Power Set and Subsets

For a finite set A, the number of elements in its power set is given by |P(A)| = 2^|A|. Since |P(A)| = 32 and 32 = 2^5, we have 2^|A| = 2^5. Equality of the powers gives |A| = 5. The other options would produce 16, 64, or an enormously larger power-set size, so they cannot be correct.

The power set P(A) consists of all subsets of A. The set A has two elements: the number 0 and the set {0}. The set {0} contains only the element 0, which belongs to A, so {0} is a subset of A and therefore an element of P(A). In contrast, 0 is an element of A but is not itself a subset of A.

Every member of \(\mathcal{P}(A)\) is a subset of \(A\). Since \(A\subseteq B\), any subset of \(A\) is automatically also a subset of \(B\). Therefore, every element of \(\mathcal{P}(A)\) belongs to \(\mathcal{P}(B)\), giving \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\). Equality occurs only when \(A=B\); in general, \(B\) may have additional elements and therefore additional subsets.

For a finite set with \(n\) elements, the power set contains \(2^n\) subsets. Therefore, \(|\mathcal{P}(B)|=2^3=8\) and \(|\mathcal{P}(A)|=2^2=4\). Because \(A\subset B\), every subset of \(A\) is a subset of \(B\), so \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\). The difference therefore has \(8-4=4\) elements. Thus option B is correct.

The even elements of \(A\) are 2 and 4, so the relevant set is \(\{2,4\}\), which has two elements. Each element can either be included or excluded independently when forming a subset. Hence the number of subsets is \(2^2=4\): \(\emptyset\), \(\{2\}\), \(\{4\}\), and \(\{2,4\}\). The empty set is included because it contains no odd elements and is a valid subset.

The odd elements of \(A\) are \(1,3,5\), so there are three odd elements. To obtain a subset with exactly two odd elements, choose two of these three odd elements: \(\binom{3}{2}=3\) choices. Each of the three even elements, \(2,4,6\), may be independently included or excluded, giving \(2^3=8\) choices. By the multiplication principle, the total is \(\binom{3}{2}2^3=3\times8=24\).

A subset with at least 4 elements can have either exactly 4 elements or exactly 5 elements. The number of 4-element subsets is C(5,4) = 5, and the number of 5-element subsets is C(5,5) = 1. Therefore, the required number is 5 + 1 = 6. Hence option B is correct. The answer 5 counts only the four-element subsets and misses the whole set.

Any subset that does not contain a can use only the remaining four elements b, c, d, and e. Each of these four elements may either be included or excluded, so the number of such subsets is 2^4 = 16. None of them can equal A because A contains a, so every one of these 16 subsets is proper. Therefore option C is correct.

For integer x, the inequality x² ≤ 4 gives −2 ≤ x ≤ 2. Hence A = {−2, −1, 0, 1, 2}, which has five elements. A set with n elements has 2^n total subsets, so A has 2^5 = 32 subsets. A proper subset is any subset other than the set itself, so the number of proper subsets is 32 − 1 = 31. Therefore option B is correct.

The power set P(A) contains every subset of A as an element. Since A = {1, 2, 3}, its elements include ∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, and A itself. In option A, ∅, {1}, and {2,3} are all subsets of A, so the whole set in option A is a subset of P(A). The other options contain 1 or 4 directly, whereas those are not necessarily elements of P(A) in the required form; 4 is not even an element of A.

The power set P(A) contains every subset of A, including the empty set and A itself. If a finite set has n elements, then it has 2ⁿ subsets because each element has two independent choices: it is either included or excluded. Here n = 3, so |P(A)| = 2³ = 8. Therefore, option C is the correct answer.

By definition, the power set P(A) is the collection whose elements are all subsets of A. It always contains the empty set and A itself, and it also contains every other possible subset. If A has n elements, P(A) has 2ⁿ elements. It is important not to confuse an element of A, such as a, with a subset of A, such as {a}.

The elements of the power set P(A) are subsets of A, not generally the individual elements of A. For A = {a, b}, the power set is {∅, {a}, {b}, {a, b}}. Hence {a} is an element of P(A) because it is a subset of A. The symbols a and b are elements of A, while ab is not a set and is not a subset of A.

The empty set has no elements, but it has exactly one subset: the empty set itself. Therefore, the power set of the empty set contains one member, namely ∅, and is written P(∅) = {∅}. The notation matters: ∅ is the empty set, whereas {∅} is a set containing the empty set as its element. Thus option B is correct.

If a finite set has n elements, its power set contains all possible subsets, and the number of these subsets is 2ⁿ. For n = 4, the number is 2⁴ = 2 × 2 × 2 × 2 = 16. This count includes the empty set and the original set itself, as well as all subsets containing one, two, or three elements. Therefore, option C, 16, is correct.

A = {x} is a singleton set, so it has exactly two subsets: the empty set ∅ and the set {x} itself. The power set is the set whose elements are these subsets, so P(A) = {∅, {x}}. Notice the different levels of braces: x is an element of A, while {x} is an element of P(A). Therefore, option C is correct.

The power set P({1, 2}) contains every subset of {1, 2}: ∅, {1}, {2}, and {1, 2}. A set belongs to the power set only when all of its elements belong to the original set. Since 3 is not an element of {1, 2}, the set {3} is not a subset of {1, 2}, and therefore it is not an element of the power set. Option D is correct.

The empty set ∅ is a subset of every set because it has no elements that could violate the requirement for being a subset. Since a power set consists of all subsets of the original set, ∅ must occur in the power set of every set, including the empty set itself. The universal set or singleton sets are not guaranteed to be subsets of every possible set. Therefore, option B is correct.

For a finite set A with n(A) elements, the number of elements in its power set is n(P(A)) = 2ⁿ. Each element of A has two choices when forming a subset: it is either included or excluded. With n(A) = 2, there are 2² = 4 possible subsets. They are ∅, the two singleton subsets, and A itself. Therefore, option C, 4, is correct.

The set {p, q} is a subset of A = {p, q, r}, because both p and q belong to A. The power set P(A) contains subsets of A as its elements. Consequently, {p, q} is an element of P(A), written {p, q} ∈ P(A). It is not equal to A because r is missing, and it is not an element of A because the elements of A are p, q, and r individually. Therefore, option A is correct.