Power Set and Subsets

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.

The governing rule is that a finite set with n elements has 2^n elements in its power set. Since A = {0, 1} has 2 elements, |P(A)| = 2^2 = 4. Applying the same rule again to P(A), which has 4 elements, gives |P(P(A))| = 2^4 = 16. Therefore option C is correct. The values 4 and 8 stop after the first stage or use an incorrect exponent.

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.

A power set P(A) is the set of all subsets of A. Since {v} contains the element v and is a subset of A = {u,v}, it is an element of P(A). In contrast, u and v are individual elements, not sets of elements, and uv is not a subset notation for A. The complete power set is {∅, {u}, {v}, {u,v}}, confirming that option B is correct.

For a finite set A, the number of elements in its power set is given by n(P(A)) = 2^n(A). Here 2^n(A) = 64. Since 64 = 2^6, the exponent must be n(A) = 6. The alternatives do not work: 2^4 = 16, 2^5 = 32, and 2^8 = 256. Therefore the set A contains six elements.

The set A = {9} has one element. Its only subsets are the empty set ∅ and the set containing its element, {9}. The power set must contain these subsets as its own elements, so P(A) = {∅, {9}}. Option B incorrectly treats 9 as an element directly inside the power set rather than the subset {9}; option A omits the empty subset, and option D mixes element and set notation.

The set \(\{3,9\}\) contains the elements 3 and 9, and both are elements of \(A\). Therefore, \(\{3,9\}\subseteq A\). The power set \(\mathcal{P}(A)\) contains every subset of \(A\), so \(\{3,9\}\) is an element of \(\mathcal{P}(A)\). Option A is false because \(A\) contains numbers, not the set \(\{3,9\}\); option C is false because 6 is missing; option D contradicts the subset relation.

The symbol \(\emptyset\) is one element of \(A\); it is not the same as saying that the set has no elements. Thus the three distinct elements of \(A\) are \(\emptyset\), 5, and 10, so \(n(A)=3\). For a finite set with \(n\) elements, the power set has \(2^n\) elements because each element can be selected or not selected. Therefore, \(n(\mathcal{P}(A))=2^3=8\), making option C correct.