I need to show that (A∩B)×C = (A×C) ∩ (B×C).
I would really appreciate some feedback on what I have so far. Thanks a bunch!!!
Proof.
First we must show that (A∩B)×C ⊆ (A×C) ∩ (B×C). To this end, let m∈(A∩B)×C. By the definition of Cartesian Product, m∈(A∩B)×C = {(x,y)| x∈(A∩B) and y∈C}. By the definition of intersection, x∈(A∩B) means x∈A and x∈B. So we can say that m = {(x,y)| x∈A and y∈C} and m = {(x,y)| x∈B and y∈C}. In other words, m∈(A×C) ∩ (B×C). Thus (A∩B)×C ⊆ (A×C) ∩ (B×C).
Now we must show that (A×C) ∩ (B×C) ⊆ (A∩B)×C. So let n∈(A×C) ∩ (B×C). This means that n∈(A×C) = {(a,c)| a∈A and c∈C}, and n∈(B×C)={(b,c)| b∈B and c∈C}. Since c∈C is mapped to both a∈A and b∈B, we can say that n∈(A∩B)×C. Thus (A×C) ∩ (B×C) ⊆ (A∩B)×C.
Hence (A∩B)×C = (A×C) ∩ (B×C). ■