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I need a counter example. I need two subsets $A, B$ of $\mathbb{R}^n$ so that $\text{Cl}(A+ B)$ is different of $\text{Cl}(A) + \text{Cl}(B)$, where $\text{Cl}(A)$ is the closure of $A$, and $A + B = \{a + b: a \in A, b \in B\}$

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I've edited the title to be a bit more informative about the content of the question (I'm assuming amWhy's interpretation of the question is correct). Feel free to change it if you prefer something else. – Zev Chonoles Aug 29 '11 at 4:11
I think by '$+$' Andres means element-wise addition which is different from the union. – Adhvaitha Aug 29 '11 at 4:11
What about $A+B=\{a+b:a\in A,\, b\in B\}$? – leo Aug 29 '11 at 4:14
"+" is not the union of sets, is the usual sum of two sets A+B={a+b:a∈A,b∈B} – Roiner Segura Cubero Aug 29 '11 at 4:24
Corrected; sorry about the confusion. – amWhy Aug 29 '11 at 4:30
up vote 7 down vote accepted

It’s also possible to find examples in $\mathbb{R}$. For instance, let $A = \{n+\frac1{2n}:n \in \mathbb{Z}^+\}$, and let $B = \{-n:n \in \mathbb{Z}^+\}$; clearly $A$ and $B$ are closed. $A$ contains no integer, so $0 \notin A+B$, but $A+B \supseteq \{\frac1{2n}:n \in \mathbb{Z}^+\}$, so $0 \in \operatorname{cl}(A+B)$.

Note that it’s never possible to take $A$ and $B$ to be compact: if $A$ and $B$ are compact subsets of $\mathbb{R}^n$, $A \times B$ is a compact subset of $\mathbb{R}^{2n}$, and its continuous image under addition is a compact and therefore closed subset of $\mathbb{R}^n$. Thus, all such examples must involve unbounded sets.

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Dave Rusin gives a counterexample in $\mathbb R^2$ at, namely $A=\{(x,1/x):x\neq 0\}$ and $B=\{(x,-1/x):x\neq 0\}$, with the comment, "Find a point not in $A+B$ which lies in the closure of $A+B$," to lead you to the proof that this is actually a counterexample. (Note that $A$ and $B$ are closed in this example.)

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Take $A=\{1,10,100,\dots\}$ and $B=\{-0.9,-9.99,-99.999,\dots\}$ in $\mathbb{R}$.

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