Equal sets and Subsets

The elements 2 and 5 must be included in every required subset, so their choices are fixed. The remaining elements 1, 3, and 4 are unrestricted; each can either be included or excluded independently. Therefore, the number of subsets is 2 × 2 × 2 = 2³ = 8. This is a standard subset-counting rule: if k elements are free, there are 2ᵏ possible choices.

The element 1 is compulsory, while 4 is forbidden. Thus, only 2 and 3 remain available for independent selection. Each of these two elements can be either included or excluded, giving 2² = 4 possible subsets. They are {1}, {1, 2}, {1, 3}, and {1, 2, 3}. Hence option B is correct. Fixed inclusion and exclusion conditions should be handled before counting the free elements.

To determine A, solve x² = 1 over the integers. Factoring gives (x − 1)(x + 1) = 0, so x = 1 or x = −1. Therefore, A = {-1, 1}. Since B has exactly the same elements, A and B are equal sets, and A = B is the only correct statement. Remember that equal sets contain precisely the same elements, regardless of their order.

Every element of A, namely 1, 2, and 3, is present in B, and B has the additional element 4. Therefore A is a proper subset of B. Similarly, both elements of C, 2 and 3, are present in A, while A contains the additional element 1, so C is a proper subset of A. Thus both statements in option A are true.

The positive multiples of 5 that are less than 20 are 5, 10, and 15. The number 20 is excluded because the condition says less than 20, not less than or equal to 20. Hence A = {5, 10, 15}. This is exactly the set B, so A = B. Equal sets must have the same elements, even if they are described in different forms.

The set A has three elements: 1, 2, and the set {1}. Therefore, {1} is directly listed as an element of A, so {1} ∈ A. Also, the only element of {1} is 1, and 1 belongs to A; hence every element of {1} belongs to A, so {1} ⊂ A as well. Both statements are true.

The outer set A has exactly two elements: the set {1,2} and the number 3. A subset must contain only elements that are themselves elements of A. The set {{1,2}} contains the single element {1,2}, which is in A, so it is a subset. The other options contain 1 or 2 individually, neither of which is an element of A.

The prime natural numbers less than 10 are 2, 3, 5, and 7, so A = {2,3,5,7}. Every element of B is in A, but B does not contain 7. Thus B is a proper subset of A and B ≠ A. The sets are not equal, and their intersection is not empty because they share 2, 3, and 5.

Because A is a subset of B, every element of A is already contained in B. If A were a proper subset, B would contain at least one additional element, and therefore n(B) would be greater than n(A). The given equality n(A) = n(B), together with finiteness, rules out any additional element. Hence the two sets contain exactly the same elements, so A = B. Therefore option A is correct.

A set with n elements has 2^n total subsets. For four elements, this gives 2^4 = 16 subsets. A proper subset is any subset that is not equal to the original set, so the original set itself must be excluded from the total. Therefore the number of proper subsets is 16 − 1 = 15. The empty set is included because it is a proper subset of every non-empty set. Hence option A is correct.

The set A has three elements: the number 1, the number 2, and the set {1, 2}. Since {1, 2} appears as one complete element inside A, the membership statement {1, 2} ∈ A is true. It is not equal to A because A has two additional elements, 1 and 2. The number 3 is absent, so neither 3 ∈ A nor {1, 2, 3} ⊆ A can be true.

Every set X must contain all elements of A, namely 1, 2, and 3. The only elements that can be chosen freely are 4 and 5, because they belong to B but not to A. Each of these two elements has two independent choices: include it in X or leave it out. Therefore the number of possible sets is 2 × 2 = 2² = 4, including A itself and B itself.

The elements 1 and 2 are required, while 5 is forbidden. Therefore the only elements whose membership is undecided are 3 and 4. Each of these two elements can independently be included or excluded, producing 2² = 4 possible subsets: {1,2}, {1,2,3}, {1,2,4}, and {1,2,3,4}. Hence option B is correct.

Solve the defining equation for A by factoring: x² − 5x + 6 = (x − 2)(x − 3) = 0. Thus x = 2 or x = 3, so A = {2, 3}. Since B is also {2, 3}, the two sets contain exactly the same elements and therefore A = B. The proper-subset options are false because neither set is strictly smaller, and their intersection is {2,3}, not empty.

Solving \(x^2=9\) over the integers gives two solutions: \(x=3\) and \(x=-3\). Therefore, \(A=\{-3,3\}\). This is exactly the same collection of elements as \(B=\{-3,3\}\), so \(A=B\). Option B omits the negative solution, option C incorrectly claims a proper subset, and option D ignores the two integer solutions. Always consider both square roots and the stated domain.

Subset inclusion is transitive. Let \(x\) be any element of \(A\). From \(A\subseteq B\), we get \(x\in B\); from \(B\subseteq C\), we then get \(x\in C\). Hence every element of \(A\) belongs to \(C\), proving \(A\subseteq C\). The reverse inclusion, equality, or emptiness of B is not forced by the given information.

Because \(A\subset B\), every element of A belongs to B, and because \(B\subset C\), every element of B belongs to C. Therefore \(A\subseteq C\). Moreover, A cannot equal C: if A equalled C, then the chain \(A\subset B\subset C=A\) would force B to be simultaneously larger than A and contained in A, which is impossible. Thus \(A\subset C\) is proper.

For finite sets, a proper subset must contain strictly fewer elements than the set containing it. Here A is contained in B, but both have the same finite cardinality. Therefore B cannot have any additional element outside A, so every element of B is already in A. Consequently, the two sets are equal: \(A=B\). This conclusion depends on finiteness.

Every set is an element of its own power set because a set is always a subset of itself. Thus \(A\in\mathcal{P}(A)\). If \(\mathcal{P}(A)=\mathcal{P}(B)\), then A also belongs to \(\mathcal{P}(B)\), so \(A\subseteq B\). Similarly, B belongs to \(\mathcal{P}(A)\), giving \(B\subseteq A\). By mutual inclusion, \(A=B\).

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 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.

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.

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 \(\{3,4\}\).