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

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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:

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

is $\tuple {d_2, d_\infty}$-continuous.


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.

$\blacksquare$


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