Projection from Cartesian Product under Chebyshev Distance is Continuous

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


Then for all $i \in \left\{ {1, 2, \ldots, n}\right\}$, $p_i$ is continuous on $\mathcal A$.


Proof

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

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

Let $B_\epsilon \left({a_i; d_i}\right)$ be the open $\epsilon$-ball of $a_i$ in $M_i$.

From Open Ball in Cartesian Product under Chebyshev Distance:

$\displaystyle B_\epsilon \left({a; d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right)$

By definition of $i$th projection:

$\forall x \in B_\epsilon \left({a; d_\infty}\right): \operatorname{pr}_i \left({x}\right) \in B_\epsilon \left({a_i; d_i}\right)$

Thus by definition of image of subset of $\mathcal A$ under $\operatorname{pr}_i$:

$\operatorname{pr}_i \left[{B_\epsilon \left({a; d_\infty}\right)}\right] \subseteq B_\epsilon \left({a_i; d_i}\right)$


Thus by definition of continuity at a point, $\operatorname{pr}_i$ is continuous at $a$.

As $a$ is arbitrary, it follows that $\operatorname{pr}_i$ is continuous at all points of $\mathcal A$.

Hence, by definition, $\operatorname{pr}_i$ is continuous from $\mathcal A$ to $A_i$.

As $i$ is arbitrary, it follows that the result holds for all $i \in \left\{ {1, 2, \ldots, n}\right\}$.

$\blacksquare$


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