Equal sets and Subsets

The symmetric difference A △ B contains elements that belong to exactly one of the two sets: elements of A not in B together with elements of B not in A. If this set is empty, neither set has an element absent from the other. Therefore A ⊆ B and B ⊆ A simultaneously, which proves A = B. A proper-subset relation would require one set to have an extra element, and disjointness is not implied.

The positive divisors of 18 are 1, 2, 3, 6, 9 and 18. Among them, the prime divisors are only 2 and 3, because both are prime and divide 18. The number 9 is a divisor but is not prime, so it cannot be included in A. Hence A = {2, 3} = B. The sets are equal, not in a proper-subset relationship, and A is certainly not empty.

The positive divisors of 24 are {1, 2, 3, 4, 6, 8, 12, 24}. Selecting the even divisors gives A = {2, 4, 6, 8, 12, 24}. Every element of option A belongs to this set, so option A is a subset. Option B contains 5, which is not a divisor of 24; option C contains 1, which is not even; and option D contains 10, which is not a divisor of 24.

Consider one particular element of A. Because X ⊆ Y, that element has exactly three permissible statuses: it may belong to neither X nor Y, it may belong to Y but not X, or it may belong to both X and Y. The status “in X but not in Y” is forbidden. Since there are 3 independent elements and 3 choices for each, the total number of ordered pairs is 3 × 3 × 3 = 3³ = 27. Thus option C is correct.

The condition A ⊆ X ⊆ B means that every element of A must be in X, while the only element that can be added beyond A is 9, because B − A = {9}. Therefore there are exactly two possibilities: X = A, when 9 is excluded, or X = A ∪ {9} = B, when 9 is included. Equivalently, the number is 2 raised to the number of optional elements: 2¹ = 2. Hence option B is correct.

For {1, k} to be a subset of {1, 2, 3}, every element of the left-hand set must belong to the right-hand set. The element 1 already satisfies this condition, so k may be any of 1, 2, or 3. When k = 1, the set {1, k} is simply {1}, because repeated elements are not counted in a set; it is still a subset. Therefore the complete set of possible values is {1, 2, 3}, making option B correct.

The subset must contain 0, so one of its two positions is already fixed. The second element can be chosen from the remaining elements of A: 1, 2, or 3. These choices produce the three distinct subsets {0, 1}, {0, 2}, and {0, 3}. No other two-element subset containing 0 is possible. Equivalently, we choose one element from the three available elements, giving C(3, 1) = 3. Hence option B is correct.

Although the symbols are related, ∅ and {∅} are different objects. The first is the empty set, containing no elements; the second is a singleton set whose only element is ∅. Thus A has exactly two distinct elements: ∅ and {∅}. A set with two elements has 2² = 4 subsets: ∅, {∅}, {{∅}}, and {∅, {∅}}. Therefore option C is correct.

A set is a subset of \(A\) only when each of its elements is also an element of \(A\). The elements of \(A\) are \(\emptyset\), \(\{\emptyset\}\), and \(1\). Options A, B, and C contain only elements from this list, so they are subsets of \(A\). Option D contains the element \(2\), but \(2\notin A\); therefore \(\{1,2\}\) is not a subset. Notice especially that \(\emptyset\) and \(\{\emptyset\}\) are different objects.

The assertion is true because every subset of \(A\) is automatically a subset of \(B\) when \(A\subset B\). Hence every element of \(\mathcal{P}(A)\) belongs to \(\mathcal{P}(B)\). The inclusion is proper, not merely equal, because \(B\subseteq B\), so \(B\in\mathcal{P}(B)\); however, \(B\notin\mathcal{P}(A)\), since \(B\) is not a subset of the smaller set \(A\). Thus both the assertion and reason are correct, and the reason explains the assertion.

To determine set \(A\), factor the quadratic equation: \(x^2-7x+12=(x-3)(x-4)=0\). Therefore, the possible values of \(x\) are \(3\) and \(4\), so \(A=\{3,4\}\). This is exactly the set given as \(B\). Since sets are equal when they contain precisely the same elements, \(A=B\). Neither set is a proper subset of the other, and their intersection is not empty; in fact, their intersection is the whole set \(\{3,4\}\).

In set notation, repeating an element does not create a new element and multiplicity is ignored. Therefore \(A=\{1,2,3\}\) after removing repeated entries. The elements of \(B\) are also \(\{1,2,3\}\), merely written in a different order. Since two sets are equal when they contain exactly the same elements, \(A=B\). Neither proper-subset option is correct because neither set has an additional element, and option D contradicts their equality.

The condition A ⊂ X ⊆ B means that X must contain p and q, may contain r and s, and must be different from A. The four possible supersets formed from r and s are {p,q}, {p,q,r}, {p,q,s}, and {p,q,r,s}. Since A itself is not allowed by the proper-subset sign, remove {p,q}. Therefore, 4 − 1 = 3 sets satisfy the condition.

The empty set ∅ is explicitly listed as an element of A, so ∅ ∈ A. Also, the empty set is a subset of every set because it has no element that could violate the subset condition; hence ∅ ⊆ A. Therefore, ∅ satisfies both requirements. Although 2 is an element of A, it is not a set and therefore is not a subset of A in this context.

A subset contains only elements that belong to A. Since 3 ∈ A, the set {3} is a subset of A. However, the listed elements of A are ∅, {∅}, and 3; the set {3} is not among them, so {3} ∉ A. The empty set, {∅}, and 3 are all elements of A, so they do not satisfy the phrase “not an element of A.”

The union \(A\cup B\) contains every element of \(B\). If this union is equal to \(A\), then every element of \(B\) must also be an element of \(A\). This is precisely the definition of \(B\subseteq A\). The reverse inclusion, \(A\subseteq B\), is not necessary; for example, \(A=\{1,2\}\) and \(B=\{1\}\) satisfy the condition. The intersection need not be empty, and a complement relation cannot be inferred.

The equality \(A\cap B=B\) means that intersecting \(B\) with \(A\) does not remove any element from \(B\). Therefore, every element of \(B\) must already belong to \(A\), which gives \(B\subseteq A\). The reverse relation \(A\subseteq B\) does not necessarily hold, because \(A\) may contain additional elements. The complement statement is unrelated, and the intersection cannot be empty unless \(B\) is empty.

The difference \(A\setminus B\) contains the elements of \(A\) that are not in \(B\). If removing the elements of \(B\) from \(A\) leaves all of \(A\) unchanged, then no element of \(A\) can belong to \(B\). Hence the two sets have no common element, so \(A\cap B=\emptyset\). This does not imply that either set is empty or that they are equal; it only establishes that they are disjoint.

The set \(A\) is itself an element of its power set \(\mathcal{P}(A)\), because every set is a subset of itself. Given \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\), it follows that \(A\in\mathcal{P}(B)\). Membership in \(\mathcal{P}(B)\) means being a subset of \(B\), so \(A\subseteq B\). The other options do not follow from the given power-set inclusion and can fail for ordinary nested sets.

The intersection A ∩ B consists of all elements that occur in both A and B. The elements 2 and 3 are common to the two sets, while 1 occurs only in A and 4 occurs only in B. Therefore A ∩ B = {2, 3}. Hence option A is correct. The intersection is not the union, so elements occurring in only one set must not be included.