Limit of Sequence in Product of Metric Spaces under Chebyshev Distance

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Theorem

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

$\blacksquare$


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