# 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.

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Their sum is not $\mathbb{Q}$: it doesn't contain $2/5$ for example. –  Chris Eagle Oct 18 '12 at 14:24
And their sum is closed. –  Brian M. Scott Oct 18 '12 at 14:25
@Ayman: That definitely won’t work: if $A$ and $B$ are finite, so is $A+B$. –  Brian M. Scott Oct 18 '12 at 14:25
In fact, the sum of a compact set and a closed set is always closed, so your two sets will both have to be unbounded. –  Chris Eagle Oct 18 '12 at 14:26
@Sebastian: What elements of $A$ and $B$ do you imagine sum to $2/5$? –  Chris Eagle Oct 18 '12 at 14:28

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.

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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 $\mathbb R$

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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.

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Very nice approach....+1 –  Babak S. Mar 23 '13 at 19:31

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 $\mathbb{R}$.

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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.

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