The Empty Set, Finite and Infinite Sets, Equal Sets

A circle is a continuous geometric curve, not a polygon made of a limited number of points. Between any two distinct points on the circle, there are further points, and this process never ends. Thus the collection of all points on a circle contains infinitely many elements and is an infinite set.

Every horizontal line has an equation of the form y = c, where c is a real number. Since there are infinitely many possible real values of c, there are infinitely many distinct lines parallel to the x-axis. The x-axis itself is one such line, and lines such as y = 1, y = 2, and y = -3 are other examples. Therefore, the set has infinitely many elements and is an infinite set.

The condition requires x to be a natural number and also to satisfy x ≤ 0. Under the definition given in the question, the natural numbers are 1, 2, 3, and so on; none of them is zero or negative. Consequently, no element satisfies both conditions simultaneously. Therefore I₁ contains no elements and is the empty set, written as ∅.

For an integer x, the inequality |x| < 3 is equivalent to -3 < x < 3. The integers strictly between -3 and 3 are -2, -1, 0, 1, and 2. Thus J₁ = {-2, -1, 0, 1, 2}, which contains five distinct elements. The endpoints -3 and 3 are excluded because the inequality is strict, so the correct answer is 5.

The absolute value |x| represents the distance of x from zero, so it can equal zero only when x = 0. Since 0 is an integer, it belongs to the stated domain. Hence K₁ = {0}. A set containing exactly one distinct element is called a singleton set. There are not two solutions, because |0| = 0 has only the single solution x = 0.

The absolute value of every real number, and therefore of every integer, is always non-negative. It can be zero or positive, but it can never equal -2. Thus there is no integer x satisfying |x| = -2. A set defined by a condition with no possible solution contains no elements, so L₁ is the empty set, denoted by ∅.

The positive perfect squares are obtained by squaring natural numbers: 1² = 1, 2² = 4, 3² = 9, and 4² = 16. The next square, 5² = 25, is not less than 20 and must be excluded. Therefore the complete set of perfect-square natural numbers below 20 is M₁ = {1, 4, 9, 16}.

The perfect-square natural numbers begin as 1, 4, 9, 16, 25, and continue as 36, 49, 64, and so on. For every natural number n, the number n² is a perfect square, and there is no largest natural number. Consequently, new perfect squares continue without end, so N₁ contains infinitely many elements and is an infinite set.

The only even prime number is 2. Every other even natural number is divisible by 2 and therefore has at least the divisors 1, 2, and itself, so it is not prime. Hence O₁ = {2}. Because this set contains exactly one element, it is a singleton set. It is neither empty nor infinite, and it certainly does not contain three elements.

Two sets are equal when they contain exactly the same elements, regardless of the order in which those elements are written. P₁ and Q₁ both contain 1, 2, and 3, so P₁ = Q₁. However, R₁ contains 4 instead of 3, so it does not have the same elements as P₁. Therefore, P₁ = Q₁ but P₁ ≠ R₁. None of these sets is empty because each contains three elements.

The natural-number factors of 8 are 1, 2, 4, and 8. The additional condition requires the factor x to be greater than 8, but none of these factors satisfies x > 8; 8 itself is equal to 8, not greater than it. Therefore no natural number meets both conditions, so A has no elements and A = ∅.

The condition -4 < x ≤ 1 requires x to be an integer greater than -4 and less than or equal to 1. Therefore, the permitted integers are -3, -2, -1, 0, and 1. The endpoint -4 is excluded because the inequality is strict, while 1 is included because the inequality is non-strict. Hence B has exactly five elements.

Since x = 2n and n ranges over the natural numbers, C contains even natural numbers such as 2, 4, 6, 8, and so on. There is no greatest natural number, so n can always be increased to produce another even member of C. Consequently, the listing never terminates and C is an infinite set. This conclusion is unchanged whether 0 is included in ℕ.

Factor the quadratic equation: x² - 5x + 6 = (x - 2)(x - 3) = 0. Thus x = 2 or x = 3. Both values are natural numbers, so D = {2, 3}. Since E is also {2, 3}, the two sets contain exactly the same elements and are equal. The order or form in which elements are written does not affect set equality.

Two sets are equal precisely when they have the same elements, meaning every element of the first set belongs to the second and every element of the second belongs to the first. The order of listing is irrelevant because sets are unordered collections. For example, {1, 2, 3} and {3, 1, 2} are equal, whereas {1, 2} and {1, 3} are not equal even though both have two elements.

Solve the defining equation: x² - 1 = 0 can be factored as (x - 1)(x + 1) = 0. Therefore, x = 1 or x = -1. Both values belong to the integers, so A = {-1, 1}. This is exactly the set B. Set elements may be written in either order, so the different possible ordering does not make the sets unequal.

Factor the equation x² - 7x + 12 = 0 as (x - 3)(x - 4) = 0. Its solutions are x = 3 and x = 4. Both are natural numbers, so both satisfy the defining condition of C. Therefore C = {3, 4}, which contains exactly two distinct elements. Repeated or equivalent descriptions would not increase the cardinality, but here the two roots are distinct.

The equation x² + 2x + 1 = 0 is (x + 1)² = 0, so its only algebraic solution is x = -1. However, the set definition requires x to be a natural number. Since -1 is not a natural number, it is rejected and no permitted element remains. Thus D contains no elements and is the empty set, written as ∅.

The condition |x + 1| ≤ 2 means that x + 1 lies between −2 and 2, inclusive. Therefore, −2 ≤ x + 1 ≤ 2. Subtracting 1 from all three parts gives −3 ≤ x ≤ 1. Since x must be an integer, the possible values are −3, −2, −1, 0, and 1. There are 5 values, so the set F has 5 elements. The endpoints are included because the inequality is ‘less than or equal to’.

The positive natural-number factors of 18 are 1, 2, 3, 6, 9, and 18. Among these, only 2 and 3 are prime numbers because each has exactly two positive divisors. The number 1 is not prime, and 6, 9, and 18 are composite. Therefore the set satisfying both conditions is G = {2, 3}.