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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.

blargen
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hafsah
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  • what exactly do you mean by 'region between two concentric circles'? I think I know but it is better if you give a clear formulation of it. – drhab Feb 15 '14 at 09:17
  • I think your approach would be ok. – Ragnar Feb 15 '14 at 09:23
  • @drhab Concentric circles are circles that share the same center point.

    Think of this as the area of one circle (r=5), minus the area of another circle inside (r=6). So it's pi(5^2) - pi(6^2). ....area between two concentric circles is anulus.

     – hafsah Feb 15 '14 at 09:29
  • What you need is two points $P,Q$ in the region between the circles (outside the smaller, inside the larger) and a convex comination of $P,Q$ which is not in the region. You certainly don't need the convex combination "not in R^2" which would be impossible. And convex combination means $Px+Qy$ where $x,y \ge 0$ and $x+y=1.$ – coffeemath Feb 15 '14 at 09:34
  • You should switch off course (your area is negative).Let the origin be the center and look at the points $(0,5.5)$ and $(0,-5.5)$. They belong to the mentioned area but the origin (a point on the linesegment that connects the points) does not. So not convex. – drhab Feb 15 '14 at 09:36
  • @coffeemath sory i can't understand what do you mean by "You certainly don't need the convex combination "not in R^2" which would be impossible. " – hafsah Feb 15 '14 at 09:39
  • What I mean is that any convex combination of two points in R^2 must always be another point in R^2. – coffeemath Feb 15 '14 at 09:41

3 Answers3

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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.

Hagen von Eitzen
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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.

drhab
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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.

Emo
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