Cartesian Product under Chebyshev Distance of Continuous Mappings between Metric Spaces is Continuous

Theorem
Let $n \in \N_{>0}$.

Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.

Let $N_1 = \struct {B_1, d'_1}, N_2 = \struct {B_2, d'_2}, \ldots, N_n = \struct {B_n, d'_n}$ be metric spaces.

Let $f_i: M_i \to N_i$ be continuous mappings for all $i \in \set {1, 2, \ldots, n}$.

Let $\ds \MM = \prod_{i \mathop = 1}^n M_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.

Let $\ds \NN = \prod_{i \mathop = 1}^n N_i$ be the cartesian product of $B_1, B_2, \ldots, B_n$.

Let $d_\infty$ be the Chebyshev distance on $\ds \AA = \prod_{i \mathop = 1}^n A_i$, and $\ds \BB = \prod_{i \mathop = 1}^n B_i$, defined as:


 * $\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$


 * $\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d'_i} {x_i, y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$ or $\BB$.

Let $F: M \to N$ be the mapping defined as:
 * $\forall x \in \AA: \map F {x_1, x_2, \ldots, x_n} = \tuple {\map f {x_1}, \map f {x_2}, \ldots, \map f {x_n} }$

Then $F$ is continuous.

Proof
Let $\epsilon \in \R_{>0}$.

Let $x \in \AA$.

Let $k \in \left\{{1, 2, \ldots, n}\right\}$.

Then as $f_k$ is continuous:
 * $(1): \quad \exists \delta_k \in \R_{>0}: \forall y_k \in A_k: \map {d_k} {x_k, y_k} < \delta_k \implies \map {d'} {\map {f_k} {x_k}, \map {f_k} {y_k} } < \epsilon$

Let $\delta = \max \set {\delta_k: k \in \set {1, 2, \ldots, n} }$.

Then:

Thus it has been demonstrated that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in X: \map {d_\infty} {x, y} < \delta \implies \map {d_\infty} {\map F x, \map F y} < \epsilon$

Hence by definition of continuity at a point, $F$ is continuous at $x$.

As $x$ is chosen arbitrarily, it follows that $F$ is continuous for all $x \in X$.

The result follows by definition of continuous mapping.