Equal sets and Subsets

Two sets are equal when they contain exactly the same elements, regardless of the order in which those elements are written. Set A contains 1, 2, and 3, and set B also contains 1, 2, and 3; only their listing order is different. Neither set is empty or infinite. Therefore, A = B, so option B is correct.

Solve the condition defining W: x² = x gives x² − x = 0, or x(x − 1) = 0. Thus the only real solutions are x = 0 and x = 1. Consequently, W = {0, 1}. Since X is also exactly the set {0, 1}, both sets contain the same elements, so W = X. Equality of sets depends on elements, not on their order.

Two sets are equal exactly when they contain the same elements. Since A = B and A contains the four distinct elements 5, 6, 7, and 8, set B must contain precisely those same four elements. The notation n(B) denotes the cardinality, or number of elements, of B. Therefore n(B) = 4, making option C correct.

The governing concept is equality of a solution set with a roster set. Solve x² − 4 = 0 by writing x² = 4; hence x = 2 or x = −2. Both solutions are real and must be included because the domain is ℝ. Therefore A = {−2, 2}, so option A is correct. Option B omits the negative root, option C confuses roots with squared values, and option D ignores the two real solutions.

The governing concept is equality of sets after interpreting an absolute-value inequality. For integers, |x| < 2 is equivalent to −2 < x < 2. The only integers in this open interval are −1, 0, and 1, so A = {−1, 0, 1}. This is exactly B; therefore A = B and option A is correct. The endpoints −2 and 2 are excluded, but that does not make A empty, and B clearly has only three elements, so it is not infinite.

A set A is a subset of B when every element of A is also an element of B. The elements of A are 1 and 2, and both occur in B. Although B has the additional element 3, that does not prevent A from being a subset of B. Thus A ⊆ B, and the inclusion is proper because A and B are not equal.

The empty set ∅ contains no elements. To check whether ∅ is a subset of A, we ask whether every element of ∅ belongs to A. Since there is no element in ∅ that could violate this requirement, the statement is vacuously true for every set A. Therefore ∅ ⊆ A always holds. This does not mean that ∅ is an element of A.

Every set is a subset of itself because each element of the set is, naturally, contained in that same set. Here the elements 2, 4, and 6 all belong to A, so A ⊆ A is true. In contrast, A ∈ A would claim that the entire set A is an element, while 2 ⊆ A incorrectly treats a number as a set.

For A to be a subset of B, every element of A must also appear in B. The elements 1 and 3 are present in B, but 5 is absent from B. A single missing element is enough to make the subset statement false. Therefore A ⊄ B because 5 ∉ B. The set A is not empty; it has three elements.

A proper subset must contain only elements of A and must not be equal to A itself. The set {7} satisfies both conditions: 7 belongs to A, and {7} has fewer elements than {7, 8}. Option B is equal to A rather than proper, while options C and D contain 9, which is not an element of A. Hence {7} is correct.

A set with n distinct elements has 2ⁿ total subsets because each element has two independent choices: it may be included or excluded. For A = {a, b}, n = 2, so the number is 2² = 4. Listing them confirms the result: ∅, {a}, {b}, and {a, b}. Thus option C is correct.

The natural numbers are 1, 2, 3, and so on, and every natural number is also a real number. Therefore, every element of ℕ belongs to ℝ, which is written as ℕ ⊆ ℝ. The reverse relation is false because real numbers also include integers, fractions, irrational numbers, and many other numbers that are not natural numbers.

The set ℝ contains all real numbers. The elements 1, 2, and 3 are real numbers, so every element of {1, 2, 3} belongs to ℝ; hence {1, 2, 3} ⊆ ℝ. The other options contain words rather than real-number elements, so they are not subsets of the real-number set in the usual mathematical interpretation.

Every integer is a real number, including negative integers, zero, and positive integers. Consequently, all elements of ℤ are contained in ℝ, so the correct relation is ℤ ⊆ ℝ. The reverse statement fails because real numbers include noninteger fractions and irrational numbers. Likewise, ℝ is not contained in ℕ, and rational numbers are not all natural numbers.

A rational number can be written in the form p/q, where p and q are integers and q is nonzero. Every such number lies on the real number line, so every element of ℚ is an element of ℝ; therefore ℚ ⊆ ℝ. The sets are not equal because irrational numbers such as √2 and π belong to ℝ but not to ℚ.

A subset contains only elements that are present in the larger set. The elements of {1, 3} are 1 and 3, and both occur in A = {1, 2, 3}. Therefore {1, 3} ⊆ A. It is not equal to A because A also contains 2. Option D is false because 2 is not an element of {1, 3}, while option C reverses the subset relation.

For a set B to be a subset of A, every element of B must belong to A. In option A, {4, 7} contains 7, but 7 is not in A = {4, 5, 6}; hence it is not a subset. The set {4} and the set {5, 6} contain only elements of A. The empty set is also a subset of every set, so option A is the unique answer.

The empty set has zero elements, so the general formula gives 2⁰ = 1 subset. That one subset is ∅ itself. Although it contains no elements, it is still a set and is considered a subset of every set, including itself. Therefore the empty set has exactly one subset, making option B correct.

The symbol A ⊆ B means that A is a subset of B. Formally, every element belonging to A must also belong to B. The sets may be equal, because the subset symbol allows equality, or A may be smaller than B. This statement is different from A ∈ B, which says that A itself is an element of B.

A proper subset contains only elements of the original set and is not equal to the original set. The set {1, 2} contains elements that are all in A, but it omits 3, so it is smaller than A and is a proper subset. Option B equals A, option C contains the extra element 4, and option D also contains 4, which is not in A.

The set A has two elements: 1 and 2. Both 1 and 2 occur as elements of B, so every element of A belongs to B; hence A ⊆ B. In addition, the complete set {1, 2}, which is exactly A, is itself listed as an element of B. Therefore A ∈ B also holds. This illustrates that being a subset and being an element are different relationships.

The closed interval [0, 3] contains every real number x satisfying 0 ≤ x ≤ 3, and both endpoints 0 and 3 are included. The elements 0, 1, 2, and 3 all satisfy this condition, so {0, 1, 2, 3} is a subset of [0, 3]. Each other option contains at least one number outside the interval: −1, 4, or −2.

The open interval (0, 3) consists of all real numbers strictly greater than 0 and strictly less than 3. Its endpoints 0 and 3 are excluded. Option A contains 0, so not every element of {0, 1} lies in (0, 3); consequently, it is not a subset. Every element in options B, C, and D lies strictly between 0 and 3.

A = [0, 2] contains all real numbers from 0 through 2, including both endpoints. B = (0, 2) contains only numbers strictly between 0 and 2, so it excludes 0 and 2. Every element of B is therefore an element of A, which proves B ⊆ A. The sets are not equal, and 0 is not an element of B because B is open at 0.

By definition, A is a subset of B when every element belonging to A also belongs to B. This relationship is written as A ⊆ B. It does not require A and B to have exactly the same elements; A may be a proper subset of B. The symbol A ∈ B would instead state that the whole set A is one element of B, which is a different claim.