Taxicab Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space

Theorem
For $n \in \N$, let $\R^n$ be a real vector space.

Let $d_1$ be the taxicab metric on $\R^n$.

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

Then
 * $\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_1} {x, y} \le n \cdot \map {d_\infty} {x, y}$

It follows that $d_1$ and $d_\infty$ are Lipschitz equivalent.

Proof
By definition of the Chebyshev distance on $\R^n$, we have:
 * $\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$

where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.

Let $j$ be chosen so that:
 * $\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$

Then: