Power Set and Subsets

For any subset S of A, its complement A \ S contains exactly the elements of A that are not in S. If S were equal to its complement, each element would have to be both in S and not in S, which is impossible. Equivalently, S and its complement are disjoint, so they cannot be equal when A is non-empty. Thus the count is 0, option A.

There are 2^5 = 32 total subsets of A. Count the opposite of the required condition: a subset fails to contain 1 or fail to omit 5 only when it does not contain 1 and does contain 5. The remaining three elements are optional, so there are 2^3 = 8 such subsets. Hence the required count is 32 - 8 = 24, option A.

The universal set contains the numbers 1 through 14. A contains all the even numbers in U, so its complement A' contains precisely the odd numbers {1,3,5,7,9,11,13}. This is exactly B. Equal sets have identical collections of subsets, so P(A') = P(B). Therefore option A is correct.

The positive divisors of 16 that lie in U are {1,2,4,8,16}, so A has 5 elements. Since U has 16 elements, B = A' has 16 - 5 = 11 elements. The power set of an 11-element set contains 2^11 subsets, which equals 2048. Thus n(P(B)) = 2048 and option A is correct.

The elements of P(A) are exactly the subsets of A. The elements of A are ∅, {1}, and 2; the number 1 by itself is not an element of A. Therefore {1, 2} is not a subset of A and cannot belong to P(A). In contrast, {∅, 2}, {{1}, 2}, and ∅ contain only elements allowed by A, so they are subsets of A.

For a finite set with \(n(A)\) elements, its power set has \(2^{n(A)}\) elements. Thus, \(2^{n(A)}=32=2^5\), so \(n(A)=5\). Since the universal set has 13 elements, the complement has \(n(A')=13-5=8\) elements. The number of 2-element subsets of an 8-element set is \(\binom{8}{2}=\frac{8\times7}{2}=28\). Therefore, option B is correct.

Within \(U\), the multiples of 4 are \(\{4,8,12,16,20\}\), so \(n(A)=5\). The multiples of 6 are \(\{6,12,18\}\), so \(n(B)=3\). Their only common element is 12, giving \(n(A\cap B)=1\). By inclusion-exclusion, \(n(A\cup B)=5+3-1=7\). Therefore, the complement contains \(20-7=13\) elements, so option D is correct.

The element a must be included, so it has only one fixed choice. The elements b and c must be excluded, so they also have no choice. Only d and e remain unrestricted; each may either be included or excluded independently. Thus the number of possible subsets is 2×2 = 2^2 = 4.

Every member of \(\mathcal{P}(A)\) is, by definition, a subset of \(A\). Since every element of \(A\) is also in \(U\), we have \(A\subseteq U\). Consequently, every subset of \(A\) is automatically a subset of \(U\). A set with four elements has \(2^4=16\) subsets, including the empty set and the set \(A\) itself. Therefore, option C is correct.

The complement is taken with respect to the given universal set. Removing the elements of \(A\) from \(U\) gives \(A'=U-A=\{b,d\}\), so \(n(A')=2\). The power set of a set containing two elements has \(2^2=4\) members: \(\varnothing\), \(\{b\}\), \(\{d\}\), and \(\{b,d\}\). Hence option B is correct.

Because \(A\subseteq U\), the difference \(U-A\) contains all elements of \(U\) that are not in \(A\). Its cardinality is \(|U-A|=|U|-|A|=9-5=4\). The power set of a four-element set contains \(2^4=16\) subsets. Therefore, \(|\mathcal{P}(U-A)|=16\), and option B is the correct answer.

The empty set has exactly one subset, namely itself, so \(\mathcal{P}(\varnothing)=\{\varnothing\}\) and its cardinality is 1. Applying the power-set operation once more gives \(|\mathcal{P}(\mathcal{P}(A))|=2^{|\mathcal{P}(A)|}=2^1=2\). Explicitly, the two members are \(\varnothing\) and \(\{\varnothing\}\). Thus, option B is correct.

The given information says that the set \(\mathcal{P}(A)\) itself has 64 elements. For any finite set \(S\), the power set \(\mathcal{P}(S)\) has \(2^{|S|}\) elements. Taking \(S=\mathcal{P}(A)\), we obtain \(|\mathcal{P}(\mathcal{P}(A))|=2^{|\mathcal{P}(A)|}=2^{64}\). It is not \(64^2\), because forming a power set counts all subsets. Therefore, option C is correct.

Using the usual school convention \(\mathbb{N}=\{1,2,3,\ldots\}\), we have \(|A|=5\) and \(|U|=8\). Hence \(|\mathcal{P}(A)|=2^5=32\), while \(|\mathcal{P}(U)|=2^8=256\). Because \(A\subseteq U\), every subset of A is also a subset of U, so \(\mathcal{P}(A)\subseteq\mathcal{P}(U)\). Therefore, \(|\mathcal{P}(U)-\mathcal{P}(A)|=256-32=224\). Option A is correct. This difference is not \(\mathcal{P}(U-A)\).

The complement of A with respect to U is A′ = U − A = {1, 3, 5}. The power set P(A′) contains every subset of {1, 3, 5}, including ∅, {1, 3}, and {5}. However, {2} contains the element 2, which is not in A′, so {2} is not a subset of A′ and therefore is not a member of P(A′). Hence, option D is correct.

The members of A are 1, {2}, and 3. A member of P(A) must be a subset of A, so every element of that member must itself be one of these three objects. The set {{2}} has the single element {2}, which belongs to A; therefore {{2}} is a subset of A and belongs to P(A). Option B is correct.

Although ∅ and {∅} look related, they are different objects: ∅ has no elements, while {∅} has one element, namely ∅. Thus A has two distinct members. For a finite set with n members, the power set has 2^n members. Therefore |P(A)|=2^2=4, so option C is correct.

For any finite set X, the number of members in its power set is |P(X)| = 2^|X|. Here, |P(A′)| = 8 = 2^3, so |A′| = 3. Since A and A′ partition the universal set U, their cardinalities add to |U|: |A| + |A′| = 7. Therefore, |A| = 7 − 3 = 4, making option C the only correct answer.

First find the union: A∪B={1,2,3}, which has three elements. Therefore its power set contains 2^3=8 subsets. A proper subset of a set is any subset other than the set itself, so exactly one subset, {1,2,3}, must be excluded. The number of proper subsets is 8−1=7; hence option B is correct.

Here A'={2,3,5}. A and A' are disjoint, so a set that is a subset of both can contain no element from A and no element from A'. The empty set is the only set with no elements, and it is a subset of every set. Thus P(A)∩P(A')={∅}, not ∅; option B is correct.