The Empty Set, Finite and Infinite Sets, Equal Sets

The empty set has no elements. The statement ∅ ⊆ A means that every element of ∅ is also an element of A. Because there are no elements in ∅, there is no counterexample to this requirement; therefore the statement is true for every set A. In contrast, A ⊆ ∅ is true only when A itself is empty, and ∅ = A is not always true. Hence option A is correct.

The order of elements does not matter in a set. Two sets are equal when they contain exactly the same elements. Both given sets contain 1, 2, and 3, even though the elements are written in different orders. Therefore, {1, 2, 3} = {3, 2, 1}; they are equal sets, and option A is correct.

The positive natural-number divisors of 9 are 1, 3, and 9. Therefore the set described by A is \(A=\{1,3,9\}\), which is exactly the given set B. Since two sets are equal when they contain precisely the same elements, A and B are equal. They are not disjoint, empty, or in a proper-subset relationship. The order in which elements are listed would not affect equality, although here the order is already the same.

Using the standard school convention N = {1, 2, 3, ...}, the natural numbers not exceeding 4 are 1, 2, 3, and 4. Hence A = {1, 2, 3, 4}. Set B contains exactly these same elements, merely written in reverse order. The order of elements does not matter in a set, so A = B. Option B is false for this reason; option C is false because equal sets are not proper subsets of one another; and option D is false because B has four elements.

Prime divisors are prime numbers that divide the given number exactly. The prime factorization of 42 is 42 = 2 × 3 × 7. Therefore, the complete set of prime divisors of 42 is A = {2, 3, 7}. This is exactly the set specified as B, so A = B. Options B and C are incorrect because equal sets are not proper subsets of one another. Option D is false because A contains three elements and is therefore not empty.

To expand B, list the positive even integers less than 8. They are 2, 4, and 6; 8 is not allowed because the inequality is strict, and neither 0 nor negative even numbers are positive. Therefore B = {2,4,6}, which has exactly the same elements as A. Hence A = B. Options B and C incorrectly claim a proper-subset relation, while option D is false because A ∩ B = {2,4,6}, not the empty set.

The positive multiples of 6 that are less than 20 are 6, 12, and 18. The next positive multiple is 24, which is not less than 20. Thus A = {6,12,18}. This is exactly the set B, so A and B are equal. The set is finite because it has only three elements, and neither set is a proper subset of the other. Therefore, option A is correct.

An even integer is divisible by 2 without leaving a remainder. Checking the elements of A one by one, 2, 4, and 6 are even, whereas 1, 3, and 5 are odd. Therefore the set containing all even elements of A is B = {2, 4, 6}. Option C misses 6, option A contains the odd elements, and option D includes every element rather than only the even ones.

The positive divisors of 15 are the positive numbers that divide 15 exactly: 1, 3, 5, and 15. Thus the rule defining A gives A = {1, 3, 5, 15}. This is precisely the list used to define B, so A and B contain the same elements and are equal. Options B and C incorrectly claim a proper-subset relation, while D omits 1 and 15.

The empty set ∅ is a subset of every set A. A subset relation requires every element of the first set to belong to the second set. Since ∅ has no elements, there is no element that can violate this condition; consequently, ∅ ⊆ A is always true. The other statements require additional conditions: A must be empty for A ⊆ ∅ or ∅ = A, and ∅ ∈ A depends on whether A contains the empty set as an element.

Having the same cardinality means only that the two finite sets contain the same number of elements; it does not mean that the elements themselves are identical. For example, \(A=\{1,2\}\) and \(B=\{3,4\}\) both have cardinality 2, but they are different sets. Equality requires every element of A to be in B and every element of B to be in A.

Two singleton sets are equal precisely when their only elements are equal. Therefore, from {2a + 1} = {7}, we obtain the equation 2a + 1 = 7. Subtracting 1 from both sides gives 2a = 6, and dividing by 2 gives a = 3. Substitution verifies the result: 2(3) + 1 = 7, so both singleton sets become {7}. Thus option B is the unique correct answer.

The notation says that \(x\) must be an integer greater than \(-3\) and less than or equal to \(2\). The integers greater than \(-3\) begin with \(-2\), and the upper endpoint \(2\) is included because of the symbol \(\le\). Thus the complete list is \(\{-2,-1,0,1,2\}\), which is option A. Option B incorrectly includes \(-3\); option C omits \(2\); and option D includes \(-3\) while omitting \(2\).

Each side is a singleton set, meaning that each set has exactly one element. Two singleton sets are equal only when their sole elements are equal, so \(a+2=9\). Subtracting 2 from both sides gives \(a=7\), which is option C. The braces indicate sets; they do not change the equality principle. A common error is to add 2 to 9 and choose 11, but the equation requires subtraction because 2 is already added to \(a\).

The statement ∅ ∈ A says that the empty set itself is an element of A. Therefore, A contains at least one element, namely ∅, and so A is non-empty. It does not mean that A equals the empty set; in fact, the empty set has no elements. Also, every set has at least the empty subset, so the remaining statements are not valid.

To list B, substitute the allowed values n = 1, 2, 3, and 4 into x = 2n. This gives x = 2, 4, 6, and 8, so B = {2, 4, 6, 8}. This roster is exactly the same as A, meaning the two sets have identical elements and therefore A = B. Neither set is a proper subset of the other, because proper inclusion requires unequal sets. Their intersection is also the whole set, not empty.

The set A has exactly two direct elements: the set {1} and the number 2. Although 1 appears inside the element {1}, it is not separately listed as an element of A. Therefore, the number of elements or cardinality of A is n(A) = 2. Nested elements must not be counted again.

Cardinality counts the elements at the outermost level of a set. In \(A=\{\{2\},4,6\}\), the first element is the set \(\{2\}\) itself, the second element is 4, and the third element is 6. The number 2 is inside the first element and is not a separate outer element of \(A\). Therefore \(A\) has three elements and \(n(A)=3\), making option B correct.

The complement A' is defined as the set of all elements in the universal set U that are not in A. If A is the empty set, it contains no elements, so no element of U is removed. Consequently every element of U belongs to A', and therefore ∅' = U. Option A confuses a set with its complement, while P(U) is a different power set.

A−B is the portion of A lying outside B. If its cardinality is zero, that portion contains no elements, so no element of A can lie outside B. Therefore every element of A belongs to B, which is exactly the definition of A⊆B. The condition n(A)>0 only confirms that A is nonempty; it does not imply B⊆A or disjointness.

The union contains all elements of both sets, so it must contain at least as many elements as the larger set. Therefore n(A ∪ B) ≥ max(n(A), n(B)) = 109. This lower bound is attainable if the smaller set B is completely contained in A. In that case A ∪ B = A and its cardinality is 109. Hence the minimum possible value is 109, option A.

The intersection consists of all elements that belong to both sets. The order is irrelevant in set notation, so \(B=\{3,2,1\}\) is the same set as \(\{1,2,3\}\). Every element 1, 2, and 3 is common to A and B. Therefore, \(A\cap B=\{1,2,3\}\), making option A correct.

The union of two sets contains every distinct element appearing in either set, but a common element is counted only once. Here, \(A\cup B=\{1,2,3,4,5,6\}\), so its cardinality is 6. Equivalently, \(n(A\cup B)=4+3-1=6\), because 4 belongs to both sets. Option B incorrectly counts 4 twice; options C and D do not represent the complete union.

The equation \(x+y=0\) is equivalent to \(y=-x\). We must ensure that the first component belongs to A and the second belongs to B. For \(x=-1\), \(y=1\); for \(x=0\), \(y=0\); and for \(x=1\), \(y=-1\). These give \((-1,1),(0,0),(1,-1)\). Values \(x=-2\) and \(x=2\) would require 2 and -2, neither of which is in B. Therefore, the answer is 3, option B.