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What interesting properties of convex sets are retained by star-convex sets?

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up vote 6 down vote accepted

"Star-convex set" is a bit of a misnomer. The prevailing term probably should have been "star domain" instead (but star-convex is so common that the ship has sailed on that one) since that is what you use it for: An open star domain is a simply connected domain - a handy fact for proving simple cases of theorems in e.g. complex analysis.

In short, they have very few of the properties of convex sets. For example:

  • The intersection of two star-convex sets need not be connected (and thus in particular not star-convex). You can take two L-shaped sets and their intersection will be two disjoint squares.
  • The interior of a star-convex set need not be connected. An example is $\mathbb{C} \setminus \lbrace x+iy \;\vert\; x = 0 \text{ and } y \neq 0 \rbrace$, i.e. the right half-plane plus the left half-plane plus the origin.
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+1 on the bit about the misnomer. Since convex $\implies$ star, it is very strange to add "star" as a qualifier over "convex". – Willie Wong Dec 12 '10 at 14:09
Thanks @kahen. Excellent answer to a vague and underdetermined question. – isomorphismes Dec 22 '10 at 7:07
Also, how does convex work in R^2? Let Y be a subset of R^2 (the plane). If you have a point x and y with x < y, then the interval (x, y) must also lie in Y (for Y to be convex in X). What is the interval (x, y) if x and y are points on the plane? – Kara May 17 '13 at 17:11
Why is a "star" in the plane not convex? I see that it is star convex, but why isn't it convex? – Kara May 17 '13 at 17:33
@Kara, instead of thinking about intervals, you should think about line segments. These can be parametrized by intervals by using vectors, e.g. $[x,y] = [3,1]t+[0,-1]$. So, a star is not convex because the line between the points is not contained inside it. – Eric Stucky Jun 26 '13 at 0:57

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