# Show by example that the Minkowski sum of two sets $X+Y$ may be convex even if neither $X$ nor $Y$ are convex

There were two parts to this question. I proved that the Minkowski sum of two sets $X+Y$ is convex whenever $X$ and $Y$ are convex, but how do I prove this second part? "Show by example that the Minkowski sum of two sets $X+Y$ may be convex even if neither $X$ nor $Y$ are convex."

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$X=\mathbb{R}\setminus\{0\}$ is not convex but $X+X=\mathbb{R}$ is.

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+1 Nice, clean example. If $a > 0$, one could also use $X = (-a,0) \cup (0,a)$. Then, $X+X = (-2a,2a)$. –  JavaMan Oct 9 '11 at 3:16
Sorry I still don't get it. Can u prove your example? Thank u so much~ –  xuan Oct 9 '11 at 5:47
What is there not to get? –  scineram Oct 9 '11 at 8:47
The example is really straight forward but I think I need a formal prove which I don't know how... –  xuan Oct 9 '11 at 9:54
@xuan: Take $x=-1\in X$, $y=1\in X$ and $t=1/2\in[0,1]$. Then $(1-t)x+ty=0\notin X$, so $X$ is not convex. $X+X=\mathbb{R}$ is the fact that any real number can be written as a sum of two nonzero real numbers; e.g. $1=1/2+1/2$ and $x=(x-1)+1$ for $x\neq 1$. That $\mathbb{R}$ is convex is an easy consequence of the definition of convexity. –  LostInMath Oct 9 '11 at 10:16

A stupid example: Let $X$ be any non-convex set and let $Y=\emptyset$. Then $X+Y=\emptyset$ is convex!

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But $Y$ is convex. –  scineram Oct 9 '11 at 8:46
If this helps you to understand the previous demonstration, you can think of the Minkowski Sum as a painting operation over the points of $X$ using $Y$ as a brush.
Now think of a convex set $X$ minus one interior point - and that is enough not to make it convex anymore.
This point or even small gaps will be painted over by the brush $Y$ standing in nearby points, if the brush is 'fat' enough. The resulting set will be made convex again.
Since a typical point of $X+Y$ will be covered by different points of the brush $Y$ standing at different points of $X$, you can remove points of the brush and it will still paint the same region.