External Direct Product Closure
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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
- $\struct {S \times T, \circ}$ 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.
$\blacksquare$