Power Set and Subsets

The power set contains all subsets of A. For A = {1,2}, the subsets are the empty set ∅, the singleton sets {1} and {2}, and the set {1,2} itself. Therefore P(A) = {∅,{1},{2},{1,2}}. In a power set, each subset is an element, so writing 1 and 2 without braces is incorrect. Thus option B is correct.

The power set P(A) is the set of all subsets of A, including the empty set, singletons, pairs, and A itself. The set {1, 2} contains only elements of A, so it is a subset of A and therefore an element of P(A). Although 1 is an element of A, it is not itself a subset in this context. Sets containing 4 are not subsets of A, so options C and D are wrong.

A subset with exactly two elements is formed by choosing any 2 of the 4 distinct elements a, b, c, and d. The order of selection does not matter, so we use combinations: C(4, 2) = 4!/(2!2!) = (4 × 3)/(2 × 1) = 6. Therefore, the six subsets are {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, and {c,d}.

Since x is a positive natural number and x² < 10, the possible values are x = 1, 2, and 3, because 1² = 1, 2² = 4, and 3² = 9, while 4² = 16 is not less than 10. Thus A = {1,2,3}. A proper subset must be contained in A but must not be equal to A. Only {1,3} satisfies both conditions. Option A equals A, option B contains 4, and option D contains 0, so the correct answer is C.

The empty set has zero elements, so |A| = 0. A set with n elements has exactly 2^n subsets, because each element may either be included or excluded from a subset. Thus the empty set has 2^0 = 1 subset. That single subset is the empty set itself. It is incorrect to say zero subsets, because every set contains itself as a subset; two subsets would apply to a one-element set.

A subset that does not contain 1 can use only the remaining elements {2, 3, 4}. Each of these three elements has two independent choices: it may either be included or excluded. Therefore, the number of permitted subsets is 2^3 = 8. Equivalently, these are all subsets of A − {1}, including the empty set and {2, 3, 4}.

A set containing n elements has 2^n subsets, because each element can either be included in or excluded from a subset. Here A has two elements, a and b. Its power set is P(A) = {∅, {a}, {b}, {a, b}}, which contains four elements. Therefore, option C is correct.

The even elements in A are 2, 4, and 6. A subset formed using only even numbers may contain any selection of these three elements, including the empty subset. Each element has two independent choices—selected or not selected—so the number of subsets is 2³ = 8. Therefore, option C is correct.

An element of \(\mathcal{P}(A)\) must be a subset of A. The set \(\{1,3\}\) contains only elements from A, so \(\{1,3\}\subseteq A\) and it belongs to \(\mathcal{P}(A)\). The object 1 is an element of A, not a subset by itself in this context; 4 is not in A, and \(\{\{4\}\}\) contains an element not belonging to A.

The odd elements of A are 1, 3 and 5, so the relevant set has three elements. Each of these three elements has two independent choices in a subset: it may be included or excluded. Therefore the number of subsets is 2³ = 8. This count includes the empty subset, the three one-element subsets, the three two-element subsets, and the full set of odd elements. Hence option C is correct.

A three-element subset is formed by choosing 3 different elements from the 5 elements of A, and the order of selection does not matter. Therefore, the required number is the combination 5C3 = 5!/(3!2!) = (5×4)/(2×1) = 10. Choosing the same three elements in another order does not create a new subset, which is why combinations rather than permutations are used.

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 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.

The complement A′ contains all elements of the universal set U that are not in A. Since A is the empty set, it contains no elements, so none of the elements of U are removed. Therefore A′ = U − ∅ = U = {1, 2, 3}. Option A would be the set A itself, while options B and C omit elements that must remain in the complement. Thus option D is correct.

The set {m, n} has two elements. Every element may either be selected or not selected when forming a subset, so the number of subsets is 2² = 4. They are the empty set ∅, the singleton sets {m} and {n}, and the complete set {m, n}. Therefore, the power set is P({m, n}) = {∅, {m}, {n}, {m, n}}, making option A correct.

A one-element subset, also called a singleton, is formed by choosing exactly one element from A. Since A contains five distinct elements, its singleton subsets are {1}, {2}, {3}, {4}, and {5}. Thus P(A) contains exactly five one-element subsets. The number 10 would count two-element subsets, because C(5, 2) = 10, not singleton subsets.