Cartesian Product of Intersections/General Case

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Theorem

Let $I$ be an index set.

Let $\family {S_i}_{i \mathop \in I}$ and $\family {T_i}_{i \mathop \in I}$ be indexed families of sets over the same index set.


Then:

$\displaystyle \paren{ \prod_{i \mathop \in I} S_i } \cap \paren{ \prod_{i \mathop \in I} T_i } = \prod_{i \mathop \in I} \paren{S_i \cap T_i}$

where $\displaystyle \prod$ denotes Cartesian product.


Proof

Let $\family {x_i}_{i \mathop \in I} \in \displaystyle \paren{ \prod_{i \mathop \in I} S_i } \cap \paren{ \prod_{i \mathop \in I} T_i }$.

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

$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} S_i$
$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} T_i$

By definition of Cartesian product, this means, for all $i \in I$:

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

Again by definition of intersection, this is equivalent to, for all $i \in I$:

$x_i \in S_i \cap T_i$

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

$\family {x_i}_{i \mathop \in I} \in \displaystyle \prod_{i \mathop \in I} \paren{S_i \cap T_i}$


The result follows by definition of set equality.

$\blacksquare$