Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A + B = \{a + b: a \in A, b \in B\}$

share|improve this question
1  
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
2  
I think by '$+$' Andres means element-wise addition which is different from the union. –  Adhvaitha Aug 29 '11 at 4:11
1  
What about $A+B=\{a+b:a\in A,\, b\in B\}$? –  leo Aug 29 '11 at 4:14
2  
"+" 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
1  
Corrected; sorry about the confusion. –  amWhy Aug 29 '11 at 4:30

3 Answers 3

up vote 6 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.

share|improve this answer

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

share|improve this answer

Dave Rusin gives a counterexample in $\mathbb R^2$ at http://www.math.niu.edu/~rusin/known-math/95/closed.djr, 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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.