Open Ball is Convex Set

Theorem
Let $V$ be a normed vector space with norm $\left\Vert{\cdot}\right\Vert$ over $\R$ or $\C$.

An open ball in the metric induced by $\left\Vert{\cdot}\right\Vert$ is a convex set.

Proof
Let $v \in V$ and $\epsilon \in \R_{>0}$.

Denote the open $\epsilon$-ball of $v$ as $B_\epsilon \left({v}\right)$.

Let $x, y \in B_\epsilon \left({v}\right)$.

Then $x + t \left({y - x}\right)$ lies on line segment joining $x$ and $y$ for all $t \in \left[{0 \,.\,.\, 1}\right]$.

The distance between $x + t \left({y - x}\right)$ and $v$ is:

Hence, $x + t \left({y - x}\right) \in B_\epsilon \left({v}\right)$.

By definition of convex set, $B_\epsilon \left({v}\right)$ is a convex set.