The Empty Set, Finite and Infinite Sets, Equal Sets

A calendar year has exactly twelve months: January, February, March, April, May, June, July, August, September, October, November, and December. These names form a set with a fixed and limited number of distinct elements. Because its number of elements is 12, not unlimited, A is a finite set. Therefore option B is correct; it is neither empty nor infinite.

The natural-number factors of 6 are 1, 2, 3, and 6, so A = {1, 2, 3, 6}. In B, the number 6 appears twice, but repeated entries are counted only once in a set. Thus B also represents {1, 2, 3, 6}. Since A and B contain exactly the same elements, A = B. Therefore option A is correct.

Solving x² + 1 = 0 gives x² = −1. For every natural number x, x² is non-negative, so it can never equal −1. Hence no natural number satisfies the defining equation. A set whose defining condition has no solution is the empty set. Therefore A = ∅, and the correct statement is that A is an empty set, represented by option A.

The months having 31 days are January, March, May, July, August, October, and December. There are exactly seven such months, and this list can be completely counted and has a final element. A set with a fixed, countable number of elements is called a finite set. Therefore, B is finite, not empty or infinite.

For B, we list the natural numbers that are multiples of 3 and less than 15. They are 3, 6, 9, and 12; 15 is excluded because the condition is x < 15. Thus B = {3, 6, 9, 12}, which contains exactly the same elements as A. Since two sets are equal when they have precisely the same elements, A = B. Therefore, option A is correct.

For every natural number n, the expression x = 2n gives an even number. If N begins with 1, the set contains 2, 4, 6, 8, and so on; if a convention includes 0, 0 is also included, without changing the conclusion. Since natural numbers continue indefinitely, there is no last generated element. Therefore C is infinite.

The prime natural numbers less than 10 are 2, 3, 5, and 7. Therefore, the set described in option A is exactly {2, 3, 5, 7}. Number 1 is not prime, so option B has an extra element. Option C contains even numbers rather than the required primes, and option D contains 11 while omitting 2. Hence only option A represents a set equal to the given set.

Evaluate the condition x² < 10 for natural numbers. For x = 1, 2, and 3, the squares are 1, 4, and 9, all less than 10. For x = 4, x² = 16, which is not less than 10; larger natural numbers also fail. Thus A = {1, 2, 3}, exactly the same as B. Therefore A = B, so option A is correct.

Solving x² = 25 over the integers gives two solutions: x = 5 and x = −5. Both numbers belong to Z, so A = {−5, 5}. These are distinct elements, and no other integer has square 25. Therefore the cardinality of A, written |A|, is 2. The negative solution must not be omitted.

The prime factorisation of 24 is 24 = 2 × 2 × 2 × 3 = 2³ × 3. The prime factors are therefore 2 and 3. A set records membership, not repeated occurrences, so 2 is written only once even though it appears three times in the factorisation. Neither 1 nor 24 is a prime factor, and 8 is not prime. Hence A = {2, 3}, making option A correct.

Factor the equation as x² + 2x + 1 = (x + 1)² = 0. Its only real and integer solution is x = −1. However, −1 is not a natural number, so it does not satisfy the restriction x ∈ N. Consequently, no permitted value belongs to A, which means A = ∅, the empty set.

List the positive factors of each number. The factors of 18 are 1, 2, 3, 6, 9, and 18, while the factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The elements appearing in both lists are 1, 2, 3, and 6. Therefore, P is the set of common factors, so P = {1, 2, 3, 6}; option A is correct.

Because x is an integer, list every integer from −3 up to but not including 2. The condition −3 ≤ x includes −3, and x < 2 excludes 2. Thus A = {−3, −2, −1, 0, 1}. Counting these elements gives 5. The set is finite, not infinite, so option B is the only correct answer.

For option A, substitute the natural-number values n = 1, 2, 3, and 4 into x = 4n. The resulting elements are 4, 8, 12, and 16, exactly the elements of A. Therefore, option A represents the same set. Equality of sets depends on having precisely the same elements; the order of elements does not matter, but missing or additional elements would make the sets unequal.

A prime number has exactly two distinct positive divisors, 1 and itself. Among natural numbers, 2 is the only number that is both even and prime. Every other even natural number greater than 2 is divisible by 2 and another number, so it is composite. Hence the set contains only one element and is A = {2}, making option B correct.

The set of natural-number factors of 12 is {1, 2, 3, 4, 6, 12}. It has six elements, so it is finite, and it has several elements, so it is not empty. Option A is empty because no integer has square −9. Options B and D are infinite. Therefore, only option C satisfies both required conditions.

The outer braces define the set A. Inside them are three distinct objects: the number 0, the empty set ∅, and the singleton set {0}. Although {0} contains 0 as its own element, it is not the same object as 0. Likewise, ∅ is a set with no elements. Thus A has three elements, so its cardinality is 3.

The natural numbers less than 40 that are divisible by 5 are 5, 10, 15, 20, 25, 30, and 35. The number 40 is excluded because the condition is x < 40, not x ≤ 40. There are seven listed elements, so the set is finite and its cardinality is 7. Therefore, option A is correct.

The natural-number multiples of 4 are 4, 8, 12, 16, 20, and so on. For every multiple listed, adding 4 produces another natural-number multiple, so the list never ends. There is no upper bound in the defining condition. Consequently, A contains infinitely many elements and is an infinite set. Option C is therefore correct.

In a set, repetition of an element does not create a new element. The repeated 2s and 3s are therefore written only once when the set is simplified. The distinct elements are 1, 2, and 3, so A = {1, 2, 3} and its cardinality is 3, not 6. It is neither empty nor infinite. Hence option B is correct.