The Empty Set, Finite and Infinite Sets, Equal Sets

Two sets are equal when they contain exactly the same elements, regardless of the order in which those elements are written. A and B both contain a, b, and c, so A = B. Set C contains only a and b; it does not contain c. Therefore, C is not equal to A, and the correct statement is A = B and A ≠ C. None of these sets is empty.

The positive natural-number factors of 18 are 1, 2, 3, 6, 9, and 18. Thus A = {1, 2, 3, 6, 9, 18}. This list is complete because every factor of 18 occurs in one of the factor pairs 1×18, 2×9, or 3×6. Since A has six countable elements, it is a finite set.

For every integer x, the square x² is non-negative: it is either zero or positive. Therefore, no integer can have a square equal to −4. Although complex numbers can solve the equation, the condition specifically restricts x to ℤ, the integers. Hence the defining condition has no allowed solution, so the set is empty and is written as ∅.

The natural numbers not exceeding 12 that are divisible by 3 are 3, 6, 9, and 12. Therefore B = {3, 6, 9, 12}, which is exactly the set A. Two sets are equal when they contain precisely the same elements; the order or the way in which the elements are described does not affect equality. Hence A = B.

The condition requires a natural number strictly greater than 5 and strictly less than 6. Since 5 and 6 are consecutive natural numbers, there is no natural number satisfying both inequalities. Thus A contains no element and is the empty set, written as ∅. It is not a singleton because no value meets the stated condition.

Since x must be an integer and must lie between −3 and 3, the elements of C are −3, −2, −1, 0, 1, 2, and 3. Counting them gives seven distinct elements. Therefore, C is a finite set and its cardinality, denoted by n(C), is 7. The inclusive signs ≤ ensure that both boundary values, −3 and 3, are included.

The odd natural numbers begin 1, 3, 5, 7, 9, and continue indefinitely. For every odd natural number, adding 2 produces another odd natural number, so there can never be a final element. The set therefore contains infinitely many elements. Option D lists only some initial elements, not the complete set.

Solving x² = 4 gives x = 2 or x = −2. However, the set specifically restricts x to natural numbers. Under the usual convention that natural numbers are positive counting numbers, 2 is natural but −2 is not. Consequently B = {2}, which has exactly one element and is a singleton set.

A set is determined by its elements, not by the order in which those elements are written. Both C and D contain exactly the four distinct elements 2, 4, 6, and 8. Reversing their display order does not create a new set. Therefore C and D are equal. The other options incorrectly treat order as important or misclassify a clearly finite set.

The positive multiples of 5 are 5, 10, 15, 20, 25, and so on. For every listed positive multiple, adding 5 gives another positive multiple, so the sequence has no final term. Because the set continues without end, it contains infinitely many elements and is an infinite set. It is neither empty nor finite.

The months in a year are January through December, giving exactly 12 distinct elements. Since this list has a fixed number of members and the counting ends after December, the set is finite. The set is not empty, not infinite, and “equal set” describes a relationship between two sets rather than the size of one set.

Repeated writing of an element does not increase the number of distinct elements in a set. In H, the symbol a appears twice, but it is still only one distinct element. Thus H can be rewritten as {a, b, c}, which is exactly G. Therefore G and H are equal sets, and H does not have four distinct elements.

The set contains integers strictly greater than 2 and strictly less than 3. The only numbers between 2 and 3 are non-integers, while there is no integer in this open interval. Therefore, no integer satisfies the condition, so the set has no elements and is the empty set, written as ∅.

By definition, an empty set is a set containing no element. Therefore, its cardinality, or number of elements, is zero, written as n(∅) = 0. It is not infinite. Also, the order in which elements are written does not matter in a set, so equal sets need not have the same arrangement.

Taking N as the natural numbers 1, 2, 3, and so on, the condition x ≤ 50 restricts J to J = {1, 2, 3, ..., 50}. This set has exactly 50 elements, so its cardinality is |J| = 50. Because the list has a fixed final element and cannot continue beyond 50, J is finite. It is not empty or infinite, while “equal set” is not a classification that answers the question.

The natural numbers satisfying x ≥ 50 are 50, 51, 52, 53, and so on. There is no greatest natural number, so more elements can always be added to the list. Consequently, K has infinitely many elements and is an infinite set, not a finite or singleton set.

For every real number x, x² ≥ 0. Consequently, x² + 1 ≥ 1, so the expression can never equal 0. Equivalently, solving the equation gives x² = −1, which has no real solution because the square of a real number cannot be negative. Therefore no real number belongs to L, and L is the empty set. The alternatives involving one, two, or infinitely many elements would require real solutions that do not exist.

The set-builder description of N selects every natural number less than 5. Thus N = {1, 2, 3, 4}, assuming natural numbers begin with 1. This is exactly the roster form of M. Two sets are equal when they contain precisely the same elements, regardless of how those elements are written.

The square of every natural number is non-negative: for every x ∈ N, x² ≥ 0. Hence the inequality x² < 0 cannot be satisfied by any natural number. There is therefore no element that belongs to Q. By definition, a set with no elements is the empty set, so Q = ∅. The choices Q = N, {0}, or {1} are impossible because each would contain elements whose squares are not negative.

To find the elements, solve the defining equation x² = 16. Taking square roots gives x = 4 or x = −4, since both 4² = 16 and (−4)² = 16. Both values are integers and therefore satisfy the domain condition x ∈ Z. Thus R = {−4, 4}. Option A omits one valid solution, option C confuses the square with the value of x, and option D wrongly claims that no solution exists.