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Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?

(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)

Thanks!

  • 0
    Does this count as a duplicate of [this](http://math.stackexchange.com/q/60452/8271)?2012-03-25
  • 0
    It's awfully similar, and it's probably quite straightforward to show that it's an equivalent question. However, I feel that the wording is different enough for it not to count as an exact duplicate.2012-03-25
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    As David shows, the answer is no. However, the sum of a closed set and a compact set is closed.2012-04-15

4 Answers 4

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Consider the sets $A=\{ n\mid n=1,2,\ldots\}$ and $B=\{- n+{1\over n}\mid n= 2,3,\ldots\}$. Note that $0$ is not in the sum, but $1\over n$ is for each $n\ge2$.

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    Clever, thanks. So $0$ is a limit point that's not in $A+B$, right?2012-03-25
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    @ro44 You're welcome. Yes to your question.2012-03-25
  • 1
    What if one of $A$ or $B$ is compact , will their sum be closed or compact?2014-01-23
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    @NeerajBhauryal It will be closed but not necessarily compact.2014-01-23
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    Perhaps it's too trivial to mention, but if both are compact, then of course their sum will be compact (the continuous image of a compact set is compact).2014-07-26
  • 0
    What happens if $A$ and $B$ are closed subsets of $[0, \infty)$? Are there such $A$ and $B$ in this case such that $A+B$ is not closed?2016-11-27
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    @PrinceKumar I don't think so. If $(a_n+b_n)$ converges to $c$, then by comparison, $(a_{n }) $ convergies to some $a\in A$ and $(b_{n }) $ converges to some $b\in B$. But then $c=a+b\in A+B$.2016-11-27
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    @DavidMitra I don't understand what you used to deduce the convergence of $(a_n)$ and $(b_n)$ from the convergence of $(a_n+b_n)$. The only way this case is different from the one is question is that we can't choose $(a_n)$ or $(b_n)$ to be negative. Will that be sufficient to deduce the convergence?2016-11-27
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    @PrinceKumar Oh, sorry, what I have above is off... But, you can choose a "common" subsequece of $(a_n)$ and $(b_n)$, both of which converge (both are bounded and non-negative).2016-11-27
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    @DavidMitra They need not have any common subsequence. I will try to find the proof. All I am getting right now is an intuition. :)2016-11-27
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    @PrinceKumar Yes they do. Take a convergent subsequence $(a_{n_k})_k$ of $(a_n)$. Then take a convergent subsequence of $(b_{n_k})_k$.2016-11-27
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    @DavidMitra Oh, you meant common indices. I thought you are talking about common terms (values) which will not be true for two arbitrary sequences. It will work. Thanks a lot!2016-11-27
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    @PrinceKumar Yes, sorry for being unclear...2016-11-27
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    How to show B is closed in your example?2017-09-03
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consider $\mathbb Z$ and $\sqrt 2 \mathbb Z$ both are closed but the sum is not...:) moreover it is dense on $\mathbb R$

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The sum $E +F$ may fail to be closed even if $E$ and $F$ are closed. For instance, set $E = \{(x, y) \in \mathbb R^2 : y > 1/x\text{ and }x > 0\}$ and $F = \{(x, y) \in\mathbb R^2 : y > -1/x\text{ and }x < 0\}$

Then $E$ and $F$ are closed, but $$E + F = \{(x, y) \in \mathbb R^2 : y > 0\}$$ is not closed.

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    hassan: I've tried to edit your question (add latex formatting for better readability). Please, check whether I did not changed the meaning unintentionally. You can find more about writing math on this site e.g [here](http://math.stackexchange.com/editing-help#latex) and [here](http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto).2012-04-15
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    No, $E$ and $F$ are not closed; you want $y \ge 1/x$ etc. And the original question was about $\mathbb R$, not ${\mathbb R}^2$.2012-04-16
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Take $A=\{(a,0):a\in\mathbb{R}$ and $B=\{(b,\frac{1}{b}):b\in \mathbb{R}-\{0\}\}$. Then both $A,B\subset \mathbb{R}^2$ are closed. But $A+B=\{(a+b,\frac{1}{b}):a\in \mathbb{R},b\in \mathbb{R}-\{0\}\}.$The sequence $\{(0,\frac{1}{n})\}=\{(n-n,\frac{1}{n})\}\subset A+B$ but the limit $(0,0)$ which is not in the sum.