Equal sets and Subsets

A set is equal to U only when it has exactly the same elements. The positive natural multiples of 3 that are less than or equal to 12 are 3, 6, 9, and 12, which gives precisely U. Option A omits 12 because it uses x < 12. Option C omits 3 and 9, while option D includes many nonmultiples of 3. Therefore option B is correct.

If two sets are equal, they contain exactly the same elements. Consequently, their cardinalities, or numbers of elements, must also be equal. However, the converse is not generally true: for example, {1, 2} and {a, b} have the same cardinality but are not equal because their elements differ. Also, ∅ has zero elements, while {0} has one element.

In a set, the order of elements does not matter. Hence {1, 2} and {2, 1} represent exactly the same inner set. The outer set A therefore contains only one distinct element, namely the set {1, 2}. Consequently, the cardinality of A is n(A) = 1, not 2. Repeated or differently ordered descriptions do not create new elements in a set.

The governing idea is equality of sets: two sets are equal when they contain exactly the same elements. For A, x² − 1 = 0 gives x² = 1, so x = 1 or x = −1. For B, |x| = 1 also has the integer solutions x = 1 and x = −1. Therefore A = {−1, 1} and B = {−1, 1}, so A = B. The other choices incorrectly omit a solution or misclassify the set as empty or infinite.

Every number strictly between 4 and 6 is also strictly between 2 and 8. Therefore each element of B belongs to A, so B ⊆ A. The intervals are not equal because A contains additional numbers such as 3 and 7. Their intersection is B, not the empty set, so option B is the only correct statement.

The closed interval A includes every number from 2 to 8, including both endpoints. The open interval B contains only numbers strictly between 2 and 8, so it excludes 2 and 8. Every element of B is nevertheless an element of A; hence B ⊆ A. The intervals are not equal and 2 is not in B.

The interval (-∞, 5) represents every real number x such that x < 5. The round bracket at 5 shows that 5 itself is excluded. The interval contains negative numbers, zero, and positive numbers less than 5, so it is certainly not restricted to positive numbers and is not empty. Infinity is not an actual endpoint, so it is always written with a round bracket.

A = {1, 2, {3}}. A subset may contain only objects that are elements of A. Both 1 and {3} belong to A, so {1, {3}} is a subset. However, 3 itself is not an element of A; only the set {3} is. Therefore the other options contain an object that is not in A or have an incorrect nested structure.

Solving x² = 4 over the real numbers gives x = 2 or x = -2. Hence A = {-2, 2}. Since B contains exactly the same two elements, A and B are equal, so A = B. Neither is a proper subset of the other. Option D is incorrect because it omits the valid solution -2. Equality of sets depends on having exactly the same elements, regardless of their order.

The symbol ∈ relates an object to a set, while ⊆ relates one set to another. Since 1 is listed in A, 1 ∈ A is true. The singleton set {1} contains only an element of A, so {1} ⊆ A is also true. Option B reverses these meanings, option C ignores 2 and 3, and option D fails because 4 is not in A.

The set A contains the two numbers 1 and 2 as its elements. Therefore 1 ∈ A is true, and both {1} ⊆ A and {1, 2} ⊆ A are true because every member of each set belongs to A. However, {1, 2} is the whole set A, not an element listed inside A. Hence {1, 2} ∈ A is false.

The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Therefore A = {1, 2, 3, 4, 6, 12}, which is exactly the set B. Since both sets contain the same elements, they are equal. Neither proper-subset statement can be true, and A is certainly not empty because 1, among other numbers, is a positive divisor of 12.

The positive divisors of 8 are 1, 2, 4, and 8, so B = {1, 2, 4, 8}. Every element of A = {2, 4, 8} is in B, but B contains the additional element 1, which is not in A. Therefore A is a proper subset of B, written A ⊂ B. The sets are not equal, and their intersection is A rather than the empty set.

When A is a subset of B, every element of A is contained in B. For finite sets, if A is a proper subset of B, then A must have fewer elements than B. Here both sets have cardinality 5, so A cannot be a proper subset of B. Since A ⊆ B and their cardinalities are equal, they must contain exactly the same elements; hence A = B. Option A is correct.

Because the element 1 must be included, it is already fixed as one member of every required subset. We therefore need to choose the remaining 2 elements from the other 5 elements, namely 2, 3, 4, 5, and 6. The number of choices is C(5, 2) = 5!/(2!3!) = 10. Hence, exactly 10 three-element subsets contain 1.

Factor the quadratic equation: x² − 7x + 12 = (x − 3)(x − 4) = 0. Therefore, the natural-number solutions are x = 3 and x = 4, so A = {3, 4}. Since B is also exactly {3, 4}, the two sets have the same elements and hence A = B. The order of elements does not matter in a set.

Use the antisymmetry property of set inclusion. The given relations produce A ⊆ B ⊆ C, while A = C changes this chain into A ⊆ B ⊆ A. Thus both A ⊆ B and B ⊆ A hold, so A = B. Since A = C is already given, all three sets are equal. Option A is correct. Options B and C incorrectly assert proper inclusion, and D contradicts the stated equality A = C.

Solving x² = 16 gives x = 4 or x = −4. Therefore A = {−4, 4}, which is exactly the set B. Sets are equal when they contain the same elements; the order in which those elements are written is irrelevant. Thus A = B, while the proper-subset and empty-intersection statements are false.

The positive even integers less than 10 are 2, 4, 6, and 8. Hence the set-builder description gives B = {2, 4, 6, 8}. Since this is exactly the roster form of A, both sets contain the same elements and A = B. No other positive even integer below 10 exists.

Factor the quadratic equation: x² - 5x + 6 = (x - 2)(x - 3). Therefore its solutions are x = 2 and x = 3, so the solution set is A = {2, 3}. Since B is also {2, 3}, the two sets have exactly the same elements and A = B. Option B confuses the constant term with the solution set. Option C is false because a proper subset must be smaller than the other set, while these sets are equal. Option D is false because their intersection is {2, 3}.

The elements 1, 2, and 3 of A all occur in B, so A is a subset of B. However, B also contains the element {1}, which is a set containing 1; it is not the same object as the number 1. Consequently, B has one additional element and A is a proper subset of B. The sets are not equal, B is not a subset of A, and their intersection is not empty because they share 1, 2, and 3.

Solving \(x^2-4=0\) gives \((x-2)(x+2)=0\), so the integer solutions are \(x=2\) and \(x=-2\). However, the definition of \(A\) asks specifically for a natural-number solution. Under the usual school convention, 2 is natural but −2 is not. Hence \(A=\{2\}\), and since \(B=\{2\}\), the two sets are equal. Therefore option A is correct.

The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Selecting only the even divisors gives \(A=\{2,4,6,8,12,24\}\). Every element of \(\{2,6,8\}\) belongs to this set, so it is a subset of \(A\). Each other option contains at least one element that is not in \(A\): 1 or 3, 5, or 9. Therefore option A is the only correct answer.

The statement A ⊆ B says that every element of A belongs to B. The additional statement that B has no element outside A says that every element of B belongs to A, or B ⊆ A. Thus both inclusions hold: A ⊆ B and B ⊆ A. By the criterion for equality of sets, A = B. The value n(A) = 8 is consistent with this and also implies n(B) = 8, not 16. Hence option A is certain.

Solve the absolute-value inequality: |x − 2| < 2 gives −2 < x − 2 < 2. Adding 2 throughout yields 0 < x < 4. Since x must be an integer, A = {1, 2, 3}. The set {1, 3} contains only elements of A, so it is a subset of A, but it does not contain 2 and therefore is not equal to A. Hence it is a proper subset. Option A equals A, while C and D contain elements not belonging to A.