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.