Remark

The proof of Proposition 5 shows that, if the topology defined by $|x|_2$ is coarser than that defined by $|x|_1$, and if $|x|_1$ is not improper, then $|x|_1$ and $|x|_2$ are equivalent, for the relation $|x|_2 < 1$ then implies $|x|_2 < 1$. Thus the topologies defined by two absolute values on $K$, neither of which is improper, cannot be comparable without being identical.