For all norms on $\mathbb{R}^n$, for all $x\in\mathbb{R}^n$, for all $r > 0$, for all $y, z \in B(x,r)$, the line segment $[y, z) \subseteq B(x,r)$.

Question

Ok, I believe that if I can prove the above statement for one norm in $\mathbb{R}^n$ then it is proven for all norms. Also, if we can show that each point on the line $y$ to $z$ is in the open ball centered at $x$ of radius $r$ then we have completed the proof. Am I on the correct path here? (bad pun intended)

Thank you.

Answer 1

Let $y,z\in B(x,r), t\in [0,1]$, $\|t(y-x)+(1-t)(z-x)\|\leq t\|(y-x)\|+(1-t)\|(z-x)\|\leq tr+(1-t)r=r$.