Power Set and Subsets

A two-element subset is formed by choosing exactly two different elements from the three elements 1, 2, and 3. The possible subsets are {1, 2}, {1, 3}, and {2, 3}. These are different subsets, while the order of elements does not matter. Therefore, A has exactly 3 two-element subsets. This also agrees with the combination formula C(3, 2) = 3.

A one-element subset, also called a singleton subset, contains exactly one member of the original set. For A = {p, q, r}, the singleton subsets are {p}, {q}, and {r}. Each of the three elements produces one different singleton subset, so the total number is 3. In general, an n-element set has n singleton subsets.

The open interval (0, 1) contains exactly the real numbers strictly greater than 0 and strictly less than 1. Both 1/4 and 3/4 satisfy these conditions, so every element of {1/4, 3/4} belongs to (0, 1). The other choices contain 0, 1, or a negative number and therefore are not subsets.

A proper subset contains only elements of the original set and is not equal to the original set. The set {1, 2, 3} contains elements that all belong to A, but it omits 4, so it is smaller than A. Option A is equal to A, while options C and D contain elements not in A. Therefore B is correct.

Taking N to mean the positive natural numbers, A = {1, 2, 3, 4}, so A has four elements. A set with n elements has exactly 2^n subsets, because each element has two independent choices: it may be included or excluded. Hence the number of subsets is 2^4 = 16. This count includes both the empty set and A itself.

The set A contains all real numbers greater than 1 and less than or equal to 6. For a proposed set to be a subset of A, every one of its elements must satisfy both conditions. In {2, 4, 6}, all elements are greater than 1 and 6 is allowed because the upper inequality is inclusive. The other choices contain 1, 0, or 7, which are outside A.

The power set of A is the set of all subsets of A. For A = {a,b}, its power set is {∅, {a}, {b}, {a,b}}. Every member of a power set must contain only elements from the original set. The set {a,c} contains c, but c is not an element of A, so {a,c} is not a subset of A and cannot belong to its power set. Hence option D is correct.

For a set with n elements, each element has two independent choices when forming a subset: it may be included or excluded. Thus the total number of subsets is 2ⁿ. With n = 4, the number is 2⁴ = 2 × 2 × 2 × 2 = 16. This count includes both the empty set, in which no element is selected, and the original set, in which all four elements are selected. Therefore C is correct.

A set with three elements has 2³ = 8 total subsets. A proper subset is a subset that is not equal to the original set itself. The only subset that must be excluded from the total is A = {a,b,c}; the empty set and all other smaller subsets are proper subsets. Therefore the number of proper subsets is 8 − 1 = 7, so option B is correct.

The power set P(A) is the set of all subsets of A. For A = {1, 2}, P(A) = {∅, {1}, {2}, {1, 2}}. Therefore, {1} is an element of P(A) because it is a subset of A. The symbols 1 and {1} must not be confused: 1 is an element of A, whereas {1} is a set and an element of P(A).

A set with n elements has exactly 2^n subsets, because each element can either be included or excluded independently. Since n(A) = 5, the number of elements in the power set is n(P(A)) = 2^5 = 32. Thus, P(A) contains 32 subsets, including the empty set and the original set A itself.

A set containing n distinct elements has 2ⁿ subsets, because each element has two independent choices: it may either be included in a subset or excluded from it. Here n = 2, so the number of subsets is 2² = 4. They are ∅, {1}, {2}, and {1, 2}. Both the empty set and the original set itself must be counted.

A set with two elements has 2² = 4 subsets. For {1, 2}, these are the empty set ∅, the singleton {1}, the singleton {2}, and the complete set {1, 2}. Option A lists all four and is therefore correct. Option B omits the empty and complete sets, option C lists only two subsets, and option D includes {3}, which contains an element not in the original set.

To form a subset that must contain 1, fix 1 as included. Each of the remaining three elements, 2, 3, and 4, has two independent choices: it may be included or omitted. Hence the number of subsets is 2 × 2 × 2 = 2^3 = 8. The value 16 counts every subset of A, including those without 1.

A subset that does not contain a must be formed only from the remaining elements {b, c, d}. A three-element set has 2^3 subsets because each element can independently be selected or not selected. Therefore the required number is 2^3 = 8. The number 16 represents all subsets of the original four-element set.

A two-element subset is formed by choosing any two different elements from the five elements of A, with order ignored. Thus the number is C(5, 2) = 5!/(2!3!) = (5 × 4)/2 = 10. For example, {1, 2} and {2, 1} represent the same subset, so they are counted only once. Hence option B is correct.

Using the usual school convention N = {1, 2, 3, ...}, the condition x ≤ 3 gives A = {1, 2, 3}. A proper subset must contain only elements of A and must not be equal to A. The set {1, 2} satisfies both conditions, so it is a proper subset. Option B equals A, option C contains 4, and option D contains 0, which is not in A under this convention.

The power set P(A) is the set of all subsets of A. Since A = {0, 1}, its power set is P(A) = {∅, {0}, {1}, {0, 1}}. Therefore {0} is an element of P(A). The number 2 is not an element of A, so {2} and {0, 2} are not subsets of A; the standalone number 2 is not a subset either. Hence option A is correct.

If a finite set has n elements, its power set has 2^n elements because each element has two independent choices: it may either be included in a subset or not included. Here A has n = 3 elements, so |P(A)| = 2^3 = 8. These eight subsets are the empty set, three one-element subsets, three two-element subsets, and A itself. Therefore option C is correct.

The elements a and b are fixed as included in every required subset. The remaining elements c, d, and e are unrestricted: each one may either be included or omitted independently. Therefore, each of the three remaining elements gives two choices, and the total number of subsets is 2 × 2 × 2 = 2³ = 8. Hence option C is correct.