Continuity of Mapping to Cartesian Product under Chebyshev Distance

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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 $\map {B_\delta} {a; d'}$.



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