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We define a set $S \subset \Bbb{R}^n$ to be star convex if there exists $a \in S$, such that the line segment connecting $a$ and any other point in $S$ lies entirely in $S$. I would like to show that it's simply connected. Can someone verify my proof?

The set $S$ is certainly path connected since given $x,y \in S$, we can construct a path from $x$ to $a$ and $a$ to $y$ , and so adjoining the paths yields a path from $x$ to $y$. Also given any loop $p(r)$, $r \in [0, 1]$, we have a straight-line homotopy

$$H(r, t) = ta + (1-t)p(r)$$

with $H((r, 0) = p(r)$ and $H(r, 1) = a$, so $p$ is homotopic to a point, meaning $S$ is simply connected.

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Looks good to me. –  Rankeya Feb 27 '13 at 5:42
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Agreed, this looks fine to me. –  Alex Youcis Feb 27 '13 at 6:24
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1 Answer 1

Yes, your proposed proof is correct.

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