The Empty Set, Finite and Infinite Sets, Equal Sets

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 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.

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 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 two-digit natural numbers are 10, 11, 12, and so on up to 99. Both the lower bound and upper bound are fixed, so only finitely many natural numbers satisfy the condition. In fact, the number of elements is 99 − 10 + 1 = 90. Therefore A is a finite set, not an empty set, an infinite set, or a singleton. Hence option B is correct.

Because x is an integer, the values in the interval are −4, −3, −2, −1, 0, 1, and 2. The lower boundary −4 is included because the symbol is ≤, whereas 3 is excluded because the symbol is <. Counting the listed integers gives 7 elements. Thus A is a finite set with cardinality 7, and option B is correct.

The natural numbers not exceeding 50 that are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. There are only ten such numbers because the condition includes the upper bound x ≤ 50. Since the list ends at 50, it cannot continue indefinitely. Therefore A is a finite set, so option B is correct.

The symbol ∅ denotes the empty set, which contains no elements at all. In contrast, {∅} is a set whose only element is the empty set itself. Therefore, {∅} has exactly one element and is a singleton set, whereas ∅ has zero elements. They are different sets, and neither statement about both being equal or infinite is correct.

An absolute value is zero only when its inside expression is exactly zero. Thus |x − 1| = 0 implies x − 1 = 0, giving x = 1. Since the variable is restricted to real numbers, the solution set contains only 1. Consequently, A = {1}, which is a singleton set. The pair {−1, 1} would arise from an equation such as |x| = 1, not from |x − 1| = 0.

The two-digit natural numbers begin at 10 and end at 99. The even members are 10, 12, 14, and so on, up to 98. This is a bounded list with a fixed first and last value, so it has only finitely many elements. In fact, there are 45 such numbers. Option D lists only one-digit even numbers and therefore does not represent A.

Factor 84 into primes: 84 = 2² × 3 × 7. Its positive divisors include 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84, but the question asks only for prime divisors. The prime divisors are therefore 2, 3, and 7, so A = {2, 3, 7}. The number 1 is not prime, and 4 is composite, so neither belongs in A.

Subtracting x from both sides of x + 5 = x gives 5 = 0, which is a contradiction. Thus no natural number, or any number at all, satisfies the condition. A set containing no elements is the empty set, so A = ∅. It is not {5}, because x = 5 gives 10 = 5.

For natural numbers, test the values around the boundary: 1³ = 1, 2³ = 8, and 3³ = 27, so 1, 2, and 3 satisfy x³ ≤ 27. The next natural number does not, because 4³ = 64 > 27. Therefore B = {1, 2, 3}, which is exactly the same collection of elements as A. Hence A = B; both sets are finite and equal.

The positive natural-number divisors of 36 are obtained from factor pairs: 1×36, 2×18, 3×12, 4×9, and 6×6. Therefore the complete divisor set is {1, 2, 3, 4, 6, 9, 12, 18, 36}, exactly the set B. Repetition is not used in a set, so 6 appears only once. Hence A = B.

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Among these, the odd numbers are 1, 3, and 9; every other divisor is even. Thus the set satisfying both conditions is A = {1, 3, 9}. Notice that 1 is odd and is also a divisor of every nonzero integer.

The equation x² = 0 has exactly one real solution, x = 0. However, the second condition requires x to be strictly greater than 0, and 0 is not greater than 0. Since both conditions must hold simultaneously, no real number satisfies them. Therefore A contains no elements and A = ∅, the empty set.

The natural-number perfect squares begin 1, 4, 9, 16, 25, and continue as n² for n = 1, 2, 3, and so on. For every natural number n, another perfect square n² can be formed, and there is no greatest natural number. Consequently, A has infinitely many elements and is an infinite set.

The governing concept is counting the elements of a finite set under an upper bound. The natural-number perfect squares below 50 are 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, and 7² = 49. The next square, 8² = 64, is not less than 50. Thus A has 7 elements, making option B correct. The bound prevents the set from being infinite.

Solving x² = 16 gives x = 4 or x = −4, because both numbers have square 16. The additional condition x < 0 eliminates 4 and retains −4. Thus the only integer satisfying both requirements is −4, so A = {−4}. This is a singleton set because it has exactly one element.