External Direct Product Closure

Theorem
Let $\left({S \times T, \circ}\right)$ be the external direct product of the two algebraic structures $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$.

If $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ are closed, then $\left({S \times T, \circ}\right)$ is also closed.

Proof
If $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ are closed, then:


 * $s_1, s_2 \in S \implies s_1 \circ_1 s_2 \in S$
 * $t_1, t_2 \in T \implies t_1 \circ_2 t_2 \in S$

Thus $\left({s_1 \circ_1 s_2, t_1 \circ_2 t_2}\right) \in S \times T$

and we see that $\left({S \times T, \circ}\right)$ is closed.