P-Product Metrics on Real Vector Space are Topologically Equivalent

Theorem
For $n \in \N$, let $\R^n$ be an Euclidean space.

Let $p \in \R_{\ge 1}$.

Let $d_p$ be the $p$-product metric on $\R^n$.

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

Then $d_p$ and $d_\infty$ are topologically equivalent.

Proof
Let $r, t \in \R_{\ge 1}$.

, assume that $r \le t$.

For all $x, y \in \R^n$, we are going to show that:


 * $\displaystyle d_r \left({x, y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_r \left({x, y}\right)$

Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows.

Let $d_r$ be the metric defined as $\displaystyle d_r \left({x, y}\right) = \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{1/r}$.

Inequality for General Case
When we combine the inequalities, we have:


 * $d_r \left({x, y}\right) \ge d_\infty \left({x, y}\right) \ge n^{-1} d_1 \left({x,y}\right) \ge n^{-1} d_r \left({x, y}\right)$

Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for all $r \in \R_{\ge 1}$.