Star-shaped set Definition.- A set $S\subset \mathbb{R}^n$ is called star-shaped if there exists a point $z_0$in $S$ such that the line segment between $z_0$ and any point $z$ in $S$ is contained in $S$, $z_0$ is called a center of $S$.
Let $S\subset \mathbb{R}^n$ be a star-shaped  open  set , I have a problem is the set $$ \color{blue}Z=\{z\in S:z \text{ is a center of S}\}$$ an open set ? 
and if $S$ is bounded ? 
Any hints would be appreciated.
 A: Take a right-angled triangle that is not isosceles.  Fit four of them around a single point, like a pinwheel.  
EDIT: Take the interior of the octagon with vertices (in order) $(0,2),(0,1),(2,0),(1,0),(0,-2),(0,-1),(-2,0),(-1,0)$
A: If you take the open unit disc in $\mathbb{R}^2$ and remove the (closure of the) first quadrant, you get an open star shaped set, with center $(-x,-x)$ for any $0 < x< \sqrt{2} $. There are, of course, many more centers for this set, for instance the points of the form $(0,y)$ with $-1<y<0$.
Any point $(x,y)$ with $0<x$ and $-1<y<0$ won't be a center, therefore the set of centers is not open.
A: The answer is no. The set
$$\{(x,y) \in \mathbb R^2 : |y| < \frac1{1+x^2}\}$$
is open and star-shaped (with $z_0 = (0,0)$, for instance). But all its centers lie on the line $y=0$, so your set $Z$ has no interior points.
Updated to answer the updated question:
If $S$ is bounded, the answer is still no. The details are a bit messy, but in essence, we can define an "infinite open daisy" with a hub equal to the open unit disk, and an infinite number of "open petals" of length 2, which get narrower and narrower as they approach the y-axis. The only center is $(0,0)$.
Updated to add: As Steve's comment and Aldo's answer show, the details don't need to be messy.
