closed balls are connected when its given the topology on $k$ given by non archimedian abosolute value?

Question

Is it true that closed balls are connected when its given the topology on $k$ given by non archimedian abosolute value ?

I've shown if $k=\mathbb{Q}_{p}$ then we can write any closed ball as disjoint union of open balls and as open balls are also closed in this topology, so closed balls are actually disconnected. Here advantage was I could actually extract maximum powers of $p$ and could show, but here I don't have any advantage like that.

Answer 1

Let $|\cdot|$ be a non-Archimedean absolute value on $k$, and define the associated metric

$$d:k\times k\to\Bbb R:\langle x,y\rangle\mapsto|x-y|\;.$$

For any $x,y,z\in k$ we have

$$\begin{align*} d(x,y)&=|x-y|\\ &=|(x-z)+(z-y)|\\ &\le\max\{|x-z|,|z-y|\}\\ &=\max\{d(x,z),d(z,y)\}\;, \end{align*}$$

i.e., $d$ is a non-Archimedean metric (or ultrametric).

For $r\in\Bbb R^+$ and $x\in k$ let

$$B(x,r)=\{y\in k:d(x,y)<r\}$$

and

$$\overline{B}(x,r)=\{y\in k:d(x,y)\le r\}\;;$$

Of course $\overline{B}(x,r)$ is closed; I’ll show that it’s also open. Suppose that $y\in\overline{B}(x,r)$. If $z\in B(y,r)$, then

$$d(x,z)\le\max\{d(x,y),d(y,z)\}\le r\;,$$

so $z\in\overline{B}(x,r)$. Thus, $B(y,r)\subseteq\overline{B}(x,r)$, and $$\overline{B}(x,r)=\bigcup_{y\in\overline{B}(x,r)}B(y,r)$$ is open.

Now suppose that $\overline{B}(x,r)\ne\{x\}$, so that there is a $y\in\overline{B}(x,r)\setminus\{x\}$. Let $s=\frac12d(x,y)$; then $B(x,s)$ is a non-empty clopen proper subset of $\overline{B}(x,r)$, and $\overline{B}(x,r)$ is therefore not connected. Thus, $\overline{B}(x,r)$ is connected if and only if it’s the singleton $\{x\}$.

Since our non-Archimedean metric arises from a non-Archimedean absolute value, we can say a little more:

• If $\inf\limits_{0\ne x\in k}|x|=0$, then for all $x\in k$ and $r>0$ the closed ball $\overline{B}(x,r)$ is not connected.