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 $\struct {T, \circ_2}$.

Then:

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

if and only if:

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

Proof

Sufficient Condition

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:

\(\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}\)

\(=\)

\(\ds \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}\)

Definition of External Direct Product

\(\ds \)

\(\in\)

\(\ds S \times T\)

as $S$ and $T$ are closed: $s_1 \circ_1 s_2 \in S, t_1 \circ_2 t_2 \in T$

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

$\Box$

Necessary Condition

Let $\struct {S \times T, \circ}$ be closed.

Let $s_1, s_2 \in S$ and $t_1, t_2 \in T$.

Then:

\(\ds \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}\)

\(=\)

\(\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}\)

Definition of External Direct Product

\(\ds \)

\(\in\)

\(\ds S \times T\)

as $\struct {S \times T, \circ}$ is closed

\(\ds \leadsto \ \ \)

\(\ds s_1 \circ_1 s_2\)

\(\in\)

\(\ds S\)

Definition of External Direct Product

\(\, \ds \land \, \)

\(\ds t_1 \circ_2 t_2\)

\(\in\)

\(\ds T\)

demonstrating that $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ are closed.

$\blacksquare$