Equal sets and Subsets

For a set B to be a subset of A, every element of B must belong to A. In option A, {4, 7} contains 7, but 7 is not in A = {4, 5, 6}; hence it is not a subset. The set {4} and the set {5, 6} contain only elements of A. The empty set is also a subset of every set, so option A is the unique answer.

The empty set has zero elements, so the general formula gives 2⁰ = 1 subset. That one subset is ∅ itself. Although it contains no elements, it is still a set and is considered a subset of every set, including itself. Therefore the empty set has exactly one subset, making option B correct.

The symbol A ⊆ B means that A is a subset of B. Formally, every element belonging to A must also belong to B. The sets may be equal, because the subset symbol allows equality, or A may be smaller than B. This statement is different from A ∈ B, which says that A itself is an element of B.

A proper subset contains only elements of the original set and is not equal to the original set. The set {1, 2} contains elements that are all in A, but it omits 3, so it is smaller than A and is a proper subset. Option B equals A, option C contains the extra element 4, and option D also contains 4, which is not in A.

The set A has two elements: 1 and 2. Both 1 and 2 occur as elements of B, so every element of A belongs to B; hence A ⊆ B. In addition, the complete set {1, 2}, which is exactly A, is itself listed as an element of B. Therefore A ∈ B also holds. This illustrates that being a subset and being an element are different relationships.

The closed interval [0, 3] contains every real number x satisfying 0 ≤ x ≤ 3, and both endpoints 0 and 3 are included. The elements 0, 1, 2, and 3 all satisfy this condition, so {0, 1, 2, 3} is a subset of [0, 3]. Each other option contains at least one number outside the interval: −1, 4, or −2.

The open interval (0, 3) consists of all real numbers strictly greater than 0 and strictly less than 3. Its endpoints 0 and 3 are excluded. Option A contains 0, so not every element of {0, 1} lies in (0, 3); consequently, it is not a subset. Every element in options B, C, and D lies strictly between 0 and 3.

A = [0, 2] contains all real numbers from 0 through 2, including both endpoints. B = (0, 2) contains only numbers strictly between 0 and 2, so it excludes 0 and 2. Every element of B is therefore an element of A, which proves B ⊆ A. The sets are not equal, and 0 is not an element of B because B is open at 0.

By definition, A is a subset of B when every element belonging to A also belongs to B. This relationship is written as A ⊆ B. It does not require A and B to have exactly the same elements; A may be a proper subset of B. The symbol A ∈ B would instead state that the whole set A is one element of B, which is a different claim.

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.