Is it possible to construct a metric in $\mathbb{R}^n $ s.t. it does not induce CONVEX balls? I'm studying point set topology and looking for a counterexample of "Balls are convex". 
We say set $K \subset \mathbb{R}^n $ is convex if $\forall x, y\in K$ implies $\lambda x + \left(1-\lambda y\right)\in K, \forall \lambda \in \mathbb{R}$. 
A counterexample has been found in $\mathbb{Q}$ (where the definition of convex has been specialized to "$\lambda\in\mathbb{Q}$") by using p-adic metric. E.g. consider ball $B_{\text{2-adic}}\left(0, 1\right)$. 

Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be any bijection, and define a metric $d'$ by$$d'(x,y)=d(f(x),f(y)),$$where $d$ is the Euclidean metric. Let $B$ denote the unit ball around zero in the Euclidean metric. Then the unit ball around zero in the new metric is$$B'=f^{-1}(B).$$The bijection $f$ can be chosen so that $f^{-1}(B)$ is a really wild subset. In particular, it can be non-convex.
Note that if $f$ is a homeomorphism, then the metrics $d,d'$ are equivalent, in the sense that both induce the standard topology on $\mathbb{R}^n$.
