# Metric induced from norm

I was trying to understand the following:

Every norm on $R^n$ is continuous (as a map from $R^n$ to $R$). Proof. We use the maximum metric on $R^n$: $d(x, y) = \max{|x_j − y_j| : j ∈ \{1, . . . ,n\}}$. Let $||\cdot||$ be a norm, and let $C = \Sigma_{j=1}^ne_j$, where $e_1, \dots , e_n$ is the canonical basis of $R^n$.

Then we have $||(x_1,\dots, x_n)|| = ||x_1e_1+...x_ne_n\le|x_1|||e_1||+...|x_n|||e_n||\le \max\{|x_1,...x_n|\}C$ From the triangle inequality, we then get $|||y|| − ||x||| ≤ ||y − x|| ≤ Cd(y, x)$ . So if $d(y, x) < ε/C$, then$|||y|| − ||x||| < \varepsilon$.

My question: The statement claims that every norm is continuous, and nothing about the metric on $R^n$ is specified, since this metric is defined via that norm, how can we take that $d(x, y) = \max{|x_j − y_j| : j ∈ \{1, . . . ,n\}}$? Is it fine to consider arbitrary metric and proceed? Please help.

$R^n$ is if not said otherwise assumed to have a distance that derives from a norm.