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
- 2000: James R. Munkres: Topology (2nd ed.): $9$: The Fundamental Group: $\S 52$: The Fundamental Group
- 2001: Christian Berg: Kompleks funktionsteori: $\S 3.1$