Questions about sum of sets Suppose $A$ and $B$ are two subsets of $\mathbb{R}^n$. Then define $A+B=\{a+b~|~a\in A,b\in B\}$. 


*

*I have proved that if $A$ and $B$ are open, then so is $A+B$. However, I need to prove that if $A$ and $B$ are closed then $A+B$ need not be closed. I can't find any suitable counterexample. 

*Moreover, another fact states that if $A$ is compact and $B$ is closed then $A+B$ is closed. I have proved that if $A$ and $B$ are both compact then so is $A+B$, but this is not helping me to prove the above fact.

*I also want to know, whether $A+B$ compact implies that $A$ and $B$ are both compact?  


*

*$A$,$B$ closed, $A+B$ not closed: Let $A=\mathbb N$, $B=\{\,-n+\frac1n\mid n\in\mathbb N, n>1\,\}$ (courtesy the answer here

*$A+B$ compact, neither $A$ nor $B$ compact: Let $A=B=[-1,1]\setminus\{0\}\subset \mathbb R^1$. Then $A+B=[-2,2]$.

*$A$ compact, $B$ closed implies $A+B$ closed:
Let $x\notin A+B$. Then for each $a\in A$, $x-a\notin B$, hence there exists $r_a>0$ with $B(r_a,x-a)\cap B=\emptyset$. The open cover
$$ A\subseteq \bigcup_{a\in A} B\left(\frac12 r_a,a\right)$$
allows a finite subcover by compactness of $A$:
$$ A\subseteq \bigcup_{1\le i\le n} B\left(\frac12 r_{a_i},a_i\right).$$
Let $r=\min_{1\le i\le n} r_{a_i}>0$.
Then for $a\in A,b\in B$ there exists $1\le i\le n$ with $\|a-a_i\|<\frac12 r$; from $b\notin B(r_{a_i},x-a_i)$ we conclude $\|b-(x-a_i)\|\ge r$, hence 
$$\|(a+b)-x\|\ge \|(a+b-x)-(a-a_i)\|-\|a-a_i\|> \frac r2. $$
Hence $B\left(\frac r2,x\right)\cap (A+B)=\emptyset$, i.e. the complement of $A+B$ is open as was to be shown.
