The Empty Set, Finite and Infinite Sets, Equal Sets

Between any two distinct real numbers, including 2 and 3, there are infinitely many real numbers such as 2.5, 2.25, and 2.125. Therefore {x ∈ ℝ : 2 < x < 3} is an infinite set. Option A contains only 2, option C contains no integer, and option D contains only 2 in ℕ. Hence option B is correct.

Two sets are equal when they contain exactly the same elements, regardless of the way they are described. In option D, the domain is the integers, so x² = 16 gives x = -4 or x = 4. Thus the first set is {-4, 4}, whereas the second set is {4}; because -4 is missing from the second set, the two sets are not equal. The other three pairs contain the same elements.

The notation 1 < x < 2 requires x to be strictly greater than 1 and strictly less than 2. Under the explicitly stated convention N = {1, 2, 3, ...}, there is no natural number satisfying both conditions. The numbers 1 and 2 are boundary values and are excluded by the strict inequalities. Therefore A contains no elements and is the empty set, written as ∅.

The empty set, written as ∅, contains no elements, so its cardinality is 0. It is also finite because a finite set has a limited number of elements, including zero elements. However, {0} is not empty: it contains the number 0 as one element and therefore has cardinality 1. Thus the statement that the empty set equals {0} is false. The symbols ∅ and {0} must not be confused.

A natural-number factor of 30 must divide 30 exactly. Every positive factor of a positive integer is less than or equal to that integer; the greatest factor of 30 is 30 itself. The additional condition x > 30 therefore cannot be satisfied by any natural factor of 30. Since no element meets both requirements, the set has no elements and A is the empty set ∅.

The positive natural-number multiples of 4 are 4, 8, 12, 16, 20, 24, and so on. The condition x ≤ 20 excludes every multiple after 20, while 20 itself is included because equality is allowed. Therefore the complete set is {4, 8, 12, 16, 20}. It is finite because the upper bound leaves only five possible multiples. Zero is not included under the stated positive-natural-number convention.

The stem defines N as the positive natural numbers {1, 2, 3, ...}. Therefore, the natural numbers satisfying x < 5 are exactly 1, 2, 3, and 4. Hence A = {1, 2, 3, 4}, which is precisely B in option C. In option A, B incorrectly includes 0; in option B, A also includes 5; and in option D, A contains numbers greater than 5. Thus only C represents equal sets.

Because the domain is the real numbers, both real square roots of 2 must be considered. Solving x² = 2 gives x = √2 or x = -√2. Therefore A = {-√2, √2}. These are two distinct real numbers, so the set has exactly two elements and is finite. A common error is to write only √2 and overlook the negative solution, even though both numbers have square 2.

There are only finitely many natural numbers less than 1000, namely the numbers from 1 through 999. The prime numbers satisfying the condition form a subset of this finite collection, so their number is also finite, even though the list may be fairly long. In contrast, there are infinitely many natural primes, infinitely many negative integers, and infinitely many real numbers between 0 and 1. Therefore option B is the finite set.

The set in option B is {3, 6, 9, 12, ...}. Every positive multiple of 3 belongs to the set, and after any listed multiple another one can be obtained by adding 3. Therefore the set has no last element and is infinite. A simple rule describes how to generate its elements, but it does not make the set finite. Option A is finite, option C has only two elements, and option D is empty under the usual convention N = {1, 2, 3, ...}.

The positive natural-number factors of 10 are A = {1, 2, 5, 10}. The positive natural-number factors of 20 are B = {1, 2, 4, 5, 10, 20}. Since 4 and 20 belong to B but not to A, the two sets are not equal. Each fixed positive integer has only finitely many factors, so both sets are finite. Thus option B is the only correct statement.

For every real number x, the square x² is greater than or equal to zero. It can equal zero only when x = 0, but it can never be negative. Consequently, the inequality x² < 0 has no real solution, so the set contains no elements and is the empty set ∅. Option B is incorrect because 0² = 0, not a negative number.

Solving x² = 0 gives x = 0, and 0 is an integer. Therefore A contains exactly one element: A = {0}. The empty set B = ∅ contains no elements. Since {0} has one element while ∅ has none, A and B are not equal. This also illustrates the important difference between a singleton set containing zero and the empty set containing nothing.

The equation x² − 1 = 0 factors as (x − 1)(x + 1) = 0, giving x = 1 or x = −1. However, A is restricted to the natural numbers. Under the usual convention N = {1, 2, 3, ...}, only 1 is admitted, so A = {1}. Set B contains both −1 and 1, and therefore A ≠ B.

A natural number divisible by both 2 and 3 is divisible by their least common multiple, 6. Hence A = {6, 12, 18, 24, ...}. This sequence continues without an end because for every member 6n, the next member 6(n + 1) is also a natural number satisfying both conditions. Therefore A is infinite, not merely the singleton {6}.

The quadratic x² − 6x + 9 is (x − 3)², so its only integer solution is x = 3. Thus A = {3}. In set notation, repetitions do not create additional elements; writing 3 three times still represents B = {3}. Consequently A and B contain exactly the same element and are equal. Repeated roots affect multiplicity in algebra, but not the number of distinct elements in a set.

There are infinitely many rational numbers strictly between 0 and 1. For example, 1/2, 1/3, and 2/3 belong to A, and for every positive integer n, the rational number 1/(n + 1) also lies between 0 and 1. Thus new distinct elements can be generated endlessly. The endpoints 0 and 1 are excluded, but excluding them does not make the set finite.

The natural numbers less than 5 are 1, 2, 3, and 4. Among these, 2 and 3 are prime. The number 1 is not prime because a prime number has exactly two distinct positive factors, while 1 has only one. The number 4 is composite, and 5 is not less than 5. Therefore A = {2, 3} = B, so option A is correct.

The expression |x − 2| + |x − 5| represents the sum of the distances from x to 2 and 5 on the real number line. By the triangle inequality, this sum is always at least the distance between 2 and 5, which is 3. Since the question requires the sum to equal 2, no real number x can satisfy it. Hence A is the empty set.

The natural numbers satisfying 1 ≤ x ≤ 4 are 1, 2, 3, and 4. Squaring these values gives 1² = 1, 2² = 4, 3² = 9, and 4² = 16. Therefore, the set described by {x² : x ∈ N, 1 ≤ x ≤ 4} contains exactly the four elements {1, 4, 9, 16}, so option A is equal to the given set. The other options either contain unsquared numbers, include an extra element such as 0, or contain many more elements.