Open Ball is Convex Set

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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:

\(\displaystyle \left\Vert{x + t \left({y - x}\right) - v}\right\Vert\) \(=\) \(\displaystyle \left\Vert{\left({1 - t}\right) \left({x - v}\right) + t \left({y - v}\right) }\right\Vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left\Vert{\left({1 - t}\right) \left({x - v}\right) }\right\Vert + \left\Vert{t \left({y - v}\right) }\right\Vert\) 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.

$\blacksquare$