Power Set and Subsets

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.

By definition, \(\mathcal{P}(A)\) contains exactly the subsets of \(A\). Thus \(\{m,n\}\in\mathcal{P}(A)\) means \(\{m,n\}\subseteq A\). Every element of this subset must belong to \(A\), so both \(m\in A\) and \(n\in A\). The statement does not imply that \(A\) has no other elements, so option C is not necessary. Option A is correct.

The set A has exactly two elements: ∅ and {∅}. Its power set therefore contains every subset of A: ∅, {∅}, {{∅}}, and {∅, {∅}}. Option A is the empty subset, option B contains the element ∅, and option C contains the element {∅}. Thus all three are subsets of A and hence all three are elements of P(A). Therefore, the correct answer is D, All of these.

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.

A set with n elements has 2^n total subsets because each element can either be included or excluded. For n = 4, the total number is 2^4 = 16. A non-empty proper subset excludes both the empty set and the complete set A. Hence the required number is 16 − 2 = 14, so option A is correct.

The power set contains all subsets of A. To form a subset with exactly 2 elements, choose any 2 of the 5 elements, without considering order. The number of choices is the combination 5C2 = 5!/(2!3!) = (5 × 4)/2 = 10. Therefore P(A) contains exactly 10 two-element subsets, making option B correct.

Let one particular element be fixed as included. The remaining 6 elements may each be either included or excluded independently. Thus they produce 2^6 possible choices. Every such choice, together with the fixed element, gives one subset containing it. Therefore the required number is 2^6 = 64, so option B is correct.

The two specified elements are forbidden, so they cannot be selected in any valid subset. The remaining 5 − 2 = 3 elements are unrestricted; each may independently be included or excluded. Therefore the number of valid subsets is 2^3 = 8. This includes the empty subset and all allowed combinations of the remaining three elements, so option B is correct.

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.

Equal sets have exactly the same elements and therefore exactly the same subsets. Since A = U, every subset of A is a subset of U and vice versa. Consequently their power sets are equal: P(A) = P(U). The other statements are false: a power set is not empty, P(U) is not generally a subset of A, and identical power sets have a non-empty intersection equal to themselves.

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.

We must choose exactly 4 elements from the 6 elements of A. The order of selection does not matter, so combinations are used. The required number is C(6,4) = 6!/(4!2!) = (6 × 5)/(2 × 1) = 15. Every four-element subset of A belongs to P(A), so there are 15 such subsets and 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.

For any finite set X, the power set has 2ⁿ(X) elements. Therefore n(P(A)) = 2⁶ = 64 and n(P(B)) = 2⁴ = 16. Their ratio is 64:16, which simplifies by dividing both terms by 16 to 4:1. The ratios 3:2 and 6:4 compare the original sets, not their power sets, so option B is correct.

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.

For a finite set with n elements, every element has two choices in a subset: it is either included or excluded. Therefore, the total number of subsets is 2^n. Here 2^n = 2048, and 2048 = 2^11. Thus n = 11, so option C is correct. The neighboring powers confirm this: 2^10 = 1024 and 2^12 = 4096.

The element 2 is required, so it is fixed as included. The element 5 is forbidden, so it is fixed as excluded. The remaining elements 1, 3, and 4 are unrestricted, and each has two choices: included or excluded. Consequently, the number of valid subsets is 2^3 = 8. Therefore, option B is correct; the restrictions leave three independent elements.