Power Set and Subsets

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 \(A\).

First find the complement of \(A\) in the universal set: \(A'=U-A=\{3,4,5,6\}\). Thus, \(A'\) has 4 elements. For any finite set with \(n\) elements, the power set has \(2^n\) elements, because each element can independently be selected or not selected. Therefore, \(|\mathcal{P}(A')|=2^4=16\).

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.

The empty set ∅ has no elements, but it has exactly one subset: the empty set itself. Therefore, its power set is 𝒫(∅) = {∅}, which contains one element. This illustrates the formula n(𝒫(S)) = 2ⁿ, where n is the number of elements in S. For S = ∅, n = 0, so n(𝒫(∅)) = 2⁰ = 1.

The set A has two distinct elements: the empty set ∅ and the number 1. The fact that one element is itself a set does not change the count; thus n(A) = 2. A finite set with n elements has 2ⁿ subsets, so n(𝒫(A)) = 2² = 4. Its subsets are ∅, {∅}, {1}, and {∅, 1}.

A singleton subset contains exactly one element. Since A has four elements, p, q, r, and s, the singleton subsets are {p}, {q}, {r}, and {s}. Thus, 𝒫(A) contains four singleton subsets. In general, an n-element set has exactly n singleton subsets, although its total number of subsets is 2ⁿ; here the total is 2⁴ = 16.

An element of the power set 𝒫(A) must be a subset of A, meaning every element of that set must belong to A. The sets {1}, {2, 3}, and ∅ are all subsets of A. However, {1, 4} is not a subset because 4 is not an element of A. Therefore, {1, 4} is not an element of 𝒫(A).

The power set P(A) is the set of every subset of A. For A = {6, 12}, the subsets are the empty set ∅, the singleton sets {6} and {12}, and the complete set {6, 12}. Therefore P(A) = {∅, {6}, {12}, {6, 12}}. Since A has two elements, it must have 2² = 4 subsets.

The power set \(\mathcal{P}(A)\) contains every subset of \(A\), including all its two-element subsets. To choose two elements from the four elements \(a,b,c,d\), use the combination formula \(\binom{4}{2}=\frac{4!}{2!2!}=6\). These subsets are \(\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\},\{c,d\}\). Therefore, option C, 6, is correct. The number \(2^4=16\) is the total number of subsets in \(\mathcal{P}(A)\), not the number of two-element subsets.

A power set contains all subsets of the original set, including the original set itself. Every set is a subset of itself because each element of A belongs to A; symbolically, A ⊆ A. Therefore, when P(A) is formed, the complete set A appears as one element of P(A), written A ∈ P(A). It is not automatically a universal set.

The power set P(A) contains subsets of A, not the individual elements of A unless those elements are themselves sets. The object {2} is a subset of A because its only element, 2, belongs to A, so {2} ∈ P(A). In contrast, 2 is an element of A, not a subset of A, and therefore 2 ∉ P(A).

A four-element subset is obtained by choosing 4 of the 5 elements of A. The number of choices is 5C4 = 5!/(4!1!) = 5. Equivalently, each four-element subset is determined by the one element left out, and there are five possible elements to omit. Therefore, P(A) has exactly five subsets containing four elements.

The empty set ∅ is a subset of every set because it has no elements that could fail to belong to that set. In particular, every element of ∅ is an element of A vacuously, so ∅ ⊆ A. Since P(A) is the collection of all subsets of A, ∅ must be one of its elements. It does not mean that ∅ equals A.

First calculate the complement relative to U: A′ = U − A = {2, 4, 5}. Thus A′ has three elements. A set with n elements has 2ⁿ subsets, so its power set has 2³ = 8 elements. Therefore, P(A′) contains 8 elements. The answer is not 16 because the complement has three, not four, elements.