# External Direct Product Closure

## Theorem

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.

Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.

Then $\struct {S \times T, \circ}$ is also closed.

### General Result

Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.

Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ all be closed algebraic structures.

Then $\struct {S, \circ}$ is also a closed algebraic structure.

## Proof

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed.

Let $\tuple {s_1, t_1} \in S \times T$ and $\tuple {s_2, t_2} \in S \times T$.

Then:

 $\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}$ $=$ $\ds \tuple {s_1 \circ_1 s_2, t_1 \circ_1 t_2}$ Definition of External Direct Product $\ds$ $\in$ $\ds S \times T$ as $S$ and $T$ are closed: $s_1 \circ_1 s_2 \in S, t_1 \circ_1 t_2 \in T$

demonstrating that $\struct {S \times T, \circ}$ is closed.

$\blacksquare$