Proposition 1

The topology $\mathscr{T}_v$ is Hausdorff and compatible with the field structure on K. The mapping $v : K^* \to G$ is continuous if G is given the discrete topology.

Let $x \in K^*$ and $a = v(x)$; then $x \notin V_a$ which shows that $\mathscr{T}_v$ is Hausdorff. For all $x_0 \in K$ and $a \in G$, there exists $\beta \in G$ such that $x_0 V_\beta \subset V_\alpha$ and $V_\beta x_0 \subset V$, (it is sufficient to take $\beta \geqslant a - v(x_0)$). On the other hand, if $a \geqslant 0$, then $V_\alpha V_\alpha \subset V_v$. The axioms (AV_I) and (AV_{II}) of General Topology, Chapter III, § 6, no. 3 being thus satisfied, $\mathscr{T}_v$ is compatible with the ring structure on $K$. Let $x_0 \in K^*$; if $x \in K^*$ satisfies $v(x - x_0) > \sup(\alpha + 2v(x_0), v(x_0))$, then $v(x^{-1} - x_0^{-1}) > a$ (Lemma 1), which shows that $x \mapsto x^{-1}$ is continuous and that $\mathscr{T}_v$ is therefore compatible with the field structure on $K$. Finally, the single condition $v(x - x_0) > v(x_0)$ implies $v(x) = v(x_0)$ ($§ 3$, no. 1, Proposition 1) and hence the mapping $v : K^* \to G$ is continuous if $G$ is given the discrete topology.

Let $\alpha \in G$ and $V'_\alpha$ be the set of $x \in K$ such that $v(x) \geq \alpha$. If $\beta < a$, then $V_\beta \supset V'_\alpha \supset V_v$. If $v$ is not improper, it is therefore seen that the $V'_\alpha$ form a fundamental system of neighbourhoods of 0 for $\mathscr{T}_v$.

The $V_v$ and the $V'_\alpha$ are open additive subgroups and therefore closed in $K$ and therefore the topological field $K$ is totally disconnected. As every non-zero ideal of the ring of $v$ contains a $V_v$, it is open and closed in $K$. The quotient topology on the residue field of $v$ is therefore discrete.

Let $A$ be the ring of $u$. If $v$ is discrete, Proposition 8 of $§ 3$, no. 6 shows that the topology induced by $\mathscr{T}_v$ on $A$ is the $m(A)$-adic topology. This is not so in general (Exercise 4).