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 of A is found relative to the universal set U: A′ = U − A. Removing 2 and 4 from U gives A′ = {1, 3, 5}. A set is a subset when every one of its elements belongs to the larger set. Thus {1, 3} is a subset of {1, 3, 5}. The other choices contain 2 or 4, which are members of A and hence cannot belong to A′.

Complements reverse the direction of inclusion. Since \(A\subseteq B\), every element outside \(B\) is certainly outside \(A\) as well. Thus, if \(x\in B^c\), then \(x\notin B\), which implies \(x\notin A\), so \(x\in A^c\). Therefore \(B^c\subseteq A^c\), making option A correct. Option B reverses this result, while option C requires \(A=B\). Option D is not generally true; it would impose an additional condition on the universal set.

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′.