Equal sets and Subsets

The elements of A are 2 and 4. Both 2 and 4 are present in B, so every element of A belongs to B; therefore A ⊆ B. B is not a subset of A because 6 and 8 are absent from A. Also, 6 is a number rather than a set, and the set A itself is not listed as an element of B, so options C and D are incorrect.

A subset may contain only elements that occur in the original set A. The empty set is a subset of every set, {a} contains an element of A, and {a, b} is A itself; therefore all three are subsets of A. Option D contains c, but c is not an element of A. Since even one outside element is enough to fail the subset condition, {a, c} is not a subset.

The natural numbers, such as 1, 2, 3 and so on, are included within the real number system. Therefore every natural number is a real number, which is written as ℕ ⊆ ℝ. The reverse inclusion is false because real numbers also include integers, fractions, irrational numbers, and decimals that are not natural numbers. ℝ is not empty, and ℕ is a set rather than one individual element of ℝ.

The two sets contain exactly the same elements: 1, 2, and 3. Since every element of A is also an element of B, A is a subset of B. Similarly, every element of B is in A, so B is a subset of A. This proves that A and B are equal sets, and both subset relations hold.

The set B contains the elements 2 and 4. Both of these elements are present in A = {1, 2, 3, 4}; therefore every element of B belongs to A, which proves B ⊆ A. The sets are not equal because A also contains 1 and 3. Also, 2 is an element of B, not a subset of B.

A set X is a subset of A when every element of X is also an element of A. Here A contains 1, 2, and 3. Both elements of {2,3} occur in A, so {2,3}⊆A. The other choices contain 4, which is not in A; in particular, C={2,3,4} is not a subset of A because of the element 4.

The first relation, A⊆B, says that every element of A belongs to B. The second relation, B⊆A, says that every element of B belongs to A. Thus the two sets contain exactly the same elements, which is precisely the definition of equal sets. They do not have to be empty; for example, A=B={1,2} satisfies both subset relations.

Subset status depends on membership, not on addition or on merely comparing the numbers of elements. The set {2,5} is a subset of A because its only elements are 2 and 5, and both of them occur in A={1,2,3,4,5}. Therefore every element of {2,5} belongs to A, so {2,5}⊆A. The size of either set alone is not sufficient evidence.

The interval (-∞,3] contains every real number extending indefinitely to the left and ending at 3. The round bracket at infinity is required because infinity is not an actual number or endpoint, while the square bracket at 3 means that 3 is included. In inequality form, the interval is {x∈R:x≤3}, so it includes negative numbers, zero, and positive numbers up to and including 3.

A subset must contain only elements that belong to the original set A. A two-element subset must also contain exactly two distinct elements. The set {2, 4} satisfies both conditions because 2 and 4 are in A and there are exactly two elements. Option B contains 8, which is not in A; option C has three elements; and option D has only one element.

The closed interval [2, 2] contains every real number x satisfying 2 ≤ x ≤ 2. The only possible value is x = 2, so the interval contains exactly one element and is therefore the singleton set {2}. It is not empty, not an ordered pair, and certainly not the set of all real numbers. Equal endpoints in a closed interval produce a singleton.

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.