Complement of a Set and Its Properties

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.

Students who do not learn music are represented by the complement Aᶜ of the set A of music learners. Assuming A is a subset of the universal group U, the number of students in the complement is n(Aᶜ) = n(U) − n(A). Substituting the given values gives 60 − 18 = 42. Therefore, 42 students do not learn music.

The complement Aᶜ contains all elements of the universal set U that do not belong to A. Therefore, n(Aᶜ) means the number of students in the class who are not included in A. Numerically, n(Aᶜ) = n(U) − n(A) = 45 − 28 = 17. Thus, option A gives the correct meaning as well as the resulting count.

A proper subset A of U is contained in U but is not equal to U. Consequently, at least one element of U is missing from A. That missing element, or those missing elements, belong to the complement Aᶜ. Hence Aᶜ must contain at least one element and cannot be empty. Also, Aᶜ is always a subset of U, so option D is false.

A complement is not determined by A alone; it is defined with respect to a specified universal set. Relative to U₁, Aᶜ = U₁ − A = {1, 3}. Relative to U₂, Aᶜ = U₂ − A = {1, 3, 5}. Thus the extra element 5 appears when the universal set changes. Set A itself remains unchanged, so option A is correct.

For B to be the complement of A in the universal set U, A and B must satisfy two conditions. First, they must be disjoint, so A ∩ B = ∅. Second, together they must contain every element of U, so A ∪ B = U. These conditions ensure that B contains exactly the elements of U that are outside A. Therefore, option A is the correct test.

The complement of a set A, written as Aᶜ, consists of all elements in the universal set U that are not elements of A. Here, U contains 10, 11, 12, 13, 14, and 15, while A contains 11, 13, and 15. Removing the elements of A from U leaves 10, 12, and 14. Therefore, Aᶜ = {10, 12, 14}.