Power Set and Subsets

A set is itself an element of its power set because every set is a subset of itself. If P(A) = P(B), then A belongs to P(B), so A ⊆ B. Similarly, B belongs to P(A), so B ⊆ A. Two sets that are subsets of each other are equal. Therefore A = B is always true, making option A the only valid conclusion.

If n(A) = n, then the power set P(A) has 2^n elements. Applying the power-set rule a second time, P(P(A)) has 2^(2^n) elements. Here 2^(2^n) = 16 = 2^4, so 2^n = 4 = 2^2. Consequently n = 2, and option B is correct. The formula is applied twice, not once.

The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19, so n(A) = 8. Because the complement is relative to U, n(A') = 20 − 8 = 12. Applying the power-set rule, a 12-element set has 2¹² = 4096 subsets. Therefore option A is correct; 256, 1024, and 8192 are incorrect powers of two for this problem.

A finite set with m elements has 2^m elements in its power set. Therefore, the given condition becomes 2^m=m+12. Testing the options, m=4 gives 2^4=16 and m+12=4+12=16, so the equation is satisfied. The other listed values do not satisfy it; hence m=4.

Since 2 must be included, the subset already has one element. To make its total size even, we must choose an odd number of elements from the remaining set {1, 3, 4}. We can choose one element in C(3,1)=3 ways or all three elements in C(3,3)=1 way. Thus the total is 3+1=4.

A set with six elements has 2^6 = 64 subsets. To count subsets containing 1 or 2, use the complement: subsets containing neither 1 nor 2 can choose freely only from {3,4,5,6}, giving 2^4 = 16. Therefore the required number is 64 - 16 = 48. Option A is correct.

The complement A' is defined relative to U, so A' = {3,4,5}. Every element of A' is therefore an element of U, which means A' ⊆ U. Whenever one set is a subset of another, every subset of the smaller set is also a subset of the larger set. Consequently P(A') ⊆ P(U), making option A correct.

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 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.

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.

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.

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.

A∩B=∅ means A and B have no common elements. Any set belonging to both P(A) and P(B) must be a subset of both A and B, so it must be a subset of A∩B. The only subset of the empty set is ∅. Hence P(A)∩P(B)={∅}, whose cardinality is 1, so option B is correct.

A set belongs to both P(A) and P(B) exactly when it is a subset of both A and B. Such sets are precisely the subsets of A∩B, so P(A)∩P(B)=P(A∩B). Since |A∩B|=2, its power set has 2^2=4 members. Therefore option B is correct; the separate sizes of A and B are not needed.

Since A and B are disjoint and each has two elements, A ∪ B = {1, 2, 3, 4}, so |P(A ∪ B)| = 2^4 = 16. Also, |P(A)| = |P(B)| = 2^2 = 4. The two power sets share only the empty set, because no nonempty subset can consist simultaneously of elements from A and from B. Thus |P(A) ∪ P(B)| = 4 + 4 − 1 = 7, and the required difference has 16 − 7 = 9 members.

For a set with n elements, exactly half of its 2^n subsets have even cardinality and half have odd cardinality. Therefore, the number of odd-cardinality subsets is 2^(n−1). With n = 7, the count is 2^6 = 64. This includes all possible odd sizes, namely 1, 3, 5, and 7. Option B is correct.

A set with 8 elements has 2^8 = 256 total subsets. For every nonempty finite set, the subsets with even cardinality and the subsets with odd cardinality occur in equal numbers. Thus the number of even-cardinality subsets is 2^8/2 = 2^7 = 128. The empty set is included because its cardinality is zero, which is even. Therefore, option B is correct.

There are 2^6 = 64 total subsets of A. It is easier to count the complement: subsets containing neither 2 nor 4 can use only the remaining four elements, so there are 2^4 = 16 of them. Therefore, subsets containing at least one of 2 or 4 equal 64 − 16 = 48. Option C is correct.

First form the union: A ∪ B = {1, 2, 3, 5, 7, 8}. Taking the complement relative to the universal set U leaves the elements of U that are not in this union: (A ∪ B)' = {4, 6}. This complement has 2 elements. A set with n elements has 2^n subsets, so the power set P((A ∪ B)') has 2^2 = 4 members. Therefore, option B is correct. The order matters: find the union first, then take its complement in U, and finally count the power set.