The ring of integers of a locally compact non-archimedean field is compact

Question

I am not looking for a solution, I want an enlightening hint.

Let $K$ be a locally compact complete non-archimedean metrizable field. Then $\mathcal{O}_K=\{x\in K: |x| \leq 1 \}$ is compact. Is $\{x\in K: |x| < 1 \}$ necessarilly compact?

Answer 1

The "enlightening hint" should probably be: What does "locally compact" mean? It means that certain small balls are compact, and if certain balls are compact, you can scale them to bigger balls which are still compact. And what you wrote down there is a closed unit ball.

Actually, it is quite easily seen that a normed vector space over a field with a non-trivial valuation is locally compact iff its closed unit ball is compact. Cf. Locally compact vector spaces over non-Archimedean field. The statement here is, if not the definition, then the case of dimension $1$, and if anything proven easier than the general statement.