# Mapping from Cartesian Product under Chebyshev Distance to Real Number Line is Continuous

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## Theorem

Let $M = \left({A, d'}\right)$ be a metric space.

Let $\displaystyle \mathcal A = A \times A$ be the cartesian product of $A$ with itself.

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

- $\displaystyle d_\infty \left({x, y}\right) = \max \left\{ {d' \left({x_1, y_1}\right), d' \left({x_2, y_2}\right)}\right\}$

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

Then the mapping:

- $d: \left({A \times A: d_\infty}\right) \to \left({\R, d}\right)$

where $d$ is the usual metric, is continuous.

## Proof

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.5$: Limits: Exercise $9$