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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!

  • 4
    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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    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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    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.