Star Convex Set is Path-Connected

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Theorem

Let $A$ be a star convex subset of a topological 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:

$\forall t \in \closedint 0 1: x_1 + \paren {1 - t} a, t x_2 + \paren {1 - t} a \in A$


Define two paths $\gamma_1, \gamma_2: t \in \closedint 0 1 \to A$ by:

$\map {\gamma_1} t = t x_1 + \paren {1 - t} a$
$\map {\gamma_2} t = t a + \paren {1 - t} x_2$


By definition of topological vector space, it follows that $\gamma_1$ and $\gamma_2$ are continuous.

Because:

$\map {\gamma_2} t = \paren {1 - t} x_2 + \paren {1 - \paren {1 - t} } a$

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 $x_2$.

It follows by definition that $A$ is path-connected.

$\blacksquare$


Sources

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