Equal sets and Subsets

The set A contains three elements: the number 1, the set {1}, and the number 2. Therefore, 1 is directly an element of A, so 1 ∈ A is true. The expression 1 ⊆ A is not the appropriate subset statement because 1 is a number, not a set. Also, {2} and {1,2} are not listed as elements of A; the presence of 2 does not mean that {2} is an element.

To determine whether A is a subset of B, check whether every element of A occurs in B. The elements of A are 2, 4, and 6, and all three are present in B. Therefore A ⊆ B is true. The reverse relation B ⊆ A is false because B also contains 1, 3, and 5. The sets are not equal because they do not have the same elements, so option A is the only correct answer.

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}.

A = {1, 2, {3}}. A subset may contain only objects that are elements of A. Both 1 and {3} belong to A, so {1, {3}} is a subset. However, 3 itself is not an element of A; only the set {3} is. Therefore the other options contain an object that is not in A or have an incorrect nested structure.

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.

Solving x² = 4 over the real numbers gives x = 2 or x = -2. Hence A = {-2, 2}. Since B contains exactly the same two elements, A and B are equal, so A = B. Neither is a proper subset of the other. Option D is incorrect because it omits the valid solution -2. Equality of sets depends on having exactly the same elements, regardless of their order.

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.