Equal sets and Subsets

A set is equal to U only when it has exactly the same elements. The positive natural multiples of 3 that are less than or equal to 12 are 3, 6, 9, and 12, which gives precisely U. Option A omits 12 because it uses x < 12. Option C omits 3 and 9, while option D includes many nonmultiples of 3. Therefore option B is correct.

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.

If two sets are equal, they contain exactly the same elements. Consequently, their cardinalities, or numbers of elements, must also be equal. However, the converse is not generally true: for example, {1, 2} and {a, b} have the same cardinality but are not equal because their elements differ. Also, ∅ has zero elements, while {0} has one element.

In a set, the order of elements does not matter. Hence {1, 2} and {2, 1} represent exactly the same inner set. The outer set A therefore contains only one distinct element, namely the set {1, 2}. Consequently, the cardinality of A is n(A) = 1, not 2. Repeated or differently ordered descriptions do not create new elements in a set.

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.

Solve the defining equation: x² − 4 = 0 gives x² = 4, so x = 2 or x = −2. Both values are real and must be included in the set because the domain is ℝ. Thus A = {−2, 2}. A set does not repeat elements, and option A represents exactly these two solutions.

For an integer x, the inequality |x| < 2 means −2 < x < 2. The integers in this interval are −1, 0, and 1, so A = {−1, 0, 1}. This is exactly the set B. Since two sets are equal when they contain precisely the same elements, A = B. Therefore, option A is correct.

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.