Power Set and Subsets

The phrase “at least 5” includes subsets having exactly 5 elements and subsets having exactly 6 elements. The number of 5-element subsets is C(6,5) = 6, while the number of 6-element subsets is C(6,6) = 1. Adding them gives 6 + 1 = 7. Hence option B is correct. The empty set and smaller subsets do not satisfy the condition.

The elements 3 and 7 must be included, so their choices are fixed. The other five elements, namely 1, 2, 4, 5, and 6, may independently be included or excluded. Each of these five elements gives two choices, so the number of valid subsets is 2^5 = 32. Therefore, option B is correct. The required elements do not add extra choices because they are already fixed.

The condition excludes 1, 2, and 3 from every permitted subset. Thus only the remaining elements 4, 5, and 6 can be selected. Each of these three elements may independently be included or excluded, so the number of possible subsets is 2^3 = 8. The empty subset is included in this count because it contains none of the forbidden elements. Therefore, option B is correct.

A has 4 elements, so its power set P(A) contains 2^4 = 16 subsets in total. A proper subset is a subset that is not equal to the original set A. Among the 16 power-set elements, only A itself is not proper. Removing it leaves 16 − 1 = 15 proper subsets. Therefore, option B is correct; the empty set is also a proper subset.

The common element 6 must be counted only once in the union. Therefore, A ∪ B = {2, 4, 6, 8, 10}, which has 5 elements. A set with n elements has 2^n subsets in its power set, including the empty set and the set itself. Hence |P(A ∪ B)| = 2^5 = 32. Therefore, option B is correct.

A has 3 elements, so its power set contains 2^3 = 8 subsets. Exactly one of these subsets is empty, namely ∅, and it does not contain at least one element. Every other subset is non-empty. Therefore, the number of subsets containing at least one element is 8 − 1 = 7. Thus option B is correct.

The complement is taken relative to U. Removing the elements of A from U gives A' = U \ A = {1, 3, 5, 7}, so A' has 4 elements. The power set of a four-element set has 2^4 subsets, because each element can either be included or excluded. Therefore n(P(A')) = 16, making option C correct.

Every subset of A is an element of the power set P(A). To form a subset B with exactly two elements, choose any 2 elements from the 4 elements of A. The number of choices is the combination 4C2 = 4!/(2!2!) = 6. Therefore, six two-element subsets belong to P(A), so option A is correct.

For a finite set A with n elements, the power set P(A) contains 2^n elements. Here 2^n = 128. Since 128 = 2^7, comparison of the powers gives n = 7. Thus |A| = 7. Notice that 2^6 = 64 and 2^8 = 256, so options 6 and 8 cannot satisfy the given condition. Therefore, option B is correct.

The elements of A are ∅, {1}, and 2. An element of P(A) must be a subset whose every member is one of these three objects. The set {∅, 2} uses two actual elements of A, so it is a subset of A and therefore belongs to P(A). In contrast, 1 is not an element of A, and neither {1,2} nor {{2}} satisfies the subset condition. Thus A is correct.

A singleton subset contains exactly one element. To form a singleton subset of A, we simply choose one element from the five elements of A. The number of choices is C(5,1) = 5. Each chosen element produces one distinct singleton subset, and all of these subsets are elements of P(A). Therefore, P(A) contains exactly 5 singleton sets, so option B is correct.

The power set P(A) contains every subset of A, including the empty set, all one-element subsets, all two-element subsets, and A itself. Options {a,b}, ∅, and {c} are subsets of A, so they belong to P(A). However, a without braces is an element of A, not a subset of A. Therefore, a is not an element of the power set. Braces are important: {a} is a subset, whereas a is only an element.

The set A contains all prime numbers from 1 to 10: 2, 3, 5, and 7. The complement A′ is taken relative to the universal set U, so it contains the elements of U that are not in A. Removing those four primes leaves {1,4,6,8,9,10}. These are the composite numbers 4, 6, 8, 9, and 10 together with 1. Since 1 is neither prime nor composite, the correct description is composite numbers and 1.

For any finite set A, the original set A is itself one of the members of its power set. In the given collection, {p,q} is the largest subset and contains both elements that occur in the singleton subsets {p} and {q}. Therefore the original set must be A = {p,q}. Notice that ∅ is a subset, not the original set here, and {{p},{q}} has different elements from A.

A subset of a four-element set can have 0, 1, 2, 3, or 4 elements. “At least 3” means exactly 3 or exactly 4 elements. The number of three-element subsets is C(4,3) = 4, because one element is omitted each time. The number of four-element subsets is C(4,4) = 1, namely A itself. Hence the required number is 4 + 1 = 5, so option B is correct.

The elements of A are 1 and the set {2}; the number 2 itself is not an element of A. Every subset of A belongs to P(A). Since {2} is an element of A, the singleton set whose only element is {2}, written {{2}}, is a subset of A and therefore belongs to P(A). Option A is an element of A, not necessarily a subset, while option C is not even an element of A.

Solving x² = 4 gives x = 2 or x = −2, so A = {−2, 2}. Because the universal set is the set of all real numbers, A′ consists of every real number that is not −2 and not 2. It can be written as ℝ \ {−2,2}, or equivalently as all real numbers except −2 and 2. It is not limited to integers, and it is certainly not empty because infinitely many other real numbers remain.

For an empty set A = ∅, its power set contains exactly one subset: P(A) = {∅}. This new set has one element, namely ∅. The power set of a one-element set has 2¹ = 2 elements: the empty subset and the set containing that one element. Thus P(P(A)) = {∅, {∅}}, so it contains two elements. The answer is not 1 because the second power-set operation is also required.

A member of P(A) is a subset of A. To count subsets that contain 1, include 1 in every subset and independently decide whether to include 2 and whether to include 3. Each of these two remaining elements has two choices, included or excluded. Therefore the number of valid subsets is 2 × 2 = 2² = 4. They are {1}, {1,2}, {1,3}, and {1,2,3}. Hence option C is correct.

If A ⊆ B, every element of A is also an element of B. Consequently, every subset of A is automatically a subset of B. Since P(A) and P(B) denote the sets of all subsets of A and B respectively, this proves P(A) ⊆ P(B). The converse is also true: because A belongs to P(A), the inclusion of power sets implies A ⊆ B. Thus option A gives the exact condition, whereas disjointness or an empty union is not generally required.