Advanced Properties, Cardinalities, and Model-Based Reasoning with Products

If A is nonempty, choose some a ∈ A. For any b ∈ B, (a,b) ∈ A × B, and since A × B ⊆ A × C, (a,b) ∈ A × C, meaning b ∈ C. Thus every b ∈ B belongs to C, so B ⊆ C. Without A nonempty this cancellation is invalid because there may be no (a,b) to test.

A counterexample: take A = C = {1}, B = {2}, D = {3}. Then (A × B) ∖ (C × D) = {(1,2)} because C × D = {(1,3)} does not remove (1,2); meanwhile (A ∖ C) × (B ∖ D) = ∅ × {2} = ∅. So they differ. In general, subtracting products does not factor as product of differences.

Explanation: Intersection of products equals product of intersections; emptiness forces one coordinate-intersection empty.

|A| = 3 and |B| = 2, so |A × B| = 3·2 = 6 by the counting rule for Cartesian product. The list of pairs would be (1,a),(1,b),(2,a),(2,b),(3,a),(3,b).

Belonging to A × B by definition means the first coordinate belongs to A and the second coordinate belongs to B. The other options swap or conflate coordinates incorrectly.

If (a,x) ∈ A × (B ∩ C) then a ∈ A and x ∈ B ∩ C, hence x ∈ B and x ∈ C, so (a,x) ∈ A × B and (a,x) ∈ A × C. Therefore (a,x) is in their intersection. Inclusion holds for all sets independent of finiteness.

If A × B is finite and nonempty, every element of A appears as the first coordinate of some pair and every element of B appears as the second coordinate of some pair (otherwise the product would omit rows or columns). More formally, projection maps from A × B onto A and B are surjective; surjective images of a finite set are finite, so A and B must be finite.

For any a ∈ A, the pair (a,a) has first coordinate a ∈ A and second coordinate a ∈ A, so (a,a) ∈ A × A. Hence the set of all such diagonal pairs is indeed a subset of the product. This holds regardless of whether A is finite or empty (if A empty both sets are empty and inclusion holds).

If A = ∅, then both sides equal ∅: A × (B ∩ C) = ∅ × (B ∩ C) = ∅, and (A × B) ∩ (A × C) = ∅ ∩ ∅ = ∅. Therefore equality holds. More broadly, the equality holds for all A (not just empty ones), but the question asked specifically about A empty where equality certainly holds.

Membership: (a,x) ∈ (A × B) ∩ (A × C) iff (a,x) ∈ A × B and (a,x) ∈ A × C, i.e., a ∈ A and x ∈ B and a ∈ A and x ∈ C; simplifying gives a ∈ A and x ∈ (B ∩ C), so (a,x) ∈ A × (B ∩ C). Hence the equality holds for all sets.

The union over all a ∈ A of the slice {a} × B collects every ordered pair (a,b) with a ∈ A and b ∈ B, which is precisely A × B. B × A would collect pairs with reversed coordinates; intersection would be too small, and ∅ is wrong unless A = ∅ and B = ∅.

For any a ∈ A and x ∈ B ∪ C, x ∈ B or x ∈ C, so (a,x) ∈ A × B or (a,x) ∈ A × C, hence in the union. Conversely, any (a,x) in the union comes from one of the products, so x ∈ B ∪ C. Therefore equality holds generally; no extra conditions needed.

For finite sets, |A × B| = |A|·|B|, so k = m·|B|, hence |B| = k/m provided m ≠ 0. The other algebraic rearrangements are incorrect. Note this assumes sets finite and m divides k.

Option A describes the diagonal set {(x,x) : x ∈ A ∩ B}. For any x ∈ A ∩ B, we have x ∈ A and x ∈ B, so the pair (x,x) has first coordinate in A and second coordinate in B, making (x,x) ∈ A × B. Therefore this diagonal set is indeed a subset of A × B. The other options don't even produce sets of ordered pairs. Options B and D are not sets of ordered pairs; option C yields unions, not pairs.

The two sides have different formal element structure ((a,b),c) vs (a,(b,c)) so they are not equal as sets; however mapping ((a,b),c) ↦ (a,(b,c)) is a natural bijection between them. Option C incorrectly asserts equality ignoring nesting; option D is false because emptiness of the nested product depends on any factor being empty, not just A.