- Problem set 10, question 2 answer is in 3 parts: conjecture, first proof, 2nd proof
- Section 5.4
- Given A = {a,b}, B = {x,y,z}, find A × B
- Set of all ordered pairs that pair element A with element of B
- { (a,x), (a,y), (a,z), (b,x), (b,y), (b,z) }
- Think of it as a Cartesian grid indexed by elements of A, B
- What about {a,b} × ∅?
- Turns out to be the empty set
- What is A × B × C? Ordered triples? Ordered pairs of ordered pairs?
- Ordered triples: {a,b} × {x,y,z} × {1,2} = { (a,x,1), (a,x,2), … }
- Thus, for example, 3D space = R3
- Note that the formal definition of an ordered triple (a,b,c) is as the ordered pair ((a,b),c)
- So A × B × C = (A × B) × C ≠ A × (B × C)
- Ordered triples: {a,b} × {x,y,z} × {1,2} = { (a,x,1), (a,x,2), … }
- Why is Cartesian product not commutative?
- Ordered pairs are ordered
- Prove (A - B) × C = (A × C) - (B × C) (Theorem 5.25 part 6)
- Proof features
- Prove set equality: A ⊆ B and B ⊆ A
- Take pairs apart, reason about parts, put together again
- First direction: (A - B) × C ⊆ (A × C) - (B × C)
- Let u = (x,y) ∈ (A - B) × C, thus x ∈ A-B = A ∩ BC, y ∈ C
- So x ∈ A, and x ∈ BC
- So (x,y) = u ∈ A × C
- And u ∈ BC × C so u ∉ B × C
- Thus u ∈ (A × C) - (B × C)
- Proof features
- Finish second direction of proof that (A - B) × C = (A × C) - (B × C)
- Indexed families of sets
- Read section 5.5