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

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Theorem

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

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 $N_1 = \left({B_1, d'_1}\right), N_2 = \left({B_2, d'_2}\right), \ldots, N_n = \left({B_n, d'_n}\right)$ be metric spaces.

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


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

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


Let $d_\infty$ be the Chebyshev distance on $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$, and $\displaystyle \mathcal B = \prod_{i \mathop = 1}^n B_i$, defined as:

$\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
$\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d'_i \left({x_i, y_i}\right)}\right\}$

where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$ or $\mathcal B$.


Let $F: M \to N$ be the mapping defined as:

$\forall x \in \mathcal A: F \left({x_1, x_2, \ldots, x_n}\right) = \left({f \left({x_1}\right), \left({x_2}\right), \ldots, \left({x_n}\right)}\right)$

Then $F$ is continuous.


Proof

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

Let $x \in \mathcal A$.

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

Then as $f_k$ is continuous:

$(1): \quad \exists \delta_k \in \R_{>0}: \forall y_k \in A_k: d_k \left({x_k, y_k}\right) < \delta_k \implies d' \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right) < \epsilon$


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

Then:

\(\, \displaystyle \forall y \in \mathcal A: \, \) \(\displaystyle d_\infty \left({x, y}\right)\) \(<\) \(\displaystyle \delta\)
\(\displaystyle \implies \ \ \) \(\displaystyle \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}\) \(<\) \(\displaystyle \delta\) Definition of Chebyshev Distance
\(\displaystyle \implies \ \ \) \(\, \displaystyle \forall k \in \left\{ {1, 2, \ldots, n}\right\}: \, \) \(\displaystyle d_k \left({x_k, y_k}\right)\) \(<\) \(\displaystyle \delta\)
\(\displaystyle \implies \ \ \) \(\displaystyle d'_k \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right)\) \(<\) \(\displaystyle \epsilon\) from $(1)$
\(\displaystyle \implies \ \ \) \(\displaystyle \max_{i \mathop = 1}^n \left\{ {d'_i \left({f_i \left({x_i}\right), f_i \left({y_i}\right)}\right)}\right\}\) \(<\) \(\displaystyle \epsilon\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_\infty \left({F \left({x}\right), F \left({y}\right)}\right)\) \(<\) \(\displaystyle \epsilon\) Definition of Chebyshev Distance

Thus it has been demonstrated that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in X: d_\infty \left({x, y}\right) < \delta \implies d_\infty \left({F \left({x}\right), F \left({y}\right)}\right) < \epsilon$

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.

$\blacksquare$


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