Equal sets and Subsets

The phrase x < 10 restricts x to the natural numbers less than 10. Under the convention N = {1, 2, 3, ...}, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}; these are exactly the one-digit natural numbers, so A = B. Even if a text includes 0 in N, both definitions include the same one-digit natural numbers, so their equality remains true. Option D is not universally correct because it depends on whether 0 is included in N.

For an integer x, the inequality |x| ≤ 2 means that x lies between -2 and 2 inclusive. The integers satisfying this condition are -2, -1, 0, 1, and 2. Hence A = {-2, -1, 0, 1, 2}, which is exactly the given set B. Therefore A = B. Option B omits the negative integers, option D omits the values between the endpoints, and option C is false because the two sets are equal rather than properly contained.

For set A, solve x² − 1 = 0. Factoring gives (x − 1)(x + 1) = 0, so x = 1 or x = −1; both are integers. Thus A = {−1, 1}. For set B, the integers strictly between −2 and 2 are −1, 0, and 1. The condition x ≠ 0 removes 0, leaving B = {−1, 1}. Since A and B contain exactly the same elements, A = B. Therefore, option A is correct; B, C, and D are false.

Two sets are equal only when they contain exactly the same elements, regardless of the order in which those elements are written. The left-hand set already contains 5, so the right-hand set must also contain 5; this requires y = 5. The right-hand set contains 3, so the left-hand set must contain 3; this requires x = 3. Therefore option A gives {3,5} = {3,5}. Each other pair produces unequal sets or a repeated element on one side.

The set A has exactly three elements: 1, 2, and the set {3}. The number 3 itself is not directly listed as an element of A; instead, the one-element set {3} is listed. Therefore, {3} ∈ A is true. Statement A is false because 3 ∉ A. Statement C is false because {1, 2, 3} has 3 as an element, whereas A has {3} as an element. Statement D is also false because {3} ⊂ A would require its element 3 to belong to A. Always distinguish membership (∈) from subset notation (⊂).

An n-element set has 2ⁿ total subsets because every element can independently be included or excluded. For n = 4, the total is 2⁴ = 16. A proper subset is a subset that is not equal to the original set, so the original set must be removed from this count. Therefore, the number of proper subsets is 16 − 1 = 15. Option B counts the original set too, while 8 and 14 do not follow from the proper-subset formula.

The statement A ⊆ B means that every element of A is also an element of B. Consequently, taking the intersection of A and B leaves every element of A and no additional element, so A ∩ B = A. Conversely, if A ∩ B = A, every element of A lies in B, which proves A ⊆ B. Option B instead represents B ⊆ A; option C means A and B are disjoint; and option D represents A ⊆ B only in a different difference condition, not the stated equivalence.

A 2-element subset is formed by choosing 2 different elements from the 4-element set A. Since the order of elements does not matter in a set, combinations are used rather than permutations. The number is C(4, 2) = 4!/(2!2!) = (4 × 3)/2 = 6. The six subsets are {p,q}, {p,r}, {p,s}, {q,r}, {q,s}, and {r,s}. Therefore, option C is correct.

The element 1 is required, so it is fixed as included in every counted subset. The remaining four elements—2, 3, 4, and 5—can each be independently included or excluded. That gives 2 choices for each of four elements, so the number of subsets is 2⁴ = 16. Equivalently, exactly half of the 2⁵ = 32 total subsets contain any specified element. Therefore, option B is correct.

Because both a and b must be present, they are fixed and do not create choices. Only the remaining three elements, c, d, and e, can be independently included or excluded. Therefore, there are 2³ = 8 possible subsets containing both required elements. They are obtained by combining {a,b} with any subset of {c,d,e}. Option D, 32, counts all subsets of A, while 4 incorrectly allows choices for only two remaining elements.

Two sets are equal only when they contain exactly the same elements. Here, \(a\in A\) but \(a\notin B\), while \(d\in B\) but \(d\notin A\), so \(A\ne B\). Since one missing element is enough to disprove a subset relation, \(A\not\subseteq B\); similarly, \(B\not\subseteq A\). Therefore, neither set is a subset of the other. The correct answer is D.

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 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.

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 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.

Taking N = {1, 2, 3, …}, the condition x ≤ 5 gives A = {1, 2, 3, 4, 5}, whereas x < 5 gives B = {1, 2, 3, 4}. Every element of B belongs to A, so B ⊆ A. Since 5 is in A but not in B, equality fails and B is a proper subset of A. Thus option B is correct; the sets are not disjoint and the inclusion direction is not reversed.

The key concept is the difference between an element and a set containing that element. Here A has exactly two elements: ∅ and the number 1. The set {∅} contains only ∅, which belongs to A, so {∅} ⊆ A. In contrast, 0 is not in A, 2 is not in A, and {{1}} contains the set {1}, not the number 1 itself. Therefore only option A is a subset of A.