Minkowski addition ($A+B$)

The Minkowski sum of subsets $A,B$ of a vector space is $A+B=\{a+b:a\in A,b\in B\}$. Minkowski addition also makes sense for nonconvex sets; translations and positive dilations commute with it.

In $\mathbb{R}$, the sets $X=\{n:n\ge 1\}$ and $Y=\{-n+1/n:n\ge 1\}$ are closed because their points have no finite accumulation point. Their Minkowski sum contains $1/n\to 0$ but does not contain zero, since $m-n+1/n=0$ has no positive-integer solution. Compactness of one summand is a useful sufficient condition missing from this counterexample.

In Euclidean space, a Minkowski sum of a compact set $X$ and a closed set $Y$ is closed. From any convergent sequence $x_j+y_j$, extract a convergent subsequence of $x_j$ using compactness. The corresponding $y_j$ then converges to the difference of the two limits, which lies in $Y$. If $Y$ is also bounded, the sum is bounded and therefore compact.

For nonempty open sets $A,B\subset {\mathbb{R}}^n$, the Lebesgue measure of their Minkowski sum satisfiesIt also holds for compact sets and in standard measurable-set formulations with the appropriate measurability qualification. Normalize both volumes to one and apply the Prékopa–Leindler inequality to indicator functions.