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Recall that a star domain is a subset $S$ of $\mathbb{R}^n$ in which there exists a point $x_0$ of $S$ such that every point of $S$ can be joined to $x_0$ by a line segment lying entirely within $S$, in which case $x_0$ is called a center.

I am interested in the case $n=2$ and where $S$ is an open subset of $\mathbb{R}^n$ (in the standard topology). My question is then: do there exists such $S$ in which the set of all points $S_0$ that are centers of $S$ is a finite set?

This clearly doesn't hold when we remove the condition for $S$ to be open, as we may take two line segments that do not lie on a common line and that intersect. This intersection point will be the unique center of the union of the two line segments.

On the other hand, there obviously exist open star domains that have infinitely-many centers, with a simple example being any open ball in $\mathbb{R}^2$.

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up vote 2 down vote accepted

The set of all centers of a given star domain is easily seen to be convex. If there are two distinct centers, then there are infinitely many. The only way you get finitely many centers is if you only have one.

An open star domain in the plane that has only one center is given by the open unit disk minus the two following segments : $\lbrace (x,0)\mid x\geq \frac12\rbrace$ and $\lbrace (0,y)\mid y\geq \frac12\rbrace$. The origin is the only center.

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That makes perfect sense that the set of centers would need to be convex, and reduces the problem to whether or not there exist open star domains that have only a single center. Do such star domains exist? – Hayden May 21 '14 at 22:57
Yes, @Hayden. $\{ (x,y) : \lvert xy\rvert < 1\}$ is one. – Daniel Fischer May 21 '14 at 22:58

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