9
$\begingroup$

Note: A subset $S\subseteq \mathbb{R}^2$ is called star-shaped if there exists an element $x\in S$ such that for every $y\in S$, the line segment connecting $x$ to $y$ is contained in $S$.

We want to give an example of a non-star shaped subeset of $\mathbb{R}^2$ and carefully prove it is not star-shaped.


I have begun by giving my example of a set that is not star shaped. It is the set of all points between the two circles, $x^2+y^2=1$ and $x^2+y^2=4$.

This set looks like an Annulus, as shown below.

enter image description here

To me, it is obvious that this set is not star shaped, because there is no point in the set that will not have to cross the circle in the middle of the set in order to make a line segment to every other point in the set.

I don't know how to prove this, however. Perhaps it is easier to prove giving a different example of a nonstar shaped set.


Any help would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ You can simplify the example by making it just a circle. Then any straight line intersects it in at most two points, and two points do not a segment make. $\endgroup$ – Brian M. Scott Nov 9 '15 at 4:14
  • 4
    $\begingroup$ A circle is not star-shaped; you’re thinking of a disk, a circle together with the region inside it. $\endgroup$ – Brian M. Scott Nov 9 '15 at 4:18
  • 2
    $\begingroup$ Ah! I am thinking of a disk. Thanks for clearing that up. So it would be easier to use just a circle as an example. $\endgroup$ – klorzan Nov 9 '15 at 4:23
  • 1
    $\begingroup$ Heck, you could just use the empty set. The definition of a star-shaped set requires it to contain at least one point. $\endgroup$ – user2357112 Nov 9 '15 at 4:31
  • 4
    $\begingroup$ If you want to be really boring, you could just take a set consisting of two points, e.g. $S = \{ (0, 0), (1, 1) \}$. After all, you did not say the set has to be connected or even represent a region. $\endgroup$ – CompuChip Nov 9 '15 at 8:27
17
$\begingroup$

Note that the figure $S$ is centered at $0$. To show that it is not star-shaped, note that $x\in S$ if and only if $-x\in S$. So the striaght line joining $x $ and $-x$ must pass the origin, which is not in $S$. So $S$ is not star-shaped.

Heuristically, everything that has a hole in between will not be star-shaped. So, as Brian suggests, you can use also something like $S = \{x^2 + y^2 = 1\}$, and the proof is the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.