Power Set and Subsets

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 {∅, {∅}}.

Set A contains two elements. For any finite set with n elements, its power set contains 2^n subsets because each element can either be included or excluded from a subset. Therefore, n(B) = n(P(A)) = 2^2 = 4. Explicitly, B = {∅, {1}, {2}, {1,2}}. The value 2 is n(A), not n(P(A)), while 8 would require a three-element original set.

The power set contains every subset of the original set, including the empty subset, both singleton subsets, and the original two-element set. Therefore, P({m,n}) = {∅, {m}, {n}, {m,n}}. The braces are important: m and n alone are elements, whereas {m} and {n} are singleton subsets. Thus option B lists all four subsets correctly.

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.

To form a three-element subset from the five elements of A, choose any 3 elements without considering their order. Therefore, the required number is C(5,3) = 5!/(3!2!) = (5×4×3)/(3×2×1) = 10. These are ten distinct members of P(A). The total number of all subsets is 2^5 = 32, but only the subsets with exactly three elements are counted here.

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.

For a finite set with n elements, the number of members of its power set is 2ⁿ. Each of the six elements of A can either be selected or not selected when forming a subset, producing 2 × 2 × 2 × 2 × 2 × 2 = 2⁶ possibilities. Since 2⁶ = 64, the power set P(A) contains 64 subsets. Therefore option D is correct; 32 is only 2⁵.

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 ∅.

A singleton subset contains exactly one element. The odd elements of A are 1 and 3, so the singleton subsets containing an odd number are {1} and {3}. There are exactly two such subsets. The even elements 2 and 4 produce {2} and {4}, but these do not satisfy the condition. Therefore option B is correct. In general, each qualifying element produces one distinct singleton subset.

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.

A set with n elements has 2^n total subsets because each element can either be included or excluded. Here, |A| = 3, so P(A) has 2^3 = 8 subsets. Exactly one of them is the empty set, so the number of non-empty subsets is 8 − 1 = 7. Therefore, option B is correct; 8 counts the empty set too.

If a is not allowed in a subset, only b and c may be selected. Each of these two elements has two independent choices: include it or leave it out. Thus the number of permitted subsets is 2^2 = 4. They are the empty set, {b}, {c}, and {b, c}. Therefore, option C is correct.

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.

The power set P(A) is the collection of all subsets of A. If a set has n elements, each element has two choices—being included or excluded—so the power set has 2^n elements. Here n = 4, giving |P(A)| = 2^4 = 16. Therefore, option C is correct. The value 4 is the size of A itself, not the size of its power set.

The complement A′ contains all elements of the universal set U that are absent from A. The set A contains the even numbers from 2 through 10, so removing them from U leaves the odd numbers 1, 3, 5, 7, and 9. Thus A′ = {1, 3, 5, 7, 9}, making option A correct. Option B is A itself, not its complement.

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.

If a finite set A has n elements, then its power set 𝒫(A) has 2ⁿ elements. Here, 2ⁿ = 32 = 2⁵, so A has 5 elements. Therefore, A has 32 subsets in total. A proper subset is any subset other than A itself, so we exclude exactly one subset: A. Hence, the number of proper subsets is 32 − 1 = 31.