Power Set and Subsets

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.

The elements of A are the set {1} and the number 2, so A has exactly two elements. The power set P(A) contains every subset of A, including the empty set, two one-element subsets, and A itself. A set with n elements has 2ⁿ subsets; therefore n(P(A)) = 2² = 4. The nested braces do not create an additional element beyond {1}.

The governing concepts are complement relative to a universal set and the power set of a set. Since A′ contains elements of U that are not in A, remove 2 and 3 from U = {1, 2, 3, 4, 5}; hence A′ = {1, 4, 5}. The power set P(A′) consists of all subsets generated from these three elements. Therefore it is based on {1, 4, 5}, making option B correct. Option C is U itself, not its complement.

A zero-element subset is a subset containing no elements. There is exactly one such subset for every set: the empty set, written as ∅. Since ∅ is always a member of the power set P(A), P(A) contains exactly one zero-element subset, regardless of the fact that A itself has four elements.

A power set P(A) is the set of all subsets of A. The set {a,c} contains only elements that already belong to A = {a,b,c}; therefore {a,c} ⊆ A. It is consequently one of the members of P(A). It is not empty, it is not necessarily a complement, and no universal set U has been specified. Hence option A is correct.

Since A is the empty set, its complement relative to U is A' = U = {1, 2, 3, 4}. This set has four elements. The power set of a finite set with four elements contains 2^4 subsets, because each element can independently be included or excluded. Thus n(P(A')) = 16.

To form a subset of A that contains 1, keep 1 fixed. The other two elements, 2 and 3, may each be either included or excluded independently. Therefore there are 2 choices for 2 and 2 choices for 3, giving 2 × 2 = 2^2 = 4 subsets. They are {1}, {1,2}, {1,3}, and {1,2,3}.

A subset that does not contain 15 can use only the remaining elements, 5 and 10. Each of these two elements has two independent choices: it may be included or excluded. Therefore, the number of permitted subsets is 2^2 = 4. They are ∅, {5}, {10}, and {5, 10}. Hence, option C is correct.

The set A has exactly three elements: a, e, and i. A three-element subset must contain all three elements, because omitting even one would leave only two elements. Thus the only three-element subset is A itself, namely {a, e, i}. Therefore, P(A) contains exactly one three-element subset, so option A is correct.

The set {1} is a subset of A because its only element, 1, belongs to A. Every subset of A is an element of the power set P(A), so {1} also belongs to P(A). In contrast, 1 is an element of A, not a subset of A, and {3} is not a subset because 3 is not in A. Hence option A is correct.

We need to count the subsets of A that contain both 2 and 4. These two elements are compulsory, while the remaining element 6 is optional. If 6 is excluded, the subset is {2, 4}; if 6 is included, the subset is {2, 4, 6}. Therefore there are 2^1 = 2 valid subsets, so option B is correct.

The conditions require 1 to be included and 4 to be excluded, so neither of these elements is free to vary. Only 2 and 3 remain, and each can independently be included or excluded. Thus the number of valid subsets is 2 × 2 = 2^2 = 4. They are {1}, {1, 2}, {1, 3}, and {1, 2, 3}; therefore, option B is correct.

The set A has two distinct elements: the empty set ∅ and the singleton set {∅}. Although one element contains the other, they are different sets, so n(A) = 2. A set with n elements has exactly 2^n subsets in its power set. Hence n(P(A)) = 2^2 = 4. The four subsets are ∅, {∅}, {{∅}}, and {∅,{∅}}.

The set A has four elements. For a finite set with n elements, the power set contains 2^n subsets, because each element can either be included or excluded. Thus P(A) has 2^4 = 16 subsets. A one-element subset is formed by choosing exactly one of the four elements, so there are 4 such subsets: {1}, {3}, {5}, and {7}. Therefore the answer is 16 and 4.

The set A contains three distinct elements. Each element has two independent possibilities when forming a subset: it may be included or excluded. Therefore the total number of subsets, and hence the number of elements in the power set, is 2^3 = 8. These include the empty set, three singleton subsets, three two-element subsets, and the original set itself. Thus option C is correct.