A+B is closed if one of them is compact 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.
 A: let $(x_n+y_n)$ a sequence that converge such that $(x_n)\subset A$, $(y_n)\subset B$. Suppose $A$ is compact. By Bolzano-Weierstrass, there is a subsequence $(x_{n_k})$ that converge to $x\in A$. Moreover, $$|y_{n_k}|\leq |x_{n_k}+y_{n_k}|+|x_{n_k}|$$
and thus $(y_{n_k})$ has a subsequence $(y_{n_{k_j}})$ that converges since it's bounded. Since $B$ is closed, $(y_{n_{k_j}})$ converges to $y\in B$. Therefore, $(x_{n_{k_j}}+y_{n_{k_j}})$ converges to $x+y\in A+B$ and thus, $(x_{n}+y_{n})$ converges to $x+y\in A+B$ which proves the claim.
A: 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.
A: Hint.
Suppose that $(c_n)$ is a convergent sequence of $A+B$ that converges to $c$. By definition, there exist sequences $(a_n)$ in $A$ and $(b_n)$ in $B$ with $c_n=a_n +b_n$ for all $n \in \mathbb N$.
By hypothesis, we can suppose $A$ compact. Therefore $(a_n)$ has a convergent subsequence $a_{\varphi(n)}$ converging to $a \in A$. $b_{\varphi(n)}=c_{\varphi(n)}-a_{\varphi(n)}$ converges to $c-a \in B$ as $ B$ is supposed to be closed.
Finally $c \in A+B$ which concludes the proof.
