# Continuity of Mapping to Cartesian Product under Chebyshev Distance

## Theorem

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

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:

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

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

For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the $i$th projection on $\AA$:

- $\forall a \in \AA: \map {\pr_i} a = a_i$

where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.

Let $M' = \struct {X, d'}$ be a metric space.

Let $f: X \to \AA$ be a mapping.

Then $f$ is continuous on $X$ if and only if each of $\pr_i \circ f: X \to A_i$ is continuous on $X$.

## Proof

Without loss of generality, let $i \in \set {1, 2, \ldots, n}$ be arbitrary.

### Necessary Condition

Let $f$ be continuous.

From Projection from Cartesian Product under Chebyshev Distance is Continuous, $\pr_i: \AA \to A_i$ is continuous on $\AA$.

From Composite of Continuous Mappings between Metric Spaces is Continuous it follows that $\pr_1 \circ f$ is continuous on $\AA$.

$\Box$

### Sufficient Condition

Let $\pr_i \circ f$ be continuous on $X$.

Let $a \in X$.

By definition of continuity at a point:

- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \pr_i \circ f \sqbrk {\map {B_\delta} {a; d'} } \subseteq \map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$

where $\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$ denotes the open $\epsilon$-ball of $\map {\pr_i \circ f} a$ with respect to the metric $d_i$, and similarly for $\map {B_\delta} {a; d'}$.

From Projection from Cartesian Product under Chebyshev Distance is Continuous, we have that:

- $\forall \epsilon \in \R_{>0}: \pr_i \sqbrk {\map {B_\epsilon} {\map f a; d_\infty} } \subseteq \map {B_\epsilon} {\map {\pr_i} {\map f a}; d_i}$

From Open Ball in Cartesian Product under Chebyshev Distance:

- $\ds \map {B_\epsilon} {\map f a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$

From Metric Space Continuity by Inverse of Mapping between Open Balls:

- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map {B_\delta} {a; d'} \subseteq \paren {\pr_i \circ f}^{-1} \sqbrk {\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i} }$

where $\map {B_\epsilon} {\map {\pr_i \circ f} a; d_i}$ denotes the open $\epsilon$-ball of $\map {\pr_i \circ f} a$ with respect to the metric $d_i$, and similarly for $\map {B_\delta} {a; d'}$.

This needs considerable tedious hard slog to complete it.In particular: not going anywhere, need to rethinkTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $7$