Questions about sum of sets

Question

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?

@DavidMitra. Quite often I see comments like 'see this' with nice links. How do people do that? Just memory, or is some sort of adequate method applied? If there is a method then it has my interest and I want to practicize that too. — drhab, Feb 08 '14 at 11:24
@DavidMitra Nice counter-example. Can you help me out eith the other problems? — QED, Feb 08 '14 at 11:25
@drhab Most of the time, I just recalled the question was asked here before. It helps if I've answered it. — David Mitra, Feb 08 '14 at 11:25
For the second bullet point: You could take a sequence $(a_n+b_n)$ in $A+B$ converging to $z$. $(a_n)$ has a convergent subsequence. Use this and the hypotheses to conclude $z\in A+B$. — David Mitra, Feb 08 '14 at 11:28
For the third: I think ${-1/n\mid n\in\Bbb N}\cup{0}$, $B={1/n\mid n\in\Bbb N}$ provides a counterexample. I can't think of a counterexample where neither $A$ nor $B$ is compact. — David Mitra, Feb 08 '14 at 11:42

score 1 · Accepted Answer · edited Apr 13 '17 at 12:19

$A$,$B$ 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.

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