# Limit of Sequence in Product of Metric Spaces under Chebyshev Distance

## Theorem

Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_k = \left({A_n, d_k}\right)$ be metric spaces.

Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^k A_i$ be the cartesian product of $A_1, A_2, \ldots, A_k$.

Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the Chebyshev distance on $\mathcal A$:

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

where $x = \left({x_1, x_2, \ldots, x_k}\right), y = \left({y_1, y_2, \ldots, y_k}\right) \in \mathcal A$.

Let $\left\langle{a_n}\right\rangle$ be a sequence of points of $\mathcal A$:

$a_n = \left({a_{n 1}, a_{n 2}, \ldots, a_{n k} }\right) \in \mathcal A$

Let $c = \left({c_1, c_2, \ldots, c_k}\right) \in \mathcal A$.

Then:

$\displaystyle \lim_{n \mathop \to \infty} a_n = c \iff \forall i \in \left\{ {1, 2, \ldots, k}\right\}: \lim_{n \mathop \to \infty} a_{n i} = c_i$

where $\displaystyle \lim_{n \mathop \to \infty}$ denotes a limit.

## Proof

 $\displaystyle \lim_{n \mathop \to \infty} a_n$ $=$ $\displaystyle c$ $\displaystyle \iff \ \$ $\displaystyle \lim_{n \mathop \to \infty} d_\infty \left({a_n, c}\right)$ $=$ $\displaystyle 0$ Definition of Convergent Sequence $\displaystyle \iff \ \$ $\displaystyle \lim_{n \mathop \to \infty} \max_{i \mathop = 1}^k \left\{ {d_i \left({a_{n i}, c_i}\right)}\right\}$ $=$ $\displaystyle 0$ Definition of Chebyshev Distance $\displaystyle \iff \ \$ $\displaystyle \max_{i \mathop = 1}^k \left\{ {\lim_{n \mathop \to \infty} d_i \left({a_{n i}, c_i}\right)}\right\}$ $=$ $\displaystyle 0$ Limit of Image of Sequence $\displaystyle \iff \ \$ $\, \displaystyle \forall i \in \left\{ {1, 2, \ldots, k}\right\}: \,$ $\displaystyle \lim_{n \mathop \to \infty} d_i \left({a_{n i}, c_i}\right)$ $=$ $\displaystyle 0$ Definition of Max Function

$\blacksquare$