Power Set and Subsets

The power set P(A) is the collection of every subset of A. A set is always a subset of itself, so A ⊆ A. Consequently, A is one of the members of P(A), which means A ∈ P(A). The other choices confuse elements with subsets or introduce irrelevant conditions. A set does not need to be universal, empty, or have a particular size to belong to its own power set.

For a finite set A with n(A) elements, the power set has n(P(A)) = 2ⁿ⁽ᴬ⁾ elements. Here 2ⁿ⁽ᴬ⁾ = 8. Since 8 = 2³, it follows that n(A) = 3. Option A would give only 2² = 4 subsets, while option C would give 2⁴ = 16 subsets. Therefore, option B is the unique correct answer.

The set \(A=\{\emptyset\}\) contains exactly one element: the empty set \(\emptyset\). It is important to distinguish \(\emptyset\), which has no elements, from \(\{\emptyset\}\), which has one element. A set with \(n\) elements has \(2^n\) subsets, so its power set has cardinality \(n(\mathcal{P}(A))=2^{n(A)}=2^1=2\). The two subsets are \(\emptyset\) and \(\{\emptyset\}\). Therefore, option C is correct.

The set A = {∅} has exactly one element: ∅. A one-element set has exactly two subsets: the empty subset ∅ and the complete subset A itself, which is {∅}. Therefore its power set is P(A) = {∅, {∅}}. The braces must be read carefully because ∅ and {∅} are different mathematical objects.

The power set P(A) contains subsets of A, not necessarily the individual elements of A. The number 1 is an element of A, but it is not a subset of A because it is not a set in this context. The set {1} is a subset of A, since its only element, 1, belongs to A. Therefore {1} belongs to P(A), and option B is correct.

The set {1, 5} contains only elements that belong to A, so {1, 5} is a subset of A. However, it is not an element of A because the elements of A are the individual numbers 1, 3, and 5. The power set 𝒫(A) contains every subset of A, including {1, 5}. Therefore, the correct answer is B, 𝒫(A).

If a finite set has n elements, its power set has 2ⁿ elements because each original element may either be included or excluded from a subset. Here 2ⁿ = 32. Since 32 = 2⁵, we obtain n = 5. Thus the original set contains five elements, so option B is correct. The other numerical choices do not satisfy 2ⁿ = 32.

A set with zero elements is the empty set, written as ∅. For a finite set with n elements, the number of elements in its power set is 2ⁿ. Here n = 0, so |𝒫(A)| = 2⁰ = 1. In fact, 𝒫(∅) = {∅}; the power set has one element, namely the empty set itself. Therefore, option B is correct.

The elements of 𝒫(A) are the subsets of A. A two-element subset is formed by choosing any two of the three elements, without considering order. The number is therefore C(3, 2) = 3. These subsets are {1, 2}, {1, 3}, and {2, 3}. Thus 𝒫(A) contains three two-element subsets, so option C is correct.

The power set 𝒫(A) is the set of all subsets of A. The empty set is a subset of every set, including A, because it has no element that could violate the subset condition. For A = {2, 3}, 𝒫(A) = {∅, {2}, {3}, {2, 3}}. Hence ∅ is an element of 𝒫(A), and option A is correct.

The power set 𝒫(A) contains all subsets of A, including A itself. Since A = {7, 8}, the set {7, 8} is equal to A, and every set is a subset of itself: A ⊆ A. Therefore {7, 8} is an element of 𝒫(A). It is not an individual element of A; the individual elements of A are 7 and 8. Thus option A is correct.

The complement A′ consists of the elements of the universal set U that are not in A. For the complement to be empty, there must be no element of U outside A. This happens exactly when A contains every element of U, meaning A = U. Therefore, A′ = ∅ when A = U, so option A is correct.

The power set P(A) contains every subset of A. A three-element subset is formed by choosing exactly 3 elements from the 4 elements of A. The number of such choices is C(4,3) = 4!/(3!1!) = 4. Therefore, P(A) contains exactly four subsets having three elements: {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}. Hence option C is correct.

A zero-element subset means the empty set ∅. Every set, including A and its power set P(A), has exactly one empty subset. Thus P(A) contains ∅ as one of its elements, and the number of zero-element subsets is exactly 1. The value 8 is the total number of subsets of A because A has three elements, not the number of empty subsets.

For any set A, its power set P(A) always contains the empty set and A itself. Here P(A) has exactly two elements: ∅ and {5}. Since the non-empty member is the whole original set, A must be {5}. This also agrees with the rule that a one-element set has 2¹ = 2 subsets: ∅ and {5}.

An element of the power set P(A) must be a subset of A, meaning every element of that set must belong to A. Although 1 and 2 belong to A, the set {1,2,3} also contains 3, and 3 is not in A. Therefore {1,2,3} is not a subset of A and cannot be an element of P(A).

If a finite set has n elements, its power set has 2ⁿ elements because each original element can either be included or excluded from a subset. Here A has n = 2 elements, book and pen. Therefore |P(A)| = 2² = 4. The four subsets are ∅, {book}, {pen}, and {book, pen}.

Both 4 and 8 are elements of A = {2,4,8}; therefore every element of {4,8} belongs to A, so {4,8} ⊆ A. By definition, every subset of A is an element of P(A), hence {4,8} ∈ P(A). The set {4,8} is not itself an element of A because A contains numbers, not this two-element set.

The power set \(\mathcal{P}(A)\) is the set of all distinct subsets of \(A\). The original set \(A\) is itself a subset of \(A\), so it is included in the power set exactly once. The number \(2^4=16\) represents the total number of different subsets, not the number of times one particular subset is repeated. Therefore, the whole set appears once.

The power set \(\mathcal{P}(A)\) contains every subset of \(A\), including the empty set, singleton subsets, two-element subsets, and \(A\) itself. Since 20 belongs to \(A\), the singleton set \(\{20\}\) is a subset of \(A\), and therefore \(\{20\}\in\mathcal{P}(A)\). The number 40 is not in \(A\), so \(\{40\}\) is not a subset of \(A\).

The set \(A=\{2,5,7\}\) contains 3 distinct elements, so \(n(A)=3\). For a finite set with \(n\) elements, every element has two choices in forming a subset: it is either included or excluded. Consequently, the number of subsets, and hence the number of elements in the power set, is \(2^n\). Therefore, \(n(\mathcal{P}(A))=2^3=8\).

The power set contains subsets of \(A\), not the individual elements of \(A\) written without braces. Since \(x\in A\), the singleton \(\{x\}\) is a subset of \(A\), so \(\{x\}\in\mathcal{P}(A)\). In fact, \(\mathcal{P}(A)=\{\emptyset,\{x\},\{y\},\{x,y\}\}\). Thus option B is the only correctly written member among the listed choices.

If a finite set \(A\) has \(n\) elements, then its power set has \(2^n\) elements. Here, \(|\mathcal{P}(A)|=16\), so \(2^n=16\). Since \(16=2^4\), it follows that \(n=4\). Therefore, the original set \(A\) contains 4 elements. The answer is not 8; 8 would be the number of subsets of a three-element set.

A power set contains all subsets of the original set. For the singleton set \(A=\{3\}\), the only subsets are the empty set \(\emptyset\) and the set \(\{3\}\) itself. Therefore, \(\mathcal{P}(A)=\{\emptyset,\{3\}\}\), which has two elements. The braces are important: 3 is an element of \(A\), whereas \(\{3\}\) is a subset and an element of the power set.

Both 1 and 9 are elements of A, so {1, 9} is a subset of A. The power set 𝒫(A) is the set of all subsets of A; therefore, {1, 9} is an element of 𝒫(A). Statement A is false because the elements of A are numbers, not the set {1, 9}. Statement D is false because {1, 9} is indeed a subset of A, and U is not defined for statement C.