Equal sets and Subsets

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 records membership, not the order in which elements are written or how many times an element is repeated. After ignoring the order and removing the repeated 4 from B, we get B = {2, 4, 6}. This is exactly the same set as A. Therefore A and B are equal. Option B is false because equality is stronger than being a proper subset; neither set has an element missing from the other. Option D is unrelated because no universal set has been specified.

The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. The condition that the divisor must be less than 5 leaves only 1, 2, 3, and 4. Hence A = {1, 2, 3, 4}. This is exactly the set given as B, so A = B. The subset options are not the best description because the sets are equal rather than proper subsets, and their intersection is the whole set, not the empty set.

The natural numbers less than 6 are 1, 2, 3, 4, and 5. Selecting only the odd numbers gives B = {1, 3, 5}. Since A is also {1, 3, 5}, both sets have exactly the same elements and are therefore equal. They are not disjoint because they share every element. Neither set has more elements than the other, and B is clearly not empty. Thus option A is correct.

To be a subset of B, every element of A must belong to B. Both a and b are in B, so A ⊆ B. In addition, B contains c, which is not in A. Therefore A is not equal to B and is a proper subset of B, written A ⊂ B. Option B reverses the relationship, option C ignores the extra element c, and option D contradicts the membership check. Hence 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.

Two sets are equal precisely when they contain exactly the same elements. This is equivalent to saying that every element of A belongs to B and every element of B belongs to A; therefore, A ⊆ B and B ⊆ A. Merely having the same number of elements does not ensure equality, because different sets can have equal cardinality. Disjointness means that the sets have no common element, not that they are equal.

The two-way subset criterion states that if A ⊆ B and B ⊆ A, then every element of A is in B and every element of B is in A. Thus the sets have exactly the same elements, so A = B. Option B is not always true because A may equal B; a proper subset must be strictly smaller. Disjoint sets need not be equal, and equal non-empty sets usually have a non-empty intersection.

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.

The elements of \(A\) are 1, 2, and 3. Each of these satisfies the condition \(x\leq 3\), so every element of \(A\) is included in \(B\). Since \(B\) is defined only using elements of \(A\), it cannot contain anything else. Thus \(B=\{1,2,3\}=A\). Neither set is a proper subset of the other, and \(B\) is certainly not empty.

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, \(A=B\).

When A ⊆ B, every element of A is also an element of B. The elements common to A and B are therefore exactly the elements of A, so A ∩ B = A. Option B would be true only if B ⊆ A as well, which is not given. Option C conflicts with A ⊆ B because A − B is empty. Option D is not necessarily true because B may contain elements that are not in A.

The equation A ∩ B = A means that intersecting A with B removes no element from A. Hence every element of A must already be in B. More formally, if x ∈ A, then x belongs to A ∩ B because that intersection equals A; therefore x ∈ B. Thus A ⊆ B. The reverse inclusion is not guaranteed, and the remaining choices require conditions that are not supplied.