Complement of a Set and Its Properties

A set and its complement contain mutually exclusive elements: an element cannot belong to A and not belong to A at the same time. Hence their intersection is always empty, A ∩ Aᶜ = ∅, so A and Aᶜ are always disjoint. The other pairs need not be disjoint because A and U, Aᶜ and U, or U and U generally share elements.

Every element of the universal set U is either an element of A or an element of its complement Aᶜ. Consequently, A ∪ Aᶜ = U. Taking cardinalities on both sides gives n(A ∪ Aᶜ) = n(U) = 25. Therefore option A is correct. The result does not depend on the particular elements or size of A.

Aᶜ consists of the elements of U that are not in A. Consequently, no element can belong to both A and Aᶜ, so their intersection is the empty set: A ∩ Aᶜ = ∅. The empty set has no elements, and therefore n(A ∩ Aᶜ) = n(∅) = 0. Thus option A is correct regardless of the size or nature of A.

The relative difference U − A contains all elements that are in U but not in A. This is exactly the definition of the complement of A with respect to U, so U − A = Aᶜ. Option B gives the removed set itself, and option C incorrectly retains the elements of A. Option D occurs only in the special case A = U, not for every subset A.

U − A means the elements that belong to U but do not belong to A. Removing 1, 2, 3, and 4 from U leaves only 5 and 6. Therefore U − A = {5, 6}, which is also Aᶜ relative to U, so option A is correct. Option B is A itself, option C removes too few elements, and option D fails to remove A.

To find Aᶜ, list the elements of U and retain only those that are absent from A. The elements 2, 3, 5, and 7 are in A, while 1, 4, and 6 are not. Therefore Aᶜ = {1, 4, 6}, making option A correct. Option B is A itself, option C incorrectly includes 2, and option D omits 1, 4, and 6.

The universal set U contains all seven days, while A contains the two weekend days, Saturday and Sunday. The complement Aᶜ therefore contains every day in U that is not in A: Monday through Friday. Hence option A is correct. Option B is A itself, option C includes the weekend days that must be excluded, and option D incorrectly claims that no day remains.

The complement Aᶜ contains all elements of the universal set U that are not in A. Therefore, no element can belong to both A and Aᶜ, so A ∩ Aᶜ = ∅, not U. The other statements are standard complement properties: A ∪ Aᶜ = U, (Aᶜ)ᶜ = A, and Aᶜ is always a subset of U. Hence option A is false.

By definition, Aᶜ consists of the elements of U that are not members of A. Thus A and Aᶜ are disjoint sets: they cannot have a common element. This gives the universal identity A ∩ Aᶜ = ∅. Although {2, 5} is Aᶜ and {1, 3, 4} is A, neither set is their intersection. Therefore option A is correct.

A set together with its complement covers every element of the universal set. Here Aᶜ = {2, 5}, so A = U − Aᶜ = {1, 3, 4}. Taking the union gives {1, 3, 4} ∪ {2, 5} = {1, 2, 3, 4, 5} = U. Thus the correct answer is option A. The other listed sets represent only one set or the empty set, not the complete union.

For a set A defined within the universal set U, an element belongs to Aᶜ precisely when it belongs to U but does not belong to A. The question states both conditions for the element 3: 3 is in U and 3 is not in A. Hence 3 must belong to Aᶜ. It cannot belong to A, and it cannot belong to A ∩ Aᶜ because the latter is empty.

Membership in the complement Aᶜ means that the element is part of the universal set U but is excluded from A. Therefore, from 4 ∈ Aᶜ, we can conclude both 4 ∈ U and 4 ∉ A. Option B contradicts the definition of a complement, option C contradicts membership in U, and option D is impossible because A ∩ Aᶜ = ∅.

The universal set is U = {1, 2, 3, 4, 5}. Since A is defined by x < 4 and x is a natural number in U, A = {1, 2, 3}. The complement contains the elements of U that are absent from A, so Aᶜ = U − A = {4, 5}. Option B is A itself, option C incorrectly includes 1, 2, and 3, and option D omits 4.

Because the natural numbers here are taken up to 8, the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8}. Removing the elements of A, namely 1, 3, 5, and 7, leaves {2, 4, 6, 8}. Thus Aᶜ = {2, 4, 6, 8}. Option B is A itself, while options C and D contain elements of A or omit required elements.

For intervals, parentheses exclude an endpoint and brackets include it. The set A = [0,4] contains every real number from 0 through 4, including both endpoints. Its complement within U = [0,10] therefore contains numbers greater than 4 and up to 10. The point 4 is excluded, while 10 is included, giving Aᶜ = (4,10]. Option B incorrectly includes 4, and option C is A itself.

The complement Aᶜ consists of the elements in U that are not in A. Because A = (-1, 2] excludes -1, the point -1 belongs to the complement. Because A includes 2, the complement must exclude 2. Thus the remaining portions of U are [-2, -1] and (2, 3], so Aᶜ = [-2, -1] ∪ (2, 3].

The complement Aᶜ contains every member of the universal set U that does not belong to A. Here U includes all students in the class, while A includes only students who play cricket. Therefore Aᶜ represents the students in the class who do not play cricket. It does not mean all students, only cricket players, or an empty group.

For a set A and its complement A^c within the universal set U, every element of U belongs to exactly one of these two disjoint sets. Therefore, n(A)+n(A^c)=n(U). Substituting the given values gives n(A)+15=40, so n(A)=40−15=25. Hence option A is correct. Option C gives the complement’s size, option B exceeds the size of U, and option D ignores the given complement.

A proper subset A of U is contained in U but is not equal to U. Thus, at least one element of U is not contained in A. By definition, all such elements form the complement A^c, so A^c must contain at least one element and cannot be empty. Also, every complement is a subset of U, which rules out option D. Therefore, option A is the only correct statement.

The complement of A is always defined relative to a specified universal set: A^c=U−A. With U_1, the complement is {3}; with U_2, the complement is {3,4}. The set A itself remains unchanged in both situations, but the collection of available elements outside A changes because the universal set changes. Hence option A correctly identifies the reason.