Let's say if we have $\displaystyle \bold{x} \in \mathbb{R}^{n} $ and we want to show that $\displaystyle B_{r}(\bold{x}) $(open ball of radius $\displaystyle r >0 $) is convex. We know that $\displaystyle d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}| $ (e.g. the norm).

We can still apply the triangle inequality right? So we have: $\displaystyle |t(\bold{y}-\bold{x})+(1-t)(\bold{z}-\bold{x})| \leq t|\bold{y}-\bold{x}| + (1-t)|\bold{z}-\bold{x}| < r $.

In other words, does $\displaystyle d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}| $ only apply to $\displaystyle \mathbb{R} $? Or does it apply to general $\displaystyle \mathbb{R}^{n} $ so that we can invoke the triangle inequality? We can use the triangle inequality to prove the general case (which is what I did) right?