# Cartesian Product of Intersections/General Case

## Theorem

Let $S_1, \ldots, S_n$ and $T_1, \ldots, T_n$ be sets.

Then:

$\displaystyle \left({ \prod_{i \mathop = 1}^n S_i }\right) \cap \left({ \prod_{i \mathop = 1}^n T_i }\right) = \prod_{i \mathop = 1}^n \left({S_i \cap T_i}\right)$

where $\displaystyle \prod$ denotes Cartesian product.

## Proof

Let $\left({x_1, \ldots, x_n}\right) \in \displaystyle \left({ \prod_{i \mathop = 1}^n S_i }\right) \cap \left({ \prod_{i \mathop = 1}^n T_i }\right)$.

By definition of intersection, this is equivalent to the conjunction of:

$\left({x_1, \ldots, x_n}\right) \in \displaystyle \prod_{i \mathop = 1}^n S_i$
$\left({x_1, \ldots, x_n}\right) \in \displaystyle \prod_{i \mathop = 1}^n T_i$

By definition of Cartesian product, this means, for all $1 \le i \le n$:

$x_i \in S_i$ and $x_i \in T_i$

Again by definition of intersection, this is equivalent to, for all $1 \le i \le n$:

$x_i \in S_i \cap T_i$

Finally, by definition of Cartesian product, this is equivalent to:

$\left({x_1, \ldots, x_n}\right) \in \displaystyle \prod_{i \mathop = 1}^n \left({S_i \cap T_i}\right)$

The result follows by definition of set equality.

$\blacksquare$