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

Theorem

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:

$I: \struct {\R^n, d_\infty} \to \struct {\R^n, d_2}$

Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.

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

Let $\delta = \dfrac {\epsilon} {\sqrt n}$.

Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_\infty} {x, a} < \delta$.

That is:

$\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } < \delta$

Then:

$\ds \map {d_2} {x, a}$

$=$

$\ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - x_i} }$

$\ds $

$\le$

$\ds \sqrt {n \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } }$

$\ds $

$<$

$\ds \sqrt {n \delta^2}$

$\ds $

$=$

$\ds \sqrt {\epsilon^2}$

$\ds $

$=$

$\ds \epsilon$

The result follows by definition of $\tuple {d_\infty, d_2}$-continuity.

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.5$