Example 3

If $p_i (1 \leq i \leq n)$ is a finite set of semi-norms on E, then clearly $p'(x) = \sup_{1 \leq i \leq n} p_i(x)$ and $p''(x) = \sum_{i=1}^n \alpha_i p_i(x)$ (where the $\alpha_i$ are $\geq 0$) are both semi-norms on E.

A mapping $p$ of E in $\mathbf{R}_+$ is called an ultra-semi-norm if it satisfies (SN$_I$) and the following axiom :

(SN$_{II}'$) If $x \in E$ and $y \in E$, then $p(x + y) \leq \sup(p(x), p(y))$.

Clearly an ultra-semi-norm is a semi-norm.

To say that the absolute value on K is ultrametric (CA, VI, § 6.2) means that it is an ultra-semi-norm on the left vector space $K_s$, which is not identically zero.