Continuity of Mapping to Cartesian Product under Chebyshev Distance

Theorem
Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be metric spaces.

Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.

Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the Chebyshev distance on $\mathcal A$:
 * $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$

where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.

For all $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $\operatorname{pr}_i: \mathcal A \to A_i$ be the $i$th projection on $\mathcal A$:
 * $\forall a \in \mathcal A: \operatorname{pr}_i \left({a}\right) = a_i$

where $a = \left({a_1, a_2, \ldots, a_n}\right) \in \mathcal A$.

Let $M' = \left({X, d'}\right)$ be a metric space.

Let $f: X \to \mathcal A$ be a mapping.

Then $f$ is continuous on $X$ iff each of $\operatorname{pr}_i \circ f: X \to A_i$ is continuous on $X$.