# Neighborhood Basis in Cartesian Product under Chebyshev Distance

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

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

For all $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $\mathcal B_{a_i}$ be a basis for the system of neighborhoods for $a_i$ in $M_i$.

Let $\mathcal B_a$ be the cartesian product of $\mathcal B_{a_1}, \mathcal B_{a_2}, \ldots, \mathcal B_{a_n}$:

$\displaystyle \mathcal B_a = \prod_{i \mathop = 1}^n \mathcal B_{a_i}$

Then $\mathcal B_a$ is a basis for the system of neighborhoods for $a$ in $M$.

## Proof

Let $N$ be a neighborhood of $a$ in $M$.

Then by definition of neighborhood:

$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a, d_\infty}\right) \subseteq N$

where $B_\epsilon \left({a}\right)$ is the open $\epsilon$-ball of $a$ under $d_\infty$.

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

From Open Ball is Neighborhood of all Points Inside, each $B_\epsilon \left({a_i; d_i}\right)$ is a neighborhood of $a_i$.

By definition of basis for system of neighborhoods:

$\exists N_i \in \mathcal B_{a_i}: N_i \subseteq B_\epsilon \left({a_i; d_i}\right)$
$\displaystyle \prod_{i \mathop = 1}^n N_i \subseteq \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right) \subseteq N$

But $\displaystyle \prod_{i \mathop = 1}^n N_i \in \mathcal B_a = \prod_{i \mathop = 1}^n \mathcal B_{a_i}$

Hence the result by definition of basis for system of neighborhoods.

$\blacksquare$