Proposition 3

Let $E$ be a left topological vector space over $K$ which is Hausdorff and of dimension 1. Suppose that $v$ is not improper. For all $x_0 \neq 0$ in $E$, the mapping $a \mapsto ax_0$ of $K_s$ onto $E$ is a topological isomorphism.

This mapping is a continuous algebraic isomorphism. It is sufficient to show that it is bicontinuous. Let $\alpha \in G$. We need to show that there exists a neighbourhood $V$ of 0 in $E$ such that the relation $ax_0 \in V$ implies $v(a) > a$. Let $a, \in K^*$ be such that $v(a_0) = a$. As $E$ is Hausdorff, there exists a neighbourhood $W$ of 0 in $E$ such that $a_0x_0 \notin W$. As $v$ is not improper, there exist a neighbourhood $W'$ of 0 in $E$ and an element $\beta$ of $G$ such that the relations $y \in W'$, $v(a) \geq \beta$ imply $ay \in W$. Let $a, \in K^*$ be such that $v(a_1) = -\beta$. The relations $ax_0 \in a_1^{-1}W'$ and $v(a) \leq a$ imply $a_1ax_0 \in W$ and $v(a_0a^{-1}a_1^{-1}) = \alpha + \beta - v(a) \geq \beta$ and hence $a_0x_0 = a_0a^{-1}a_1^{-1}(a_1ax_0) \in W$, which is absurd; in other words, the relation $ax, \in a_1^{-1}W$ implies $v(a) > a$.