Sets and their representations

The square bracket in [3, 7) means that the left endpoint 3 belongs to the interval. The round bracket after 7 means that 7 is excluded. In inequality form, the interval is 3 ≤ x < 7, so it contains 3 but not 7. Consequently, option A is correct. The other choices reverse or ignore the meanings of the brackets.

The symbol 0 ≤ x allows x to equal 0, so the lower endpoint is included and must use a square bracket. The symbol x < 6 excludes 6, so the upper endpoint must use a round bracket. Therefore the interval is [0, 6), making option B correct. The other forms incorrectly include or exclude one or both endpoints.

The inequality 2 < x excludes 2, so the interval must begin with a round parenthesis. The condition x ≤ 9 includes 9, so the interval must end with a square bracket. Combining these endpoint rules gives (2, 9], which is option A. Option B includes 2 and excludes 9; C includes both endpoints, while D excludes both.

The inequality x ≥ 5 includes 5 because equality is allowed, so a square bracket is used at 5. It also includes every real number greater than 5, continuing without upper bound toward infinity. Infinity is not an actual real endpoint, so a round bracket is always used with it. Thus the correct interval is [5, ∞).

The power set of a set is the set of all its subsets. Therefore, an element of P({1, 2, 3}) must itself be a set whose every element comes from {1, 2, 3}. The set {1, 3} satisfies this condition, so it belongs to the power set. The single number 1 is not written as a subset here, and 4 is not in the original set.

The condition requires x to be an integer between -2 and 2, with both endpoints included because the inequalities use ≤. Listing the integers in order gives -2, -1, 0, 1, and 2. Thus A = {-2, -1, 0, 1, 2}. Fractions and decimals are not considered, and no endpoint may be omitted.

An interval includes an endpoint when a square bracket is used and excludes it when a round bracket is used. In (1,4], the round bracket at 1 means that 1 is excluded, while the square bracket at 4 means that 4 is included. Option A excludes both endpoints, option B includes 1 but excludes 4, and option D includes both endpoints.

An open interval excludes both of its finite endpoints. Round brackets indicate exclusion, so (2,6) represents all real numbers x satisfying 2<x<6; neither 2 nor 6 belongs to the interval. Option A is closed because both endpoints are included, while options C and D are half-open because exactly one endpoint is included.

The condition 1<x<2 is strict at both ends. Therefore, x may be any real number between 1 and 2, but it cannot equal 1 or 2. Strict inequalities are represented with round brackets, so the interval notation is (1,2). Square brackets would incorrectly include the corresponding endpoint; hence A, C, and D do not match the given set-builder description.

The inequality 1 ≤ x ≤ 5 includes both endpoints because equality is allowed at 1 and at 5. In interval notation, an included endpoint is written with a square bracket. Therefore the correct interval is [1, 5]. Option A excludes both endpoints, option C excludes 5, and option D excludes 1, so none of them represents the given double inequality exactly.

The condition x > 7 includes all real numbers greater than 7 but excludes 7 itself. Exclusion of a finite endpoint is shown by a round parenthesis, so the interval is (7, ∞). Infinity is never an actual endpoint and is always written with a round parenthesis. The other choices either include 7 or describe numbers less than 7.

The endpoint −1 is included, so a square bracket is used at the left side: [−1. The endpoint 2 is excluded, so a round bracket is used at the right side: 2). Therefore, the required interval is [−1, 2). This is called a left-closed, right-open or half-open interval. The corresponding inequality is −1 ≤ x < 2.

The inequality −4 < x means that x is greater than −4, but x cannot equal −4; therefore, −4 is excluded and receives a round bracket. The inequality x ≤ 1 means that x may equal 1; therefore, 1 is included and receives a square bracket. Hence A is written in interval notation as (−4, 1].

Both inequalities are strict: 2 < x excludes the endpoint 2, and x < 7 excludes the endpoint 7. An excluded endpoint is represented by a round bracket in interval notation. Since neither endpoint is included, the set is the open interval (2, 7). Equivalently, it contains every real number strictly between 2 and 7, but not 2 or 7 themselves.

The square bracket at 3 shows that 3 is included, so the inequality must contain 3 ≤ x. The round bracket at 8 shows that 8 is excluded, so the inequality must contain x < 8. Combining these conditions gives {x : x ∈ ℝ, 3 ≤ x < 8}. Thus option B is the correct set-builder form of [3, 8).

The inequality x ≤ 3 includes every real number less than 3 as well as 3 itself. The interval extends without bound to the left, represented by −∞, and ends at 3. Since 3 is included because of the symbol ≤, a square bracket is used at 3. Infinity is never included as an endpoint, so a parenthesis is used at −∞. Therefore, the interval is (−∞,3].

The condition x > −2 represents all real numbers strictly greater than −2. The number −2 itself is not allowed because the inequality is strict, so the left endpoint must use a parenthesis. The solution continues indefinitely to the right, which is shown by ∞; infinity is not an included real endpoint, so it also uses a parenthesis. Hence the interval is (−2,∞).

The condition 0 ≤ x means that x may equal 0, so the left endpoint is included and receives a square bracket. The condition x < 4 means that x must be strictly less than 4, so 4 is excluded and receives a parenthesis. Combining these endpoint rules gives the half-open interval [0,4). Both boundaries must be checked separately in a mixed inequality.

The closed interval [2,2] represents all real numbers x satisfying 2 ≤ x ≤ 2. A number cannot be both less than and greater than 2 under this condition; the only possible value is exactly x = 2. Since the endpoint is included at both sides, the interval contains one element only. Therefore, [2,2] is the singleton set {2}, not the empty set.

The set of all real numbers extends without bound in both the negative and positive directions. In interval notation it is written as (-∞, ∞), which represents every real number x satisfying -∞ < x < ∞. Infinity is not an actual real number or endpoint, so it is always written with round brackets, never square brackets. Therefore, option A is correct.