Why is the unit ball open in a $\mathbb Q_p$-vector space $V$ with a norm?

Question

Suppose $V$ is a $\mathbb Q_p$-vector space. After giving $\mathbb Q_p$ an absolute value $|\cdot|_{\mathbb Q_p}$, we can equip $V$ with a norm by defining a map:$|\cdot|_V:V\to \mathbb R_{\ge 0}$ which satisfies : $$|\lambda v|_{V}=|\lambda|_{\mathbb{Q}_{p}} \cdot|v|_{V} \quad|v+w|_{V} \leq \max \left(|v|_{V},|w|_{V}\right) \quad v=0 \Leftrightarrow|v|_{V}=0$$

Such a norm gives a metric topology on $V$. If we define $V_0=\{v\in V:|v|_V\leq 1\}$, can we prove $V_0$ is open?

Answer 1

I see someone else gave a reference but let's describe a proof because it's quick, and we even give a more general statement: if you have a topological group $(G,+)$ and two subgroups $H\subseteq K\subseteq G$, then $H$ open implies $K$ open. One sees this by writing $K=\bigcup_{x\in K}(x+H)$ and using that $x+H$ is open for each $x$.

In the scenario you've described you're taking $K=V_0$ and $H=\{v:|v|_V<1\}$, and easily verifying that $H$ and $K$ are both subgroups.

Answer 2

You did not define $|\lambda|_{\Bbb Q_p}$ but if it is the usual norm, then $V$ is ultrametric, so yes, the closed balls are open, see Section "Properties" in the Wikipedia article.