Question

Given sets A and B, when will A×B be equal to B×A ? Write N×R using set-builder notation and then describe this set grometrically by interpreting it as a subset of R².

We now turn our attention to subsets of Cartesian products.
Let A,B,C, and D be sets. If A⊆C and B⊆D, then A×B⊆C×D. Is it true that if A×B⊆C×D, then A⊆C and B⊆D ?
Do not forget to think about cases involving the empty set.
Is every subsct of C×D of the form A×B, where A⊆C and B⊆D ? If so, prove it. If not, find a counterexample.
If A,B, and C are nonempty sets, is A×B a subset of A×B×C ? Let A=[2,5],B=[3,7],C=[1,3], and D=[2,4].
Compute each of the following.

Answer 1

The sets A × B and B × A are equal when the sets A and B are commutative. N × R can be written using set-builder notation. The statement A × B ⊆ C × D does not imply that A ⊆ C and B ⊆ D.

Step-by-step explanation:

The sets A × B and B × A are equal when the sets A and B are commutative, meaning that the order of elements does not matter. For example, if A = {1, 2} and B = {3, 4}, then A × B = {(1, 3), (1, 4), (2, 3), (2, 4)} and B × A = {(3, 1), (4, 1), (3, 2), (4, 2)}. These two sets are equal.

The set N × R can be written using set-builder notation as n ∈ N, r ∈ R, which represents the Cartesian product between the set of natural numbers and the set of real numbers. Geometrically, this set can be interpreted as a subset of the Euclidean plane, R², meaning that each ordered pair (n, r) represents a point on the plane.

The statement A × B ⊆ C × D does not imply that A ⊆ C and B ⊆ D. For example, if A = {1} and B = {2}, and C and D are empty sets, then A × B is also an empty set, which is a subset of C × D. However, A is not a subset of C and B is not a subset of D.

Not every subset of C × D is of the form A × B. For example, the subset {(1, 2)} is a subset of C × D = {[1, 3] × [2, 4]}, but it cannot be written as A × B for any subsets A and B of C and D, respectively.

If A, B, and C are nonempty sets, then A × B is not a subset of A × B × C. This is because A × B × C contains ordered triples of the form (a, b, c), where a ∈ A, b ∈ B, and c ∈ C, whereas A × B only contains ordered pairs.

Using the given sets A = [2, 5], B = [3, 7], C = [1, 3], and D = [2, 4], we can compute the following:

1. A × B = {[2, 3], [2, 4], [2, 5], [2, 6], [2, 7], [3, 3], [3, 4], [3, 5], [3, 6], [3, 7], [4, 3], [4, 4], [4, 5], [4, 6], [4, 7], [5, 3], [5, 4], [5, 5], [5, 6], [5, 7]}
2. A × B × C = {[2, 3, 1], [2, 3, 2], [2, 4, 1], [2, 4, 2], [2, 5, 1], [2, 5, 2], [2, 6, 1], [2, 6, 2], [2, 7, 1], [2, 7, 2], [3, 3, 1], [3, 3, 2], [3, 4, 1], [3, 4, 2], [3, 5, 1], [3, 5, 2], [3, 6, 1], [3, 6, 2], [3, 7, 1], [3, 7, 2], [4, 3, 1], [4, 3, 2], [4, 4, 1], [4, 4, 2], [4, 5, 1], [4, 5, 2], [4, 6, 1], [4, 6, 2], [4, 7, 1], [4, 7, 2], [5, 3, 1], [5, 3, 2], [5, 4, 1], [5, 4, 2], [5, 5, 1], [5, 5, 2], [5, 6, 1], [5, 6, 2], [5, 7, 1], [5, 7, 2]}