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 $y, z \in \map {B_\epsilon^-} x$.

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 $\iota : \map {B_\epsilon^-} x \to V$ be the inclusion mapping of $\map {B_\epsilon^-} x$ in to $V$

From Continuity of Composite with Inclusion: Inclusion on Mapping, to show that $f$ is continuous it is sufficient to show that $\iota \circ f$ is continuous.

Let $g = \iota \circ f$.

Then $g: \closedint 0 1 \to V$ is the mapping defined by:
 * $\forall t \in \closedint 0 1 : \map g t = t y + (1 - t) y$