## Minkowski Sums in Euclidean Space

Suppose $A,B$ are subsets of the Euclidean space $\mathbb{R}^{k}$. We define the Minkowski sum of $A$ and $B$ by

$\displaystyle A+B:=\left\{z\in\mathbb{R}^{k}:z=x+y,x\in A\text{ and }y\in B\right\}$

Of course, we could define the Minkowski sum on any (topological) vector space, but it’s the weekend, so let’s keep things simple.

With no topological restrictions on $A$ and $B$, the set $A+B$ is not so interesting. But if we impose the condition $A, B$ be open, closed, or possibly even compact, we become curious about whether these topological properties are preserved.

Suppose $A$ and $B$ are both open subsets of $\mathbb{R}^{k}$. Translation is a homeomorphism, so for each point $x\in A$, the set $x+B$ is open. Since we can write the Minkowksi sum $A+B$ as the union,

$\displaystyle A+B=\bigcup_{x\in A}x+B$,

we see that $A+B$ is the union of open sets and therefore open.

Now suppose that $A$ and $B$ are both closed. Is it necessarily true that the set $A+B$ is closed? The answer is no. Pick your favorite irrational real number $\alpha$. Let $C_{1}$ denote the set of integers, and let $C_{2}$ denote the set of integer multiples of $\alpha$. Then $C_{1}$ is closed, as for any noninteger $x$, the open interval

$\displaystyle\left(x-\delta,x+\delta\right),\indent\delta:=\min\left\{\left|x-\lfloor{x}\rfloor\right|,\left|x-\lceil{x}\rceil\right|\right\}$

does not intersect $C_{1}$. To see that $C_{2}$ is closed, let $n_{k}\alpha$ be a sequence in $C_{2}$ converging to a point $x$. If $x\in C_{2}$, then $n_{k}$ must be constant for all but finitely many indices $k$. Suppose that $x\notin C_{2}$. Then we can write

$\displaystyle\left|n_{k}\alpha-x\right|=\left|n_{k}-\frac{x}{\alpha}\right|\cdot\left|\alpha\right|$

Since $\frac{x}{\alpha}$ is not an integer, we have the lower bound

$\displaystyle\left|n_{k}-\frac{x}{\alpha}\right|>\min\left\{\left|\frac{x}{\alpha}-\left\lfloor{\frac{x}{\alpha}}\right\rfloor\right|,\left|\frac{x}{\alpha}-\left\lceil{\frac{x}{\alpha}}\right\rceil\right|\right\}>0,\indent\forall k\in\mathbb{N}$

which contradicts convergence. I claim that the set $C_{1}+C_{2}$ is not closed. To prove this claim, it suffices to show that $C_{1}+C_{2}$ is countable, dense subset of $\mathbb{R}$, which I have done here in an earlier post.

If we require that at least one of the sets $A,B$ are compact, then the Minkowski sum is closed. Indeed, suppose $A$ is compact. Suppose $z_{n}$ is a sequence in $A+B$ with limit $z\in\mathbb{R}^{k}$. We can write $z_{n}$ as the sum

$\displaystyle z_{n}=x_{n}+y_{n},\indent x_{n}\in A\text{ and }y_{n}\in B \ \forall n\in\mathbb{N}$

Since $A$ is compact, there exists a convergent subsequence $x_{n_{k}}\rightarrow x\in A$, as $k\rightarrow\infty$. Since the subsequence $z_{n_{k}}$ converges to $z$, we conclude that

$\displaystyle\lim_{k\rightarrow\infty}y_{n_{k}}=z_{n_{k}}-x_{n_{k}}=z-x=:y$

Since $B$ is closed, we see that $y\in B$ and therefore $z\in A+B$.

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### One Response to Minkowski Sums in Euclidean Space

1. leo says:

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