Complement of a Set and Its Properties

By definition, Aᶜ = U − A, so we list the elements of U and exclude every element belonging to A. Removing 1, 4, and 9 from U = {1, 2, 3, 4, 5, 6, 7, 8, 9} leaves {2, 3, 5, 6, 7, 8}. Option B is A itself, not its complement, while the other options either include elements of A or omit valid elements.

The complement of A contains every element of the universal set U that is not an element of A. Starting with U, remove 4, 8, and 12, because these are the elements of A. The elements left are 1, 2, 3, 5, 6, 7, 9, 10, and 11. Therefore, Aᶜ = U − A = {1, 2, 3, 5, 6, 7, 9, 10, 11}, so option A is correct. The universal set must always be used as the boundary for a complement.

The complement Aᶜ contains elements that are in U but absent from A. Starting with U = {0, 1, 2, 3, 4, 5, 6}, remove 0, 3, and 6 because they belong to A. The elements left are 1, 2, 4, and 5. Therefore Aᶜ = {1, 2, 4, 5}, so option A is the only correct answer.

Since U contains the natural numbers from 1 through 10, we can write U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The even elements form A = {2, 4, 6, 8, 10}. The complement contains the elements of U that are not even; these are precisely the odd numbers {1, 3, 5, 7, 9}. Hence Aᶜ = {1, 3, 5, 7, 9}, making option A correct. The answer depends on the stated universal set.

Because N is the set of positive natural numbers and x < 7, the universal set is U = {1, 2, 3, 4, 5, 6}. The condition x ≤ 3 gives A = {1, 2, 3}. The complement Aᶜ consists of elements in U that do not belong to A. Removing 1, 2, and 3 from U leaves {4, 5, 6}. Therefore, option A is correct. Option B is A itself, while option D is the complete universal set.

First form the union by including every element that occurs in A or B: A ∪ B = {1, 2, 4, 5, 6}. The complement of this union contains the elements of U that are absent from it. From U = {1, 2, 3, 4, 5, 6, 7}, only 3 and 7 remain. Thus (A ∪ B)ᶜ = {3, 7}, so option A is correct.

First find the intersection: A ∩ B contains elements common to both A and B, so A ∩ B = {3, 4}. To find its complement, select all elements of U that are not in {3, 4}. From U = {1, 2, 3, 4, 5, 6, 7}, the remaining elements are {1, 2, 5, 6, 7}. Thus (A ∩ B)ᶜ = {1, 2, 5, 6, 7}, and option A is correct. Option B gives the intersection itself, not its complement.

If every element of A is also an element of B, then any element outside B must certainly be outside A. Therefore the complement of B is contained in the complement of A: Bᶜ ⊆ Aᶜ. This reversal of inclusion is called the complement or order-reversing property. The other statements are not generally true because A and B may be unequal and need not exhaust U.

The complement \(A^c\) contains all elements of the universal set \(U\) that are not in \(A\). Taking the complement once more selects exactly the elements that were originally in \(A\), so \((A^c)^c=A\). In contrast, \(A\cap A^c=\varnothing\) and \(A\cup A^c=U\). Thus, only option A is always true.

The complement \(A^c\) is defined relative to the universal set \(U\). It contains every student in the school who is not a member of \(A\), where membership in \(A\) means coming by bus. Therefore, \(A^c\) represents students who use another mode of travel or otherwise do not come by bus. It does not mean that there are no students.

The complement of \(A\) is found by taking all elements of \(U\) and removing the elements that belong to \(A\). The universal set contains red, blue, green, and yellow, while \(A\) contains red and yellow. The remaining elements are blue and green. Hence, \(A^c=\{\text{blue},\text{green}\}\), so option A is correct.

The universal set of digits is \(U=\{0,1,2,3,4,5,6,7,8,9\}\). The complement \(A^c\) consists of every element of the universal set that does not belong to \(A\). Since \(A\) contains the even digits 0, 2, 4, 6, and 8, removing them from \(U\) leaves the odd digits 1, 3, 5, 7, and 9. Hence, \(A^c=\{1,3,5,7,9\}\), so option A is correct. The complement always depends on the stated universal set.

For a finite universal set, the set \(A\) and its complement \(A^c\) are disjoint and together contain every element of \(U\). Therefore, their cardinalities satisfy \(n(A)+n(A^c)=n(U)\). Substituting the given values gives \(n(A)+19=48\), so \(n(A)=48-19=29\). Thus, option A is correct.

A set and its complement are disjoint, and their union is the universal set. Hence, \(n(A)+n(A^c)=n(U)=24\). Since the problem states that \(n(A)=n(A^c)\), let each cardinality be \(x\). Then \(x+x=24\), so \(2x=24\) and \(x=12\). Therefore, \(n(A)=12\).

Every element of the universal set belongs either to \(A\) or to its complement \(A^c\). Therefore, their union covers the entire universal set: \(A\cup A^c=U\). Taking cardinalities gives \(n(A\cup A^c)=n(U)=31\). The empty set is the intersection \(A\cap A^c\), not the union, so option B is incorrect.

By definition, \(A^c\) contains precisely the elements of the universal set that are not in \(A\). Consequently, no element can belong to both \(A\) and \(A^c\) at the same time. Their intersection is therefore empty: \(A\cap A^c=\varnothing\). Although the given complement allows us to identify \(A=\{1,3,5,7\}\), it does not change this identity.

The complement Aᶜ contains exactly those elements of the universal set U that are not in A. Therefore, A and Aᶜ together contain every element of U and have no element outside U. Hence, the union identity is A ∪ Aᶜ = U. Here Aᶜ = {b, e}, so A = {a, c, d}; their union is {a, b, c, d, e}, which is U. Option B is only the complement, not the union.

The complement Aᶜ of A with respect to U is defined as Aᶜ = U \ A. Thus, an element belongs to Aᶜ precisely when it belongs to U but does not belong to A. The given facts state that 6 is in U and is not in A, so 6 must be in Aᶜ. It cannot simultaneously belong to A and Aᶜ, because A ∩ Aᶜ = ∅.

For an element of the universal set U, membership is divided between A and its complement Aᶜ: every element of U is in exactly one of these two sets. Since 5 belongs to U but is explicitly not in Aᶜ, it must belong to A. Option C contradicts the given condition, option B contradicts 5 ∈ U, and option D is impossible because A ∩ Aᶜ is empty.

The complement is taken relative to U, so Aᶜ = U \ A. The interval A = [1, 5) contains 1 and all numbers up to but not including 5. Therefore, 5 is excluded from A and included in its complement. The endpoint 9 is included in U, and A contains no numbers beyond 5, so 9 remains included. Thus Aᶜ = [5, 9].

The complement Aᶜ consists of elements in U that are not in A, so Aᶜ = U \ A. The number 1 is included in A because the interval ends with a square bracket at 1; therefore, 1 is excluded from Aᶜ. The number 4 is included in U and is not in A, so it remains included. Hence, Aᶜ = (1, 4].

Because x is an integer and -2 ≤ x ≤ 3, the universal set is U = {-2, -1, 0, 1, 2, 3}. The complement Aᶜ contains the elements of U that are absent from A. Removing {-1, 0, 1} from U leaves {-2, 2, 3}. Therefore, Aᶜ = {-2, 2, 3}; option B is A itself, not its complement.

The complement Aᶜ contains every element of U that is not an element of A. The numbers 2, 3, 5, and 7 are already in A, so none of them belongs to Aᶜ. Since 9 is present in U but absent from A, it satisfies the definition of complement and therefore belongs to Aᶜ. Thus, option A is the only correct choice.

The complement of A with respect to U is Aᶜ = U \ A = {1, 3, 4}. Therefore, 4 belongs to Aᶜ, making option A correct. Statement B is false because 2 belongs to A, statement C is false because 5 belongs to A, and statement D is false because 1 is not in A. The result follows directly by removing the elements of A from U.

The complement of A is formed relative to the stated universal set U. It contains the directions in U that are not listed in A. Since A contains north and south, removing these two directions from U leaves east and west. Therefore, Aᶜ = {east, west}. The complement is not the same as A, U, or the empty set.