Star convex set is simply connected.
Olivia Zamora 
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.
$\endgroup$ 43 Answers
$\begingroup$Yes, your proposed proof is correct.
$\endgroup$ $\begingroup$Not quite correct, you need to check that $H(x,t)$ always lies in the set $S$ (because star convex set need not be convex).
$\endgroup$ $\begingroup$Doesn't look right because to show something is Simply Connected, you need to show that the fundamental group is trivial alongside X being path connected. This means that you would want to create a path homotopy and not just a homotopy.
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