Power Set and Subsets

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 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.

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.

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.

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.

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.

In set theory, repeated elements do not create new elements. Thus the given collection is the set A = {1,2,3}, whose cardinality is 3 rather than 5. A set with n elements has 2^n subsets, so n(P(A)) = 2^3 = 8. The repeated 1s and 2s must therefore be ignored when determining the number of elements and the size of the power set.

If a finite set has n elements, then its total number of subsets, including the empty set and the set itself, is 2^n. Here 2^n = 256. Since 256 = 2^8, it follows that n = 8. The neighboring choices do not work: 2^7 = 128 and 2^9 = 512. Therefore, A has exactly eight elements.

The integers from −2 through 2 are −2, −1, 0, 1, and 2, so A has 5 elements. A set with n elements has exactly 2ⁿ subsets because each element can independently be included or excluded. Therefore, |P(A)| = 2⁵ = 32. The answer is consequently option C, not 16 or 64, which would result from counting the elements of A incorrectly.

The elements 2 and 5 must be present in every allowed subset, so their choices are fixed. The remaining elements 1, 3, 4, and 6 are unrestricted. Each of these four elements can independently be included or excluded, giving 2⁴ possible choices. Therefore the number of subsets containing both 2 and 5 is 16, which is option B.

A subset that contains neither 1 nor 2 can use only the remaining elements 3, 4, and 5. Each of these three elements may independently be selected or not selected. Therefore the number of such subsets is 2³ = 8. This includes the empty set, single-element subsets, larger permitted subsets, and {3,4,5}; hence option B is correct.

A has three elements, so its power set contains 2³ = 8 subsets. Every subset of A belongs to P(A), but a proper subset must not be equal to A itself. Exactly one subset, namely {x,y,z}, is not proper. Removing it from the eight subsets leaves 8 − 1 = 7 proper subsets. Therefore option C is correct.

The union contains each distinct element from A and B only once. Thus A ∪ B = {1,2,3,4}, which has four elements; the repeated element 3 is counted only once. A set with four elements has 2⁴ subsets. Therefore P(A ∪ B) contains 16 elements, so option B is correct.

To form a subset of A that must contain 3, include 3 first. The remaining elements, 1 and 5, can each be either included or excluded independently. Thus there are 2 choices for 1 and 2 choices for 5, giving 2 × 2 = 2² = 4 subsets. They are {3}, {1,3}, {3,5}, and {1,3,5}. Therefore, option B is correct.

An element of P(A) must be a subset of A. Option B, {r,t}, contains only r and t, and both are elements of A, so it is a subset and therefore belongs to P(A). Option A is an element of A rather than a set of elements; options C and D contain v, which is not in A. Hence option B is correct.

The power set is the set of all subsets. For the singleton \(A=\{k\}\), the only subsets are the empty set \(\varnothing\) and the set itself \(\{k\}\). Therefore, \(\mathcal{P}(A)=\{\varnothing,\{k\}\}\). Notice the braces: \(k\) is an element, whereas \(\{k\}\) is a subset. Thus option C is correct.

The empty set is a subset of every set, including A. By definition, P(A) is the set of all subsets of A. Therefore ∅ is always an element of P(A), so option C is universally true. Options A and B are not necessary because A may not contain itself and may not contain the empty set. Option D is not guaranteed because U need not be a subset of A.

The empty set has no elements, but it does have one subset: itself. Therefore the power set of the empty set contains exactly one element, namely the empty set: P(∅) = {∅}. It is important to distinguish ∅ from {∅}; the former has zero elements, while the latter has one element. Thus option B is correct.

The set A has 4 elements. A set with n elements has 2ⁿ subsets because each element can either be included or excluded. Therefore, P(A) has 2⁴ = 16 subsets in total. Exactly one of these subsets is the empty set, ∅. Hence, the number of non-empty subsets is 16 − 1 = 15, so option B is correct.

A singleton subset contains exactly one element. Since A contains five distinct elements— a, e, i, o, and u—each element produces one singleton subset: {a}, {e}, {i}, {o}, and {u}. Thus P(A) contains exactly 5 singleton subsets. The total number of subsets is 2⁵ = 32, but the question asks only for subsets having one element, so option B is correct.

The power set is the set of all subsets, including the empty set and the original set itself. For A = {2, 3}, the subsets are ∅, {2}, {3}, and {2, 3}. Notice that 2 and 3 must appear inside braces when they are considered as singleton subsets. Therefore P(A) is option A, which correctly lists all four subsets.

The natural numbers strictly between 5 and 11 are 6, 7, 8, 9, and 10. Thus A = {6,7,8,9,10} and n(A) = 5. The number of elements in the power set of an n-element set is 2ⁿ. Consequently, n(P(A)) = 2⁵ = 32. Therefore option B is correct; the endpoint values 5 and 11 are not included.

A set records each distinct element only once. Although the word MOON has four positions, the letter O is repeated, so the distinct-letter set is A = {M,O,N}. Thus n(A) = 3. The power set of a three-element set contains 2³ = 8 subsets, including the empty set and A itself. Hence option B is correct.

Repeated entries do not create new elements in a set. Therefore, A = {4,4,5,6,6,7} is the same set as {4,5,6,7}, which has 4 distinct elements. A set with 4 elements has 2⁴ subsets in its power set. Hence n(P(A)) = 16, so option B is correct. Counting the displayed entries as six different elements would be an error.