Is it possible to construct a metric in $\mathbb{R}^n $ s.t. it does not induce CONVEX balls?

Question

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 $B_{\text{2-adic}}\left(0, 1\right)$.

Answer 1

Denote with $||\cdot||$ the usual euclidean metric on $\Bbb{R}^n$, and define the following metric on $\Bbb{R}^n$: $$d(x,y) = \left\{\begin{matrix} ||x-y|| & \mbox{if $x,y$ are linearly dependent} \\ ||x||+||y|| &\mbox{otherwise} \end{matrix} \right.$$

In this metric there exist non-convex balls.

For example, in $\Bbb{R}^2$, consider the ball $B=\{ v \in \Bbb{R}^2 : d(v, (0,1)) < 2\}$. You can check that $$B=\{ (x,y) \in \Bbb{R}^2 : x^2+y^2 < 1\} \cup \{ 0 \} \times (0,2)$$ is not convex.

However in this metric all balls are star-shaped. It would be interesting to find some "nice" example of metrics with non-star-shaped balls. Why do I say "nice"? Well, let $f : \Bbb{R}^n \longrightarrow \Bbb{R}^n$ be an ugly permutation (a bijection different from the identity, possibly non-continuous, nonn-measurable, non-anything). Then $$d_f(x,y) = ||f(x)-f(y)||$$ would be an ugly distance, and I suppose that you are able to find some non-convex balls.

Answer 2

Stereographically project $\mathbb R^n$ to the sphere missing its north pole, and use the usual metric on the sphere. A spherical cap which contains the north pole will correspond to a non-convex set in $\mathbb R^n$.

Answer 3

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 $\mathbb{R}^n$.

Answer 4

Simply consider the distance induced on $\mathbb{R}^n$ by an $\ell_p$-norm with $0 < p < 1$. It's not too hard to show that balls here are non-convex :)