Real Vector Space under Chebyshev Distance is Homeomorphic to that under Euclidean Metric

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

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

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

Let $M_1 = \left({\R^n, d_\infty}\right)$ and $M_2 = \left({\R^n, d_2}\right)$ be the corresponding metric spaces.

Then $M_1$ and $M_2$ are homeomorphic.

Proof
From Relation between $p$-Product Metric and Chebyshev Distance on Real Vector Space:
 * $\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_p \left({x, y}\right) \le n^{1/p} d_\infty \left({x, y}\right)$

The Euclidean metric $d_2$ is a special case of the $p$-product metric $d_p$ for $p = 2$.

It follows by definition that $d_\infty$ and $d_2$ are Lipschitz equivalent.

It follows from Lipschitz Equivalent Metrics are Topologically Equivalent that $d_\infty$ and $d_2$ are topologically equivalent.

The result follows from Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic.