Equal sets and Subsets

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.

Every number strictly between 4 and 6 is also strictly between 2 and 8. Therefore each element of B belongs to A, so B ⊆ A. The intervals are not equal because A contains additional numbers such as 3 and 7. Their intersection is B, not the empty set, so option B is the only correct statement.

The closed interval A includes every number from 2 to 8, including both endpoints. The open interval B contains only numbers strictly between 2 and 8, so it excludes 2 and 8. Every element of B is nevertheless an element of A; hence B ⊆ A. The intervals are not equal and 2 is not in B.

The interval (-∞, 5) represents every real number x such that x < 5. The round bracket at 5 shows that 5 itself is excluded. The interval contains negative numbers, zero, and positive numbers less than 5, so it is certainly not restricted to positive numbers and is not empty. Infinity is not an actual endpoint, so it is always written with a round bracket.