A subset \(A\) of \(\mathbb{R}^n\) is said to be star convex if for some point \(a_0\) of \(A\), all the line segments joining \(a_0\) to other points of \(A\) lie in \(A\).
Find a star convex set that is not convex.
Show that if \(A\) is star convex, \(A\) is simply connected.
Solution
For example, \[ A=\{(x,y)\in\mathbb R^2:0\le x\le1,\ |y|\le x\}\cup\{(x,0):1\le x\le2\} \] is star convex with respect to \((0,0)\): every radial segment from the origin to a point of \(A\) stays in \(A\). It is not convex, since \((1,1),(2,0)\in A\) but their midpoint \((3/2,1/2)\notin A\).
Let \(a_0\) be a star center. Define \[ H:A\times I\to A,\qquad H(x,t)=(1-t)x+t a_0. \] By star convexity the whole segment from \(x\) to \(a_0\) lies in \(A\), so \(H\) is well defined and continuous. Moreover \[ H(x,0)=x,\qquad H(x,1)=a_0. \] Thus \(A\) is contractible. A contractible space is path connected, and every loop \(f:I\to A\) contracts by \[ F(s,t)=H(f(s),t). \] Hence \(\pi_1(A,a_0)=0\), so \(A\) is simply connected.