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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 $Y$)


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Does this count as a duplicate of this? – leo Mar 25 '12 at 6:37
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. – user22805 Mar 25 '12 at 6:44
As David shows, the answer is no. However, the sum of a closed set and a compact set is closed. – Robert Israel Apr 15 '12 at 8:05
up vote 41 down vote accepted

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? – ro44 Mar 25 '12 at 3:26
@ro44 You're welcome. Yes to your question. – David Mitra Mar 25 '12 at 3:32
What if one of $A$ or $B$ is compact , will their sum be closed or compact? – Neeraj Bhauryal Jan 23 '14 at 5:00
@NeerajBhauryal It will be closed but not necessarily compact. – David Mitra Jan 23 '14 at 9:02
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). – Marcel T. Jul 26 '14 at 6:49

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 and here. – Martin Sleziak Apr 15 '12 at 8:34
No, $E$ and $F$ are not closed; you want $y \ge 1/x$ etc. And the original question was about $\mathbb R$, not ${\mathbb R}^2$. – Robert Israel Apr 16 '12 at 7:31

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