# Open Ball is Convex Set

## Theorem

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

## 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:

 $\displaystyle \left\Vert{x + t \left({y - x}\right) - v}\right\Vert$ $=$ $\displaystyle \left\Vert{\left({1 - t}\right) \left({x - v}\right) + t \left({y - v}\right) }\right\Vert$ $\displaystyle$ $\le$ $\displaystyle \left\Vert{\left({1 - t}\right) \left({x - v}\right) }\right\Vert + \left\Vert{t \left({y - v}\right) }\right\Vert$ by triangle inequality $\displaystyle$ $=$ $\displaystyle \left({1 - t}\right) \left\Vert{x - v}\right\Vert + t \left\Vert{y-v}\right\Vert$ by norm axiom $(N2)$ $\displaystyle$ $<$ $\displaystyle \left({1 - t}\right) \epsilon + t \epsilon$ as $x, y \in B_\epsilon \left({v}\right)$ $\displaystyle$ $=$ $\displaystyle \epsilon$

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.

$\blacksquare$