# 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 $t \in \left[{0 \,.\,.\, 1}\right]$, we have $t x_1 + \left({1 - t}\right) a, t x_2 + \left({1 - t}\right) a \in A$.

Define two paths $\gamma_1, \gamma_2: t \in \left[{0 \,.\,.\, 1}\right] \to A$ by $\gamma_1 \left({t}\right) = t x_1 + \left({1 - t}\right) a$, and $\gamma_2 \left({t}\right) = t a + \left({1 - t}\right) x_2$.

As $\gamma_2 \left({t}\right) = \left({1 - t}\right) x_2 + \left({1 - \left({1 - t}\right) }\right) a$, and $\left({1 - t}\right) \in \left[{0 \,.\,.\, 1}\right]$, it follows that $\gamma_2 \left({t}\right) \in A$.

Note that $\gamma_1 \left({0}\right) = x_1$, $\gamma_1 \left({1}\right) = \gamma_2 \left({0}\right) = a$, and $\gamma_2 \left({1}\right) = x_2$.

Define $\gamma: \left[{0 \,.\,.\, 1}\right] \to A$ 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.

$\blacksquare$