Equal sets and Subsets

The condition 1 ≤ n ≤ 3 with n a natural number allows n = 1, 2, and 3 only. Substituting these values into x = 2n gives x = 2, 4, and 6. Therefore B = {2, 4, 6}. Since A contains exactly the same elements as B, the two sets are equal, so option A is correct. The order in which elements are listed does not matter.

Because x must be an integer, we list the integers beginning at −2 and less than 2: −2, −1, 0, and 1. The symbol ≤ includes −2, whereas the symbol < excludes 2. Hence A = {−2, −1, 0, 1}, which is exactly option A. The order of elements does not affect equality of sets.

In ordinary set theory, repeated listings of an element do not create new elements. Thus B simplifies to {1, 2, 3}, which is exactly A. The sets have the same distinct elements, so A = B. Repetition would matter in a list or multiset, but not in a standard set.

First list the positive divisors of 12: 1, 2, 3, 4, 6, and 12. Selecting only the even divisors removes 1 and 3, leaving A = {2, 4, 6, 12}. This is exactly the set specified as B, so the two sets have identical membership and A = B. Option A is correct. Options B and C wrongly claim proper inclusion, while D includes the odd divisors 1 and 3.

By definition, B is a subset of A when every element belonging to B also belongs to A. Here the elements of B are 2 and 4, and both occur in A. Having fewer elements, being unequal, or containing only even numbers is not by itself the definition of a subset. Therefore the precise reason is that every element of B is in A.

The set {1, 2} contains only elements that are also in A, and it is not equal to A because A additionally contains 3. Therefore, {1, 2} is a proper subset of A, written {1, 2} ⊂ A. The notation {1, 2} ∈ A would incorrectly treat a set as one of A’s listed elements.

Solve the defining equation x² = 4 by taking both square roots: x = 2 or x = −2. Thus the solution set is A = {−2, 2}. Since B is given as the same two-element set, A and B have exactly identical members, so A = B. Option A is correct. Option B omits the negative solution, C incorrectly calls equal sets a proper inclusion, and D contradicts their common elements.

The governing concept is the comparison of sets by listing their elements. The positive natural-number divisors of 15 are 1, 3, 5, and 15, so A = {1, 3, 5, 15}. Every element of B = {1, 3, 5} belongs to A, but A also contains 15. Therefore B is a proper subset of A, written B ⊂ A, and the sets are not equal. Thus option B is correct; option A ignores 15, while option D is false because the sets overlap.

The governing idea is translating set-builder notation into roster form and then comparing the resulting elements. Whole numbers are W = {0, 1, 2, 3, ...}. Since n < 4, the possible values are n = 0, 1, 2, 3. Substitution in x = 2n gives x = 0, 2, 4, 6, so B = {0, 2, 4, 6}. This is exactly A; hence A = B. Option C omits 0, and option D contradicts n = 0.

In a set, the order in which elements are written has no significance, and repeated listing does not create new elements. A contains 1, 2, and 3, and B contains exactly the same three elements. Therefore A = B because equality of sets depends on identical membership, not on order or appearance.

The governing concept is equality of sets: two sets are equal when they contain exactly the same elements, regardless of how they are described. The odd natural numbers less than 9 are 1, 3, 5, and 7. The endpoint 9 is excluded because the condition is strictly less than 9. Thus A = {1, 3, 5, 7}, which is precisely B. Neither set is a proper subset of the other, so option A is correct.

A proper subset must be contained in the original set but must not be equal to the original set. The sets {2,5}, {7}, and ∅ are all contained in A and are smaller than A, so they are proper subsets. However, {2,3,5,7} is exactly A itself. It is a subset, but not a proper subset. Therefore option C 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.

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 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.

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.

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.

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.

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.