Power Set and Subsets

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.

Set A has two elements, so its power set contains 2^2 = 4 subsets: the empty set, the two singleton subsets, and A itself. The set P(A) therefore has 4 elements. Taking the power set once more gives |P(P(A))| = 2^4 = 16. The exponent in the second step is 4 because P(A), rather than A, is the set whose subsets are being counted.

The condition requires 2 to be included and 3 to be excluded, so those two choices are fixed. Only 1 and 4 remain freely selectable. Each can be included or omitted independently, giving 2 × 2 = 2^2 = 4 valid subsets: {2}, {1,2}, {2,4}, and {1,2,4}. Therefore, option B is correct.

The three listed objects are distinct elements: ∅ is the empty set, {∅} is a one-element set whose element is ∅, and 0 is a number. Since A has 3 distinct elements, the number of its subsets is 2^3 = 8. Therefore n(P(A)) = 8. It is important not to confuse ∅ with {∅}; they are different sets.

If the original set A has n elements, then its power set has 2^n elements. The question gives 2^n = 32. Since 32 = 2^5, it follows that n = 5. Therefore, the original set A contains five elements. Choosing 4 would give only 16 subsets, while choosing 6 would give 64 subsets. Hence option B is correct.

Although A contains the empty set, A itself is not empty. The set A = {∅} has exactly one element: the empty set. A set with one element has 2^1 = 2 subsets. They are ∅ and {∅}. Therefore, P(A) = {∅, {∅}} and contains two elements. Confusing ∅ with {∅} would incorrectly lead to one, so option B is correct.

A set containing n distinct elements has exactly 2ⁿ subsets, because each element can either be included in or excluded from a subset. Here A = {k,l,m} has three distinct elements. Therefore n(P(A)) = 2³ = 8. The power set includes the empty set, the three singleton subsets, the three two-element subsets, and A itself. Hence option C is correct.

The set A has exactly two elements: the number 1 and the set {2}. The inner set {2} counts as one element of A; the number 2 is not separately an element of A. A set with n elements has 2ⁿ subsets, so here the number is 2² = 4. Explicitly, the subsets are ∅, {1}, {{2}}, and {1,{2}}. Hence option C is correct.

First find the complement of A in U: A′ = U − A = {b,c}. Thus A′ has two elements. The power set of any finite set containing n elements has 2ⁿ elements, because every element has two choices—present or absent—in a subset. Therefore n(P(A′)) = 2² = 4. Option A gives the size of A′, not the size of its power set, so option B is correct.

The power set of A contains every subset of A, including the empty set and A itself. The given power set has exactly two elements: ∅ and {x}. Since the non-empty member must be A itself, A = {x}. This also agrees with |P(A)| = 2^|A|: because |P(A)| = 2, we get |A| = 1. Notice that {∅} would produce a different power set, namely {∅, {∅}}.

Both elements of the set {0,2} occur in A = {0,1,2}, so {0,2} is a subset of A, written {0,2} ⊆ A. Every subset of A is, by definition, an element of the power set P(A). Hence {0,2} ∈ P(A). It is not an element of A because the elements of A are the numbers 0, 1, and 2, not the set {0,2} itself.

The power set P(A) consists of every subset of A, including the empty set and A itself. A four-element subset must contain exactly four distinct elements. Among the choices, ∅ has zero elements, {a, b} has two, and {a, b, c} has three. The set {a, b, c, d} contains all four elements and is A itself, so it is the required member of P(A).

If a finite set A has n elements, then its power set P(A) has 2ⁿ elements because each element independently has two choices: it may be included in a subset or excluded. We need 2ⁿ = 16. Since 16 = 2⁴, n = 4. Thus P(A) has exactly 16 members when A contains four elements. The other choices give 4, 8, and 256 subsets respectively.

The empty set has no elements, so its only subset is the empty set itself. Therefore P(∅) = {∅}. The notation is important: ∅ is the sole member of the power set, whereas {∅} is the power set as a whole, not a member of that power set. The number 0 and the singleton {0} are not subsets of ∅ because 0 is not an element of ∅.

To form a subset containing x, x must be included, while y and z may each independently be included or excluded. Thus there are 2 choices for y and 2 choices for z, giving 2 × 2 = 4 subsets. They are {x}, {x, y}, {x, z}, and {x, y, z}. The value 8 is the total number of all subsets of A, including those that do not contain x.

At least three elements means that a subset may contain exactly three elements or exactly four elements. The number of three-element subsets is C(4,3) = 4, and the number of four-element subsets is C(4,4) = 1. Adding them gives 4 + 1 = 5. Hence option B is correct; counting only three-element subsets would miss A itself.

For a set A with n elements, its power set P(A) contains 2^n elements. Since |P(A)| = 8, we have 2^n = 8 = 2^3, so |A| = 3. The power set therefore has eight total subsets, including exactly one empty set. Consequently, the number of non-empty subsets is 8 − 1 = 7, so option C is correct.

First find the complement relative to U: A′ = U − A = {3}. The power set of a one-element set contains exactly two subsets: the empty set and the set itself. Hence P(A′) = P({3}) = {∅, {3}}. Option B is incorrect because 3 is an element, not a subset written as {3}; the other options do not represent this power set.

An element of P(A) must be a subset of A. The elements of A are a, b, and the set {c}; importantly, c itself is not an element of A. Option B, {a, {c}}, uses two actual elements of A and is therefore a subset of A. Option A contains c rather than {c}, option C is not a subset, and option D also contains c directly, so B is the only correct answer.

The elements of A are 1, 2, and the set {3}. In particular, 3 itself is not an element of A; only {3} is. A member of 𝒫(A) must be a subset whose every element belongs to A. Options A, B, and C satisfy this condition. Option D contains 3 as an element, so it is not a subset of A and therefore is not an element of 𝒫(A).

First evaluate the inner power set. Since ∅ has no elements, 𝒫(∅) = {∅}, which has one element. Now take the power set of this one-element set. A one-element set has 2¹ = 2 subsets: ∅ and {∅}. Therefore, 𝒫(𝒫(∅)) contains two elements. The two power-set operations must be applied successively from the inside outward.

The positive divisors of 18 are 1, 2, 3, 6, 9, and 18, so A has 6 elements. For any finite set with n elements, its power set P(A), the set of all subsets of A, has 2ⁿ elements because each element can either be included or excluded from a subset. Hence n(P(A)) = 2⁶ = 64. Therefore, option C is correct.

The power set P(A) is the set of all subsets of A. For A = {∅, 1}, the empty set ∅ is a subset, {∅} is a subset because ∅ belongs to A, and {1} is a subset because 1 belongs to A. Therefore all three listed sets are elements of P(A), so option D is correct. Notice that ∅ itself and {∅} are different sets.

A set with n elements has 2^n subsets. When n = 3, A has 2^3 = 8 subsets in total. A proper subset cannot be the set A itself, and a non-empty subset cannot be ∅. Removing these two subsets from the total gives 8 − 2 = 6 non-empty proper subsets. Therefore option B is correct.

Elements of P(A) are subsets of A. To form a subset with exactly two elements from the four elements 1, 2, 3, and 4, choose any two of them. The number of choices is the combination C(4,2) = 4!/(2!2!) = 6. Hence P(A) contains six two-element subsets: {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, and {3,4}.