Power Set and Subsets

A and its complement A' are disjoint and together form the universal set U. Therefore |U| = |A| + |A'| = 4 + 5 = 9. The power set of an n-element set has 2ⁿ elements, so |P(U)| = 2⁹ = 512. Option B is correct. The other choices are 2⁸, 2¹⁰, and 2¹¹, which use an incorrect size for U.

Let the fixed 4-element subset be F. A subset of U is disjoint from F precisely when it contains none of the elements of F. Therefore, its elements can be selected only from the remaining 10 − 4 = 6 elements. Each of these six elements has two independent choices: included or excluded. Hence the number of such subsets is 2^6 = 64. This is a direct application of the rule that an n-element set has 2^n subsets, with the forbidden elements removed.

For a finite set A, the power set has 2^|A| members. Since |P(A)| = 256 = 2^8, we get |A| = 8. The universal set U has 16 elements, so its complement A' has |A'| = 16 − 8 = 8 elements. A set with 8 elements has 2^8 = 256 subsets, including the empty set. Therefore, the number of non-empty subsets is 256 − 1 = 255, so option B is correct.

The common elements of A and B are 4 and 10, so A ∩ B = {4,10}. The power set contains every subset of this two-element set. These are the empty set, the two singleton subsets {4} and {10}, and the whole set {4,10}. Therefore P(A ∩ B) = {∅,{4},{10},{4,10}}, which is option C. The universal set U is not needed after finding the intersection.

We are counting 4-element subsets of A, not subsets of the power set as objects of a second-level selection. The element a must be included and b must be excluded. Thus, one of the four positions is already occupied by a, while the remaining three elements must be chosen from c,d,e,f,g, which gives five available elements. The number of choices is C(5,3) = 5!/(3!2!) = 10. Hence option A is correct.

The complement of A relative to U consists of the elements of U that are not in A. Therefore A' = U − A = {5,6}, so |A'| = 2. Every element of A' can independently be either included in or excluded from a subset. Consequently, the power set P(A') contains 2^2 = 4 subsets: ∅, {5}, {6}, and {5,6}. Thus the correct answer is option B.

Let |A| = n. The first power set P(A) has 2^n elements. Taking the power set again gives |P(P(A))| = 2^(2^n). The given value is 256 = 2^8, so 2^n = 8. Since 8 = 2^3, it follows that n = 3. Therefore |A| = 3, and option B is correct. The nested power set must be handled by applying the power-set formula twice.

The universal set contains the integers from 1 through 10. Removing the elements of A leaves A' = {1,3,5,7,9}, which has five elements. A singleton subset of A' contains exactly one element, so there is one singleton subset for each element of A': {1}, {3}, {5}, {7}, and {9}. Hence P(A') contains exactly five singleton sets, making option B correct. The total size of P(A') is 32, but that is not the number of singleton members.

Because A has 4 elements and U has 9 elements, its complement A' has 9 − 4 = 5 elements. Therefore P(A') has 2^5 = 32 elements. A set with 32 elements has 2^32 total subsets, including itself. Proper subsets exclude the set itself, so their number is 2^32 − 1 = 4,294,967,295. Thus option D is correct. The value 1023 would incorrectly correspond to proper subsets of a 10-element set, not a 32-element set.

A set is always an element of its own power set, so A ∈ P(A). If P(A) = P(B), then every subset of A is also a subset of B and vice versa. In particular, each element of A, viewed as a singleton subset, belongs to P(B), which implies that it belongs to B; thus A ⊆ B. By the same argument B ⊆ A. Therefore A = B, so option B is the only valid conclusion. Equal power sets cannot arise from different original sets.

The power set of the empty set contains exactly one subset, namely the empty set itself: P(∅) = {∅}. Now take the power set of the one-element set {∅}. Its subsets are the empty set and the set containing its only element, namely {∅}. Therefore P(P(A)) = P({∅}) = {∅,{∅}}. Notice that ∅ and {∅} are different: the first has no elements, while the second has one element. Hence option D is correct.

First find the union: A ∪ B = {a, c, d}. The complement is taken relative to U, so (A ∪ B)' = U − {a, c, d} = {b, e}. This set has two elements. A set with n elements has 2ⁿ subsets in its power set; therefore, n(P((A ∪ B)')) = 2² = 4. Hence option B is correct.

Using De Morgan’s law, A' ∩ B' = (A ∪ B)'. The union A ∪ B is {1, 2, 3, 4, 5, 6}; therefore its complement in U is {7, 8}. Thus A' ∩ B' has two elements. The power set of any two-element set contains 2² = 4 subsets, including the empty set and the set itself. Therefore the correct answer is option B, 4.

An element of P(A) is a subset of A. To form a subset containing exactly three elements from the five elements of A, choose any 3 of them. The number of such choices is the combination 5C3 = 5!/(3!2!) = (5 × 4)/(2 × 1) = 10. Therefore, exactly 10 elements of P(A) have cardinality 3, so option B is correct.

The condition requires 1 to be included and 4 to be excluded. These two elements therefore have fixed statuses. The remaining elements 2 and 3 are unrestricted; each can be included or omitted independently. Consequently, the number of valid subsets is 2² = 4. They are {1}, {1, 2}, {1, 3}, and {1, 2, 3}. Hence option B is the only correct answer.

A has four elements, so its power set contains 2⁴ = 16 subsets in total. Exactly one of these subsets is the empty set, ∅. The question asks for subsets that are not empty, so remove that one subset: 16 − 1 = 15. Therefore P(A) has 15 non-empty elements, and option A is correct.

The power set P(A) contains every subset of A, including A itself. A proper subset must be strictly smaller than the original set, so A is not a proper subset of itself. Every other subset of A has fewer than four elements and is proper. Thus exactly one element of P(A), namely A, is not a proper subset. The answer is option B.

A six-element set has 2⁶ = 64 subsets in total. The subsets can be paired with their complements. Because the set has six elements, the complement of an even-sized subset is also even-sized, while the standard binomial identity shows that even- and odd-sized subsets occur equally often. Therefore half of 64, namely 2⁵ = 32, have even cardinality. Option C is correct.

For every non-empty finite set with n elements, the numbers of even-cardinality and odd-cardinality subsets are equal. Since A has five elements, P(A) has 2⁵ = 32 subsets in total. Therefore the number with an odd number of elements is half of 32, or 2⁴ = 16. Equivalently, 5C1 + 5C3 + 5C5 = 5 + 10 + 1 = 16. Thus option C is correct.

An element of P(A) must be a subset of A, so all of its elements must be 1, 2, or 3. The sets {1, 2}, ∅, and {1, 2, 3} satisfy this requirement. However, {{1}} has the single element {1}, which is itself a set and is not an element of A. Therefore {{1}} is not a subset of A and is not an element of P(A).