Closed Ball is Path-Connected

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

A closed ball in the metric induced by $\norm {\,\cdot\,}$ is path-connected.

Proof
Let $\map {B_\epsilon^-} x$ be a closed ball in the metric induced by $\norm {\,\cdot\,}$ with center $x \in V$ and radius $\epsilon \in \R_{>0}$

Let $y, z \in \map {B_\epsilon^-} x$ such that $y \neq z$.

Let $f: \closedint 0 1 \to \map {B_\epsilon^-} x$ be the mapping defined by:
 * $\forall t \in \closedint 0 1 : \map f t = t y + (1 - t) z$

From Closed Ball is Convex Set, it follows that $f$ is well-defined.

Furthermore:
 * $\map f 0 = y$

and
 * $\map f 1 = z$

It remains to show that $f$ is continuous.

Let $s \in \closedint 0 1$.

Let $\epsilon \in \R_{> 0}$.

Let:
 * $\delta = \dfrac \epsilon {2 \max \set{\norm y, \norm z}}$.

Let:
 * $t \in \openint {s - \delta} {s + \delta} \cap \closedint 0 1$.

We have:

It follows that $f$ is continuous at $s$ by definition.

Since $s \in \closedint 0 1$ was arbitrary, then $f$ is continuous on $\closedint 0 1$ by definition.

It follows that there exists a path from $y$ to $z$ by definition.

Since $y, z \in \map {B_\epsilon^-} x$ were arbitrary, then $\map {B_\epsilon^-} x$ is path-connected by definition.