Equal sets and Subsets

The set A contains three elements: the number 1, the number 2, and the set {1, 2}. The inner set {1, 2} is counted as one single element of A, not as two additional elements. Therefore, n(A) = 3. This illustrates that a set itself can be an element of another set, and elements must be counted according to how they are listed at the outermost level.

To determine A, solve x^2 = 9. The integer solutions are x = -3 and x = 3, so A = {-3, 3}. Set B contains these two elements and one additional element, 9: B = {-3, 3, 9}. Hence every element of A belongs to B, but A and B are not equal. Therefore, A is a proper subset of B.

Set B has exactly three elements: 2, 3, and 5. A subset with three elements must therefore contain all elements of B, so the only such subset is B itself, namely {2, 3, 5}. Since every element of B is also an element of A, B is a subset of A; in fact, it is a proper subset because A contains additional elements 1, 4, and 6.

Assuming N denotes the positive natural numbers, A = {1, 2, 3, 4, 5, 6}. The positive divisors of 6 are B = {1, 2, 3, 6}. Every element of B is in A, so B is a subset of A. However, A also contains 4 and 5, which are not in B. Thus B is not equal to A; it is a proper subset of A, making option B correct.

Solve the defining equation: x² − 2x = x(x − 2) = 0. Therefore, x = 0 or x = 2. Both values are integers, so A = {0,2}. Since B is also defined as {0,2}, the two sets contain exactly the same elements and therefore A = B. Option B omits 0, option C incorrectly claims a proper subset relationship, and option D ignores the two valid solutions.

A prime number is a natural number greater than 1 that has exactly two positive divisors: 1 and itself. The prime numbers from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Number 1 is not prime because it has only one positive divisor. Thus A and B contain exactly the same elements, so A = B. Option D incorrectly includes 1.

The positive multiples of 5 are 5, 10, 15, 20, 25, 30, and so on. The strict condition x < 30 excludes 30, leaving exactly 5, 10, 15, 20, and 25. These are precisely the elements listed in B. Therefore A and B contain the same elements, so A = B. Option B wrongly includes the boundary value 30; options C and D do not describe the actual relationship.

A set records membership, not the number of times an element is written. Thus the repeated 1 in the notation for A is counted only once, so A simplifies to {1,2,3}. The order of elements is also irrelevant, meaning B = {3,2,1} represents the same set {1,2,3}. Since both sets contain exactly the same distinct elements, A = B. Neither set is a proper subset of the other.

The elements a, b, and c of A all occur in B, so A is a subset of B. However, B also contains d, which is not in A. Therefore the two sets are not equal, and A is a proper subset of B, written A ⊂ B. Statement A is false because of the extra element d in B. Statement C reverses the inclusion, and statement D is false because every element of A is indeed in B.

To find A, solve x² = 4 over the integers. Taking square roots gives x = 2 or x = -2, and both values belong to Z. Therefore A = {-2,2}. This is exactly the roster description used for B, so A = B. Option B incorrectly omits the negative solution. Option C is false because equal sets are not proper subsets of one another, and option D is false because A has two elements rather than none.

The positive factors of 12 are the positive integers that divide 12 without a remainder. Checking the divisors gives 1, 2, 3, 4, 6, and 12; numbers such as 5 and 8 do not divide 12 exactly. Hence the set-builder description produces A = {1,2,3,4,6,12}, which is precisely the roster form of B. Therefore A = B. It is finite, so option D is also false, and neither proper-subset option applies.

A set is a subset of A only when every one of its elements belongs to A. The empty set is a subset of every set, so option A is valid. Both 1 and 3 belong to A, so {1, 3} is a subset. The set {1, 2, 3} is A itself and is therefore also a subset of A. However, 4 is not an element of A, so {2, 4} cannot be a subset. Hence option C is the only correct answer.

The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, A = {2,3,5,7}, which contains exactly the same elements as B. Sets are equal when they have the same elements, regardless of order. Option B and option C incorrectly claim a proper-subset relationship, while option D incorrectly includes 1; 1 is neither prime nor composite. Hence option A is correct.

To determine B, solve x² = x. Rearranging gives x² − x = 0, or x(x − 1) = 0. Hence x = 0 or x = 1, so B = {0,1}. Since A also contains exactly 0 and 1, the two sets have identical elements and are equal. Thus A = B. The other choices wrongly describe a proper subset, omit 0, or claim that the intersection is empty.

When A ⊆ B, every element of A is also contained in B. The intersection A ∩ B consists of elements common to both sets. Because all elements of A are common to A and B, and B cannot contribute an element that is not in A to the intersection, the result is exactly A. Hence A ∩ B = A, so option A is correct.

The empty set ∅ has no elements, and it is a subset of every set. Since B = {0} is nonempty, A and B are not equal; therefore, A is a proper subset of B. The statement 0 ∈ A is false because A has no elements, and B cannot be a subset of the empty set because B contains 0. Hence option B is correct.

Every element of B, namely 2 and 4, is present in A, so B is a proper subset of A and also a subset of A. Every set is a subset of itself, so A ⊆ A is true. However, A has four elements while B has only two, and 1 and 3 are not in B. Therefore A cannot be a proper subset of B, making option D false.

The positive odd natural numbers less than 8 are 1, 3, 5, and 7. Thus the rule-based description of B gives B = {1, 3, 5, 7}, which is exactly the listed set A. Since two sets are equal when they contain precisely the same elements, A = B. Neither is a proper subset of the other, and their intersection is not empty.

For any set \(A\), every element of \(A\) is automatically an element of \(A\). This satisfies the definition of a subset, so \(A\subseteq A\) is always true. The statement does not depend on whether the set is empty, finite, or infinite. For the empty set, there is no element that violates the condition, so \(\varnothing\subseteq\varnothing\) is also true. Therefore, option A is correct.

Here \(A\subset B\) is understood as a proper-subset relation: every element of \(A\) belongs to \(B\), and at least one element of \(B\) is not in \(A\). Consequently, the two sets cannot have exactly the same elements, so \(A\ne B\). This conclusion applies to finite, infinite, and empty-set examples whenever the proper-subset relation is valid. Therefore, the statement is true and option A is correct.

A positive prime number has exactly two positive divisors. The positive prime numbers less than 4 are 2 and 3; 1 is not prime, and no other positive integer less than 4 qualifies. Thus the descriptive set-builder form becomes A = {2,3}. Since B is also {2,3}, both sets contain precisely the same elements and are equal. Therefore option B is correct.

W denotes the set of whole numbers: 0, 1, 2, 3, and so on. The condition x < 3 restricts x to 0, 1, and 2. Therefore B = {0,1,2}, which is exactly the same set as A. The order of elements does not affect a set, and no additional element is present in either set. Hence A = B, so option A is correct.

A set is a subset of A only when every element of that set belongs to A. The elements 1 and 3 are in A, so {1,3} is a subset. The empty set is a subset of every set, and A itself is also a subset of A. However, {2,4} contains 4, which is not in A. Therefore {2,4} is not a subset, making option C correct.

For a finite set with n elements, the total number of subsets is 2^n. A proper subset is a subset that is not equal to the original set. Among all 32 subsets, exactly one subset is the original set itself. Hence the number of proper subsets is 32 − 1 = 31. The empty set is included among the proper subsets.

The positive divisors of 6 are the positive integers that divide 6 without leaving a remainder. They are 1, 2, 3, and 6. Therefore A = {1, 2, 3, 6}, which is exactly the same collection of elements as B. Since sets are equal when they contain precisely the same elements, A = B is necessary.