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