# When $(-)+M$ sends closed sets to closed sets?

Let $$M$$ be a subset of $$\mathbb R^n$$. I wonder when $$M+S$$ remains closed for arbitrary closed set $$S\subseteq \mathbb R^n$$?

I thought of this question in the study of topological group; but unfortunately, I still have few clues about the simple case of $$\mathbb R^n$$. Here are some of my attemps.

So far I have proved that $$M$$ has the closed-set-preserving property if ($$\Leftarrow$$)

1. $$M$$ is finite (this is trivial);
2. $$M$$ is compact (this is well-known);
3. $$M$$ is the intersection of finite many sets which are isometrically homeomorphic to $$\mathbb R^{n-1}\times \mathbb R_{\geq 0}$$;
4. $$M$$ is the finite sum of 1.-3. above;
5. $$M$$ is the finite union of 1.-4. above.

It is also clear that $$M$$ has the closed-set-preserving property only if ($$\Rightarrow$$) $$M$$ is closed, since [$$M$$ is closed] $$\Leftrightarrow$$ [$$M+\{\mathrm{pt}\}$$ is closed].

And I have proved that $$M$$ doesnot have to be convex. For $$N=2$$, set $$\Gamma:=\{(x,y)\mid xy\geq 1,x,y\geq 0\}$$, and its reflection $$\Gamma':=\{(x,-y)\mid (x,y)\in \Gamma\}$$. We see that $$\Gamma+\Gamma'=\mathbb R_{>0}\times \mathbb R$$ is not closed. One can also generalise it to $$\mathbb R^n$$.

• A concrete example disproving 3 (see Apass.Jack's nice answer below): $\{(x,y): y=\tan(x), -\pi/2<x<\pi/2\}+\{(x,y): x \ge 0\} = \{(x,y): x >-\pi/2\}$ Jun 6, 2023 at 11:52

A very nice question! This answer will show that a set is closed-set-preserving iff it is closed with bounded boundary.

#### Notations

Fix our universe $$\Bbb R^n$$. Either $$0$$ or the origin will denote $$(0,0,\cdots, 0)$$. For point $$x$$, let $$\|x\|$$ be the Euclidean distance from $$x$$ to the origin. Let $$M+S$$ be $$\{m+s\mid m\in M, s\in S\}$$ for set $$M,S$$. Call $$M$$ closed-set-preserving if $$M+S$$ is closed whenever $$S$$ is closed.

#### $$M$$ is a closed set with bounded boundary $$\implies$$$$M$$ is closed-set-preserving.

Proof. Let $$S$$ be a closed set and $$a$$ be a limit point of $$M+S$$, i.e, $$a=\lim_{n\to\infty}(m_i+s_i)$$ for $$m_i\in M$$ and $$s_i\in S$$. There are two cases.

• $$m_1, m_2, \cdots$$ is a bounded sequence.
Then a subsequence of $$m_1, m_2, \cdots$$ converges. WLOG, assume $$m_1, m_2,\cdots$$ converges to some point $$\mu$$, which must be in $$M$$ since $$M$$ is closed. Hence $$a-\mu=\lim_{n\to\infty}s_i$$, which must be in $$S$$ since $$S$$ is closed. Hence, $$a = \mu+(a-\mu)\in M+S$$.
• $$m_1, m_2, \cdots$$ is not bounded.
Since the boundary of $$M$$ is bounded, there exists arbitrarily large $$i$$ such that every point less than $$1$$ away from $$m_i$$ is in $$M$$.
Since $$a=\lim_{n\to\infty}(m_i+s_i)$$, we can assume that $$\|a-(m_i+s_i)\|<1$$. Let $$d=a-(m_i+s_i)$$. Then $$a = (m_i+d)+s_i,$$ where $$m_i+d\in M$$ and $$s_i\in S$$.

In all cases, $$a$$ is in $$M+S$$. Hence $$M+S$$ is closed.

#### $$M$$ is a closed set with bounded boundary $$\impliedby$$$$M$$ is closed-set-preserving.

Proof. $$M=M+\{\text{the origin}\}$$ is closed since $$\{\text{the origin}\}$$ is a closed set.

Towards a contradiction suppose the boundary of $$M$$ is not bounded. Then there are points $$p_1, p_2, \cdots$$ in the boundary such that $$\|p_i\|\ge i+1$$. Let $$q_i\notin M$$ such that $$\|q_i-p_i\|\le\frac1i$$. Then $$0<\text{distance}(q_i,M)\le\frac1i$$

Let $$S=\{-q_1, -q_2, -q_3, \dots\}$$.

• Since $$q_i=p_i+(q_i-p_i)$$, we have $$\|q_i\|\ge \|p_i\|-\|q_i-p_i\|\ge i+1-\frac1i\ge i.$$ So $$S$$ is a closed set.
• For all $$m\in M$$, $$\|m+(-q_i)\| \ge\text{distance}(q_i, M)>0$$. So $$0\notin M+S$$
• $$\text{distance}(M+\{-q_i\}, 0)=\text{distance}(q_i,M)\le\frac1i.$$ Hence, $$0$$ is a limit point of $$M+S$$.

We see that $$M+S$$ is not closed while $$S$$ is closed. This is a contradiction. Hence the boundary of $$M$$ is bounded.

For $$n\ge2$$, since $$\Bbb R^{n-1}\times \Bbb R_{\geq 0}$$ has unbounded boundary, $$M$$ of type 3 in the question is not necessarily closed-set-preserving. Neither is $$M$$ of type 4 or 5.

• (+1) Very nice answer! It's great to see that there is such a simple description of OP's sets. I do believe you made a typo saying "suppose the boundary of $M$ is not closed" and "hence the boundary of $M$ is closed" when you meant bounded, but the word closed is used so many times it is very much understandable :) Jun 6, 2023 at 8:58
• @BrunoB Thanks for spotting my typo. Thank Kolja for correcting it. Jun 6, 2023 at 13:34