Equal sets and Subsets

The closed interval [2,4] contains every real number from 2 through 4, including both endpoints 2 and 4. The open interval (2,4) contains only numbers strictly greater than 2 and strictly less than 4, so it excludes both endpoints. Since 2 and 4 belong to the first set but not the second, [2,4] is not a subset of (2,4). One counterexample is enough to disprove a subset statement.

The set-builder condition −1 < x < 3 describes all real numbers strictly between −1 and 3. Because both inequalities are strict, neither boundary is included. In interval notation, an excluded endpoint is represented by a parenthesis, so the equivalent interval is (−1,3). The square-bracket alternatives incorrectly include at least one endpoint.

In interval notation, a square bracket at an endpoint means that endpoint belongs to the set. The interval [a,b] has a square bracket on both sides, so it contains a, b, and every real number between them. The condition a < b ensures that the interval has positive length, but it does not change the inclusion rule. Therefore, both endpoints are included.

The open interval (2,2) consists of real numbers x satisfying 2 < x < 2. This condition is impossible because no real number can be strictly greater than 2 and strictly less than 2 at the same time. The endpoints are excluded as well, so 2 is not present. Consequently, the interval contains no elements and is equal to the empty set, written as ∅.

The statement A ⊆ A follows directly from the definition of a subset. A set X is a subset of Y if every element of X belongs to Y. When both sets are A, every element of A is certainly an element of A itself. This remains true whether A is empty, finite, or infinite. Therefore, every set is a subset of itself, and option A is correct.

Subset inclusion is transitive. From A ⊆ B, every element of A belongs to B. From B ⊆ C, every element of B belongs to C. Therefore, each element of A, being an element of B, must also be an element of C. Hence A ⊆ C. The conditions do not imply that A and C are equal or that the reverse inclusions hold, so option A is the only justified conclusion.

Every number x in the open interval (0, 1) satisfies 0 < x < 1. Since every such number is also less than 2, it satisfies 0 < x < 2 and therefore belongs to (0, 2). Hence (0, 1) is a subset of (0, 2). The reverse inclusion is false because numbers such as 1.5 belong to (0, 2) but not to (0, 1); the intervals are not equal. Option A is correct.

The interval [0, 1] includes both endpoints, so it contains 0 and 1. The interval (0, 1] excludes 0 because it has a round bracket at the left endpoint, but it includes 1 because it has a square bracket there. Since 0 is an element of the first interval but not of the second, the first interval cannot be a subset of the second. Therefore, option A is correct.

For a set to be a subset of A, each of its elements must belong to A = {1, 2, 3, 4}. Options A, B, and C contain only elements from A, so they are subsets. Option D contains 3, which is in A, but also contains 5, which is not in A. A single element outside A is enough to disqualify the set as a subset. Therefore, option D is correct.

The set A contains the elements a and b. The set {a} is a singleton set containing only a. Since a is an element of A, every element of {a} belongs to A; therefore {a} ⊆ A is true. In contrast, {c} and {a, c} contain c, which is not in A. Also, a is an element, not a set, so writing a ⊆ A is not correct here.

To test whether P is a subset of Q, check every element of P. The elements 2, 3, 5, and 7 all occur in Q, so every element of P belongs to Q. Hence P ⊆ Q. The reverse relation Q ⊆ P is false because Q contains 1, 4, and 6, which are absent from P. The two sets are therefore not equal, and 1 is not an element of P.

A proper subset is a set whose every element belongs to another set, while the two sets are not equal. The condition A ⊆ B establishes that all elements of A are in B. The additional condition A ≠ B ensures that B has at least one element not contained in A. Thus A is a proper subset of B. Nothing in the statement requires A to be empty or B to be a universal set.

Each member of S—namely a, e, i, o, and u—is an alphabet letter, so each one belongs to V. Therefore S ⊆ V. The reverse statement V ⊆ S is false because V also contains consonants and possibly other letters. S is not equal to V for the same reason, and a ∉ V is false because a is explicitly an alphabet letter.

The inequality 5 < x < 9 describes all real numbers greater than 5 and less than 9. Because both inequalities are strict, neither endpoint is included. In interval notation, an excluded endpoint is shown with a round parenthesis, so the correct interval is (5, 9). Square brackets would incorrectly include the corresponding endpoint.

The left square bracket in [1, 6) means that 1 is included, so the left condition is 1 ≤ x. The right round parenthesis means that 6 is excluded, so the right condition is x < 6. Combining both conditions gives {x ∈ ℝ : 1 ≤ x < 6}. The variable is restricted to real numbers because interval notation describes real values.

For an interval to be a subset of [0, 10], every one of its numbers must lie between 0 and 10, including the permitted endpoint conditions. Every number in (2, 8] is greater than 0 and no greater than 8, so it lies completely inside [0, 10]. The other intervals contain numbers below 0 or above 10, so they cannot be subsets.

The open interval (1, 4) contains every real number strictly greater than 1 and strictly less than 4. The number 3/2 equals 1.5, so it lies between 1 and 4 and belongs to A. The endpoints 1 and 4 are excluded because the interval uses round parentheses, while 5 lies outside the interval. Therefore B is the only correct choice.

Assuming the school convention ℕ = {1, 2, 3, ...}, the condition x ≤ 5 gives A = {1, 2, 3, 4, 5}. Every element of A occurs in B, so A ⊆ B. The sets are not equal because B also contains 6 and 7. Moreover, 7 does not satisfy x ≤ 5, so 7 ∉ A. Thus A is the only correct statement.

The elements listed directly in A are ∅, {1}, and 2. Therefore, the empty set ∅ is an element of A, so ∅ ∈ A is true. The number 1 is not directly listed; {1} is listed instead. Similarly, {2} is not listed because 2, not the set {2}, is an element. The number 3 is also absent. This question tests the important difference between an element and a set containing an element.

The interval A = [-1, 3] contains every real number from -1 through 3, including both endpoints. The interval B = (-2, 4) contains all real numbers strictly greater than -2 and strictly less than 4. Since -1 > -2 and 3 < 4, every number in A lies in B, including -1 and 3. Thus A ⊆ B. Statement D is false because 4 is excluded from B by the round bracket.

A = [2, 5) contains 2 and every real number greater than 2 but less than 5; it does not contain 5. B = (1, 5) contains every real number greater than 1 and less than 5. Every element of A is therefore also an element of B, so A ⊆ B. The reverse inclusion is false because values such as 1.5 belong to B but not to A. The intervals are not equal, and 1 is not in A.

A ⊆ B says that every element of A belongs to B. B ⊆ A says that every element of B belongs to A. Thus the two sets contain exactly the same elements, which is the definition of equality of sets. Therefore A = B. This is commonly called the two-way subset criterion for proving set equality.

The standard number-system inclusion chain is N ⊆ Z ⊆ Q ⊆ R. Every natural number is an integer; every integer can be written as a rational number with denominator 1; and every rational number is real. The reverse containments are false because, for example, negative integers are not natural numbers and irrational real numbers are not rational. Therefore option A is the only valid relation.

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.