TITLE: A+B is closed if one of them is compact
QUESTION [8 upvotes]: Q. Show that if A and B are closed subsets of $R^n$ and one of them is compact then A+B is closed.
My doubt:
A+B is not necessarily closed given A and B are closed. I need hints to start this question since it involves the concept of compactness.

REPLY [4 votes]: Let $c\notin A+B$. Then for each $a\in A$, $c-a\notin B$. As $B$ is closed, there is an open ball $B_{r(a)}(c-a)$ of radius $r(a)$ around $c-a$ that is disjoint from $B$. Then $B_{r(a)}(c)$ is disjoint from $a+B$, and $B_{r(a)/2}(c)$ is disjoint from $a'+B$ for all $a'\in A$ with $d(a,a')<\frac {r(a)}2$. The open balls $B_{r(a)/2}(a)$, $a\in A$, cover $A$. By compactness of $A$, a finite subcover suffices, say $A\subseteq \bigcup_{i=1}^n B_{r(a_i)/2}(a_i)$. Let $r=\min\{\,r(a_i):1\le i\le n\,\}$. Then $B_r(c)\cap (a+B)=\emptyset$ for all $a\in A$, i.e., $B_r(c)\cap (A+B)=\emptyset$.
We conclude that the complement of $A+B$ is open, in other words, $A+B$ is closed.