Star Convex Set is Path-Connected

Theorem
Let $A$ be a star convex subset of a vector space $V$ over $\R$ or $\C$.

Then $A$ is path-connected.

Proof
Let $x_1, x_2 \in A$.

Let $a \in A$ be a star center of $A$.

By definition of star convex set, it follows that for all $t \in \closedint 0 1$, we have $t x_1 + \paren {1 - t} a, t x_2 + \paren {1 - t} a \in A$.

Define two paths $\gamma_1, \gamma_2: t \in \closedint 0 1 \to A$ by $\map {\gamma_1} t = t x_1 + \paren {1 - t} a$, and $\map {\gamma_2} t = t a + \paren {1 - t} x_2$.

As $\map {\gamma_2} t = \paren {1 - t} x_2 + \paren {1 - \paren {1 - t} } a$, and $\paren {1 - t} \in \closedint 0 1$, it follows that $\map {\gamma_2} t \in A$.

Note that $\map {\gamma_1} 0 = x_1$, $\map {\gamma_1} 1 = \map {\gamma_2} 0 = a$, and $\map {\gamma_2} 1 = x_2$.

Define $\gamma: \closedint 0 1 \to A$ as the concatenation $\gamma_1 * \gamma_2$.

Then $\gamma$ is a path in $A$ joining $x_1$ and $x_2$, so $A$ is path-connected.