Three points in a metric space that satisfy the triangle inequality with equality, but are not collinear.

Question

I'm looking for three points $x,y,z$ in a metric space $X$ that satisfy the triangle inequality with equality, so $$ \mathrm{d}(x,z)=\mathrm{d}(x,y)+\mathrm{d}(y,z), $$ but are not collinear.

Three points $x,y,z$ are collinear if there is a continuous, distance-preserving mapping $ \gamma : I \to X $ with $x,y,z \in \gamma(I)$ and $ I \subseteq \mathbb{R}$ is an interval.

I have no idea right now. Maybe someone has a tip?

Answer 1

Hint: A subspace of any given metric space keep all the same distances, but a lot of previously collinear points may be non-collinear.

Answer 2

Let $X=\{0,1,2\}$, and define a metric $d$ on $X$ by setting $d(0,1)=d(1,2)=1$ and $d(0,2)=2$. (Of course $d(x,x)=0$ for all $x\in X$.)