The Empty Set, many more. Step 3: Hence the set is infinite.

The braces in A = {0} show that 0 is an element of the set. There is exactly one distinct element, so A is a singleton set. A set containing one element is also finite. It is not empty, because the empty set contains no elements, and {0} is not equal to ∅. Therefore, option B is correct.

There are irrational numbers strictly between 0 and 1; for example, √2/2 is approximately 0.707 and is irrational. Moreover, infinitely many distinct irrational numbers lie in this interval. For instance, the density of irrational numbers in the real numbers guarantees irrational numbers between any two different real numbers. Thus the set is infinite, and option D is correct.

Taking n = 1, 2, 3, 4, and so on produces the distinct values 1, 1/2, 1/3, 1/4, and so on. Although these positive fractions become smaller and approach zero, they never become zero and the process of choosing a new natural number never ends. Thus the set has infinitely many elements, so option C is correct.

The first set has exactly two elements: the number 1 and the set {2, 3}. The second set has three elements: the numbers 1, 2, and 3. Although the inner set contains 2 and 3, it is itself one element of the first set; it is not automatically replaced by its members. Therefore the sets are not equal, and option B is correct.

Find the natural-number solutions by factoring the quadratic: x² − 5x + 6 = (x − 2)(x − 3). Hence x = 2 or x = 3. Both values are natural numbers, so the set defined by the condition is A = {2, 3}. Therefore, option B is equal to A. Option A uses incorrect roots, option C gives unrelated negative values, and option D wrongly treats the solution set as empty.

The closed interval [1, 1] contains all real numbers x satisfying 1 ≤ x ≤ 1. The only real number that satisfies both inequalities is x = 1, and the square brackets indicate that the endpoint is included. Therefore [1, 1] contains exactly one element and is equal to the singleton set {1}. Option A is correct.

The open interval (1, 1) consists of real numbers x satisfying 1 < x < 1. No real number can be simultaneously greater than 1 and less than 1. The parentheses also exclude both endpoints, so 1 is not included. Consequently the interval contains no elements and is the empty set ∅. Hence option A is correct.

The symbol ≤ includes equality, so both boundary values are part of the set. The condition −1 ≤ x includes −1, and x ≤ 4 includes 4. In interval notation, an included endpoint is written with a square bracket. Since both endpoints are included, the set is the closed interval [−1, 4].

Set A contains all numbers greater than 1 up to and including 5. Set B contains all numbers from 5, including 5, up to but not including 9. The only number common to both sets is 5. Hence their intersection is the singleton set {5}, not an interval extending on either side of 5.

The complement of A ∪ B consists of all elements in the universal set U that are not in the union. For a finite universal set, n((A ∪ B)') = n(U) − n(A ∪ B). Substituting the values gives 60 − 45 = 15. Thus, option A is correct. In a Venn diagram, this is the area of the rectangle outside both circles.