5. Hence A′ = U − A = {1, 3, 5}.

The complement A′ and the set A together divide the universal set U into two non-overlapping parts. Therefore, A is obtained by removing the elements of A′ from U: A = U − A′ = {1, 2, 3, 4} − {1, 4} = {2, 3}. Option A gives A′ itself, not A. The answer follows from the complement relationship A ∪ A′ = U.

The complement A′ is defined relative to U and consists of every element of U that is not in A. From U = {sun, moon, star}, remove sun because it is the only element of A. The remaining elements are moon and star, so A′ = U − A = {moon, star}. Thus option B is correct; the other options omit a required element or repeat A.