Power Set and Subsets

The power set lists all subsets of A: P(A) = {∅, {1}, {2}, {1, 2}}. Membership in P(A) therefore requires an object to be a subset of A. The number 1 is an element of A, but 1 alone is not a subset; {1} is the subset. Hence 1 ∉ P(A), making option A false. The empty set and A itself are always subsets of A, so options C and D are true.

To form a subset with exactly two elements from a four-element set, choose 2 elements at a time without regard to order. The number is given by the combination formula C(4, 2) = 4!/(2!2!) = 6. These six pairs are all members of P(A), because every subset of A belongs to its power set.

The universal set U contains 15 elements, while A contains the five multiples of 3 from 1 to 15. Therefore, the complement A' contains the remaining 15 − 5 = 10 elements. A set with n elements has 2^n subsets, so its power set contains 2^10 = 1024 elements. Hence, n(P(A')) = 1024, which is option A.

The notation distinguishes an element from a subset. Here the set {1,2} is explicitly listed as one element of A, so {1,2}∈A. It is also a subset of A because both 1 and 2 are individually elements of A. Every subset of A belongs to P(A), so {1,2}∈P(A) as well. Therefore option A is the only correct statement.

A subset with at least four elements must have either exactly four elements or exactly five elements, since A itself has only five elements. The number of four-element subsets is C(5,4)=5, and the number of five-element subsets is C(5,5)=1. Adding them gives 5+1=6. Therefore P(A) has exactly six subsets with at least four elements, making option A correct.

For a three-element set, an odd-sized subset has either one element or three elements. There are C(3,1)=3 one-element subsets: {a}, {b}, and {c}. There is C(3,3)=1 three-element subset, namely {a,b,c}. Thus the total number is 3+1=4, so option A is correct. The empty set and two-element subsets have even size and are excluded.

The statement A⊆B means every element of A is also an element of B. Take any set X in P(A). By definition, X⊆A. Since every element of A belongs to B, all elements of X also belong to B, so X⊆B. Therefore X∈P(B). This proves P(A)⊆P(B), making option A always true. Equality occurs only in special cases, such as A=B.

Let S be a subset of U. If its complement U − S is a singleton, then exactly one element of U is outside S, so S must contain exactly three elements. The number of three-element subsets of a four-element set is C(4,3) = 4. Therefore, exactly four elements of P(U) have singleton complements.

The number of two-element subsets of a five-element set is C(5,2) = 10. The number of three-element subsets is C(5,3) = 10. These two classes are disjoint because a subset cannot simultaneously have two and three elements. Therefore, the required total is 10 + 10 = 20.

Solve the defining equation: \(x^2=x\) implies \(x^2-x=0\), or \(x(x-1)=0\). Hence \(x=0\) or \(x=1\), so \(A=\{0,1\}\). Its complement in \(U\) is \(A'=\{2,3,4,5\}\), which has four elements. A set with \(k\) elements has \(2^k\) subsets, so \(n(\mathcal{P}(A'))=2^4=16\). Therefore option A is correct; 4 is only the size of \(A'\), not of its power set.

The elements a and b must be included in every required subset, so their choices are fixed. The remaining elements c, d, and e are each optional: each can either be included or excluded. Thus there are 2 choices for each of three independent elements, giving 2³ = 8 subsets. Therefore, option A is correct.

P(A) consists of all subsets of A: ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and {a,b,c}. A subset is disjoint from A only when it has no element in common with A. Since every non-empty subset listed contains elements of A, only the empty set qualifies. Hence the number is 1, option A.

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.

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 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.

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.

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.

The set A has 3 distinct elements. A two-element subset is formed by choosing any 2 of these 3 elements, so the required number is C(3, 2) = 3!/(2!1!) = 3. Explicitly, the subsets are {p, q}, {p, r}, and {q, r}. Each of these is a member of P(A), because the power set contains every subset of A. Therefore, option B is correct.

The power set P(A) consists of all subsets of A. A member of P(A) having exactly 4 elements is therefore a 4-element subset of A. Since A has 6 elements, the number of such subsets is C(6, 4) = 6!/(4!2!) = (6 × 5)/(2 × 1) = 15. Equivalently, choosing 4 elements to include is the same as choosing the 2 elements to exclude, so C(6, 4) = C(6, 2) = 15. Hence option B is correct.

Element 1 is compulsory, so it has only one status: it must be included. Element 5 is forbidden, so it also has only one status: it must be excluded. The remaining elements 2, 3, and 4 may each be included or excluded independently, giving 2^3 = 8 possible subsets. Hence option B is correct.

The elements a and b are required, so they are fixed as included in every valid subset. The remaining four elements, c, d, e, and f, are unrestricted. Each of these four elements has two independent choices: include it or exclude it. Therefore, the number of valid subsets is 2^4 = 16, so option B is correct.