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Is it true that if $F$ is a locally compact topological field with a proper nonarchimedean absolute value $A$, then $F$ is totally disconnected? I am aware of the classifications of local fields, but I can’t think of a way to prove this directly.

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Yes: the non-Archimedean absolute value yields a non-Archimedean metric (also known as an ultrametric), and every ultrametric space is totally disconnected. In fact, every ultrametric space is even zero-dimensional, as it has a base of clopen sets.

**Proof:** Let $\langle X,d\rangle$ be an ultrametric space, meaning that $d$ is a metric satisfying $$d(x,y)\le\max\{d(x,z),d(y,z)\}$$ for any $x,y,z\in X$. Let $B(x,r)=\{y\in X:d(x,y)<r\}$; by definition $B(x,r)$ is open. Suppose that $y\in X\setminus B(x,r)$; then $d(x,y)\ge r$, and I claim that $B(x,r)\cap B(y,r)=\varnothing$. To see this, suppose that $z\in B(x,r)\cap B(y,r)$; then $$d(x,y)\le\max\{d(x,z),d(y,z)\}<r\;,$$ which is impossible. Thus, $y\notin\operatorname{cl}B(x,r)$, and $B(x,r)$ is closed. Thus, every open ball is clopen, and $X$ is zero-dimensional (and hence totally disconnected). $\dashv$

The metric associated with the non-Archimedean absolute value $\|\cdot\|$ is of course $d(x,y)=\|x-y\|$.

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