Open Ball is Convex Set

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

An open ball in the metric induced by $\norm {\,\cdot\,}$ is a convex set.

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

Denote the open $\epsilon$-ball of $v$ as $\map {B_\epsilon} v$.

Let $x, y \in \map {B_\epsilon} v$.

Then $x + t \paren {y - x}$ lies on line segment joining $x$ and $y$ for all $t \in \closedint 0 1$.

The distance between $x + t \paren {y - x}$ and $v$ is:

Hence, $x + t \paren {y - x} \in \map {B_\epsilon} v$.

Thus, by definition, $\map {B_\epsilon} v$ is a convex set.