Equal sets and Subsets

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.

For natural numbers, the condition \(x\le4\) selects \(1,2,3,4\) under the usual convention \(\mathbb N=\{1,2,3,\ldots\}\). The condition \(x<5\) selects exactly the same natural numbers, because a natural number less than 5 must be one of 1, 2, 3, or 4. Thus both sets have the same elements: \(A=B=\{1,2,3,4\}\). Therefore, option A is correct.

Solve the quadratic condition defining \(A\): \(x^2-4x+3=(x-1)(x-3)=0\). Hence the integer solutions are \(x=1\) and \(x=3\), so \(A=\{1,3\}\). This is exactly the set \(B\). The order in which elements are written does not matter in a set, and no element is repeated. Therefore, \(A=B\), making option A correct.

Equal sets contain exactly the same elements, although the order of listing does not matter. The two numbers in the left-hand set differ by 2, just as 4 and 6 do. Taking m = 4 gives m + 2 = 6, so the left side becomes {4,6}, which is exactly the right-hand set. The alternative m = 6 would give {6,8}, not {4,6}. Therefore, m = 4 and option B is correct.

The set A has three distinct elements: the empty set ∅, the singleton set {2}, and the number 2. Since {2} is explicitly listed, {2} ∈ A is true. Also, the only element of {2} is 2, and 2 is explicitly in A; therefore {2} is a proper subset of A, so {2} ⊂ A is true. Hence option A is correct. Notice that 2 and {2} are different objects.

Because A ⊆ B, every element of A is already an element of B. If A were a proper subset of B, then B would have at least one additional element and would therefore contain more than seven elements. But both sets have cardinality 7. This is impossible, so no extra element exists in B and the two sets must be equal. Therefore A = B, making option A 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.

A set treats a number and a set containing that number as different elements. In A, the third element is the singleton set {3}; in B, the third element is the number 3. Although 3 belongs to {3}, the objects 3 and {3} are not identical. Consequently, A and B do not contain exactly the same elements, so A ≠ B. Option C is correct.

The notation {} is another way to write the empty set ∅. Therefore B = {{}} means that B contains one element, namely the empty set. Similarly, A = {∅} also contains one element, namely ∅. Thus A and B both represent the singleton set whose only element is the empty set. They are equal, but neither A nor B is itself empty. Hence option A is correct.

Factor the quadratic equation: x² − 5x + 6 = (x − 2)(x − 3) = 0. Hence x = 2 or x = 3, so the set of solutions is A = {2,3}. The given set B is also {2,3}. Since two sets are equal when they have exactly the same elements, A = B. Therefore option A is the correct relation.

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.

Both elements of A, namely 1 and 2, are also elements of B, so A is a subset of B. However, B contains one additional element, the empty set ∅, which is not an element of A. Thus the two sets are not equal, and A is a proper subset of B. Notice that ∅ as an element differs from the empty set B itself.

Subset inclusion is transitive. From A ⊆ B and B ⊆ C, we obtain A ⊆ C. The additional condition C ⊆ A gives inclusion in both directions between A and C, so A = C. Similarly, B is contained in C and C is contained in A, which forces B to contain no extra elements. Therefore all three sets are equal: A = B = C. They may all be empty.

The condition 1 ≤ n ≤ 3 with n a natural number allows n = 1, 2, and 3 only. Substituting these values into x = 2n gives x = 2, 4, and 6. Therefore B = {2, 4, 6}. Since A contains exactly the same elements as B, the two sets are equal, so option A is correct. The order in which elements are listed does not matter.

The set A has three elements: the number 1, the set {1, 2}, and the number 2. Therefore {1, 2} is itself an element of A, so {1, 2} ∈ A is true. In fact, {1, 2} is also a subset of A because both 1 and 2 belong to A. The symbols ∈ and ⊆ must not be confused: one describes an element, while the other describes a set contained in another set.

Since p must be present, it is fixed in every counted subset. The remaining elements q and r are optional, and each can independently be included or excluded. Therefore there are 2 choices for q and 2 choices for r, giving 2 × 2 = 4 subsets. They are {p}, {p, q}, {p, r}, and {p, q, r}. Equivalently, a three-element set has 2^3 subsets, and fixing p reduces the free choices to two elements, giving 2^2 = 4.

Because x must be an integer, we list the integers beginning at −2 and less than 2: −2, −1, 0, and 1. The symbol ≤ includes −2, whereas the symbol < excludes 2. Hence A = {−2, −1, 0, 1}, which is exactly option A. The order of elements does not affect equality of sets.