Power Set and Subsets

The complement of A relative to U is A′ = U − A = {1,3,5,7}. It has 4 elements. A member of P(U) that is a subset of A′ can be any subset of A′, including the empty set and A′ itself. Therefore, the number of such members equals |P(A′)| = 2⁴ = 16. Thus option B is correct.

A singleton set is a set containing exactly one element. The singleton members of P(A) are {1}, {2}, {3}, and {4}, one for each element of A. Therefore, a set with 4 elements has exactly 4 singleton subsets. The empty set is not a singleton because it contains no elements, while larger subsets contain more than one element. Hence option C is correct.

If a set A has 4 elements, then its power set P(A) has 2^4 = 16 elements. A singleton member of P(P(A)) is a one-element subset whose only element is one member of P(A). There is exactly one singleton subset for each element of P(A), so P(P(A)) contains 16 singleton members. Therefore, option C is correct.

The power set P(A) contains every subset of A. Since {1} ⊆ A and {2} ⊆ A, both {1} and {2} are elements of P(A). The set {{1}, {2}} has exactly these two elements, and each of them belongs to P(A). Therefore every element of {{1}, {2}} is an element of P(A), which proves that {{1}, {2}} ⊆ P(A). The statement is therefore true. It is important to distinguish {1} from 1: the power set contains sets such as {1}, not necessarily the number 1 as an element.

Because {1} is a subset of A = {1, 2}, it is an element of P(A); therefore {1} ∈ P(A) is true. However, {1} ⊆ P(A) means that the element 1 itself must belong to P(A). The elements of P(A) are ∅, {1}, {2}, and {1, 2}; the number 1 is not one of them. Thus only the first statement is true, so option A is correct.

The universal set U contains the integers 1 through 15, so it has 15 elements. The numbers divisible by 3 are 3, 6, 9, 12, and 15; hence |A| = 5. Its complement A' therefore contains 15 − 5 = 10 elements. A set with 10 elements has 2^10 subsets, so |P(A')| = 2^10. Therefore option C is correct.

The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19, so A has 8 elements. Since U has 20 elements, the complement A' has 20 − 8 = 12 elements. Its power set contains 2^12 = 4096 subsets, including the empty set. Therefore the number of non-empty subsets is 4096 − 1 = 4095, so option B is correct.

Because the endpoints are included, the integers in A are −2, −1, 0, 1, 2, and 3. Thus A contains 6 elements. For every finite set with n elements, the number of subsets in its power set is 2^n, because each element can either be selected or not selected. Therefore |P(A)| = 2^6 = 64. Hence option C is correct.

The odd elements form A = {1, 3, 5, 7, 9, 11}, so the complement A' consists of the six even elements {2, 4, 6, 8, 10, 12}. Members of P(A') are subsets of A'. The number of members containing exactly 3 elements is the number of ways to choose 3 elements from 6, namely C(6,3) = 6!/(3!3!) = 20. Thus option C is correct.

An n-element set has 2^n total subsets. A proper subset is any subset other than the set itself, so the number of proper subsets is 2^n − 1. Given 2^n − 1 = 31, adding 1 gives 2^n = 32 = 2^5. Therefore n = 5. For comparison, n = 4 would give 15 proper subsets and n = 6 would give 63, so option B is the only correct answer.

Since A is a subset of U, its complement has cardinality |A'| = |U| − |A| = 2m − m = m. The power set of any set with m elements has 2^m elements. Therefore |P(A)| = 2^m and |P(A')| = 2^m, so the two cardinalities are equal for every allowed value of m, not just for a special value. Hence option A is correct.

For every finite set S, the number of elements in its power set is |P(S)| = 2^|S|. Since |P(A′)| = 1024 = 2^10, we obtain |A′| = 10. The complement A′ is taken relative to U, so A and A′ partition U and |A| + |A′| = |U|. Hence |A| = 18 − 10 = 8. Applying the power-set formula again gives |P(A)| = 2^8. Therefore option B is correct. Option C, 2^10, is the size of P(A′), not the size of P(A), which is the key distinction in this problem.

A set with n elements has 2^n subsets because each element has two choices: it may be included or excluded. Here |A| = 4, so P(A) contains 2^4 = 16 subsets. The set A itself is one of these subsets, so excluding A gives 16 - 1 = 15. Therefore, option B is correct.

An n-element set has equally many subsets of even and odd cardinality. Since A has 6 elements, its total number of subsets is 2^6 = 64. Exactly half have even cardinality and half have odd cardinality, so the number of even-cardinality subsets is 64/2 = 32. This includes the empty set, whose cardinality is zero and is even.

The elements 1 and 2 are compulsory, while 5 is forbidden. Thus only 3 and 4 remain optional. Each optional element can independently be included or excluded, giving 2 choices for each and therefore 2^2 = 4 valid subsets. They are {1,2}, {1,2,3}, {1,2,4}, and {1,2,3,4}. Hence option B is correct.

The total number of subsets of A is 2^5 = 32. It is easier to count the complement: subsets containing neither a nor b. Such subsets can use only c, d, and e, giving 2^3 = 8 subsets. Therefore, subsets containing at least one of a or b equal 32 - 8 = 24. Option C is correct.

Use De Morgan’s law or calculate directly. A ∪ B = {1,2,3,4,5,6}, so its complement in U is {7,8}; hence A' ∩ B' = {7,8}. This set has two elements, and the power-set rule gives |P(A' ∩ B')| = 2² = 4. Therefore option B is correct. The other values correspond to using an incorrect number of elements in the final set.

By De Morgan’s law, A' ∪ B' = (A ∩ B)'. The intersection A ∩ B is {4,5}, containing 2 elements. Its complement in the 9-element universal set U therefore contains 9 - 2 = 7 elements. The power set of a 7-element set has 2^7 = 128 members. Hence option D is correct.

For a set A with 5 elements, the power set contains 2^5 = 32 subsets. Every power set contains A itself and the empty set ∅. Since these are two different subsets and both must be excluded, the number remaining is 32 - 2 = 30. Thus option B is correct.

The union is A ∪ B = {1,2,3,4,5,6}, so it has 6 elements and its power set has 2^6 = 64 elements. The intersection is A ∩ B = {3,4}, so it has 2 elements and its power set has 2^2 = 4 elements. Hence the required difference is 64 − 4 = 60, so option C is correct.