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:


 * $\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$

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

Let $d_r$ be the metric defined as:
 * $\ds \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$

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


 * $\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y} \ge n^{-1} \map {d_r} {x, y}$

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