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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.
in the case of $\mathbb R$ :- $\mathbb Z$ is closed so is $\mathbb a \mathbb Z$ where a is irrational number then $\mathbb Z + a \mathbb Z$ is not closed in $\mathbb R$ it is a dense set in $\mathbb R$
It’s sufficient to find an example in $\Bbb R$: if $A$ and $B$ are an example in $\Bbb R$, then $A\times\{0\}$ and $B\times\{0\}$ are an example in $\Bbb R^2$.
Here’s a big hint for $\Bbb R$. Let $A=\Bbb Z^+$ and $B=\left\{n+\frac1{n+1}:n\in\Bbb Z^+\right\}$. Show that $A$ and $B$ are closed, but that $B-A$ is not. (It may help to make a rough sketch of $A$ and $B$.) Now modify the example slightly to get a pair of closed sets whose sum is not closed.
Let $A=\mathbb{Z}$ and $B=p\mathbb{Z}:=\{pn: n\in\mathbb{Z}\}$ where $p$ is any irrational number. So $A$ and $B$ are two closed subsets of $\mathbb{R}$, but, $A+B :=\{m+pn: m,n\in\mathbb{Z}\}$ is not closed in $\mathbb{R}$.
Let $F=\{(x,y)\in \mathbb{R}^2 : xy\geq 1\}$ be the set of points $(x,y)$ which lie on and above the arm of hyperbola $xy=1$ in the first quadrant and C be the Y-axis. then
$F+C=\{ (x,y):x>0\}$ which is not closed.
See the above figure in first picture $F$ and $C$ both are closed but in second picture $F+C$ 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.
