Basis for Product of Metric Spaces 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 $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $U_i$ be open in $M_i$.


Then $\left\{ {\displaystyle \prod_{i \mathop = 1}^n U_i}\right\}$ is a basis for the open sets of $M$.


Proof


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