Equal sets and Subsets

The statement B ⊆ A means that every element of B must also be an element of A. The elements of B are 2 and 4, and both occur in A = {1, 2, 3, 4}. It is not necessary for every element of A to belong to B; 1 and 3 show that B is a proper subset of A. Thus option A gives the correct definition.

The symbol ∈ relates an object to a set, while ⊆ relates one set to another. Since 1 is listed in A, 1 ∈ A is true. The singleton set {1} contains only an element of A, so {1} ⊆ A is also true. Option B reverses these meanings, option C ignores 2 and 3, and option D fails because 4 is not in A.

The set A contains the two numbers 1 and 2 as its elements. Therefore 1 ∈ A is true, and both {1} ⊆ A and {1, 2} ⊆ A are true because every member of each set belongs to A. However, {1, 2} is the whole set A, not an element listed inside A. Hence {1, 2} ∈ A is false.

The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Therefore A = {1, 2, 3, 4, 6, 12}, which is exactly the set B. Since both sets contain the same elements, they are equal. Neither proper-subset statement can be true, and A is certainly not empty because 1, among other numbers, is a positive divisor of 12.

The positive divisors of 8 are 1, 2, 4, and 8, so B = {1, 2, 4, 8}. Every element of A = {2, 4, 8} is in B, but B contains the additional element 1, which is not in A. Therefore A is a proper subset of B, written A ⊂ B. The sets are not equal, and their intersection is A rather than the empty set.

The prime natural numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19, which are exactly the elements listed in B. Therefore A and B contain the same elements, so A = B. The number 1 is neither prime nor composite, and 20 is not less than 20. Also, 2 is the smallest and the only even prime number.

The odd natural numbers less than 6 are 1, 3, and 5, so the set-builder description gives A = {1, 3, 5}. This is exactly the set listed as B. Since two sets are equal when they have precisely the same elements, A = B. Neither A nor B is a proper subset of the other because a proper subset must be strictly smaller and cannot be equal to the original set. Thus option A is correct.

The even natural numbers less than 10 are 2, 4, 6, and 8. Therefore B = {2,4,6,8}, which is exactly A, so A = B. Options B and C incorrectly claim a proper-subset relationship. Option D is false because 10 is not an element of A, and the condition is x < 10 rather than x ≤ 10.

Two sets are equal if and only if they contain exactly the same elements. The names used for the sets do not matter, and the order in which elements are written is irrelevant because sets are unordered. Having only one common element is insufficient; every element of each set must belong to the other set.

A proper subset is a subset that is strictly smaller than the containing set. Thus, A ⊂ B requires two conditions: every element of A must also be an element of B, and A must not be equal to B. For example, {1, 2} is a proper subset of {1, 2, 3}. Therefore, option A gives the complete and correct definition.

For A to be a subset of B, every element of A must occur in B. This fails because 1 and 2 belong to A but not to B. Similarly, B is not a subset of A because 5 belongs to B but not to A. The sets have common elements, 3 and 4, but sharing some elements is not sufficient for a subset relation. Hence option D is correct.

The set-builder description says that x is an integer from −1 through 2, including both endpoints. Listing these integers gives A = {−1, 0, 1, 2}. This is exactly the roster form used to define B. Since both sets contain precisely the same elements, they are equal; neither is a proper subset of the other. Thus option A is correct.

Although the word LEVEL has five positions, its distinct letters are only L, E, and V because L and E are repeated. Sets do not record repetition, so A = {L, E, V}. This is exactly the set B, meaning A and B have the same elements and are equal. Therefore, option A is correct, while option D incorrectly counts repeated occurrences.

The prime factorization of 18 is 18 = 2 × 3 × 3 = 2 × 3². When prime factors are written as a set, a repeated factor is listed only once. Thus A = {2, 3}, which is exactly the set B. The sets are equal, not proper subsets of one another, so option A is correct.

The positive divisors of 16 are 1, 2, 4, 8, and 16. The condition requires x to be even, so the odd divisor 1 must be removed. The remaining elements are 2, 4, 8, and 16; therefore A = {2, 4, 8, 16}, which is option A. Option B includes 1, option C includes 6, which is not a divisor of 16, and option D omits 2.

The natural-number divisors of 15 are A = {1, 3, 5, 15}. Every element of C = {1, 3, 5} is present in A, so C is a subset of A. However, C does not contain 15, which belongs to A, so the two sets are not equal. Hence C is a proper subset of A, written C ⊂ A and C ≠ A. Therefore option A is correct.

When A is a subset of B, every element of A is contained in B. For finite sets, if A is a proper subset of B, then A must have fewer elements than B. Here both sets have cardinality 5, so A cannot be a proper subset of B. Since A ⊆ B and their cardinalities are equal, they must contain exactly the same elements; hence A = B. Option A is correct.

The symbol A ⊂ B denotes a proper subset, so A is contained in B but is not equal to B. For finite sets, a proper subset must have strictly fewer elements than the original set. Since B has 7 elements, n(A) may be 0, 4, or 6, depending on the subset, but it cannot be 7. If n(A) were 7, A and B would have equal finite cardinality and would be equal, contradicting A ⊂ B. Thus option D is impossible.

The elements of B are 4 and 8, and both of them occur in A. Therefore B is a subset of A. Since A also contains 2, 6, and 10, which are not in B, the two sets are not equal; B is specifically a proper subset of A. Statement A is therefore correct. Statement B reverses the inclusion, statement C ignores the extra elements of A, and statement D is false because 10 is not in B.

The condition 3 ≤ x < 7 gives A = {3, 4, 5, 6}. A proper subset must contain only elements of A and must omit at least one element of A. Option A, {3, 5, 6}, satisfies both conditions: all its elements belong to A, but it omits 4, so it is smaller than A. Option B equals A, while options C and D contain elements outside A. Hence option A is correct.