# 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:

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

## Proof

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

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

Let $\delta = \epsilon$.

Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_2} {x, a} < \delta$.

That is:

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

Then:

 $\ds \sum_{i \mathop = i}^n \paren {a_i - x_i}$ $<$ $\ds \delta^2$ $\ds \leadsto \ \$ $\ds \forall i \in \set {1, 2, \ldots, n}: \,$ $\ds \paren {a_i - x_i}^2$ $<$ $\ds \delta^2$ $\ds \leadsto \ \$ $\ds \forall i \in \set {1, 2, \ldots, n}: \,$ $\ds \size {a_i - x_i}$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} }$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \map {d_\infty} {x, a}$ $<$ $\ds \delta$

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

$\blacksquare$