Continuity of Mapping to Cartesian Product under Chebyshev Distance

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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 $X$ if and only if each of $\operatorname{pr}_i \circ f: X \to A_i$ is continuous on $X$.


Proof

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


Necessary Condition

Let $f$ be continuous.

From Projection from Cartesian Product under Chebyshev Distance is Continuous, $\operatorname{pr}_i: \mathcal A \to A_i$ is continuous on $\mathcal A$.

From Composite of Continuous Mappings between Metric Spaces is Continuous it follows that $\operatorname{pr}_1 \circ f$ is continuous on $\mathcal A$.

$\Box$


Sufficient Condition

Let $\operatorname{pr}_i \circ f$ be continuous on $X$.

Let $a \in X$.


By definition of continuity at a point:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \operatorname{pr}_i \circ f \left[{B_\delta \left({a; d'}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$

where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the open $\epsilon$-ball of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the metric $d_i$, and similarly for $B_\delta \left({a; d'}\right)$.


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

$\forall \epsilon \in \R_{>0}: \operatorname{pr}_i \left[{B_\epsilon \left({f \left({a}\right); d_\infty}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \left({f \left({a}\right)}\right); d_i}\right)$


From Open Ball in Cartesian Product under Chebyshev Distance:

$\displaystyle B_\epsilon \left({f \left({a}\right); d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$


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

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: B_\delta \left({a; d'}\right) \subseteq \left({\operatorname{pr}_i \circ f}\right)^{-1} \left[{B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)}\right]$

where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the open $\epsilon$-ball of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the metric $d_i$, and similarly for $B_\delta \left({a; d'}\right)$.



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