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My try:

Consider points $a_0,a \in A$ where A is star convex. Now by the definition of star convex there exists a path between $a, a_0$ lying in A. That is, $F(a,t)=(1-t)a+t(a_0)$. Consider a loop $p$ . Then the map defined by $f:I \times I \to A$, $$f(s,t)=F(p(s),t)$$ such that $f(s,0)=p(s), f(s,1)=a_0$ is a path homotopy between the constant map and $p$ . Since the straight line homotopy is the path homotopy here, $\pi_{1}(A,a_0)$ is trivial. Also A is path connected . These two show that A is simply connected.

Is my proof correct?

  • But by definition in a star convex set there is some point $a_0$ in A such that lines joining $a_0$ with other points of the set A are in A.@JohnMa –  Oct 04 '15 at 12:24
  • Yes, I see now (I thought your $a_0$ is changing) –  Oct 04 '15 at 12:24
  • And your proof is correct. –  Oct 04 '15 at 12:26

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