Power Set and Subsets

The set A has exactly two elements: the empty set ∅ and the singleton set {1}. The fact that ∅ itself is an element does not change the count. For any finite set with n elements, its power set contains 2ⁿ subsets, because each element can either be selected or not selected. Therefore, n(P(A)) = 2² = 4.

The power set P(A) is the set of all subsets of A. Every set is a subset of itself, so A itself must be an element of P(A). Here A = {∅, {∅}}, where its two distinct elements are ∅ and {∅}. Therefore, the element {∅, {∅}} must occur in P(A), making option B correct.

The universal set is U = {−2, −1, 0, 1, 2}. The condition x² = 1 gives x = −1 or x = 1, so A = {−1, 1}. Therefore, the complement relative to U is A′ = {−2, 0, 2}, which has 3 elements. A set with n elements has 2ⁿ subsets in its power set. Hence n(P(A′)) = 2³ = 8, so option B is correct. The complement must always be taken with respect to the stated universal set.

The prime numbers in U are 2, 3, 5, and 7, so A = {2, 3, 5, 7}. Notice that 1 is not prime. The complement is therefore A′ = U − A = {1, 4, 6, 8, 9}, containing 5 elements. The power set of a finite set with n elements contains 2ⁿ subsets. Thus n(P(A′)) = 2⁵ = 32. Therefore, option B is correct. The main steps are identifying the primes, forming the complement, and then applying the power-set formula.

If A has 4 elements, then its power set P(A) has 2⁴ = 16 elements. The question does not ask for the number of 2-element subsets of A; it asks for the number of 2-element subsets of P(A), which is a new set containing 16 elements. Therefore, we choose 2 elements from 16: C(16, 2) = 16 × 15 ÷ 2 = 120. Hence option C is correct. The key idea is to calculate the size of the power set first and then use combinations.

For finite sets, n(P(A)) = 2ⁿ(A) and n(P(B)) = 2ⁿ(B). If these two powers are equal, their nonnegative integer exponents must be equal, so n(A) = n(B). This does not imply A = B, because different sets can have the same number of elements; for example, {1,2} and {3,4}.

P(A) = {∅, {1}, {2}, {1,2}} and P(B) = {∅, {3}, {4}, {3,4}}. The only common subset is ∅, because no nonempty subset can simultaneously contain only elements from A and only elements from B when A and B are disjoint. Therefore, their intersection is the set containing ∅, namely {∅}.

An element of P(A) ∩ P(B) is a set that is a subset of both A and B. A set is a subset of both precisely when every one of its elements belongs to A ∩ B. Thus the common elements of the two power sets are exactly all subsets of A ∩ B, giving the identity P(A) ∩ P(B) = P(A ∩ B).

The identity P(A) ∪ P(B) = P(A ∪ B) is generally false. The left side contains sets that are subsets entirely of A or entirely of B, whereas the right side also contains mixed subsets containing elements from both A and B. For example, if A={1} and B={2}, {1,2} belongs to P(A∪B) but not to P(A)∪P(B).

First find the power sets: P(A) = {∅, {1}, {2}, {1,2}} and P(B) = {∅, {2}, {3}, {2,3}}. Each power set has 4 elements. Their common elements are ∅ and {2}, so the intersection has 2 elements. Using the union formula, n(P(A) ∪ P(B)) = 4 + 4 − 2 = 6. Therefore, option B is correct. It is important to compare the subsets themselves, not merely the original elements of A and B, when removing duplicates.

The complement of A in U is A′ = {4, 5, 6}. Since A = {1, 2, 3} and A′ = {4, 5, 6} have no common element, the only subset common to both sets is the empty set ∅. Therefore, the only common member of their power sets is ∅ itself, so P(A′) ∩ P(A) = {∅}. It is {∅}, not ∅, because the power set contains the empty set as an element.

A ⊆ U means that every element of A is also an element of U. Now take any element X of P(A). By definition, X is a subset of A, so every element of X belongs to A and therefore also belongs to U. Hence X is a subset of U, which means X ∈ P(U). Since this holds for every X in P(A), we conclude that P(A) ⊆ P(U).

For a finite set X, the number of elements in its power set is n(P(X)) = 2ⁿ⁽ˣ⁾. Since n(P(A)) = 16 = 2⁴, we get n(A) = 4. The complement therefore has n(A′) = n(U) − n(A) = 7 − 4 = 3 elements. Hence n(P(A′)) = 2³ = 8. Option B is correct. The value 16 belongs to P(A), while 4 is n(A), not the requested power-set cardinality.

For every finite set X, n(P(X)) = 2ⁿ⁽ˣ⁾. Thus 64 = 2⁶ gives n(A′) = 6. Because A′ is the complement of A within U, A and A′ partition U, so n(A) + n(A′) = n(U). Therefore n(A) = 10 − 6 = 4. Option A is correct. The value 6 is the cardinality of the complement, not of A; the other values do not satisfy the power-set relation.

For any sets X and Y, P(X) = P(Y) implies X = Y, because each set can be recovered as the union of all members of its power set. Therefore P(A′) = P(A) requires A′ = A. In an ordinary nonempty universal set, a set cannot equal its own complement, because A and A′ are disjoint; thus this equality is possible only in the degenerate case U = ∅. Among the given conditions, C states the required condition.

A six-element set has 2⁶ = 64 total subsets. The subsets with fewer than two elements are the empty subset, counted by C(6,0)=1, and the six one-element subsets, counted by C(6,1)=6. Thus 7 subsets have fewer than two elements. Subtracting them from all subsets gives 64 − 7 = 57 subsets containing at least two elements. Therefore option B is correct.

“At most 2” means the subset may contain 0, 1, or 2 elements. The numbers of such subsets are C(5,0)=1, C(5,1)=5, and C(5,2)=10. Adding these mutually exclusive cases gives 1 + 5 + 10 = 16. Therefore, 16 elements of P(A) have at most two elements, so option C is correct. The count must include the empty set because zero is allowed.

The elements a and b must be included in every desired subset, so they are fixed choices. The remaining three elements c, d, and e may each either be included or omitted independently. Hence there are 2³ = 8 possible choices for the remaining elements. Every such choice produces one subset containing both a and b, so the required number of elements of P(A) is 8. Option B is correct.

There are 2⁵ = 32 total subsets of A. It is easier to count the complement: subsets containing neither a nor b. Such subsets can use only c, d, and e, and each of these three elements may be chosen or omitted, giving 2³ = 8 subsets. Therefore the number containing at least one of a or b is 32 − 8 = 24. This includes subsets containing a only, b only, or both, so option B is correct.

Only the odd elements 1 and 3 affect whether a subset sum is odd or even; the even elements 2 and 4 do not change parity. The sum is even when the subset contains an even number of odd elements: either neither 1 nor 3, or both 1 and 3. That gives 2 choices for the odd elements. The even elements are independently optional, giving 2² choices. Thus the total is 2 × 4 = 8, so option B is correct.