the family set of{Ai,iEI}is a partition of A,
the family set of{Bj,jEJ}is a partition of B,
than prove {AixBj,iEI and jEJ}is a parititon of AxB.
thanks for your helps.
First we show that each cell is not empty.
Given $\displaystyle A_c \times B_d $ because of partitions $\displaystyle \left( {\exists x \in A_{\,c} } \right)\left( {\exists y \in B_d } \right) \Rightarrow \quad \left( {x,y} \right) \in A_c \times B_d $.
Does that prove the first requirement?
Now disjoint cells.
If $\displaystyle \left( {x,y} \right) \in \left( {A_c \times B_d } \right) \cap \left( {A_j \times B_k } \right)$ then $\displaystyle x \in A_c \cap A_j \,\& \,y \in \left( {B_d \times B_k } \right)
$ WHY?
WHY does mean that $\displaystyle \left( {A_c = A_j } \right)\,\& \,\left( {B_d = B_k } \right)$?
Now can you explain how this shows that different cells are disjoint?
Now the covering property.
If $\displaystyle \left( {a,b} \right) \in \left( {A \times B} \right) \Rightarrow \quad \left( {\exists j} \right)\left[ {a \in A_j } \right]\,\& \,\left( {\exists k} \right)\left[ {b \in B_k } \right]\,$. How are we sure of this?
This means that $\displaystyle \left( {a,b} \right) \in \left( {A_j \times B_k } \right)$.
How does that show that $\displaystyle \left( {A \times B} \right) = \bigcup\limits_{j,k} {\left( {A_j \times B_k } \right)}$?