The Empty Set, Finite and Infinite Sets, Equal Sets

For every real number x, x² is nonnegative, so x² + 1 is at least 1 and can never equal zero. Alternatively, the equation would give x² = −1, which has no real solution; its solutions are imaginary numbers. Because the domain is R, no solution is admitted, and the set is empty.

A calendar month may have 28, 29, 30, or 31 days, but no month has 32 days. Therefore no month satisfies the stated property. A set whose defining condition is satisfied by no object is the empty set. It is finite as well, with cardinality zero, but the most specific answer here is empty set.

The symbol ∅ denotes a set with no elements, so its cardinality is 0. In contrast, {∅} is a set whose sole element is the empty set itself; therefore its cardinality is 1. The braces change the role of ∅ from representing a set to being an element inside another set. Hence A is correct.

A finite set has a fixed and limited number of distinct elements. Therefore, its elements can be listed or counted completely, even if the number is large. For example, {2, 4, 6, 8} has four elements. The empty set is also finite because it has zero elements. A set does not need to contain natural numbers to be finite.

The set contains all positive even numbers from 2 through 50. These can be written as 2 × 1, 2 × 2, ..., 2 × 25, so each integer from 1 to 25 gives exactly one element. Alternatively, for an arithmetic sequence, the number of terms is (last term − first term) ÷ common difference + 1 = (50 − 2) ÷ 2 + 1 = 25. Hence the set has 25 elements.

The integers strictly between 20 and 30 are 21 through 29. Among them, 23 and 29 are prime: each has exactly two positive divisors, 1 and itself. The numbers 21 and 27 are composite, so they cannot be included. Thus A = {23, 29}, which has two elements and is therefore finite.

The set is {101, 102, 103, ...}. After every listed positive integer, another larger positive integer exists, so the sequence never reaches a final element. Consequently, the elements cannot be completely counted or listed in a finite list. The set therefore has infinitely many elements and is an infinite set.

The multiples of 7 in the natural numbers are 7, 14, 21, 28, 35, and so on. For every natural number n, 7n is another multiple of 7, so the list never ends. Therefore, the set has infinitely many elements and is an infinite set. It is not empty, a singleton, or a set containing only seven elements.

Whole numbers begin with 0 and continue as 1, 2, 3, and so on. The whole numbers less than 10 are exactly {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. This is a complete list containing ten distinct elements, so the set is finite. It is not empty, because it contains elements, and it is not infinite because the list ends at 9.

In set notation, repetition does not create a new element. Thus A = {1, 2, 3} after removing the repeated 1, and B = {1, 2, 3} after removing the repeated 2 and 3. Since two sets are equal when they contain exactly the same elements, A and B are equal. A has three distinct elements, and B also has three distinct elements.

We need natural-number values of x satisfying x² = 4. The equation has the two integer solutions x = 2 and x = −2, but under the usual school convention N contains positive natural numbers, so −2 is excluded. The only allowed value is x = 2. Therefore, the solution set is A = {2}, making option A correct.

A straight line is continuous and contains points between every pair of its points. No matter how many points are identified, more points can always be found on the same line. Therefore the collection cannot be completely listed as a finite set. The set of all points on a straight line is infinite.

Through one fixed point, a line can be drawn in every possible direction. Different directions produce different lines, and there is no finite limit on the number of directions or lines. Hence the collection contains infinitely many lines. It is therefore an infinite set, not an empty, finite, or two-element set.

The only even prime number is 2. Every other even number is divisible by 2 and therefore has at least two positive divisors, so it is not prime. Since the question asks for even prime numbers greater than 2, no number satisfies the condition. Hence the set has no elements and is the empty set, written as ∅.

The notation {0} means a set whose single element is the number 0. It is important not to confuse the set {0} with the empty set ∅. The empty set contains no elements, whereas {0} contains exactly one element. Therefore, {0} is a non-empty singleton set and option A is correct.

The condition x² = 9 means that x can be 3 or −3, because both (3)² and (−3)² equal 9. Both values belong to the set of integers ℤ, so both must be included in A. Therefore A = {−3, 3}. The values −9 and 9 are not solutions because the equation involves the square of x.

Taking ℕ = {1, 2, 3, ...}, test the natural numbers in order. We have 1² = 1 < 10, 2² = 4 < 10, and 3² = 9 < 10, while 4² = 16 is not less than 10. Thus A = {1, 2, 3}, so it contains exactly three elements. If a convention includes 0 in ℕ, 0² also satisfies the condition; however, the options and standard school convention here use ℕ beginning with 1.

The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. This list contains eight distinct numbers and is complete because no other positive integer divides 24 without a remainder. Since the set can be listed completely and has a fixed number of elements, it is finite. Therefore, option A is correct.

For any number x, the equation x + 3 = x would require subtracting x from both sides, giving 3 = 0. This is impossible, so no natural number satisfies the stated condition. A set defined by a condition has no elements when the condition has no solution. Therefore, the given solution set is the empty set ∅.

A real number has square zero only when the number itself is zero. Thus the equation x² = 0 has the single solution x = 0, and 0 belongs to ℝ. The set is therefore {0}, which contains exactly one element. A one-element set is called a singleton and is finite, so option A is correct.