Equal sets and Subsets

The statement A ⊆ B requires every element of A to be an element of B. Although 1, 3, and 5 occur in both sets, the element 7 belongs to A and does not belong to B. A single element of A missing from B is enough to make the subset statement false. Therefore the precise reason is that 7 is in A but not in B, which makes option A correct.

Every set is a subset of itself because each element of the set is, by definition, contained in that same set. Here the elements 5, 6, and 7 all belong to A, so A ⊆ A is true. The set is not empty, so A = ∅ is false. Also, A ∈ A does not follow from A ⊆ A; being a subset and being an element are different relationships. Therefore option A is correct.

In the interval [3, 9], square brackets are used at both ends. A square bracket means that the endpoint is included. Hence x may be equal to 3 and may also be equal to 9, while every value between them is allowed. The equivalent inequality is 3 ≤ x ≤ 9, so option B is correct.

In set notation, repeated elements are written only once for purposes of membership. Thus, {a, a, b} represents exactly the same set as {a, b}; both contain the distinct elements a and b. They are therefore equal sets. Repetition can appear in an expression, but it does not create an additional element or change the set.

Set A is given in roster form as {2, 4, 6}. Set B is given in set-builder form: it contains every x that is one of 2, 4, or 6. Therefore, B also contains exactly 2, 4, and 6, so B = {2, 4, 6}. Since both sets have the same elements, A = B. Neither is a proper subset of the other.

A is a subset of B because every element of A—1, 3, and 5—is also present in B. The sets are not equal because B contains one additional element, 7, which is absent from A. Therefore A is a proper subset of B, written as A ⊂ B. A proper subset must be contained in the larger set and must not be equal to it.

The set A = {0} contains one element, namely 0, so A is not empty. The empty set ∅ is a subset of every set, including {0}; therefore ∅ ⊆ A is true. The expression 0 ⊆ A is not the correct subset statement because 0 is an element, not a set in this context. Also, {1} is not a subset because 1 does not belong to A.

A set X is a subset of A only when every element of X is also an element of A. The sets {p, q} and {r} satisfy this condition because their elements occur in A. The empty set is a subset of every set by definition. However, {p, s} contains s, and s is not an element of A, so {p, s} is not a subset of A.

For a set to be a subset of A, every one of its elements must belong to A. Both 1 and 2 belong to A, so option A is true, and 3 belongs to A, so option B is true. The empty set is a subset of every set, making option D true. Since 4 is not in A = {1, 2, 3}, the statement {4} ⊆ A is false.

The positive even numbers less than 10 are 2, 4, 6, and 8. Therefore the set-builder description gives A = {2, 4, 6, 8}. This is exactly the roster form used to define B. Since two 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.

Taking natural numbers as 1, 2, 3, …, the condition x < 5 gives A = {1, 2, 3, 4}. Every element of A is in B, so A ⊆ B. The inclusion is proper because B also contains 5, while 5 is not in A. Hence A ⊂ B. The strict inequality is important: x < 5 excludes 5.

Two sets are equal precisely when each set is a subset of the other. The condition A ⊆ B says that every element of A is in B, while B ⊆ A says that every element of B is in A. Together, these statements show that the sets contain exactly the same elements. Therefore A = B; neither set is required to be empty.

Every element of A—2, 3, and 5—is also found in B, so A ⊆ B. However, B contains the additional element 7, which is not in A; therefore A ≠ B. Thus A is a proper subset of B. Option B fails because equality requires identical elements, option C fails because 7 is not in A, and option D is false for the same membership reason.

The set A has exactly three elements: 1, 2, and the set {3}. Therefore, {3} is one complete element of A. The number 3 itself is not listed separately as an element; it is contained inside the element {3}. Also, {1, 2, 3} is not an element of A, and although ∅ is a subset of every set, it is not necessarily an element of A. Hence, the correct answer is {3}.

The number of subsets of a finite set with n elements is 2^n. The empty set has n = 0 elements, so it has 2^0 = 1 subset. That single subset is the empty set itself: P(∅) = {∅}. It is important not to confuse the number of elements in ∅, which is zero, with the number of its subsets, which is one.

A set formed from the letters of MATH contains each distinct letter as a separate element. The word has four different letters: M, A, T, and H. Therefore A = {M, A, T, H}. The complete word MATH is not the same as the set of its letters; treating it as one element would produce the singleton set {MATH}, which is different.

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 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.