Algebra of Cartesian Products: Unions, Intersections, and Differences

1) If f: A → B is a function, its graph Γ(f) ⊆ A × B consists of:

All pairs (a,b) where b = f(a)

All pairs (b,a) where b=f(a)

All pairs (a,a)

None of the above

The graph of a function f is the set of ordered pairs (a,f(a)) for each a in the domain A. It records the association of each input a with its unique output f(a). Writing pairs (b,a) would reverse the order; (a,a) would only apply to identity-type functions; “none” is false.

Explanation

The graph of a function f is the set of ordered pairs (a,f(a)) for each a in the domain A. It records the association of each input a with its unique output f(a). Writing pairs (b,a) would reverse the order; (a,a) would only apply to identity-type functions; “none” is false.

2) For sets A,B, which holds: (A × B) ∖ (A × C) = ?

A × (B ∖ C)

(A∖A) × (B∖C)

(A×B)∖A

(A∖C) × B

Remove from A × B those pairs whose second coordinate lies in C. Formally, (a,b) ∈ (A × B) ∖ (A × C) iff a ∈ A, b ∈ B, and (a,b) ∉ A × C. The last condition means not (a ∈ A and b ∈ C) which simplifies—since a ∈ A is known—to b ∉ C. Thus b ∈ B ∖ C, giving A × (B ∖ C).

Explanation

Remove from A × B those pairs whose second coordinate lies in C. Formally, (a,b) ∈ (A × B) ∖ (A × C) iff a ∈ A, b ∈ B, and (a,b) ∉ A × C. The last condition means not (a ∈ A and b ∈ C) which simplifies—since a ∈ A is known—to b ∉ C. Thus b ∈ B ∖ C, giving A × (B ∖ C).

3) Which describes A × {b} for fixed b?

A set of ordered pairs with second component b

Set of singletons {b}

Union of A and {b}

Equal to A

A × {b} = {(a,b) : a ∈ A}. Each element is an ordered pair whose second coordinate is the fixed element b. It is not a set of singletons {b}, not a union, and not equal to A unless one intentionally identifies (a,b) with a which is not standard.

Explanation

A × {b} = {(a,b) : a ∈ A}. Each element is an ordered pair whose second coordinate is the fixed element b. It is not a set of singletons {b}, not a union, and not equal to A unless one intentionally identifies (a,b) with a which is not standard.

4) If A = {1,2} and B = {1,2}, is (1,2) ∊ A × B?

Yes

No

Only if A≠B

Only if ordered pairs unordered

Since 1 ∈ A and 2 ∈ B, the ordered pair (1,2) is in A × B by definition. Options C and D bring irrelevant constraints.

Explanation

Since 1 ∈ A and 2 ∈ B, the ordered pair (1,2) is in A × B by definition. Options C and D bring irrelevant constraints.

5) What is (A × B) ∩ (C × D) equal to?

(A ∩ C) × (B ∩ D)

(A ∪ C) × (B ∪ D)

A × (B ∩ D)

C × (A ∩ B)

A pair (x,y) belongs to both A × B and C × D iff x ∈ A and y ∈ B and x ∈ C and y ∈ D, which is equivalent to x ∈ A ∩ C and y ∈ B ∩ D. Hence the product of intersections. The union or mixed forms are incorrect.

Explanation

A pair (x,y) belongs to both A × B and C × D iff x ∈ A and y ∈ B and x ∈ C and y ∈ D, which is equivalent to x ∈ A ∩ C and y ∈ B ∩ D. Hence the product of intersections. The union or mixed forms are incorrect.

6) If A × B ⊆ C × D, does A ⊆ C necessarily hold?

Yes always

No, not necessarily

Only if B nonempty

Only if sets finite

If B is nonempty, pick b ∈ B. For any a ∈ A, (a,b) ∈ A × B; since A × B ⊆ C × D, (a,b) ∈ C × D, so a ∈ C and thus A ⊆ C. If B were empty then A × B is empty and the subset relation holds vacuously while A might not be a subset of C. Finiteness is irrelevant.

Explanation

If B is nonempty, pick b ∈ B. For any a ∈ A, (a,b) ∈ A × B; since A × B ⊆ C × D, (a,b) ∈ C × D, so a ∈ C and thus A ⊆ C. If B were empty then A × B is empty and the subset relation holds vacuously while A might not be a subset of C. Finiteness is irrelevant.

7) Which is a valid element of 𝒫(A) × 𝒫(B) (cartesian product of the power sets)?

({1},{a}) if A={1},B={a}

{1,a}

(1,a)

Elements of 𝒫(A) × 𝒫(B) are ordered pairs where the first element is a subset of A and the second is a subset of B. If A = {1} and B = {a}, then {1} ∈ 𝒫(A) and {a} ∈ 𝒫(B), so ({1},{a}) is an element. Option B is a set combining elements from both A and B, not a pair of subsets; C is a pair of elements not subsets; D is a singleton set containing the union.

Explanation

Elements of 𝒫(A) × 𝒫(B) are ordered pairs where the first element is a subset of A and the second is a subset of B. If A = {1} and B = {a}, then {1} ∈ 𝒫(A) and {a} ∈ 𝒫(B), so ({1},{a}) is an element. Option B is a set combining elements from both A and B, not a pair of subsets; C is a pair of elements not subsets; D is a singleton set containing the union.

8) True or False: (A × B) ∪ (C × D) = (A ∪ C) × (B ∪ D).

True

False

True iff A⊆C

True iff B⊆D

The right-hand side contains pairs (a,d) where a ∈ A and d ∈ D, even if that pair was never in A × B or C × D; for example take a ∈ A\C and d ∈ D\B, then (a,d) is in (A ∪ C)×(B ∪ D) but not in either A×B or C×D. Thus the union of products is generally a subset of but not equal to the product of unions. Special cases (like A ⊆ C and B ⊆ D) can make the equality hold, but in general it is false.

Explanation

The right-hand side contains pairs (a,d) where a ∈ A and d ∈ D, even if that pair was never in A × B or C × D; for example take a ∈ A\C and d ∈ D\B, then (a,d) is in (A ∪ C)×(B ∪ D) but not in either A×B or C×D. Thus the union of products is generally a subset of but not equal to the product of unions. Special cases (like A ⊆ C and B ⊆ D) can make the equality hold, but in general it is false.

9) If (a,b) ∊ A × B and (a,c) ∊ A × C, then what can we infer?

B = c

A ∊ A and b ∊ B and c ∊ C

(b,c) ∊ A × C

A ∊ B

From the membership statements we immediately know the coordinates’ set memberships: (a,b) ∈ A × B implies a ∈ A and b ∈ B; (a,c) ∈ A × C implies a ∈ A and c ∈ C. So overall a ∈ A, b ∈ B, c ∈ C. Nothing forces b and c to be equal, (b,c) in B × C is true but is a separate assertion not implied directly from given pairs (though b ∈ B and c ∈ C makes (b,c) ∈ B × C true—this is derivable but the safest immediate inference is the coordinate memberships). Option D is wrong because a need not be in B.

Explanation

From the membership statements we immediately know the coordinates’ set memberships: (a,b) ∈ A × B implies a ∈ A and b ∈ B; (a,c) ∈ A × C implies a ∈ A and c ∈ C. So overall a ∈ A, b ∈ B, c ∈ C. Nothing forces b and c to be equal, (b,c) in B × C is true but is a separate assertion not implied directly from given pairs (though b ∈ B and c ∈ C makes (b,c) ∈ B × C true—this is derivable but the safest immediate inference is the coordinate memberships). Option D is wrong because a need not be in B.

10) Which mapping is a bijection between A × {b} and A?

(a,b) ↦ a

A ↦ (b,a)

(a,b) ↦ b

(a,b) ↦ (a,b)

The map π: A × {b} → A defined by π((a,b)) = a is injective because if π((a1,b)) = π((a2,b)) then a1 = a2, and surjective because for each a ∈ A we have (a,b) ∈ A × {b} mapping back to a. Thus it is a bijection. Option B has the coordinates reversed and does not even map from A × {b} to A. Option C maps every pair to the same b, not a bijection unless A is singleton; option D is the identity map on A × {b}, not a map to A.

Explanation

The map π: A × {b} → A defined by π((a,b)) = a is injective because if π((a1,b)) = π((a2,b)) then a1 = a2, and surjective because for each a ∈ A we have (a,b) ∈ A × {b} mapping back to a. Thus it is a bijection. Option B has the coordinates reversed and does not even map from A × {b} to A. Option C maps every pair to the same b, not a bijection unless A is singleton; option D is the identity map on A × {b}, not a map to A.

11) If set A is nonempty, is there a bijection between A and A × {0}?

Yes

No

Only if A finite

Only if A infinite

Define f: A → A × {0} by f(a) = (a,0). This map is injective (f(a1)=f(a2) implies (a1,0)=(a2,0), hence a1=a2) and surjective because every element of A × {0} is of the form (a,0) for some a ∈ A. Hence A and A × {0} are in bijection regardless of finiteness, provided A is nonempty (if A empty both sides are empty and there is a trivial bijection between empty sets).

Explanation

Define f: A → A × {0} by f(a) = (a,0). This map is injective (f(a1)=f(a2) implies (a1,0)=(a2,0), hence a1=a2) and surjective because every element of A × {0} is of the form (a,0) for some a ∈ A. Hence A and A × {0} are in bijection regardless of finiteness, provided A is nonempty (if A empty both sides are empty and there is a trivial bijection between empty sets).

12) Which statement holds for all sets: ∅ × A = A × ∅ ?

True

False, unless A is empty

False, they differ

True only if A empty

Both ∅ × A and A × ∅ are empty sets since either coordinate has no element, so both products equal ∅ and are therefore equal. This equality holds for all sets A, regardless of whether A is empty or not. Option B’s phrasing is confusing but the core claim is that both are empty so equality holds—thus True is correct without extra conditions.

Explanation

Both ∅ × A and A × ∅ are empty sets since either coordinate has no element, so both products equal ∅ and are therefore equal. This equality holds for all sets A, regardless of whether A is empty or not. Option B’s phrasing is confusing but the core claim is that both are empty so equality holds—thus True is correct without extra conditions.

13) Let A={1}, B={2}, C={3}. Which is an element of (A × B) × C?

((1,2),3)

(1,(2,3))

(1,2,3)

(3,(1,2))

First A × B = {(1,2)}. Then (A × B) × C consists of ordered pairs whose first coordinate is an element of A × B (i.e., (1,2)) and second coordinate is an element of C (3), hence ((1,2),3). Option B is of the other nesting type A × (B × C); C is a shorthand triple not formal as an ordered pair nesting; D reverses the coordinates.

Explanation

First A × B = {(1,2)}. Then (A × B) × C consists of ordered pairs whose first coordinate is an element of A × B (i.e., (1,2)) and second coordinate is an element of C (3), hence ((1,2),3). Option B is of the other nesting type A × (B × C); C is a shorthand triple not formal as an ordered pair nesting; D reverses the coordinates.

14) If A × B = B × A for all sets, what would follow?

Ordered pairs symmetric always

(a,b) = (b,a) for all a,b

Sets must be equal with identification of pairs

Nothing special

For two arbitrary singletons A = {a} and B = {b}, their products would be {(a,b)} and {(b,a)} respectively. If these sets are equal for all possible choices of sets, then (a,b) must equal (b,a) as ordered pairs for arbitrary a and b. That implies an equality of ordered-pair objects that abolishes the order distinction entirely. Option C is vague; the precise logical consequence is pairwise equality.

Explanation

For two arbitrary singletons A = {a} and B = {b}, their products would be {(a,b)} and {(b,a)} respectively. If these sets are equal for all possible choices of sets, then (a,b) must equal (b,a) as ordered pairs for arbitrary a and b. That implies an equality of ordered-pair objects that abolishes the order distinction entirely. Option C is vague; the precise logical consequence is pairwise equality.

15) If A finite nonempty and B finite nonempty, is A × B finite?

Yes

No

Only if A=B

Only if sets disjoint

The product of two finite sets is finite; specifically, |A × B| = |A|·|B|, which is finite when both factors are finite and nonempty. Neither equality of A and B nor disjointness is required.

Explanation

The product of two finite sets is finite; specifically, |A × B| = |A|·|B|, which is finite when both factors are finite and nonempty. Neither equality of A and B nor disjointness is required.

16) If A ⊆ C and B ⊆ D, then A × B ⊆ C × D ?

True

False

Only if A=C and B=D

Only if sets finite

If (a,b) ∈ A × B, then by definition a ∈ A and b ∈ B. Since A ⊆ C and B ⊆ D, it follows a ∈ C and b ∈ D, so (a,b) ∈ C × D. Therefore A × B ⊆ C × D. No finiteness assumption is needed.

Explanation

If (a,b) ∈ A × B, then by definition a ∈ A and b ∈ B. Since A ⊆ C and B ⊆ D, it follows a ∈ C and b ∈ D, so (a,b) ∈ C × D. Therefore A × B ⊆ C × D. No finiteness assumption is needed.