Equal sets and Subsets

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.

Taking N = {1, 2, 3, …}, the condition x ≤ 5 gives A = {1, 2, 3, 4, 5}, whereas x < 5 gives B = {1, 2, 3, 4}. Every element of B belongs to A, so B ⊆ A. Since 5 is in A but not in B, equality fails and B is a proper subset of A. Thus option B is correct; the sets are not disjoint and the inclusion direction is not reversed.

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.

The key concept is the difference between an element and a set containing that element. Here A has exactly two elements: ∅ and the number 1. The set {∅} contains only ∅, which belongs to A, so {∅} ⊆ A. In contrast, 0 is not in A, 2 is not in A, and {{1}} contains the set {1}, not the number 1 itself. Therefore only option A is a subset of A.

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.

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.