Suppose are subsets of the Euclidean space . We define the Minkowski sum of and by
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 and , the set is not so interesting. But if we impose the condition be open, closed, or possibly even compact, we become curious about whether these topological properties are preserved.
Suppose and are both open subsets of . Translation is a homeomorphism, so for each point , the set is open. Since we can write the Minkowksi sum as the union,
we see that is the union of open sets and therefore open.
Now suppose that and are both closed. Is it necessarily true that the set is closed? The answer is no. Pick your favorite irrational real number . Let denote the set of integers, and let denote the set of integer multiples of . Then is closed, as for any noninteger , the open interval
does not intersect . To see that is closed, let be a sequence in converging to a point . If , then must be constant for all but finitely many indices . Suppose that . Then we can write
Since is not an integer, we have the lower bound
which contradicts convergence. I claim that the set is not closed. To prove this claim, it suffices to show that is countable, dense subset of , which I have done here in an earlier post.
If we require that at least one of the sets are compact, then the Minkowski sum is closed. Indeed, suppose is compact. Suppose is a sequence in with limit . We can write as the sum
Since is compact, there exists a convergent subsequence , as . Since the subsequence converges to , we conclude that
Since is closed, we see that and therefore .