Why is it that if we have a compact set $X$ and a closed set $Y$ then the Minkowski sum $X+Y$ is necessarily closed? (Sorry for keep asking questions about the Minkowski sum, I am trying to figure out how these things work.) Thanks.


3 Answers 3


Here's a proof that works in any topological vector space.

If $z \notin X + Y$, then for every $x \in X$ we have $z - x \notin Y$; since $Y$ is closed and addition is continuous there are open sets $B_x$ containing $x$ and $U_x$ containing $z$ such that $(U_x - B_x) \cap Y = \emptyset$. By compactness of $X$, a finite collection $B_{x_1}, \ldots, B_{x_n}$ covers $X$. Thus $U = U_{x_1} \cap \ldots \cap U_{x_n}$ is an open set containing $z$ which does not intersect $X + Y$.


(I assume the setting for this question is a normed vector space $V$)

Suppose $z_n$ is a sequence in $X+Y$, and assume it converges to some point $z$ in $V$. We want to show that $z$ is in $X+Y$. By definition of the Minkowski sum, there must exist $x_n \in X, y_n \in Y$ such that $z_n=x_n+y_n$. Since $X$ is compact, there is a subsequence $x_{n_i}$ such that $x_{n_i} \rightarrow x$ as $i \rightarrow \infty$, where $x$ is in $X$. Thus $y_{n_i}=z_{n_i}-x_{n_i} \rightarrow z-x$. Since $Y$ is closed, we have that $z-x$ is in $Y$, and so $z=x+(z-x)$ is in $X+Y$ as desired.

It's worth noting that it's not enough to simply have $X$ and $Y$ both closed. For example, suppose $V=\mathbb{R}$, $x=\mathbb{Z}$ and $y=\sqrt{2}\mathbb{Z}$. Then $X$ and $Y$ are both closed but $X+Y$ is a countable dense set.

  • $\begingroup$ I wonder, is there a way to relax the condition of compactness of one of the spaces? $\endgroup$
    – Simon
    May 12, 2017 at 14:54

Here it goes : if we denote $\overline{A}$ for the adherence of $A$, if $X$ is compact and $Y$ is another set, we have $$ \overline{X+Y} = \overline X + \overline Y. $$ $(\subseteq)$ Let $z \in \overline{X+Y}$. Thus there exists a sequence $\{z_n\} \in X+Y$ such that $z_n \to z$. But $z_n = x_n + y_n$ with $x_n \in X$ and $y_n \in Y$, and since $X$ is compact, there exists a subsequence $x_{n_k}$ such that $x_{n_k} \to x \in \overline X$. But that means $$ y_{n_k} = x_{n_k} + y_{n_k} - x_{n_k} = z_{n_k} - x_{n_k} \to z - x \in \overline Y $$ because $y_{n_k} \in Y$. Hence $z = z-x + x \in \overline X + \overline Y$, thus the inclusion.

$(\supseteq)$ Let $z \in \overline X + \overline Y$. Thus $z = x+y$ and $x_n \to x \in \overline X$, $y_n \to y \in \overline Y$ and so $x_n + y_n \to x+y = z \in \overline {X+Y}$.

I needed the compacity of $X$ to show equality but I didn't need the fact that $Y$ is. Since compact sets are closed, $X = \overline X$ and $Y = \overline Y$, so $X+Y = \overline X + \overline Y = \overline { X+Y}$.

Hope that helps,

  • $\begingroup$ Chris Eagle's proof is shorter, but I thought this would give you some more properties of Minkowski sums and make you understand things a little easier by having things worked out here. $\endgroup$ Nov 11, 2011 at 0:18

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