region between two concentric circles are not convex set in eucledian space of order 2. I want a counterexample to show that the region between two concentric circles in $\mathbb{R}^2$ is not a convex set. I think we need to find two points in the common region of concentric circles and then will show that their convex linear combination is not in $\mathbb{R}^2$, but I am unsure of this.
 A: The counteraxample is straightforward: Any straight line through the common center passes (in this order) through big circle, small circle, centre, small circle, big circle. This makes the center a convex combinatoin of points in the first and the second annulus segment encountered.
A: As an example take the concentric circles $C_{1}=\left\{ \left(x,y\right)\mid x^{2}+y^{2}=1\right\} $
and $C_{3}=\left\{ \left(x,y\right)\mid x^{2}+y^{2}=3\right\} $.
Denote the disks by $D_{1}=\left\{ \left(x,y\right)\mid x^{2}+y^{2}\leq1\right\} $
and $D_{3}=\left\{ \left(x,y\right)\mid x^{2}+y^{2}\leq3\right\} $.
Then $\left(0,2\right),\left(0,-2\right)\in D_{3}\backslash D_{1}$
and $\frac{1}{2}\left(0,2\right)+\frac{1}{2}\left(0,-2\right)=\left(0,0\right)\notin D_{3}\backslash D_{1}$.
This proves that $D_{3}\backslash D_{1}$ is not convex.
A: Let $k_1$ and $k_2$ be two concentric circles. Wlog we take $k_1: x^2+y^2<=1$ and $k_2: x^2+y^2<=4$. Than take the points (2,0) and (-2,0). These points are in $k_2$ but not in $k_1$. Than the segment $(2,0)*t+(-2,0)*(1-t)$ where $0<=t<=1$ goes out of the region between $k_1$ and $k_2$ (because it passes through the center (0,0) (see for t=0,5)). By definition this means that the region between two concentric circles is not convex.
