Equal sets and Subsets

The open interval (2, 2) consists of real numbers x satisfying 2 < x < 2. No real number can be simultaneously greater than 2 and less than 2. Therefore, the interval contains no elements and is the empty set, written as ∅. The endpoint 2 is excluded by open brackets, and in any case it cannot satisfy both strict inequalities.

A set X is a subset of a set Y when every element of X is also an element of Y. The set {1} has only one element, namely 1, and 1 belongs to {1, 2}. Therefore {1} ⊆ {1, 2}. The sets are not equal because the second set also contains 2, and {1} is not empty.

The interval [1, 3) contains 1 and every real number less than 3, while the interval (0, 4) contains all real numbers strictly between 0 and 4. Every element of [1, 3) is greater than 0 and less than 4, including the endpoint 1 because 1 lies inside (0, 4). The fact that 3 is excluded from the first interval does not violate the subset relation.

The subset {3, 7} is formed by selecting the elements 3 and 7 from the original set A = {2, 3, 5, 7}. Both selected numbers appear in A, so {3, 7} is indeed a subset of A. The elements 2 and 5 are not selected in this particular subset, but their absence does not affect the subset relationship. Therefore both 3 and 7 are taken from A.

The set A has exactly three elements: 1, 2, and the set {3}. The braces around 3 are important because they make {3} one complete element of A. The number 3 by itself is not listed as an element. Also, {1, 3} is not a subset because 3 is not an element of A, and A is not a subset of {1, 2, 3} because {3} differs from 3.

A proper subset must contain only elements of the original set and must not be equal to the original set. The set {2, 4} contains elements that are both in A, and it omits 6, so it is smaller than A. Option B is equal to A and is therefore not proper. Options C and D contain 8, which is not an element of A.

The empty set has no elements. To be a subset of A, every element of the first set must belong to A; because the empty set has no elements, there is no possible violation of this condition. Thus ∅ ⊆ A for every set A. The other statements are not always true: A may be nonempty, A need not equal ∅, and ∅ has no elements at all.

Interval notation uses round parentheses for excluded endpoints and square brackets for included endpoints. In (5, 10], the parenthesis at 5 excludes 5, while the square bracket at 10 includes 10. Thus only 10 is an included endpoint. Option B is correct; option C wrongly includes 5, and A reverses the bracket meanings. D excludes both and is also incorrect.

The standard inclusion chain is N ⊆ Z ⊆ Q ⊆ R. Natural numbers are contained in the integers; integers can be written as rational numbers, so they are contained in Q; and every rational number is real, so Q is contained in R. The reverse statements are false because each larger set contains numbers absent from the smaller one. Therefore A is correct.

Q is the set of rational numbers, which can be written as p/q where p and q are integers and q is nonzero. The number 3/4 has this form, so it is rational. Every rational number is also real, meaning 3/4 belongs to both Q and R. In contrast, √2 and π are irrational, while i is non-real. Therefore A is correct.

The notation A ⊆ B means that A is a subset of B. Formally, for every element x, if x belongs to A, then x also belongs to B. A may be smaller than B or equal to B, and it may even be empty, but none of those possibilities is required. Therefore the defining statement is option A.

The interval (2, 5) is an open interval. It contains every real number that is greater than 2 and less than 5, but it does not contain either endpoint because both endpoints are written with round brackets. Among the given choices, 3 satisfies 2 < 3 < 5, whereas 2 and 5 are excluded and 6 lies outside the interval. Therefore, the correct answer is 3.

The closed interval [2, 5] contains every real number x satisfying 2 ≤ x ≤ 5. Square brackets show that both endpoints, 2 and 5, are included. The numbers 2 and 4 are inside the interval, and 5 is included as the right endpoint. The number 6 is greater than 5, so it lies outside the interval. Therefore D is correct.

The open interval (2, 4) contains all real numbers strictly between 2 and 4. The closed interval [2, 4] contains those same interior numbers and also includes the endpoints 2 and 4. Therefore every element of (2, 4) is an element of [2, 4], so (2, 4) ⊆ [2, 4]. The reverse inclusion and equality are false, and [2, 4] is not contained in (4, 6). Hence A is correct.

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.