The Empty Set, Finite and Infinite Sets, Equal Sets

There are infinitely many rational numbers strictly between 0 and 1. Examples include 1/2, 1/3, 2/3, and 1/n for every integer n greater than 1. More generally, between any two distinct rational numbers there are further rational numbers. Therefore, the set of rational numbers in the interval (0, 1) is infinite.

The integers are ..., −2, −1, 0, 1, 2, ... . The numbers 0 and 1 are consecutive integers, so there is no integer strictly between them. Because the interval is strict, 0 and 1 themselves are not included either. Consequently, no integer satisfies the condition, and the set is empty: ∅.

The interval (0, 1) contains infinitely many real numbers. It includes rational numbers such as 1/2 and 3/4, as well as irrational numbers such as √2/2. In fact, between any two distinct real numbers there are infinitely many other real numbers. Thus the set of real numbers strictly between 0 and 1 is infinite.

A number must satisfy both conditions to belong to A: it must be even and prime. The only even prime number is 2, because every other even natural number greater than 2 is divisible by 2 and is therefore composite. Thus A contains exactly one element and is the singleton set {2}. Option B is incorrect because the set is not empty; C includes composite numbers, and D includes 1, which is not prime.

A prime number has exactly two distinct positive factors, whereas a composite natural number has more than two positive factors. These definitions are mutually exclusive, so no natural number can be prime and composite at the same time. Consequently, the set contains no elements and is the empty set, written as ∅.

Using the convention N = {1, 2, 3, ...}, there is no natural number less than 1. Therefore B has no elements and B = ∅. Since A is also defined as the empty set, both sets contain exactly the same elements. Hence A = B. The answer depends on the stated convention that natural numbers begin with 1.

For option A, assuming N = {1, 2, 3, ...}, the condition x ≤ 3 gives the elements 1, 2, and 3. Thus the first set is exactly {1, 2, 3}, which is the second set. In the other options, one element is missing or an extra element is present. Equality of sets depends on having precisely the same elements, not on their order.

The empty set is defined as the set that contains no elements. The number of elements of a set is called its cardinality, so the cardinality of the empty set is n(∅) = 0. It is important not to confuse ∅ with {0}: the former has no elements, while the latter contains the element 0 and therefore has one element.

The notation {∅} means a set whose single element is the empty set itself. Therefore its cardinality is 1. This differs from ∅, which has no elements and cardinality 0. Braces are significant: ∅ denotes nothing inside the set, whereas {∅} contains one object, namely the empty set. Thus n({∅}) = 1.

Equal sets contain exactly the same elements. Consequently, they must have the same cardinality, regardless of the order in which their elements are written. Since A = B, we can write n(A) = n(B). The given value n(A) = 5 therefore immediately gives n(B) = 5. No additional information is needed.

Set order does not matter, so B contains the elements 1, 2, and 3. For A to equal B, it must contain exactly these same elements. Since 1 and 2 are already present in A, the remaining element represented by a must be 3. Values 1 or 2 would create repetition rather than supply the missing element, and 0 would make the sets unequal.

Assuming N = {1, 2, 3, ...}, the set A contains all natural numbers from 1 through a. The set B contains exactly 1, 2, 3, and 4, so its largest element is 4. For A and B to be equal, the upper bound must be a = 4. If a were 3, an element would be missing; if a were 5, an extra element would appear.

The prime natural numbers less than 10 are 2, 3, 5, and 7. Each of these numbers has exactly two positive factors, while 1 is neither prime nor composite and 4, 6, 8, and 9 are composite. Thus the descriptive set has precisely the elements listed in A, so the two sets are equal.

Factor the quadratic expression: x² − 5x + 6 = (x − 2)(x − 3). Therefore, the equation is satisfied when x = 2 or x = 3. Since x is restricted to real numbers and both values are real, the solution set contains exactly these two elements. Hence the given set is {2, 3}. The order of elements does not affect a set.

For an integer x, the inequality |x| < 3 means that x lies strictly between −3 and 3. The integers in this interval are −2, −1, 0, 1, and 2. The endpoints −3 and 3 are excluded because the inequality is strict. Thus A = {−2, −1, 0, 1, 2}, which contains five distinct elements.

A positive divisor of a number is a positive integer that divides it exactly. The only positive integer that divides 1 is 1 itself, because 1 ÷ 1 = 1. Thus the set of positive divisors of 1 is {1}. A set containing exactly one element is called a singleton, and because its elements can be counted, it is also finite.

The set ℕ denotes the natural numbers, which are non-negative in the convention used here, or positive numbers in conventions that begin with 1. In either convention, no negative number belongs to ℕ. Therefore there is no element satisfying the condition “x is a negative natural number.” The required set has no elements and is consequently the empty set, written ∅.

Using the standard school convention, the vowels in the English alphabet are a, e, i, o, and u. Sometimes y is treated as a vowel in particular words, but it is not included in the usual basic alphabet classification. In either interpretation, the number of vowels is limited. Therefore, the set of vowels is finite, not empty or infinite.

By definition, a triangle is a polygon with exactly three sides. A figure with four sides is a quadrilateral, not a triangle. Consequently, no geometric object can satisfy both conditions “is a triangle” and “has four sides.” Since the defining condition produces no elements, the set of such triangles is empty, denoted by ∅.

We need natural numbers n such that n² = 25. Over the integers, the equation has two solutions, n = 5 and n = −5. However, the question restricts n to natural numbers, and −5 is not natural. Since 5² = 25, the only allowed element is 5. Therefore, A = {5}, a singleton set.