Power Set and Subsets

The power set of a set is the set containing all its subsets. The empty set has exactly one subset: the empty set itself. Therefore, 𝒫(∅) = {∅}. Notice the distinction between ∅, which has no elements, and {∅}, which has one element—the empty set. This distinction is essential when determining power sets and their cardinalities.

The set A contains five distinct elements: 2, 4, 6, 8, and 10. For every element, a subset can either include it or exclude it, giving two choices independently. Therefore, a set with n elements has 2ⁿ subsets. Here n = 5, so |𝒫(A)| = 2⁵ = 32. The requested number is the cardinality of the power set, not the number of elements in A.

The value n(𝒫(B)) = 128 tells us that B has 128 subsets in total. Every set is a subset of itself, but it is not a proper subset of itself. Therefore, exactly one subset—the set B—is removed when counting proper subsets. The number of proper subsets is 128 − 1 = 127. Although 128 = 2⁷ also shows that B has seven elements, that value is not the requested answer.

An element of the power set 𝒫(A) must be a subset of A. The set A contains only p, q, and r. The object {p, q} contains only elements of A, so it is a subset and therefore belongs to 𝒫(A). Option A is the element p rather than the subset {p}; options C and D contain s, which is not an element of A, so they are not subsets of A.

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).

The empty set has zero elements, but it has one subset: itself. The general rule says that a set with n elements has 2ⁿ subsets. Taking n = 0 gives |𝒫(∅)| = 2⁰ = 1. Thus 𝒫(∅) = {∅}, which is a set containing one element. It is important not to confuse the empty set with its power set.

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.

For every set A, the empty set ∅ is a subset of A because it has no element that can violate the subset condition. The set A itself is also a subset of A because every element of A is contained in A. Since P(A) contains all subsets of A, both ∅ and A are always elements of P(A). The universal set U need not be a subset of A.

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}.

Let one particular element be fixed and required to appear in every selected subset. That element has no choice because it must be included. Each of the remaining four elements can independently be included or excluded, giving two choices for each. Therefore the number of subsets is 2^4 = 16. This is half of the total 2^5 subsets, since exactly half contain the chosen element.

Exclude the specified element from every allowed subset. The remaining set has 6 − 1 = 5 elements, and each of these five elements may be included or excluded independently. Therefore the number of permitted subsets is 2^5 = 32. Equivalently, the 2^6 = 64 subsets of A split equally into those containing the chosen element and those not containing it. Thus option B is correct.

The power set of a three-element set contains 2^3 = 8 subsets. Exactly one of these subsets is empty, namely ∅. Every other subset is non-empty, so the number of non-empty elements of P(A) is 8 − 1 = 7. These include the three one-element subsets, three two-element subsets, and A itself. Therefore option C is correct.

A singleton subset contains exactly one element. Since A has four distinct elements, the singleton subsets of A are {m}, {n}, {o}, and {p}. Therefore, P(A), which contains every subset of A, contains exactly four singleton sets. The formula 2^n counts all subsets, but singleton subsets specifically are counted by choosing one element from n elements, namely C(4,1) = 4.

A subset with exactly three elements is formed by choosing three elements from the five elements of A. The number of such choices is the combination C(5,3) = 5!/(3!2!) = 10. Each distinct choice gives one distinct subset, and every subset of A is an element of P(A). Hence, P(A) contains 10 subsets having exactly three elements.

The power set contains every subset of A, including the empty set and A itself. For A = {0,1}, the subsets are ∅, {0}, {1}, and {0,1}. Therefore P(A) = {∅, {0}, {1}, {0,1}}. Notice that 0 and 1 alone are elements of A, whereas {0} and {1} are subsets; the braces are essential in a power-set representation.

For a finite set X with n elements, the power set P(X) has 2^n elements because each original element may either be included or excluded from a subset. Therefore, n(P(A)) = 2^4 = 16 and n(P(B)) = 2^3 = 8. Their ratio is 16:8, which simplifies by dividing both terms by 8 to 2:1. Thus option A is correct.

A set records each distinct element only once. Although LEVEL has five letter positions, the distinct letters are L, E, and V, so A = {L,E,V} and n(A) = 3. The power set of a finite set with n elements contains 2^n subsets. Therefore, n(P(A)) = 2^3 = 8. Repeated occurrences of L and E do not increase the set's cardinality.