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 $\mathbb R^n$.

  • $\begingroup$ A concrete example disproving 3 (see Apass.Jack's nice answer below): $\{(x,y): y=\tan(x), -\pi/2<x<\pi/2\}+\{(x,y): x \ge 0\} = \{(x,y): x >-\pi/2\}$ $\endgroup$
    – Dunnò000
    Jun 6, 2023 at 11:52

1 Answer 1


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


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
    $\begingroup$ (+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 :) $\endgroup$
    – Bruno B
    Jun 6, 2023 at 8:58
  • $\begingroup$ @BrunoB Thanks for spotting my typo. Thank Kolja for correcting it. $\endgroup$
    – Apass.Jack
    Jun 6, 2023 at 13:34

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