The meaning of $\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$? Definition $\{A_i\}_{i\in I}$ be an indexed family of classes; Let
$$A=\bigcup_{i\in I} A_i.$$
The $product$ of the classes $A_i$ is defined to be the class
$$\prod_{i\in I}A_i\{f:f:I\rightarrow A\ is\ a\ function,\ and\ f(i)\in A_i,\forall i\in I \}$$
And I want to prove 
Let $\{A_i\}_{i\in I}$ and $\{B_j\}_{j\in J}$ be families of classes. Prove the following:
$$(\prod_{i\in I}A_i)\cap(\prod_{j\in J}B_j)=\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$$
if $$\bigcup_{i\in I} A_i=\bigcup_{j\in J} B_j=X$$
is satisfying.
But I don't know the meaning of $\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$.
If there are someone who know meaning can you give me some explain using example 
$I=\{a,b\},J=\{x,y,z\},A_a=\{1,2\},A_b=\{2,3,4\},B_x=\{1,4\},B_y=\{1,3\},B_z=\{1,2\}.$
and if you can please explain 
how $$(\prod_{i\in I}A_i)\cap(\prod_{j\in J}B_j)=\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$$ is satisfying?
 A: $I$ and $J$ are sets. Define a family $\{ X_k \}_{k \in \{ 1, 2 \}}$ by $X_1 = I, X_2 = J$ and form the product $P = \prod_{k \in \{ 1, 2 \}} X_k$. We write $P = I \times J$. Then $I \times J$ is again a set and it easy to see that is nothing else than the "naive" construct $\{ (i,j) \mid i \in I, j \in J \}$ (each function $f : \{ 1, 2 \} \to X_1 \cup X_2 = I \cup J$ such that $f(1) \in X_1 = I, f(2) \in X_2 = J$ can be identified with the pair $(f(1),f(2))$).
For any $(i,j) \in I \times J$ you obtain the intersection class $C_{(i,j)} = A_i \cap B_j$ and you can form the product
$$\prod_{(i,j) \in I \times J} C_{(i,j)} = \prod_{(i,j) \in I \times J} (A_i \cap B_j) .$$
This shows that writing
$$(\prod_{i\in I}A_i)\cap(\prod_{j\in J}B_j)=\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$$
is incorrect because $\prod_{i\in I}A_i$ consists of functions defined on $I$, $\prod_{j\in J}B_j$ of functions defined on $J$ and $\prod_{(i,j)\in{I\times J}}(A_i\cap B_j)$ of functions defined on $I \times J$. The intersection on the left side is empty if $I \ne J$. But even if $I = J$, we can never have an equation between the left and the right side. However, the range of all the functions is $X$. For the first two products this is the assumption in your question, and for the third product you have to verify that $\bigcup_{(i,j)\in{I\times J}} (A_i \cap B_j) = X$. This is an easy exercise.
Let us consider the case that $I = J$ in which $\prod_{i\in I}A_i, \prod_{i\in I}B_i$ are both subsets of the set of all functions $I \to X$. We shall show that in general there does not even exist a bijection
$$\beta : (\prod_{i\in I}A_i)\cap(\prod_{i\in I}B_i) \to \prod_{(i,j)\in{I\times I}}(A_i\cap B_j) .$$
Let $X$ be a set and $A_i = B_i = X$ for all $i \in I$. Then $\prod_{i\in I}A_i = \prod_{i\in I}B_i = \prod_{i\in I}X $, hence $(\prod_{i\in I}A_i)\cap(\prod_{i\in I}B_i) = \prod_{i\in I}X $. But $\prod_{(i,j)\in{I\times I}}(A_i\cap B_j) = \prod_{(i,j)\in{I\times I}}X$. It is clear that if $I, X$ are finite sets with $n, m$ elements, respectively, such that $n, m > 1$, then $\prod_{i\in I}X $ has $m^n$ elements whereas $\prod_{(i,j)\in{I\times I}}X$ has $m^{n^2}$ elements.
We have seen that the statement in your question is wrong. So what can be said positively? The only thing which makes sense is this:
If $A'_i \subset A_i$, then we have a natural injection $\iota : \prod_{i\in I}A'_i \to \prod_{i\in I}A_i$ so that we can write by a little abuse of notation $\prod_{i\in I}A'_i \subset \prod_{i\in I}A_i$. In fact, each element of $\prod_{i\in I}A'_i$ is a function $f : I \to \bigcup_{i \in I}A'_i$, the latter being a subset of $\bigcup_{i \in I}A_i$ so that we can identity $f$ with $\iota(f) : I \stackrel{f}{\rightarrow} \bigcup_{i \in I}A'_i \hookrightarrow \bigcup_{i \in I}A_i$.
With $C_i = A_i \cup B_i$ we therefore obtain
$$\prod_{i\in I}A_i, \prod_{i\in I}B_i, \prod_{i\in I}(A_i \cap B_i) \subset \prod_{i\in I}C_i$$
and for these subsets we get
$$(\prod_{i\in I}A_i) \cap (\prod_{i\in I}B_i) = \prod_{i\in I}(A_i \cap B_i) .$$
A: This question appears in Pinter, Exercises 2.5, 6 a). I've been mulling over if for a while. I believe it's not correct as you state above, but as it appears in a textbook I've been trying to work out a way to prove it (perhaps by taking some liberty in notation or defintion!).
If  $f\in\prod\limits_{i\in I}A_i\cap\prod\limits_{j\in J}B_j$ then dom $f = I$ and dom $f = J$ and so either

*

*$I=J$ and $\ f:I \to X$ or

*$I\ne\ J$ and $\prod\limits_{i\in I}A_i\cap\prod\limits_{j\in J}B_j=\ \phi$
And if $f\in\prod\limits_{<i,j>\in I\mathrm{x}J}A_i\cap\ B_j$ then $\ f:I\mathrm{x}J \to X$. For the case $I=J$ I can't see how these can be reconciled as the functions will have different domains. Still, it bothers me!
By the way, $\prod\limits_{i\in I}A_i\cap\prod\limits_{i\in I}B_i=\prod\limits_{i\in I}A_i\cap\ B_i$ appears in 2.5 exercise 5 and yes, I have proved this as well!
