0

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 $S$.

Shuichi
  • 461
  • 2
  • 9

2 Answers2

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 $\rho(a,s)=r$.

user10354138
  • 33,887
  • No, it's really the Euclidean metric. – Shuichi Sep 22 '21 at 12:59
  • Huh? Where is this "Euclidean metric" defined? – user10354138 Sep 22 '21 at 13:01
  • 2
    @shuichi-saihara Then your question does not make any sense. – J.-E. Pin Sep 22 '21 at 13:02
  • @J.-E.Pin actually $C(a,r)$ is not induced by the metric $\rho$. It describes the circle that we know, and we show that although it's closed in Euclidean metric, it is open in ultrametric space. It didn't make sense at first. – Shuichi Sep 23 '21 at 05:33
  • @ShuichiSaihara What is this mysterious "Euclidean metric" you kept referring to? $S$ isn't embedded in any Euclidean space. You are only given "$(S,\rho)$ is an ultrametric space". – user10354138 Sep 23 '21 at 05:40
0

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 $C(a,r)$.

Shuichi
  • 461
  • 2
  • 9