Closed subsets $A,B\subset\mathbb{R}^2$ so that $A+B$ is not closed I am looking for closed subsets $A,B\subset\mathbb{R}^2$ so that $A+B$ is not closed.
I define
$A+B=\{a+b:a\in A,b\in B\}$
I thought of this example, but it is only in $\mathbb{R}$. Take:
$A=\{\frac{1}{n}:n\in\mathbb{Z^+}\}\cup\{0\}$ and $B=\mathbb{Z}$
both of these are closed (is this correct?). But their sum $A+B=\mathbb{Q}$ which is not closed.

For a specific example in $\mathbb{R}^2$, let $A$ be the epigraph of $t \mapsto e^t$ (ie. $A= \{ (t,y) \mid y \geq e^t \}$) and $B= [0,+ \infty) \times \{0\}$. Then $A+B= \mathbb{R} \times (0,+ \infty)$ is not closed.
Notice that if $A$ or $B$ is bounded, then $A+B$ is necessarily closed. 
Indeed, suppose $B$ bounded (and so compact): if $(x_n+y_n)$ is a sequence in $A+B$ converging to some $z \in \mathbb{R}^2$, there exists a subsquence $y_{\sigma(n)}$ converging to some $y \in B$ by compactness; moreover, $x_{\sigma(n)}$ converges to $x=y-z \in A$ ($A$ is closed). You deduce that $z=x+y \in A+B$: $A+B$ is closed.
