Identity Mapping on Real Vector Space from Euclidean to Chebyshev Distance is Continuous

Theorem
Let $\R^n$ be an $n$-dimensional real vector space.

Let $d_2$ be the Euclidean metric on $\R^n$.

Let $d_\infty$ be the Chebyshev distance on $\R^n$.

Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself.

Then the mapping:
 * $I: \left({\R^n, d_2}\right) \to \left({\R^n, d_\infty}\right)$

is $\left({d_2, d_\infty}\right)$-continuous.

Proof
Let $a = \left({a_1, a_2, \ldots, a_n}\right) \in \R^n$.

Let $\epsilon \in \R_{>0}$.

Let $\delta = \epsilon$.

Let $x = \left({x_1, x_2, \ldots, x_n}\right)$ be such that $d_2 \left({x, a}\right) < \delta$.

That is:
 * $\displaystyle \sqrt{\sum_{i \mathop = i}^n \left({a_i - x_i}\right)} < \delta$

Then:

The result follows by definition of $\left({d_2, d_\infty}\right)$-continuity.