Proposition 13

— For every $M > 0$, let $V_M$ be the set of $x \in K$ such that $\mathrm{mod}(x) \leq M$. If K is nondiscrete, the $V_M$ form a fundamental system of compact neighborhoods of 0 in K.

The $V_M$ are closed neighborhoods of 0 by Prop. 12. Let us show that they are compact. Let U be a compact neighborhood of 0. There exists an $r \neq 0$ in K such that $\mathrm{mod}(r) < 1$ and $r^n \in U$ for all $n > 0$: for, let W be a neighborhood of 0 such that $WU \subset U$; by Prop. 12, there exists an $r \neq 0$ in K such that $\mathrm{mod}(r) < 1$ and $r \in U \cap W$; then $r^2 \in WU \subset U$, and $r^n \in U$ for all $n > 0$ by induction on $n$. We are going to show that $V_M$ is contained in a finite union of sets $r^{-q}U$ ($q$ an integer $\geq 0$), which will prove that the $V_M$ are indeed compact. If $x$ is a cluster point of the sequence $(r^n)$, then $\mathrm{mod}(x)$ is a cluster point of the sequence $(\mathrm{mod}(r)^n)$, therefore $\mathrm{mod}(x) = 0$, $x = 0$; since U is compact, it follows (GT, I, §9, No. 1, Cor. of Th. 1) that $\lim_{n \to \infty} r^n = 0$. Now let $a \in V_M$. Since the sequence $(r^n a)_{n \geq 0}$ tends to 0, there exists a smallest integer $n \geq 0$ such that $r^n a \in U$. If $n > 0$ then $r^{n-1} a \notin U$, therefore $r^n a \in U \cap C(rU)$; the closure X of $U \cap C(rU)$ is compact since U is compact, and it does not contain 0 since $rU$ is a neighborhood of 0; therefore, in X, $\mathrm{mod}(x)$ is bounded below by a number $m > 0$. Thus, if $n > 0$, we have $m \leq \mathrm{mod}(r^n a)$, whence $\mathrm{mod}(r^{-1})^n \leq M/m$. Since $\mathrm{mod}(r^{-1}) > 1$, the integer $n$ can only take on a finite number of values, a number not depending on $a$, which completes the proof of our assertion.

This being so, since the intersection of the $V_M$ reduces to $\{0\}$ the $V_M$ form a fundamental system of neighborhoods of 0 (GT, I, §9, No. 2, Prop. 1).