Equal sets and Subsets

A set records membership, not the number of times an element is written. Thus the repeated 1 in the notation for A is counted only once, so A simplifies to {1,2,3}. The order of elements is also irrelevant, meaning B = {3,2,1} represents the same set {1,2,3}. Since both sets contain exactly the same distinct elements, A = B. Neither set is a proper subset of the other.

The elements a, b, and c of A all occur in B, so A is a subset of B. However, B also contains d, which is not in A. Therefore the two sets are not equal, and A is a proper subset of B, written A ⊂ B. Statement A is false because of the extra element d in B. Statement C reverses the inclusion, and statement D is false because every element of A is indeed in B.

Two sets are equal only when they contain exactly the same elements, regardless of the order in which those elements are written. The left-hand set already contains 5, so the right-hand set must also contain 5; this requires y = 5. The right-hand set contains 3, so the left-hand set must contain 3; this requires x = 3. Therefore option A gives {3,5} = {3,5}. Each other pair produces unequal sets or a repeated element on one side.

To find A, solve x² = 4 over the integers. Taking square roots gives x = 2 or x = -2, and both values belong to Z. Therefore A = {-2,2}. This is exactly the roster description used for B, so A = B. Option B incorrectly omits the negative solution. Option C is false because equal sets are not proper subsets of one another, and option D is false because A has two elements rather than none.

The positive factors of 12 are the positive integers that divide 12 without a remainder. Checking the divisors gives 1, 2, 3, 4, 6, and 12; numbers such as 5 and 8 do not divide 12 exactly. Hence the set-builder description produces A = {1,2,3,4,6,12}, which is precisely the roster form of B. Therefore A = B. It is finite, so option D is also false, and neither proper-subset option applies.

The set A has exactly three elements: 1, 2, and the set {3}. The number 3 itself is not directly listed as an element of A; instead, the one-element set {3} is listed. Therefore, {3} ∈ A is true. Statement A is false because 3 ∉ A. Statement C is false because {1, 2, 3} has 3 as an element, whereas A has {3} as an element. Statement D is also false because {3} ⊂ A would require its element 3 to belong to A. Always distinguish membership (∈) from subset notation (⊂).

A set is a subset of A only when every one of its elements belongs to A. The empty set is a subset of every set, so option A is valid. Both 1 and 3 belong to A, so {1, 3} is a subset. The set {1, 2, 3} is A itself and is therefore also a subset of A. However, 4 is not an element of A, so {2, 4} cannot be a subset. Hence option C is the only correct answer.

An n-element set has 2ⁿ total subsets because every element can independently be included or excluded. For n = 4, the total is 2⁴ = 16. A proper subset is a subset that is not equal to the original set, so the original set must be removed from this count. Therefore, the number of proper subsets is 16 − 1 = 15. Option B counts the original set too, while 8 and 14 do not follow from the proper-subset formula.

The statement A ⊆ B means that every element of A is also an element of B. Consequently, taking the intersection of A and B leaves every element of A and no additional element, so A ∩ B = A. Conversely, if A ∩ B = A, every element of A lies in B, which proves A ⊆ B. Option B instead represents B ⊆ A; option C means A and B are disjoint; and option D represents A ⊆ B only in a different difference condition, not the stated equivalence.

A 2-element subset is formed by choosing 2 different elements from the 4-element set A. Since the order of elements does not matter in a set, combinations are used rather than permutations. The number is C(4, 2) = 4!/(2!2!) = (4 × 3)/2 = 6. The six subsets are {p,q}, {p,r}, {p,s}, {q,r}, {q,s}, and {r,s}. Therefore, option C is correct.

The element 1 is required, so it is fixed as included in every counted subset. The remaining four elements—2, 3, 4, and 5—can each be independently included or excluded. That gives 2 choices for each of four elements, so the number of subsets is 2⁴ = 16. Equivalently, exactly half of the 2⁵ = 32 total subsets contain any specified element. Therefore, option B is correct.

Because both a and b must be present, they are fixed and do not create choices. Only the remaining three elements, c, d, and e, can be independently included or excluded. Therefore, there are 2³ = 8 possible subsets containing both required elements. They are obtained by combining {a,b} with any subset of {c,d,e}. Option D, 32, counts all subsets of A, while 4 incorrectly allows choices for only two remaining elements.

The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, A = {2,3,5,7}, which contains exactly the same elements as B. Sets are equal when they have the same elements, regardless of order. Option B and option C incorrectly claim a proper-subset relationship, while option D incorrectly includes 1; 1 is neither prime nor composite. Hence option A is correct.

To determine B, solve x² = x. Rearranging gives x² − x = 0, or x(x − 1) = 0. Hence x = 0 or x = 1, so B = {0,1}. Since A also contains exactly 0 and 1, the two sets have identical elements and are equal. Thus A = B. The other choices wrongly describe a proper subset, omit 0, or claim that the intersection is empty.

When A ⊆ B, every element of A is also contained in B. The intersection A ∩ B consists of elements common to both sets. Because all elements of A are common to A and B, and B cannot contribute an element that is not in A to the intersection, the result is exactly A. Hence A ∩ B = A, so option A is correct.

The empty set ∅ has no elements, and it is a subset of every set. Since B = {0} is nonempty, A and B are not equal; therefore, A is a proper subset of B. The statement 0 ∈ A is false because A has no elements, and B cannot be a subset of the empty set because B contains 0. Hence option B is correct.

Every element of B, namely 2 and 4, is present in A, so B is a proper subset of A and also a subset of A. Every set is a subset of itself, so A ⊆ A is true. However, A has four elements while B has only two, and 1 and 3 are not in B. Therefore A cannot be a proper subset of B, making option D false.

The positive odd natural numbers less than 8 are 1, 3, 5, and 7. Thus the rule-based description of B gives B = {1, 3, 5, 7}, which is exactly the listed set A. Since two sets are equal when they contain precisely the same elements, A = B. Neither is a proper subset of the other, and their intersection is not empty.

Two sets are equal only when they contain exactly the same elements. Here, \(a\in A\) but \(a\notin B\), while \(d\in B\) but \(d\notin A\), so \(A\ne B\). Since one missing element is enough to disprove a subset relation, \(A\not\subseteq B\); similarly, \(B\not\subseteq A\). Therefore, neither set is a subset of the other. The correct answer is D.

For any set \(A\), every element of \(A\) is automatically an element of \(A\). This satisfies the definition of a subset, so \(A\subseteq A\) is always true. The statement does not depend on whether the set is empty, finite, or infinite. For the empty set, there is no element that violates the condition, so \(\varnothing\subseteq\varnothing\) is also true. Therefore, option A is correct.