Operations on Sets (Union, Intersection, Difference)

To prove \(B=C\), consider any element \(x\). If \(x\in A\), equality of intersections gives \(x\in B\) exactly when \(x\in C\). If \(x\notin A\), equality of unions gives the same equivalence, because membership in either union must then come from \(B\) or \(C\). Thus every element has identical membership in \(B\) and \(C\), so \(B=C\).

First find each difference separately. Elements in \(A\) but not \(B\) are \(a,c\), so \(A\setminus B=\{a,c\}\). Elements in \(B\) but not \(A\) are \(e\), so \(B\setminus A=\{e\}\). Their union is therefore \(\{a,c\}\cup\{e\}=\{a,c,e\}\). This is the symmetric difference, so option A is correct.

For any two finite sets, \(|A\cup B|=|A|+|B|-|A\cap B|\). Here \(A\cap B=\varnothing\), so the intersection has cardinality zero. Substitution gives \(|A\cup B|=|A|+|B|-0=|A|+|B|\). The sets are disjoint, so no element is counted twice. Hence option A is correct.

Every element of the union belongs to exactly one of three disjoint regions: \(A\setminus B\), \(A\cap B\), or \(B\setminus A\). Combining the given regions gives \(\{1,4\}\cup\{2,3\}\cup\{5,6,7\}=\{1,2,3,4,5,6,7\}\). Therefore option A is correct. The other options omit one of the three regions.

To find each set, solve its defining quadratic equation. For A, x² − 5x + 6 = (x − 2)(x − 3) = 0, so A = {2, 3}. For B, x² − 7x + 12 = (x − 3)(x − 4) = 0, so B = {3, 4}. The intersection contains only elements present in both sets. Since 3 is common to A and B, A ∩ B = {3}. The set {2, 3, 4} would be the union, not the intersection.

The sets A \ B, B \ A, and A ∩ B represent three mutually disjoint regions in the Venn diagram. Every element of A ∪ B belongs to exactly one of these regions. Therefore, |A ∪ B| = |A \ B| + |B \ A| + |A ∩ B| = 9 + 4 + 11 = 24. Equivalently, |A| = 9 + 11 = 20 and |B| = 4 + 11 = 15, so |A ∪ B| = |A| + |B| − |A ∩ B| = 20 + 15 − 11 = 24.

The equation \(A\cap B=B\) says that every element of \(B\) is also an element of \(A\), so \(B\subseteq A\). The equation \(A\cup B=A\) gives exactly the same conclusion: adding all elements of \(B\) to \(A\) does not enlarge \(A\). Thus option A is necessary. The reverse inclusion, disjointness, and complement relation do not necessarily follow.

First form the union: \(B\cup C=\{0,3,4,5,6,9\}\), because every element appearing in either set is included once. Now intersect this result with \(A=\{1,3,5,7,9\}\). The common elements are 3, 5, and 9, so \(A\cap(B\cup C)=\{3,5,9\}\). Therefore option A is correct; option D is only the union, not the final intersection.

Every set can be decomposed into two disjoint parts relative to \(C\): the elements outside \(C\), namely \(A\setminus C\), and the elements inside \(C\), namely \(A\cap C\). The two corresponding parts of \(A\) and \(B\) are given equal. Their unions therefore are equal: \(A=(A\setminus C)\cup(A\cap C)=(B\setminus C)\cup(B\cap C)=B\). Hence option A must hold.

The statement A ∩ B ⊆ C means that every element belonging to both A and B also belongs to C. Therefore, there cannot be any element of A ∩ B that lies outside C. The difference set (A ∩ B) \ C contains exactly those elements of A ∩ B that are not in C, so it must be the empty set, ∅.

We have \(A\cap B=\{3,4\}\) and \(A\setminus B=\{1,2\}\). These two sets are disjoint. A set that belongs to both power sets must be a subset of both \(\{3,4\}\) and \(\{1,2\}\). The only common subset of disjoint sets is the empty set. Since the empty set is an element of every power set, the intersection of the two power sets is \(\{\varnothing\}\), not \(\varnothing\).

Use the standard decomposition \(B=(B\setminus A)\cup(A\cap B)\). From \(A\cup B=A\cup C\), the parts outside \(A\) are equal: \(B\setminus A=C\setminus A\). The additional condition \(A\cap B=A\cap C\) makes the parts inside \(A\) equal as well. Thus both disjoint components of \(B\) and \(C\) match, so \(B=C\). The other statements do not generally determine equality.

From A ∩ B = A \ C, every element of A ∩ B belongs to A but does not belong to C. Hence no element can simultaneously belong to A ∩ B and C. Since A ∩ B ∩ C consists precisely of elements common to A, B, and C, it follows that A ∩ B ∩ C = ∅. The first given equality is consistent with this result but is not needed for the final conclusion.

The inequality \(|x|\le4\) gives \(-4\le x\le4\), so \(A=\{-4,-3,-2,-1,0,1,2,3,4\}\). The condition \(x^2\le9\) is equivalent to \(|x|\le3\), giving \(B=\{-3,-2,-1,0,1,2,3\}\). Removing all elements of \(B\) from \(A\) leaves only \(-4\) and 4. Therefore \(A\setminus B=\{-4,4\}\).

An element of \(A\setminus(B\setminus C)\) is in \(A\), but it is not in \(B\setminus C\). Being outside \(B\setminus C\) means either being outside \(B\), or being in \(C\). Therefore the result consists of elements in \(A\setminus B\) together with elements in \(A\cap C\). Hence \(A\setminus(B\setminus C)=(A\setminus B)\cup(A\cap C)\), making option A correct.

A set with cardinality zero is the empty set, so \(n(A)=0\) implies \(A=\varnothing\). The intersection of the empty set with any set is empty because there is no element that can belong to both sets. Therefore \(A\cap B=\varnothing\). The cardinality equation is also consistent with this: \(n(A\cup B)=n(B)=0+n(B)\).

The sets A ∩ B and A \ B are always disjoint: an element cannot be both in B and outside B. If two disjoint sets are equal, their common set must contain no elements. Therefore A ∩ B = A \ B = ∅. Since A \ B = ∅ alone does not always imply A = ∅, the equality with A ∩ B is essential here, and option A is the only necessary conclusion.

Because A ⊆ B, every element of A is also an element of B. The condition B ∩ C = ∅ says that B and C have no common elements. Consequently, no element of A can belong to C either, because any such element would also belong to B and would contradict the disjointness of B and C. Hence A ∩ C = ∅.

Assuming ℕ includes the positive natural numbers, A = {1, 2, 3, ..., 12}. The even elements are B = {2, 4, 6, 8, 10, 12}, and the prime elements are C = {2, 3, 5, 7, 11}. The union contains every element appearing in either set: B ∪ C = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}. The common element 2 is counted only once, so the union has 9 elements. Equivalently, |B ∪ C| = |B| + |C| − |B ∩ C| = 6 + 5 − 1 = 10? Careful: the displayed union omits no elements; it contains 10 elements, so the correct answer is B.

Every element of B belongs to A ∪ B, so the left side contains all elements of B. However, A \ B is defined as the set of elements in A that are not in B; it contains no element of B. If the two sides are equal, B can have no elements at all. Hence B = ∅. The other options are not forced by the given equality.