The Empty Set, Finite and Infinite Sets, Equal Sets

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.

The condition 1 < x < 2 describes numbers strictly between 1 and 2. Although real numbers exist in that interval, no natural number lies strictly between two consecutive natural numbers 1 and 2. Therefore no natural number satisfies the defining condition, so A has no elements and is the empty set, A = ∅.

The governing concept is describing a set by solving its defining equation over the stated domain. From x³ = x, move all terms to one side: x³ − x = 0. Factorisation gives x(x − 1)(x + 1) = 0, so x = 0, 1, or −1. All three values belong to ℤ, and no other roots arise from the factors. Therefore A = {−1, 0, 1}, making option A correct. The other finite options omit one solution, while ℤ includes many values that do not satisfy the equation.

For option A, x² = x becomes x² − x = 0, or x(x − 1) = 0. Thus the integer solutions are x = 0 and x = 1, so the described set is exactly {0, 1}. Option B gives only {−1, 1} if interpreted over integers, but its stated natural domain gives {1}; option C gives {−1, 1}, and option D is infinite.

The only natural number less than 2 is 1, assuming the usual positive-natural-number convention. However, 1 is not prime: a prime number has exactly two distinct positive divisors, whereas 1 has only one positive divisor. Thus no natural number satisfies both conditions, so A is the empty set, ∅.

The natural numbers less than 10 that are even are 2, 4, 6, and 8. Among these numbers, only 2 is prime, because a prime number has exactly two distinct positive divisors, 1 and itself. Therefore, the conditions “even,” “prime,” and “less than 10” are satisfied simultaneously only by 2. Hence A is the singleton set {2}, making option A correct.

Two sets are equal precisely when every element of the first set belongs to the second set and every element of the second belongs to the first. In other words, their elements must be exactly the same. Equal cardinality alone does not guarantee equality; for example, {1, 2} and {a, b} have the same size but different elements. The order and names used to write sets are irrelevant.

Using the usual school convention ℕ = {1, 2, 3, ...}, the condition x ≤ 4 gives A = {1, 2, 3, 4}. For B, the integers satisfying 1 ≤ x ≤ 4 are also {1, 2, 3, 4}. Since two sets are equal when they contain exactly the same elements, A = B. The different stated domains do not matter after the conditions are applied.

For option B, x² = 9 gives x = 3 or x = −3. The additional condition x > 0 excludes −3 and leaves only x = 3, so the set is {3}. It is non-empty and has exactly one element, making it a singleton set. Option A is empty, option C has two elements, and option D is infinite.

The three-digit natural numbers begin at 100 and end at 999, with both endpoints included. The number of integers in this inclusive interval is 999 − 100 + 1 = 900. Therefore, A is a finite set with 900 elements. The plus-one is necessary because both 100 and 999 are counted.

The natural numbers satisfying 100 < x < 999 are 101, 102, 103, ..., 998. This is a bounded list of consecutive natural numbers, so it contains only finitely many elements. In fact, its cardinality is 998 − 101 + 1 = 898. The strict inequalities exclude 100 and 999, but they do not make the set empty or infinite.

The set in option A is {0, 1, 2, 3, ...}, which continues without end and is therefore infinite. It is not equal to ℤ because ℤ also contains negative integers such as −1 and −2, while option A does not. Option B is finite, option C equals {−1, 1}, and option D is empty because no integer lies strictly between 0 and 1.

Factor the expression as x² − 4x + 4 = (x − 2)². The equation is zero only when x − 2 = 0, so x = 2. Consequently, A = {2} and has exactly one element. Although 2 is a repeated root of the polynomial, a set lists an element only once; repeated occurrence does not increase the cardinality.

Solving x² − 1 = 0 gives (x − 1)(x + 1) = 0, so x = 1 or x = −1. Both solutions belong to ℤ, hence A = {−1, 1}. Under the usual convention ℕ = {1, 2, 3, ...}, only 1 belongs to the natural numbers, so B = {1}. The domain in a set-builder description determines which solutions are included.

A natural number divisible by 11 must be a positive multiple of 11, such as 11, 22, 33, and so on. The smallest positive multiple is 11 itself, but the condition x < 11 excludes 11 and every larger multiple. Therefore, no natural number satisfies both conditions, so A is the empty set, written as ∅.

In a set, the order in which elements are written is irrelevant. The sets {p, q, r} and {r, p, q} both contain exactly the same elements: p, q, and r. Therefore, they are equal. Option B is false because the second set has one additional element, option C is false because {0} contains 0 whereas the empty set contains nothing, and option D is false because the elements are different.

The natural numbers not exceeding 8 are 1, 2, 3, 4, 5, 6, 7, and 8. Among these, the prime numbers are 2, 3, 5, and 7. Removing the primes leaves 1, 4, 6, and 8, so A = {1, 4, 6, 8}. Notice that 1 is neither prime nor composite, but it is certainly not prime, so it must be included.