A circle is open in an ultrametric space.

Question

I've been stuck with this problem for hours. The claim is that every circle, although closed in Euclidean space, is open in an ultrametric space. I tried using all the properties of the ultrametric space to construct a radius for the open ball, but I can't. I'm desperate. I would appreciate it if someone can give me a hint. Here's the formal description of the problem:

Given an ultrametric space $(S, \rho)$, show that the circle $C(a,r):=\{s\in S:|s-a|=r\}$ is open in $S$.

Answer 1

You mean $\rho(a,s)=r$ not $\lvert s-a\rvert=r$.

Hint: Prove that if $p\in C(a,r)$ and $s\in S$ with $\rho(p,s)<r$, then $\rho(a,s)=r$.

Answer 2

Thank you for the responses. There was a mistake in my professor's part. It should be $C(a,r):=\{s\in S: \rho(s,a)=r\}$. To prove that this is open, I just took $r$ to be the radius of the open ball of an element in $C(a,r)$.