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:

    I do believe so. I do believe your report will give those people a good showing. And they will express thanks to you later

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